In 2 Physics

Stephen Bosi

John O’Byrne

Peter Fletcher

Joe Khachan

Jeff Stanger

Sandra Woodward

PHYSICS@ HSC

Sydney, Melbourne, Brisbane, Perth, Adelaide and associated companies around the world

iii

ContentsSeries features viHow to use this book viiiStage 6 Physics syllabus grid x

Module 1 Space Module 1 Introduction 2

Chapter 1 Cannonballs, apples, planets and gravity 41.1 Projectile motion 41.2 Gravity 101.3 Gravitational potential energy 16Practical experiences 20Chapter summary 22Review questions 22

Chapter 2 Explaining and exploring the solar system 262.1 Launching spacecraft 262.2 Orbits and gravity 352.3 Beyond Kepler’s orbits 412.4 Momentum bandits: the slingshot effect 442.5 I’m back! Re-entry 46Practical experiences 52Chapter summary 53Review questions 54

Chapter 3 Seeing in a weird light: relativity 583.1 Frames of reference and classical relativity 583.2 Light in the Victorian era 613.3 Special relativity, light and time 643.4 Length, mass and energy 69Practical experiences 75Chapter summary 76Review questions 76

Module 1 Review 80

Module 2 Motors and Generators Module 2 Introduction 82

Chapter 4 Electrodynamics: moving charges and magnetic fields 844.1 Review of essential concepts 844.2 Forces on charged particles in magnetic fields 894.3 The motor effect 904.4 Forces between parallel wires 93Practical experiences 97Chapter summary 98Review questions 98

Chapter 5 Induction: the influence of changing magnetism 1005.1 Michael Faraday discovers electromagnetic

induction 1005.2 Lenz’s law 1045.3 Eddy currents 106Practical experiences 109Chapter summary 110Review questions 110

Chapter 6 Motors: magnetic fields make the world go around 1146.1 Direct current electric motors 1146.2 Back emf and DC electric motors 1206.3 Alternating current electric motors 121Practical experiences 126Chapter summary 127Review questions 127

Chapter 7 Generators and electricity supply: power for the people 1307.1 AC and DC generators 1307.2 Transformers 1367.3 Electricity generation and transmission 141Practical experiences 148Chapter summary 149Review questions 149

Module 2 Review 152

Module 3 From Ideas to Implementation Module 3 Introduction 154

Chapter 8 From cathode rays to television 1568.1 Cathode ray tubes 1568.2 Charges in electric fields 1608.3 Charges moving in a magnetic field 1648.4 Thomson’s experiment 1658.5 Applications of cathode rays 167Practical experiences 170Chapter summary 171Review questions 171

Chapter 9 Electromagnetic radiation: particles or waves? 1749.1 Hertz’s experiments on radio waves 1749.2 Black body radiation and Planck’s hypothesis 1789.3 The photoelectric effect 1829.4 Applications of the photoelectric effect 184Practical experiences 185Chapter summary 186Review questions 187

Contents

iv

Chapter 10 Semiconductors and the electronic revolution 18810.1 Conduction and energy bands 18910.2 Semiconductors 19010.3 Semiconductor devices 19310.4 The control of electrical current 197Practical experiences 201Chapter summary 202Review questions 202

Chapter 11 Superconductivity 20411.1 The crystal structure of matter 20411.2 Wave interference 20511.3 X-ray diffraction 20711.4 Crystal structure 20811.5 Electrical conductivity and the crystal

structure of metals 20911.6 The discovery of superconductors 21111.7 The Meissner effect 21211.8 Type-I and type-II superconductors 21211.9 Why is a levitated magnet stable? 21311.10 BCS theory and Cooper pairs 21511.11 Applications of superconductors 217Practical experiences 220Chapter summary 221Review questions 221

Module 3 Review 224

Module 4 Quanta to Quarks Module 4 Introduction 226

Chapter 12 From Rutherford to Bohr 22812.1 Atomic timeline 22812.2 Rutherford’s model of the atom 22912.3 Planck’s quantised energy 23112.4 Spectral analysis 23212.5 Bohr’s model of the atom 23512.6 Bohr’s explanation of the Balmer series 23612.7 Limitations of the Rutherford–Bohr model 239Practical experiences 241Chapter summary 242Review questions 243

Chapter 13 Birth of quantum mechanics 24713.1 The birth 24713.2 Louis de Broglie’s proposal 24813.3 Diffraction 25013.4 Confirming de Broglie’s hypothesis 25113.5 Electron orbits revisited 25213.6 Further developments of atomic theory

1924–1930 253Practical experiences 256Chapter summary 256Review questions 257

Chapter 14 20th century alchemists 26014.1 Discovery of the neutron 26014.2 The need for the strong force 26114.3 Atoms and isotopes 26214.4 Transmutation 26314.5 The neutrino 26514.6 Was Einstein right? 26614.7 Binding energy 26814.8 Nuclear fission 26914.9 Chain reactions 27014.10 Neutron scattering 272Practical experiences 273Chapter summary 274Review questions 275

Chapter 15 The particle zoo 27915.1 The Manhattan Project 27915.2 Nuclear fission reactors 28015.3 Radioisotopes 28215.4 Particle accelerators 28615.5 The Standard Model 292Practical experiences 295Chapter summary 296Review questions 297

Module 4 Review 300

Module 5 Medical Physics Module 5 Introduction 302

Chapter 16 Imaging with ultrasound 30416.1 What is ultrasound? 30416.2 Principles of ultrasound imaging 30516.3 Piezoelectric transducers 30816.4 Acoustic impedance 31016.5 Types of scans 31216.6 Ultrasound at work 315Practical experiences 317Chapter summary 318Review questions 318

Chapter 17 Imaging with X-rays 32017.1 Overview and history: types of X-ray images 320 17.2 The X-ray tube 32117.3 Types of X-rays 32217.4 Production of X-ray images 32417.5 X-ray detector technology 32617.6 Production of CAT X-ray images 32617.7 Benefits of CAT scans over conventional

radiographs and ultrasound 329Practical experiences 330Chapter summary 331Review questions 331

Contents

v

Chapter 18 Imaging with light 33318.1 Endoscopy 33318.2 Medical uses of endoscopes 336Practical experiences 338Chapter summary 339Review questions 339

Chapter 19 Imaging with gamma rays 34019.1 Isotopes and radioactive decay 34019.2 Half-life 34319.3 Radiopharmaceuticals: targeting tissues

and organs 34419.4 The gamma camera 34619.5 Positron emission tomography 347Practical experiences 350Chapter summary 351Review questions 351

Chapter 20 Imaging with radio waves 35420.1 Spin and magnetism 35420.2 Hydrogen in a magnetic field 35520.3 Tuning in to hydrogen 35720.4 It depends on how and where you look 35920.5 The MRI scanner 36020.6 Applications of MRI 362Practical experiences 363Chapter summary 364Review questions 364

Module 5 Review 366

Module 6 Astrophysics Module 6 Introduction 368

Chapter 21 Eyes on the sky 37021.1 The first telescopes 37021.2 Looking up 37321.3 The telescopic view 37421.4 Sharpening the image 37721.5 Interferometry 38021.6 Future telescopes 382Practical experiences 383Chapter summary 384Review questions 384

Chapter 22 Measuring the stars 38822.1 How far? 38822.2 Light is the key 38922.3 The stellar alphabet 39422.4 Measuring magnitudes 39722.5 Colour matters 400Practical experiences 403Chapter summary 405Review questions 405

Chapter 23 Stellar companions and variables 40723.1 Binary stars 40723.2 Doubly different 41123.3 Variable stars 41323.4 Cepheid variables 415Practical experiences 418Chapter summary 418Review questions 419

Chapter 24 Birth, life and death 42224.1 The ISM 42224.2 Star birth 42324.3 Stars in the prime of life 42524.4 Where to for the Sun? 42824.5 The fate of massive stars 43024.6 How do we know? 433Practical experiences 435Chapter summary 436Review questions 436

Module 6 Review 438

Module 7 Skills Module 7 Introduction 440

Chapter 25 Skills stage 2 44225.1 Metric prefixes 44225.2 Numerical calculations 44325.3 Sourcing experimental errors 44525.4 Presenting research for an exam 44625.5 Australian scientist 44725.6 Linearising a formula 447

Chapter 26 Revisiting the BOS key terms 44826.1 Steps to answering questions 449

Numerical answers 452Glossary 454Index 465Acknowledgements 471Formulae and data sheets 473Periodic table 474

vi

PHYSICS@ HSC

in2 Physics is the most up-to-date physics package written for the NSW Stage 6 Physics syllabus. The materials comprehensively address the syllabus outcomes and thoroughly prepare students for the HSC exam.

Physics is presented as an exciting, relevant and fascinating discipline. The student materials provide clear and easy access to the content and theory, regular review questions, a full range of exam-style questions and features to develop an interest in the subject.

in2 Physics @ HSC student book

• ThestudentbookcloselyfollowstheNSWStage6Physicssyllabus and its modular structure.

• Itclearlyaddressesboththecontextsandtheprescribedfocusareas (PFAs).

• Modulesconsistofchaptersthatarebrokenupinto manageable sections.

• Checkpointquestionsreviewkeycontentatregular intervals throughout each chapter.

• PhysicsPhilespresentshort,interestingsnippets of relevant information about physics or physics applications.

• PhysicsFeatureshighlightimportantreal-lifeexamples of physics.

• PhysicsForFun—TryThis!providehands-onactivities that are easy to do.

• PhysicsFocusbringstogetherphysicsconceptsin the context of one or more PFAs and provides students with a graded set of questions to develop their skills in this vital area.

Each student book includes an interactive student CD containing:• anelectronicversionofthestudentbook.• allofthestudentmaterialsonthecompanionwebsitewithlive

links to the website.

From cathode rays to television8

168

169

From ideas toimplementation

The vertical deflection plates cause the beam to move up or down in

synchronisation with an input voltage. For example, a sinusoidal voltage will

display a sinusoidal waveform (known as a trace) on the screen.

TelevisionCathode ray tube (CRT) television sets used the principles of the cathode ray

tube for most of the 20th century. These are now being superseded by plasma

and liquid crystal display television sets, which use different operating principles

and allow a larger display area with a sharper image. However, the CRT

television holds quite a significant historical place in this form of

communication. A schematic diagram of a colour CRT television set is shown in Figure 8.5.5.

Its basic elements are similar to those of the CRO. The main difference is the

method of deflecting the electrons. Magnetic field coils placed outside the tube

produce horizontal and vertical magnetic fields inside it. The magnitude and

direction of the current determine the degree and direction of electron beam

deflection. Recall your right-hand palm rule for the force on charged particles

in a magnetic field. The vertical magnetic field will deflect the electrons

horizontally; the horizontal field will deflect them vertically.

The picture on the screen is formed by scanning the beam from left to right

and top to bottom. The electronics in the television switches the beam on and

off at the appropriate spots on the screen in order to reproduce the transmitted

picture. However, to reproduce colour images, colour television sets need to

control the intensity of red, blue and green phosphors on the screen. Three

separate electron guns are used, each one aimed at one particular colour. The

coloured dots on the screen are clustered in groups of red, blue and green dots

that are very close to each other and generally cannot be distinguished by eye

without the aid of a magnifying glass. For this reason a method of guiding the

different electron beams to their respective coloured dots was devised. A metal

sheet, known as a shadow mask (Figure 8.5.6) and consisting of an array of

holes, is placed behind the phosphor screen. Each hole guides the three beams to

their respective coloured phosphor as the beams move horizontally and vertically.

Black and white television sets did not need the shadow mask since they had

only one beam.

heater

cathode (negative)

electrons'boil' offthe heatedcathode

anode (positive)

electronbeam

electrons attractedto the positive anode

collimator

Figure 8.5.3 The components of an electron gun used in both cathode ray

oscilloscopes and CRT televisions

V TimeV Time

sawtooth voltage for timebase sinusoidal vertical voltage

Figure 8.5.4 A sawtooth voltage waveform on the horizontal

deflection plates of a CRO sweeps the electron beam

across the screen to display the sinusoidal waveform

on the vertical deflection plates.

electron gun

magneticcoilselectron

beam

fluorescent screen

Figure 8.5.5 A television picture tube

showing the electron gun,

deflection coils and

fluorescent screen

electronguns

deflectingcoils

focusingcoils

glass

fluorescent screen

vacuum

mask

phosphor dots on screen

fluorescentscreen

mask

holes inmask

bluebeam

redbeam greenbeam

RGB

electronbeams

Figure 8.5.6 A colour CRT television set has three electron guns

that will only strike their respective coloured

phosphor dots with the aid of a shadow mask.

try this!Do not aDjust your horizontal!If you have access to an old

black and white TV set or an old

style monochrome computer

monitor, try holding a bar

magnet near the front of the

screen and watch how the

image distorts. This occurs

because the magnetic field

deflects the electrons that strike

the screen. DO NOT do this with

a colour TV set. This can

magnetise the shadow mask and

cause permanent distortion of

the image and its colour. You

can move a bar magnet near the

back of a colour TV set to

deflect the electrons from the

electron gun and therefore

distort or shift the image

without causing permanent

damage to the TV set.

Can an osCillosCope be used

as a television set?

The similarity between the cathode ray oscilloscope (CRO) and CRT

television suggests that a CRO can be used as a television set. In

fact, there have been some devices that have made use of the CRO as

you would a computer monitor. So, in principle, it can be used as a

television. One is then forced to ask ‘why did they need to deflect the

beam in a television set with magnetic fields rather than with electric

fields as in the CRO?’

In principle all television sets could be made in the same design as

a CRO; however, it is much easier and cheaper to deflect the beam with

a magnetic field on the outside of the tube rather than embed electrodes

in the glass and inside the vacuum—this is a little trickier. So now

another question arises: ‘why not deflect the beam of the CRO using

magnetic fields, wouldn’t it result in cheaper CROs?’.

Cathode ray oscilloscopes are precision instruments. The horizontal

sweep rate must be able to be increased to very high frequencies in order

to detect signals that change very quickly. Electric fields can be made to

change very quickly without significant extra power requirements.

However, a magnetically deflected system requires higher and higher

voltages with increasing horizontal and vertical deflection frequencies in

order to maintain the same current in the coils, and therefore, the same

angle of beam deflection – thus having a significantly greater power

requirement. Cathode ray tube television sets, however, only operate at

fixed and relatively low scanning horizontal and vertical frequencies.

Thus it is simpler and cheaper for the mass market to deflect with a

magnetic field.

CheCkpoint 8.51 Outline the purpose of a CRO.

2 List the main parts of a CRO.

3 Describe the role of each of these parts in the CRO.

4 State the similarities and differences between the cathode ray tube CRO and CRT TV.

THE COMPLETE PHYSICS PACKAGE FOR NSW STUDENTS

vii

in2 Physics @ HSC Activity Manual

• Awrite-inworkbookthat provides a structured approach to the mandatory practical experiences, both first-hand and secondary-source investigations.

• Dotpointandskillsfocused.

in2 Physics @ HSC Teacher Resource

• Editableteachingmaterials,includingteachingprograms, so that teachers can tailor lessons to suit their classroom.

• Answerstostudentbookandactivitymanualquestions, with fully worked solutions and extended answers and support notes.

• Riskassessmentsforallfirst-handinvestigations.

in2 Physics @ HSC companion website

Visit the companion website in the student lounge and teacher lounge of Pearson Places

• Reviewquestions— auto-correcting multiple-choice questions for each chapter.

• Webdestinations—alist of reviewed websites that support further investigation.

68

3MODULE from ideas toimplementation

Chapter 8 from cathode rays to television

69

Method1 Set up the equipment as shown in Figure 8.1.1.

2 Observe the patterns and note the pressure in the tube.

3 Replace the tube with the next in the series.

4 Repeat the process of observing the patterns and

noting the pressure for each of the tubes in your set.

HAZARD

High voltages are produced by induction coils and may produce unwanted X-rays. The voltages necessary to operate the

tubes depend upon the dimensions of the tube and the pressure of the gas in the tube. Generally, the higher the voltage used,

the greater the danger of the production of unwanted X-rays.

Use the lowest possible voltage and stand a minimum of 1 m away from the equipment.

Chapter 8 from Cathode rays to television

Changing pressure of discharge tubes

Perorm an investigation and gather first-hand information to observe the occurrence of different striation

patterns for different pressures in discharge tubes.

Physics skillsThe skills outcomes to be practised in this activity include:

12.1 perform first-hand investigations

12.2 gather first-hand information

14.1 analyse information

The complete statement of these skills outcomes can be found in the syllabus grid on pages vi–viii.

AimTo observe the striation patterns for different pressures in discharge tubes.

Hypothesis

TheoryEver since Heinrich Geissler and Julius Plücker collaborated to create a tube in which the pressure could be reduced

substantially, our understanding of the atom and developing uses for the cathode ray tube have advanced tremendously.

Because of this development, it was found that electric current could be passed through air. The different patterns that

could be seen depended on the pressure.

Normally air is considered to be an insulator, but it can be made to conduct by ionising the air molecules. To do this,

the very small fraction of free electrons that are always in air are accelerated (with an electric field). At high pressures these

electrons collide frequently with the air, losing their energy and, as a result, do not gain sufficient energy to ionise the air

atoms. As pressure is reduced, these electrons travel further before colliding with air molecules, thereby acquiring enough

energy to ionise the air molecules. This will produce more free electrons that, in turn, can ionise other atoms. When they are

able to travel far enough to gain the energy to be absorbed by atoms, we see a light show (known as a discharge). The lower

the pressure, the further the electrons can travel before colliding with gas molecules and producing a discharge.

The light that is emitted is a result of the electrons around the gas atom becoming excited (increasing in energy) and

re-emitting the photon of light as they return to the ground state (the lowest energy they can have in an atom). Light will also

be produced when free electrons recombine with ions and the electrons return to the ground state, emitting photons. As every

element has a distinct set of energy levels, the colour of light seen will vary with the element with which the electron collides.

EquipmentIf you have the apparatus at school, you can carry out the experiment first hand. The patterns are hard to see unless the room

is very dark.

• induction coil • discharge tubes at different pressures

• connecting wires • DC power supply

Alternatively, you can use the simulations in Part B and make observations from them.

Risk assessmentaCtivity 8.1first-hand investigation

DC power supply

Figure 8.1.1 Induction coil and discharge tube

For more information on the in2 Physics series, visit www.pearsonplaces.com.au

viii

How to use this bookin2 Physics @ HSC is structured to enhance student learning and their enjoyment of learning. It contains many outstanding and unique features that will assist students succeed in Stage 6 Physics. These include:

• Moduleopeningpagesintroducearangeofcontextsforstudy, as well as an inquiry activity that provides immediate activities for exploration and discussion.

2 Motors and Generators

83

Figure 4.0.1 A generator produces electricity in each of these wind turbines.

82

The first recorded observations of the relationship between electricity and magnetism date back more than 400 years. Many unimagined discoveries followed, but progress never waits. Before we understood their nature, inventions utilising electricity and magnetism had changed our world forever.

Today our lives revolve around these forms of energy. The lights you use to read this book rely on them and the CD inside it would be nothing but a shiny coaster for your cup. We use magnetism to generate the electricity that drives industry, discovery and invention. Electricity and magnetism are a foundation for modern technology, deeply seated in the global economy, and our use impacts heavily on the environment.

The greatest challenge that faces future generations is the supply of energy. As fossil fuels dry up, electricity and magnetism will become even more important. New and improved technologies will be needed. Whether it’s a hybrid car, a wind turbine or a nuclear fusion power plant, they all rely on applications of electricity and magnetism.

Context

InQUIRY ACtIVItY

BUIld YoUR own eleCtRIC motoR

Many of the devices you use every day have electric motors. They spin your DVDs, wash your clothes and even help cook your food. Could you live without them, and how much do you know about how they work?

The essential ingredients for a motor are a power source, a magnetic field and things to connect these together in the right way. It’s not as hard as you think. All you need is a battery, a wood screw, a piece of wire and a cylindrical or spherical magnet. Put these things together as shown in Figure 4.0.2 and see if you can get your motor to spin. Be patient and keep trying. Then try the following activities.1 Test the effects of changing the voltage you use. You could add another

battery in series or try a battery with a higher voltage.2 Try changing the strength of the magnet by using a different magnet or

adding another. What does this affect?3 Try changing the length of the screw, how sharp its point is or the material

it is made from. Does it have to be made of iron?

Figure 4.0.2 A simple homopolar motor

11

204

Superconductivity

205

from ideaS toimplementation

crystal, constructive interference, destructive interference, path length,

diffraction grating, Bragg law, phonons, critical temperature,

type-I superconductors, type-II superconductors,

critical field strength, vortices, flux pinning, BCS theory, Cooper pair,

coherence length, energy gap, spin

Surprising discoveryJust as an improved understanding of the conducting properties of semiconductors led to the wide variety of electronic devices, research into the conductivity of metals produced quite a surprising discovery called superconductivity. This is the total disappearance of electrical resistance below a certain temperature, which has great potential applications ranging from energy transmission and storage to public transport. An understanding of this phenomenon required a detailed understanding of the crystal structure of conductors and the motion of electrons through them.

of interference of electromagnetic radiation, and examine how this was applied to crystals using X-rays. Then we will see how the BCS theory of superconductivity made use of the crystal structure of matter.

11.1 The crystal structure of matterA crystal is a three-dimensional regular arrangement of atoms. Figure 11.1.1 shows a sodium chloride crystal (ordinary salt also called rock salt when it comes as a large crystal). The crystal is made from simple cubes repeated many times, with sodium and chlorine atoms at the corners of the cubes. Crystals of other materials may have different regular arrangements of their atoms. There are 14 types of crystal arrangements that solids can have.

The regular arrangement of atoms in crystals was a hypothesis before Max Von Laue and his colleagues confirmed it by X-ray diffraction experiments. William and Lawrence Bragg took this method one step further by measuring the spacing between the atoms in the crystal. Let us first look at the phenomenon

Figure 11.1.1 Crystal structure of sodium chloride. The red spheres represent positive sodium ions, and the green spheres represent negative chlorine ions.

try thiS!Crystals in the kitChenLook at salt grains through a magnifying lens. Each grain is a single crystal that is made from the basic arrangement of sodium and chlorine atoms shown in Figure 11.1.1. Although the grains mostly look irregular due to breaking and chipping during the manufacturing process, occasionally you will see an untouched cubic or rectangular prism that reflects the underlying crystal lattice structure.

CheCkpoInT 11.1Explain what is meant by the crystal structure of matter.

11.2 Wave interferenceThe wave nature of light can be used to measure the size of very small spaces. Recall that two identical waves combine to produce a wave of greater amplitude when their crests overlap, as shown in Figure 11.2.1a (see in2 Physics @ Preliminary sections 6.4 and 7.4). The overlapping waves will cancel to produce a resulting wave of zero amplitude when the crest of one wave coincides with the trough of the other (Figure 11.2.1b). This addition and subtraction is called constructive and destructive interference respectively and is a property of all wave phenomena.

As an example, two identical circular water waves in a ripple tank overlap (see Figure 11.2.2). The regions of constructive and destructive interference radiate outwards along the lines as shown. Increasing the spacing between the sources causes the radiating lines to come closer together (Figure 11.2.2b).

Figure 11.2.1 Two identical waves (red, green) travelling in opposite directions can add (blue) (a) constructively or (b) destructively.

Figure 11.2.2 Interference of water waves for two sources that are (a) close together and (b) further apart

t = 0 s

t = 1 s

t = 3 s

t = 4 s

t = 5 s

t = 6 s

t = 7 s

t = 0 s

t = 1 s

t = 3 s

t = 4 s

t = 5 s

t = 6 s

t = 7 s

a

b

lines of constructiveinterference

lines of destructiveinterference

ba

The interference of identical waves from two sources can also be represented by outwardly radiating transverse waves (see Figure 11.2.3). The distance that a wave travels is known as its path length. Constructive interference occurs when the difference in the path length of the two waves is equal to 0, λ, 2λ, 3λ, 4λ or any other integer multiple of the wavelength λ. Destructive interference occurs when the two waves are half a wavelength out of step. This corresponds to a path length difference of λ/2, 3λ/2, 5λ/2 etc.

constructiveinterference


destructiveinterference

wavesin phase

Figure 11.2.3 Constructive and destructive interference between identical transverse waves from two sources

3

72

Seeing in a weird light: relativity

73

Space

PHYSICS FEATURE

TwISTIng SPACETImE ... And YoUR mInd

There are two more invariants in special relativity. Maxwell’s equations (and hence relativity)

requires that electrical charge is invariant in all frames. Another quantity invariant in all inertial frames is called the spacetime interval.

You may have heard of spacetime but not know what it is. One of Einstein’s mathematics lecturers Hermann Minkowski (1864–1909) showed that the equations of relativity and Maxwell’s equations become simplified if you assume that the three dimensions of space (x, y, z) and time t taken together form a four‑dimensional coordinate system called spacetime. Each location in spacetime is not a position, but rather an event—a position and a time.

Using a 4D version of Pythagoras’ theorem, Minkowski then defined a kind of 4D ‘distance’ between events called the spacetime interval s given by:

s 2 = (c × time period)2 – path length2 = c 2t 2 – ((∆x)2 + (∆y)2 + (∆z)2)

Observers in different frames don’t agree on the 3D path length between events, or the time period between events, but all observers in inertial frames agree on the spacetime interval s between events.

In general relativity, Einstein showed that gravity occurs because objects with mass or energy cause this 4D spacetime to become distorted. The paths of objects through this distorted 4D spacetime appear to our 3D eyes to follow the sort of astronomical trajectories you learned about in Chapter 2 ‘Explaining and exploring the solar system’. However, unlike Newton’s gravitation, general relativity is able to handle situations of high gravitational fields, such as Mercury’s precessing orbit around the Sun and black holes. General relativity also predicts another wave that doesn’t require a medium: the ripples in spacetime called ‘gravity waves’.

Figure 3.4.6 One of the four ultra-precise superconducting spherical gyroscopes on NASA’s Gravity Probe B, which orbited Earth in 2004/05 to measure two predictions of general relativity: the bending of spacetime by the Earth’s mass and the slight twisting of spacetime by the Earth’s rotation (frame-dragging)

1. The history of physics

Mass, energy and the world’s most famous equationThe kinetic energy formula K = 1

2mv 2 doesn’t apply at relativistic speeds,

even if you substitute relativistic mass mv into the formula. Classically, if you apply a net force to accelerate an object, the work done equals the increase in kinetic energy. An increase in speed means an increase in kinetic energy. But in relativity it also means an increase in relativistic mass, so relativistic mass and energy seem to be associated. Superficially, if you multiply relativistic mass by c 2 you get mv c

2, which has the same dimensions and units as energy. But let’s look more closely at it.

Solve problems and analyse information using:

E = mc2

l lv

cv = −0

2

21

tt

v

c

v =

−

0

2

21

mm

v

c

v =

−

0

2

21

How does this formula behave at low speeds (when v 2/c 2 is small)?

m cm c

v

c

m cv

cv

2 02

2

2

02

2

2

12

1

1=

−

= −

−

Using a well-known approximation formula that you might learn at university, (1 – x )n ≈ 1 – nx for small x:

m cv

c0

22

2

12

1−

−

≈ m cv

c0

22

21

12

+ ×

= m0c 2 + 1

2m0v 2

Rearrange: mvc 2 – m0c 2 = (mv – m0)c 2 ≈ 1

2m0v 2

In other words, at low speeds, the gain in relativistic mass (mv – m0)multiplied by c 2 equals the kinetic energy—a tantalising hint that at low speed mass and energy are equivalent. It can also be shown to be true at all speeds, using more sophisticated mathematics. In general, mass and energy are equivalent in relativity and c 2 is the conversion factor between the energy unit (joules) and the mass unit (kg). In other words:

E = mc 2

where m is any kind of mass. In relativity, mass and energy are regarded as the same thing, apart from the change of units. Sometimes the term mass-energy is used for both. m0 c 2 is called the rest energy, so even a stationary object contains energy due to its rest mass. Relativistic kinetic energy therefore:

m c m cm c

v

c

m cv2

02 0

2

2

2

02

1

− =

−

−

Whenever energy increases, so does mass. Any release of energy is accompanied by a decrease in mass. A book sitting on the top shelf has a slightly higher mass than one on the bottom shelf because of the difference in gravitational potential energy. An object’s mass increases slightly when it is hot because the kinetic energy of the vibrating atoms is higher.

Because c 2 is such a large number, a very tiny mass is equivalent to a large amount of energy. In the early days of nuclear physics, E = mc 2 revealed the enormous energy locked up inside an atom’s nucleus by the strong nuclear force that holds the protons and neutrons together. It was this that alerted nuclear physicists just before World War II to the possibility of a nuclear bomb. The energy released by the nuclear bomb dropped on Hiroshima at the end of that war (smallish by modern standards) resulted from a reduction in relativistic mass of about 0.7 g (slightly less than the mass of a standard wire paperclip).

Worked exampleqUESTIonWhen free protons and neutrons become bound together to form a nucleus, the reduction in nuclear potential energy (binding energy) is released, normally in the form of gamma rays. Relativity says this loss in energy is reflected in a decrease in mass of the resulting atom.

Discuss the implications of mass increase, time dilation and length contraction for space travel.

evil tWinS

The most extreme mass–energy conversion involves antimatter.

For every kind of matter particle there is an equivalent antimatter particle, an ‘evil twin’, bearing properties (such as charge) of opposite sign. Particles and their antiparticles have the same rest mass. When a particle meets its antiparticle, they mutually annihilate—all their opposing properties cancel, leaving only their mass‑energy, which is usually released in the form of two gamma‑ray photons. Matter–antimatter annihilation has been suggested (speculatively) as a possible propellant for powering future interstellar spacecraft.

PRACTICAL EXPERIENCES

350

19 Imaging with gamma rays

351

Chapter summary mEdICALPhySICS

Activity 19.2: HeAltHy or diseAsed?Typical images of healthy bone and cancerous bone are shown. The tumours show up as hot-spots. Use the template in the activity manual to research and compare images of healthy and diseased parts of the body.

Discussion questions1 Examine Figure 19.4.2. There is a hot-spot that is not cancerous near the

left elbow. Explain.2 In the normal scan (Figure 19.6.2a), the lower pelvis has a region of high

intensity. Why is this? (Hint: It may be soft tissue, not bone. Looking at Figure 19.6.2b might help you with this question.)

3 State the differences that can be observed by comparing an image of a healthy part of the body with that of a diseased part of the body.

Gather and process secondary information to compare a scanned image of at least one healthy body part or organ with a scanned image of its diseased counterpart.

Review questions

ChAPTER 19This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 19.1: Bone scAnsA bone scan is performed to obtain a functional image of the bones and so can be used to detect abnormal metabolism in the bones, which may be an indication of cancer or other abnormality. Because cancer mostly involves a higher than normal

Perform an investigation to compare a bone scan with an X-ray image.

Figure 19.6.1 Comparison of an X-ray and bone scan of a hand

Figure 19.6.2 Bones scans of (a) a healthy person and (b) a patient with a tumour in the skeleton

• Thenumberofprotonsinanucleusisgivenbytheatomic number, while the total number of nucleons is given by the mass number.

• Atomsofthesameelementwithdifferentnumbersofneutrons are called isotopes of that element.

• Manyelementshavenaturallyoccurringunstableradioisotopes.

• Inalphadecayanunstablenucleusdecaysbyemittingan alpha particle (α-particle).

• Inbetadecay,aneutronchangesintoaprotonand a high-energy electron that is emitted as a beta particle (β-particle).

• Inpositrondecay,apositron—theantiparticleoftheelectron—isemitted.

• Whenapositronandanelectroncollide,theirtotalmass is converted into energy in the form of two gamma-ray photons.

• Ingammadecayagammaray (g) is emitted from a radioactive isotope.

• Thetimeittakesforhalfthemassofaradioactiveparent isotope to decay into its daughter nuclei is the half-life of the isotope.

• Artificialradioisotopesareproducedintwomainways:in a nuclear reactor or in a cyclotron.

• Agammacameradetectsgammaraysemittedby a radiopharmaceutical in the patient’s body.

• PETimagingusespositron-emittingradiopharmaceuticals to obtain images using gamma rays emitted from electron–positron annihilation.

PHysicAlly sPeAkingBelow is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

Radioactive decay

Radiation Radioisotope Nucleon

Neutron Proton Isotope Alpha decay

Beta decayGamma decay

Antimatter PET

Half-life Bone scanPositron decay

Scintillator

reviewing 1 Recall how the bone scan produced by a radioisotope

compares with that from a conventional X-ray.

2 Analyse the relationship between the half-life of a radiopharmaceutical and its potential use in the human body.

3 Explain how it is possible to emit an electron from the nucleus when the electron is not a nucleon.

4 Assess the statement that ‘Positrons are radioactive particles produced when a proton decays’.

5 Discuss the impact that the production and use of radioisotopes has on society.

6 Describe how isotopes such as Tc-99m and F-18 can be used to target specific organs to be imaged.

7 Use the data in Table 19.6.1 to answer the questions:a Which radioactive isotope would most likely be

used in a bone scan? Justify your choice.b Propose two reasons why cesium-137 would not

be a suitable isotope to use in medical imaging.

Table 19.6.1 Properties of some radioisotopes

Radioactive souRce Radiation emitted Half-lifeC-11 β+, g 20.30 minutes

Tc-99m g 6.02 hours

TI-201 g 3.05 days

I-131 β, g 8.04 days

Cs-137 α 30.17 years

U-238 α 4.47 × 109 years

rate of cell division (thus producing a tumour), chemicals involved in metabolic processes in bone tend to accumulate in higher concentrations in cancerous tissue. This produces areas of concentration of gamma emission, indicating a tumour.

Compare the data obtained from the image of a bone scan with that provided by an X-ray image.

Discussion questions1 Identify the best part of the body for each of these

diagnostic tools to image.2 Compare and contrast the two images in terms of

the information they provide.

a b

• Chapteropeningslistthekeywordsofeachchapterandintroduce the chapter topic in a concise and engaging way.

• Keyideasareclearlyhighlightedwitha and Syllabus flags indicate where domain dot points appear in the student book. The flags are placed as closely as possible to where the relevant content is covered. Flags may be repeated if the dot point has multiple parts, is complex or where students are required to solve problems.

• Eachchapterconcludeswith:– a chapter summary– review questions, including literacy-based questions

(Physically Speaking), chapter review questions (Reviewing) and physics problems (Solving Problems). Syllabus verbs are clearly highlighted as and where appropriate

– Physics Focus—a unique feature that places key chapter concepts in the context of one or more prescribed focus areas.

• Chaptersaredividedintoshort,accessiblesections— the text itself is presented in short, easy-to-understand chunks of information. Each section concludes with a Checkpoint—a set of review questions to check understanding of key content and concepts.

ix

How to use this book

• Modulereviewsprovideafullrangeofexam-stylequestions, including multiple-choice, short-response and extended-response questions.

224 225

from ideas toimplementation

3 The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 25 marks. It should take you approximately 45 minutes to complete this review.

multiple choice(1 mark each) 1 Predict the direction of the electron in Figure 11.13.1

as it enters the magnetic field.A Straight upB LeftC RightD Down

2 The diagrams in Figure 11.13.2 represent semiconductors, conductors and insulators. The diagrams show the conduction and valence bands, and the energy gaps. Which answer correctly labels each of the diagrams?

I II IIIA Conductor Insulator Semiconductor

B Insulator Conductor Semiconductor

C Insulator Semiconductor Conductor

D Semiconductor Conductor Insulator

3 The graph in Figure 11.13.3 shows how the resistance of a material varies with temperature. Identify each of the parts labelled on the graph.

I II IIIA Critical

temperatureSuperconductor material

Normal material

B Superconductor material

Critical temperature

Normal material

C Critical temperature

Normal material Superconductor material

D Normal material Superconductor material


Figure 11.13.1 An electron in a magnetic field

Figure 11.13.2 Energy bands

Figure 11.13.3 Resistance varies with temperature

–

I II III

Temperature (K)

Res

ista

nce

(Ω)

I

IIIII

4 Experimental data from black body radiation during Planck’s time showed that predicted radiation levels were not achieved in reality. Planck best described this anomaly by saying that:A classical physics was wrong.B radiation that is emitted and absorbed is

quantised.C he had no explanation for it.D quantum mechanics needed to be developed.

5 Figure 11.13.4 shows a cathode ray tube that has been evacuated. Which answer correctly names each of the labelled features?

I II IIIA Striations Cathode Anode

B Faraday’s dark space

Striations Cathode

C Crooke’s dark space

Anode Faraday’s dark space

D Cathode Faraday’s dark space

Striations

extended response 6 Explain, with reference to atomic models, why

cathode rays can travel through metals. (2 marks)

7 Outline how the cathode ray tube in a TV works in order to produce the viewing picture. (2 marks)

8 Give reasons why CRT TVs use magnetic coils and CROs use electric plates in order to deflect the beams, given that both methods work. (2 marks).

9 In your studies you were required to gather information to describe how the photoelectric effect is used in photocells.a Explain how you determined which material was

relevant and reliable.b Outline how the photoelectric effect is used in

photocells. (3 marks)

10 Justify the introduction of semiconductors to replace thermionic devices. (4 marks)

11 Magnetic levitation trains are used in Germany and Japan. The trains in Germany use conventional electromagnets, whereas the one in Japan uses superconductors. Compare and contrast the two systems. (3 marks)

12 a Determine the frequency of red light, which has a wavelength λ = 660 nm. (Speed of light c = 3.00 × 108 m s–1)b Calculate the energy of a photon that is emitted

with this wavelength. (Planck’s constant h = 6.63 × 10–34 J s) (4 marks)

Figure 11.13.4 An evacuated cathode ray tube

IIIII I

48

MO

DU

LE

motors andgenerators2

49

Chapter 6motors: magnetic fields make the world go around

Risk assessment

Method1 Cut a length of cotton-covered wire so that the wire is long enough

to wrap around the exterior of a matchbox three times (as shown in Figure 6.2.2).

2 Leave a straight piece (approx. 10 cm long) hanging out and then wind the remainder of the wire around the box 2½ times. Leave another straight piece the same length as at the start, on the opposite side.

3 Wrap the straight pieces around the loops so that they tie both ends.

4 Fan out the loops so that you get equally spaced loops and that it looks like a bird cage (see Figure 6.2.3).

5 Push out the middle of the paper clip as shown and Blu-Tack to the bench.

6 Slip the straight pieces of wire through the paper clip supports. Unwrap the cotton from these parts.

7 Connect an AC power supply to the paper clips.

8 Place two magnets so that a north pole and a south pole face on opposing sides of the cage.

9 Turn on. You may need to give the cage a tap to get it spinning.

Results1 Record your observations of the motor.

2 How did adding more magnets affect how the motor ran?

3 When the current is increased, what changes occurred?

Motors and torque

Solve problems and analyse information about simple motors using:

τ = nBIA cos θ

Physics skillsThe skills outcomes to be practised in this activity include:

12.4 process information14.1 analyse information

The complete statement of these skills outcomes can be found in the syllabus grid on pages vii–viii.

Aim

Hypothesis

TheoryThe motor effect means that a current-carrying wire experiences a force when placed in a magnetic field. This is the basis for the workings of a motor.

For a motor to work as needed, the motion resulting from the motor effect needs to be circular and the force needs to be adjusted so the direction of rotation does not change.

QuestionFigure 6.2.1 shows the simplified workings of a motor that you will be making. Label all the parts of the motor.

Equipment• insulatedwirefromwhichinsulationcanberemovedeasily • Blu-Tack• magnets • connectingwireswithalligatorclips• magneticfieldsensoranddatalogger(ifavailable) • powersupply• paperclips

matchbox

wire

loop wirethrough

a

b

alligator clipwires

paper clip

cage fanned out

power source

Figure 6.2.2 Equipment set-up 1

Figure 6.2.3 Equipment set-up 2

First-hand investigationaCtIVItY 6.2

A:

C:

D:

B:

N S

Figure 6.2.1 Simplified motor

Other features• PhysicsPhilespresentshort,interestingitemsto

support or extend the text.

• PhysicsforFun—TryThis!activitiesareshort,hands-on activities to be done quickly, designed to provoke discussion.

• PhysicsFeaturesareakeyfeatureastheyhighlightcontextual material, case studies or prescribed focus areas of the syllabus.

• Acompleteglossaryofallthekeywordsisincludedatthe end of the student book.

• Thefinaltwochaptersprovideessentialreferencematerial: ‘Skills stage 2’ and ‘Revisiting the BOS key terms’.

• Inallquestionsandactivities,exceptmodulereviewquestions, the BOS key terms are highlighted.

in2 Physics @ HSC Student CDThis is included with the student book and contains:

• anelectronicversionofthestudentbook

• interactivemodulesdemonstratingkeyconcepts

Practical experiencesThe accompanying activity manual covers all of the mandatory practical experiences outlined in the syllabus.

in2 Physics @ HSC Activity Manual is a write-in workbook that outlines a clear, foolproof approach to success in all the required practical experiences.

Within the student book, there are clear cross-references to the activity manual: Practical Experiences icons refer to the activity number and page in the activity manual. In each chapter, a summary of possible investigations is provided as a starting point to get students thinking. These include the aim, a list of equipment and discussion questions. Activity 10.2


Activity Manual, Page 94

• thecompanionwebsiteonCD

• alinktothelivecompanionwebsite(Internetaccessrequired) to provide access to the latest information and web links related to the student book.

The complete in2 Physics @ HSC package Remember the other components of the complete package:

• in2 Physics @ HSC companion website at Pearson Places

• in2 Physics @ HSC Teacher Resource.

x

Stage 6 Physics syllabus grid

Prescribed focus areas1. The history of physics H1. evaluates how major advances in scientific understanding and

technology have changed the direction or nature of scientific thinkingFeature: pp. 12, 29, 72

Focus: pp. 25, 246, 299

2. The nature and practice of physics H2. analyses the ways in which models, theories and laws in physics have been tested and validated

Focus: p. 79

3. Applications and uses of physics H3. assesses the impact of particular advances in physics on the development of technologies

Feature: pp. 12, 29, 307, 334, 346

Focus: pp. 57, 79, 129, 173, 223, 246, 259, 278

4. Implications for society and the Environment

H4. assesses the impacts of applications of physics on society and the environment

Feature: pp. 29, 307, 344

Focus: pp. 113, 173, 353

5. Current issues, research and developments in physics

H5. identifies possible future directions of physics research Feature: pp. 391, 410

Focus: pp. 79, 113, 173, 223, 353, 386

Module 1 Space

1. The Earth has a gravitational field that exerts a force on objects both on it and around it

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdefine weight as the force on an object due to a gravitational field

13 perform an investigation and gather information to determine a value for acceleration due to gravity using pendulum motion or computer-assisted technology and identify reason(s) for possible variations from the value 9.8 m s–2

Act. 1.2

explain that a change in gravitational potential energy is related to work done

16 gather secondary information to predict the value of acceleration due to gravity on other planets

Act. 1.3

define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field:

m mr= 1 2

PE G

16 analyse information using the expression:

F = mg

to determine the weight force for a body on Earth and for the same body on other planets

Act. 1.3

2. Many factors have to be taken into account to achieve a successful rocket launch, maintain a stable orbit and return to Earth

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components

5 solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using:

vx2 = ux

2

v = u + at

vy2 = uy

2 + 2ay ∆y

∆x = ux t

∆y = uyt + 12ay t

2

7, 9, 23, 24

describe Galileo’s analysis of projectile motion

5 perform a first-hand investigation, gather information and analyse data to calculate initial and final velocity, maximum height reached, range and time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis

Act. 1.1

explain the concept of escape velocity in terms of the:– gravitational constant– mass and radius of the planet

18 identify data sources, gather, analyse and present information on the contribution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun

29 Act. 2.1

xi


outline Newton’s concept of escape velocity

18

identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch

31

discuss the effect of the Earth‘s orbital motion and its rotational motion on the launch of a rocket

34

analyse the changing acceleration of a rocket during launch in terms of the:– Law of Conservation of Momentum– forces experienced by astronauts

30, 33

analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth

25, 32, 34, 37, 54, 55

solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using:

F mv 2

r=

37, 54, 55 Act. 2.2

compare qualitatively low Earth and geo-stationary orbits

43

define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods

36, 40, 56

solve problems and analyse information using:r

T

GM3

2 24=

π

39, 43, 56

account for the orbital decay of satellites in low Earth orbit

46

discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface

47

identify that there is an optimum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle

47

3. The solar system is held together by gravity

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe a gravitational field in the region surrounding a massive object in terms of its effects on other masses in it

13 present information and use available evidence to discuss the factors affecting the strength of the gravitational force

Act. 1.3

define Newton’s Law of Universal Gravitation:

m m

d= 1 2

2F G

11 solve problems and analyse information using:m m

d= 1 2

2F G

23, 24, 25, 37, 54, 55

discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites

35, 38

identify that a slingshot effect can be provided by planets for space probes

44

xii


4. Current and emerging understanding about time and space has been dependent upon earlier models of the transmission of light

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEoutline the features of the aether model for the transmission of light

61

describe and evaluate the Michelson-Morley attempt to measure the relative velocity of the Earth through the aether

62 gather and process information to interpret the results of the Michelson-Morley experiment

62 Act. 3.2

discuss the role of the Michelson-Morley experiments in making determinations about competing theories

62

outline the nature of inertial frames of reference

58 perform an investigation to help distinguish between non-inertial and inertial frames of reference

60 Act. 3.1

discuss the principle of relativity 58 analyse and interpret some of Einstein’s thought experiments involving mirrors and trains and discuss the relationship between thought and reality

66

describe the significance of Einstein’s assumption of the constancy of the speed of light

65 analyse information to discuss the relationship between theory and the evidence supporting it, using Einstein’s predictions based on relativity that were made many years before evidence was available to support it

78

identify that if c is constant then space and time become relative

65

discuss the concept that length standards are defined in terms of time in contrast to the original metre standard

79

explain qualitatively and quantitatively the consequence of special relativity in relation to:– the relativity of simultaneity– the equivalence between mass and

energy– length contraction– time dilation– mass dilation

64, 69 solve problems and analyse information using:

E = mc 2

l lv

cv = −0

2

21

tv

cv

t0

= −2

21

mv

cv

m 0

= −2

21

66, 69, 72, 77, 78

discuss the implications of mass increase, time dilation and length contraction for space travel

70, 73

Module 2 Motors and Generators

1. Motors use the effect of forces on current-carrying conductors in magnetic fields

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdiscuss the effect on the magnitude of the force on a current-carrying conductor of variations in:– the strength of the magnetic field in

which it is located– the magnitude of the current in the

conductor– the length of the conductor in the

external magnetic field– the angle between the direction

of the external magnetic field and the direction of the length of the conductor

92 perform a first-hand investigation to demonstrate the motor effect Act. 4.1

xiii


describe qualitatively and quantitatively the force between long parallel current-carrying conductors:

Fl

kI Id

= 1 2

94 solve problems using:Fl

kI Id

= 1 2

94

define torque as the turning moment of a force using:t = Fd

115 solve problems and analyse information about the force on current-carrying conductors in magnetic fields using: F = BIl sin θ

92 Act. 4.1

identify that the motor effect is due to the force acting on a current-carrying conductor in a magnetic field

90, 116

solve problems and analyse information about simple motors using: t = nBIA cos θ

117 Act. 6.2

describe the forces experienced by a current-carrying loop in a magnetic field and describe the net result of the forces

117 identify data sources, gather and process information to qualitatively describe the application of the motor effect in:– the galvanometer– the loudspeaker

91, 119 Act. 6.1

describe the main features of a DC electric motor and the role of each feature

115

identify that the required magnetic fields in DC motors can be produced either by current-carrying coils or permanent magnets

115

2. The relative motion between a conductor and magnetic field is used to generate an electrical voltage

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEoutline Michael Faraday’s discovery of the generation of an electric current by a moving magnet

100 perform an investigation to model the generation of an electric current by moving a magnet in a coil or a coil near a magnet

101 Act. 5.1

define magnetic field strength B as magnetic flux density

101 plan, choose equipment or resources for, and perform a first-hand investigation to predict and verify the effect on a generated electric current when:– the distance between the coil and magnet is varied– the strength of the magnet is varied– the relative motion between the coil and the magnet is varied

Act. 5.1

describe the concept of magnetic flux in terms of magnetic flux density and surface area

101 gather, analyse and present information to explain how induction is used in cooktops in electric ranges

108 Act. 5.2

describe generated potential difference as the rate of change of magnetic flux through a circuit

103 gather secondary information to identify how eddy currents have been utilised in electromagnetic braking

Act. 5.2 113

account for Lenz’s Law in terms of conservation of energy and relate it to the production of back emf in motors

105, 120

explain that, in electric motors, back emf opposes the supply emf

120

explain the production of eddy currents in terms of Lenz’s Law

106

3. Generators are used to provide large scale power production

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe the main components of a generator

131 plan, choose equipment or resources for, and perform a first-hand investigation to demonstrate the production of an alternating current

Act. 5.1

compare the structure and function of a generator to an electric motor

135 gather secondary information to discuss advantages/disadvantages of AC and DC generators and relate these to their use

135 Act. 7.1

describe the differences between AC and DC generators

135 analyse secondary information on the competition between Westinghouse and Edison to supply electricity to cities

141 Act. 7.2

discuss the energy losses that occur as energy is fed through transmission lines from the generator to the consumer

144 gather and analyse information to identify how transmission lines are:– insulated from supporting structures– protected from lightning strikes

146 Act. 7.3

assess the effects of the development of AC generators on society and the environment

147

xiv


4. Transformers allow generated voltage to be either increased or decreased before it is used

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe the purpose of transformers in electrical circuits

136 perform an investigation to model the structure of a transformer to demonstrate how secondary voltage is produced

Act. 7.3

compare step-up and step-down transformers

137 solve problems and analyse information about transformers using:V

V

n

np

s

p

s

=

137 Act. 7.3

identify the relationship between the ratio of the number of turns in the primary and secondary coils and the ratio of primary to secondary voltage

137 gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome

139 Act. 7.3

explain why voltage transformations are related to conservation of energy

139 gather and analyse secondary information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use

145 Act. 7.3

explain the role of transformers in electricity substations

142

discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer

136, 144

discuss the impact of the development of transformers on society

147

5. Motors are used in industries and the home usually to convert electrical energy into more useful forms of energy

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe the main features of an AC electric motor

124 perform an investigation to demonstrate the principle of an AC induction motor Act. 6.3

gather, process and analyse information to identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry

124, 153 Act. 7.3

Module 3 From Ideas to Implementation

1. Increased understandings of cathode rays led to the development of television

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEexplain why the apparent inconsistent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves

157 perform an investigation and gather first-hand information to observe the occurrence of different striation patterns for different pressures in discharge tubes

Act. 8.1

explain that cathode ray tubes allowed the manipulation of a stream of charged particles

157 perform an investigation to demonstrate and identify properties of cathode rays using discharge tubes:– containing a Maltese cross– containing electric plates– with a fluorescent display screen– containing a glass wheel

analyse the information gathered to determine the sign of the charge on cathode rays

Act. 8.2

Act. 8.2

identify that moving charged particles in a magnetic field experience a force

164 solve problem and analyse information using:F = qvB sin θF = qEand

EVd

=

162, 164

identify that charged plates produce an electric field

161

xv


describe quantitatively the force acting on a charge moving through a magnetic field: F = qvB sin θ

164

discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates

160

describe quantitatively the electric field due to oppositely charged parallel plates

161

outline Thomson’s experiment to measure the charge/mass ratio of an electron

165

outline the role of:– electrodes in the electron gun– the deflection plates or coils– the fluorescent screen– in the cathode ray tube of

conventional TV displays and oscilloscopes

167

2. The reconceptualisation of the model of light led to an understanding of the photoelectric effect and black body radiation

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but failed to investigate

182 perform an investigation to demonstrate the production and reception of radio waves

Act. 9.1

outline qualitatively Hertz’s experiments in measuring the speed of radio waves and how they relate to light waves

175 identify data sources, gather, process and analyse information and use available evidence to assess Einstein’s contribution to quantum theory and its relation to black body radiation

Act. 9.2

identify Planck’s hypothesis that radiation emitted and absorbed by the walls of a black body cavity is quantised

179 identify data sources, gather, process and present information to summarise the use of the photoelectric effect in photocells

184 Act. 9.3

identify Einstein’s contribution to quantum theory and its relation to black body radiation

179 solve problems and analyse information using:E = hfandc = f λ

181 Act. 9.3

explain the particle model of light in terms of photons with particular energy and frequency

179 process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces

Act. 9.4

identify the relationships between photon energy, frequency, speed of light and wavelength:E = hfandc = f λ

179

xvi


3. limitations of past technologies and increased research into the structure of the atom resulted in the invention of transistors

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEidentify that some electrons in solids are shared between atoms and move freely

189 perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor

Act. 10.1

describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance

189 gather, process and present secondary information to discuss how shortcomings in available communication technology lead to an increased knowledge of the properties of materials with particular reference to the invention of the transistor

Act. 10.2

identify absences of electrons in a nearly full band as holes, and recognise that both electrons and holes help to carry current

191 identify data sources, gather, process, analyse information and use available evidence to assess the impact of the invention of transistors on society with particular reference to their use in microchips and microprocessors

Act. 10.2

compare qualitatively the relative number of free electrons that can drift from atom to atom in conductors, semiconductors and insulators

190 identify data sources, gather, process and present information to summarise the effect of light on semiconductors in solar cells

Act. 10.3

identify that the use of germanium in early transistors is related to lack of ability to produce other materials of suitable purity

199

describe how ‘doping’ a semiconductor can change its electrical properties

193

identify differences in p and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes

193

describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices

199

4. Investigations into the electrical properties of particular metals at different temperatures led to the identification of superconductivity and the exploration of possible applications

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEoutline the methods used by the Braggs to determine crystal structure

208 process information to identify some of the metals, metal alloys and compounds that have been identified as exhibiting the property of superconductivity and their critical temperatures

211

identify that metals possess a crystal lattice structure

209 perform an investigation to demonstrate magnetic levitation Act. 11.1

describe conduction in metals as a free movement of electrons unimpeded by the lattice

209 analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting

Act. 11.1

identify that resistance in metals is increased by the presence of impurities and scattering of electrons by lattice vibrations

209 gather and process information to describe how superconductors and the effects of magnetic fields have been applied to develop a maglev train

Act. 11.1

describe the occurrence in superconductors below their critical temperature of a population of electron pairs unaffected by electrical resistance

215 process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors and transmission of electricity through power grids

219 Act. 11.1

discuss the BCS theory 215

discuss the advantages of using superconductors and identify limitations to their use

217

xvii


Module 4 From Quanta to Quarks

1. Problems with the Rutherford model of the atom led to the search for a model that would better explain the observed phenomena

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdiscuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits

230, 244

perform a first-hand investigation to observe the visible components of the hydrogen spectrum

Act. 12.1

analyse the significance of the hydrogen spectrum in the development of Bohr’s model of the atom

236 process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series

236 Act. 12.1

define Bohr’s postulates 236 solve problems and analyse information using:

1 112λ

R

nf

2ni

= −

233, 245 Act. 12.1

discuss Planck’s contribution to the concept of quantised energy

231 analyse secondary information to identify the difficulties with the Rutherford-Bohr model, including its inability to completely explain:– the spectra of larger atoms– the relative intensity of spectral lines– the existence of hyperfine spectral lines– the Zeeman effect

Act. 12.2

describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum:

1 112λ

R

nf

2ni

= −

237, 244

discuss the limitations of the Bohr model of the hydrogen atom

239

2. The limitations of classical physics gave birth to quantum physics

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties

250, 259

solve problems and analyse information using:

λh

mv=

249, 258

define diffraction and identify that interference occurs between waves that have been diffracted

250, 257

gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory

255 Act. 13.1

describe the confirmation of de Broglie’s proposal by Davisson and Germer

251, 257

explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis

253, 257

xviii


3. The work of Chadwick and Fermi in producing artificial transmutations led to practical applications of nuclear physics

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdefine the components of the nucleus (protons and neutrons) as nucleons and contrast their properties

261, 278

perform a first-hand investigation or gather secondary information to observe radiation emitted from a nucleus using Wilson Cloud Chamber or similar detection device

Act. 14.1

discuss the importance of conservation laws to Chadwick’s discovery of the neutron

261, 275

solve problems and analyse information to calculate the mass defect and energy released in natural transmutation and fission reactions

267, 277

define the term ‘transmutation’ 263

describe nuclear transmutations due to natural radioactivity

263

describe Fermi’s initial experimental observation of nuclear fission

269

discuss Pauli’s suggestion of the existence of neutrino and relate it to the need to account for the energy distribution of electrons emitted in β-decay

266, 276

evaluate the relative contributions of electrostatic and gravitational forces between nucleons

261

account for the need for the strong nuclear force and describe its properties

262

explain the concept of a mass defect using Einstein’s equivalence between mass and energy

267

describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942

270, 275

compare requirements for controlled and uncontrolled nuclear chain reactions

271, 275

4. An understanding of the nucleus has led to large science projects and many applications

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEexplain the basic principles of a fission reactor

280, 298

gather, process and analyse information to assess the significance of the Manhattan Project to society

280 Act. 15.1

describe some medical and industrial applications of radioisotopes

283, 298

identify data sources, and gather, process, and analyse information to describe the use of:– a named isotope in medicine– a named isotope in agriculture– a named isotope in engineering

284, Act. 15.2

describe how neutron scattering is used as a probe by referring to the properties of neutrons

272, 298

identify ways by which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter

286, 299

discuss the key features and components of the standard model of matter, including quarks and leptons

292, 298

xix


Module 5 Medical Physics

1. The properties of ultrasound waves can be used as diagnostic tools

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEidentify the differences between ultrasound and sound in normal hearing range

305 solve problems and analyse information to calculate the acoustic impedance of a range of materials, including bone, muscle, soft tissue, fat, blood and air and explain the types of tissues that ultrasound can be used to examine

312

describe the piezoelectric effect and the effect of using an alternating potential difference with a piezoelectric crystal

308 gather secondary information to observe at least two ultrasound images of body organs

Act. 16.1

define acoustic impedance:Z = ρυand identify that different materials have different acoustic impedances

310, 311

identify data sources and gather information to observe the flow of blood through the heart from a Doppler ultrasound video image

Act. 16.2

describe how the principles of acoustic impedance and reflection and refraction are applied to ultrasound

311 identify data sources, gather, process and analyse information to describe how ultrasound is used to measure bone density

315 Act. 16.3

define the ratio of reflected to initial intensity as:

II

Z Z

Z Zr

o

=− +

2 1

2 1

2

2

310 solve problems and analyse information using:Z = ρυand

II

Z Z

Z Zr

o

=− +

2 1

2 1

2

2

310, 311

identify that the greater the difference in acoustic impedance between two materials, the greater is the reflected proportion of the incident pulse

310

describe situations in which A scans, B scans and sector scans would be used and the reasons for the use of each

312

describe the Doppler effect in sound waves and how it is used in ultrasonics to obtain flow characteristics of blood moving through the heart

315

outline some cardiac problems that can be detected through the use of the Doppler effect

316

2. The physical properties of electromagnetic radiation can be used as diagnostic tools

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe how X-rays are currently produced

321 gather information to observe at least one image of a fracture on an X-ray film and X-ray images of other body parts

Act. 17.1

compare the differences between ‘soft’ and ‘hard’ X-rays

322 gather secondary information to observe a CAT scan image and compare the information provided by CAT scans to that provided by an X-ray image for the same body part

Act. 17.1

explain how a computed axial tomography (CAT) scan is produced

326 perform a first-hand investigation to demonstrate the transfer of light by optical fibres

Act. 18.1

describe circumstances where a CAT scan would be a superior diagnostic tool compared to either X-rays or ultrasound

329 gather secondary information to observe internal organs from images produced by an endoscope

Act. 18.1

explain how an endoscope works in relation to total internal reflection

334

discuss differences between the role of coherent and incoherent bundles of fibres in an endoscope

336

explain how an endoscope is used in:– observing internal organs– obtaining tissue samples of internal

organs for further testing

337

xx


3. Radioactivity can be used as a diagnostic tool

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEoutline properties of radioactive isotopes and their half-lives that are used to obtain scans of organs

340, 343, 344

perform an investigation to compare an image of bone scan with an X-ray image Act. 19.1

describe how radioactive isotopes may be metabolised by the body to bind or accumulate in the target organ

344 gather and process secondary information to compare a scanned image of at least one healthy body part or organ with a scanned image of its diseased counterpart

Act. 19.2

identify that during decay of specific radioactive nuclei positrons are given off

342

discuss the interaction of electrons and positrons resulting in the production of gamma rays

342

describe how the positron emission tomography (PET) technique is used for diagnosis

349

4. The magnetic field produced by nuclear particles can be used as a diagnostic tool

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEidentify that the nuclei of certain atoms and molecules behave as small magnets

355 perform an investigation to observe images from magnetic resonance image (MRI) scans, including a comparison of healthy and damaged tissue

Act. 20.1

identify that protons and neutrons in the nucleus have properties of spin and describe how net spin is obtained

354 identify data sources, gather, process and present information using available evidence to explain why MRI scans can be used to:– detect cancerous tissues– identify areas of high blood flow– distinguish between grey and white matter in the brain

Act. 20.1

explain that the behaviour of nuclei with a net spin, particularly hydrogen, is related to the magnetic field they produce

355 gather and process secondary information to identify the function of the electromagnet, radio frequency oscillator, radio receiver and computer in the MRI equipment

Act. 20.1

describe the changes that occur in the orientation of the magnetic axis of nuclei before and after the application of a strong magnetic field

355 identify data sources, gather and process information to compare the advantages and disadvantages of X-rays, CAT scans, PET scans and MRI scans

Act. 20.2

define precessing and relate the frequency of the precessing to the composition of the nuclei and the strength of the applied external magnetic field

356 gather, analyse information and use available evidence to assess the impact of medical applications of physics on society

Act. 20.3

discuss the effect of subjecting precessing nuclei to pulses of radio waves

357

explain that the amplitude of the signal given out when precessing nuclei relax is related to the number of nuclei present

359

explain that large differences would occur in the relaxation time between tissue containing hydrogen bound water molecules and tissues containing other molecules

360

xxi


Module 6 Astrophysics

1. Our understanding of celestial objects depends upon observations made from Earth or from space near the Earth

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdiscuss Galileo’s use of the telescope to identify features of the Moon

371 Act. 21.1

identify data sources, plan, choose equipment or resources for, and perform an investigation to demonstrate why it is desirable for telescopes to have a large diameter objective lens or mirror in terms of both sensitivity and resolution

377 Act. 21.2

discuss why some wavebands can be more easily detected from space

373

define the terms ‘resolution’ and ‘sensitivity’ of telescopes

375

discuss the problems associated with ground-based astronomy in terms of resolution and absorption of radiation and atmospheric distortion

373, 378

outline methods by which the resolution and/or sensitivity of ground-based systems can be improved, including:– adaptive optics – interferometry– active optics

378, 380

2. Careful measurement of a celestial object’s position in the sky (astrometry) may be used to determine its distance

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdefine the terms parallax, parsec, light-year

388 solve problems and analyse information to calculate the distance to a star given its trigonometric parallax using:

d1p

=

Act. 22.1

explain how trigonometric parallax can be used to determine the distance to stars

388 gather and process information to determine the relative limits to trigonometric parallax distance determinations using recent ground-based and space-based telescopes

Act. 22.2

discuss the limitations of trigonometric parallax measurements

389

3. Spectroscopy is a vital tool for astronomers and provides a wealth of information

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEaccount for the production of emission and absorption spectra and compare these with a continuous black body spectrum

390 perform a first-hand investigation to examine a variety of spectra produced by discharge tubes, reflected sunlight, or incandescent filaments

Act. 22.3

describe the technology needed to measure astronomical spectra

390 analyse information to predict the surface temperature of a star from its intensity/wavelength graph

Act. 22.4

identify the general types of spectra produced by stars, emission nebulae, galaxies and quasars

393

describe the key features of stellar spectra and describe how these are used to classify stars

395

describe how spectra can provide information on surface temperature, rotational and translational velocity, density and chemical composition of stars

393

xxii


4. Photometric measurements can be used for determining distance and comparing objects

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdefine absolute and apparent magnitude

398 solve problems and analyse information using:

M md

10= −

5 log

andII

100A

B

= (mB – mA)/5

to calculate the absolute or apparent magnitude of stars using data and a reference star

400

explain how the concept of magnitude can be used to determine the distance to a celestial object

399 perform an investigation to demonstrate the use of filters for photometric measurements

Act. 22.5

outline spectroscopic parallax 401 identify data sources, gather, process and present information to assess the impact of improvements in measurement technologies on our understanding of celestial objects

Act. 22.6

explain how two-colour values (i.e. colour index, B – V) are obtained and why they are useful

401

describe the advantages of photoelectric technologies over photographic methods for photometry

397

5. The study of binary and variable stars reveals vital information about stars

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe binary stars in terms of the means of their detection: visual, eclipsing, spectroscopic and astrometric

411 perform an investigation to model the light curves of eclipsing binaries using computer simulation

Act. 23.1

explain the importance of binary stars in determining stellar masses

408 solve problems and analyse information by applying:

m + mGT1 2 2

r 324=

π420

classify variable stars as either intrinsic or extrinsic and periodic or non-periodic

413

explain the importance of the period–luminosity relationship for determining the distance of cepheids

416

xxiii


6. Stars evolve and eventually ‘die’

STuDEnTS lEARn TO: PAGE STuDEnTS: PAGEdescribe the processes involved in stellar formation

423 present information by plotting Hertzsprung–Russell diagrams for: – nearby or brightest stars– stars in a young open cluster– stars in a globular cluster

Act. 24.1

outline the key stages in a star’s life in terms of the physical processes involved

428 analyse information from an HR diagram and use available evidence to determine the characteristics of a star and its evolutionary stage

437

describe the types of nuclear reactions involved in Main-Sequence and post-Main Sequence stars

425, 430

present information by plotting on a HR diagram the pathways of stars of 1, 5 and 10 solar masses during their life cycle

437

discuss the synthesis of elements in stars by fusion

425, 430

explain how the age of a globular cluster can be determined from its zero-age main sequence plot for a HR diagram

433

explain the concept of star death in relation to:– planetary nebula– supernovae– white dwarfs– neutron stars/pulsars– black holes

429, 431

Space

Context

2

Figure 1.0.1 The knowledge of how things move through space, influenced by gravity, has transformed the way we work, play and think.

1Modern physics was born twice. The first time (arguably) was in the 17th century when Newton used his three laws of motion and his law of universal gravitation to connect Galileo’s equations of motion with Kepler’s laws of planetary motion. Then early in the 20th century, when many thought physics had almost finished the job of explaining the universe, it was unexpectedly born again. Einstein, in trying to understand the nature of light, proposed the special and general theories of relativity (and simultaneously helped launch quantum mechanics).

Space was the common thread—Kepler, Galileo, Newton and Einstein were all trying to understand the motion of objects (or light) through space.

Newton’s laws of mechanics and his theory of gravitation led to space exploration and artificial satellites for communication, navigation and monitoring of the Earth’s land, oceans and atmosphere. Einstein’s theory of relativity showed that mass and energy are connected, and that length, mass and even space and time are rubbery. Relativity has come to underlie most new areas of physics developed since then, including cosmology, astrophysics, radioactivity, particle physics, quantum electrodynamics, anything involving very precise measurements of time and the brain-bending ‘string theory’.

So, whenever you use the global positioning system (GPS), consult Google maps, check the weather report or make an international call on your mobile phone, remember that the technology involved can be traced directly back to physics that started 400 years ago.

3

Figure 1.0.2 The revolution in our understanding of the universe started with the humble question of how projectiles move.

InquIry aCtIvIty

Go ballIstIC!

The path through the air of an object subject only to gravity and air resistance, is called a ballistic trajectory. If the object is compact and its speed is low, then air resistance is negligible and its trajectory is a parabola.

Investigate parabolic trajectories using a tennis ball, an A4 piece of paper, a whiteboard or a blackboard and a digital camera.1 On a board about 2 m wide, draw an accurate grid of horizontal and vertical lines

10 cm apart.2 With a firmly mounted camera, take a movie of a tennis ball thrown slowly in

front of the board. Try different angles and speeds to get eight or more frames with the ball on screen, and get as much of a clear parabolic shape (including the point of maximum height) as you can.

3 Using video-editing software, view the best movie, frame by frame, on a computer. If your software allows it, create a single composite image with all the ball’s positions shown on one image, to show the parabolic trajectory.

4 If you can’t do that, then for each frame, on the board, and using the grid, estimate the x- and y-coordinates of the ball’s centre to the nearest 5 cm or better. Some video software allows you to read the x- and y-coordinates (in pixels) by clicking on the image.

5 Plot a graph of x versus y to produce a graph of the parabolic trajectory. The graph might be a bit irregular because of random error in reading the blackboard scale.

6 Video the trajectory of a loosely crumpled-up piece of A4 paper. Now air resistance is NOT negligible. Does the trajectory still look like an ideal parabola?

1 cannonballs, apples, planets and gravity

4

projectile, trajectory, parabola, ballistics, vertical and horizontal

components, Galilean transformation, range, launch angle, time of flight,

inverse square law, law of universal gravitation, universal gravitation

constant G, gravitational field g, test mass, central body, density, gravimeter, low earth orbit,

gravitational potential energy, escape velocity, gravitationally bound

1.1 Projectile motionUp and down, round and roundBefore Galileo Galilei (1564–1642), it was a common belief that an object such as a cannonball projected through open space (a projectile) would follow a path (trajectory) through the air in a nearly straight line until it ran out of ‘impetus’ and then drop nearly straight down in agreement with the ideas of Aristotle. However, through experiments (Figure 1.1.1) in which he rolled balls off the edge of a table at different speeds and then marked the position of collisions with the ground, Galileo demonstrated that the trajectory of a falling ball is actually part of a parabola (see Figure 1.1.2). Remember that a parabola is the shape of the graph of a quadratic equation. The immediate result of Galileo’s discovery was that the art of firing cannonballs at your enemies became a science (ballistics). However, there were also more far-reaching, constructive consequences.

What goes up must come downOne of the powers of physics is that it enables us to find connections between seemingly unconnected things and then use those connections to predict new and unexpected phenomena. What started as separate questions about the shape of the path of cannonballs through the air and the speed of the Moon’s orbit around the Earth eventually led to the law of gravitation. This explained how the solar system works, but also led to the development of artificial satellites and spacecraft for the exploration of the solar system.

Figure 1.1.1 Galileo’s laboratory notes on his experiments showing that projectiles follow parabolic paths

5

Space

Opponents of Copernicus’ heliocentric universe claimed that if the Earth was rotating and orbiting the Sun, then a person jumping vertically into the air would have the ground move under their feet, so that they would land very far away from where they started.

Galileo argued that a person jumping from a moving Earth is like a projectile dropped by a rider on a horse (representing the Earth) moving with a constant velocity (Figure 1.1.3). From the rider’s point of view, the projectile would appear to drop vertically, straight to the ground, accelerating downwards the whole time. A bystander who is stationary relative to the ground would see the rider, horse and projectile whoosh past and, like any other projectile, the dropped object would appear to follow a parabolic trajectory.

Galileo argued that the parabolic motion of the projectile was made up of two separable parts: its accelerating vertical motion as seen by the rider, and its constant horizontal velocity (which is the same as that of the horse). Recall from your Preliminary physics course that these two contributions to velocity are called vertical and horizontal components (see in2 Physics @ Preliminary section 2.2, p 26).

Galileo then argued that the Earth doesn’t zoom away under your feet because at the moment you jump upwards you already have the same horizontal component of velocity as the Earth’s surface. Relative to the Earth’s surface, your horizontal velocity is zero and so you land on the same spot.

In connecting the two problems of projectile motion and a moving Earth, Galileo developed two important new concepts. The first is the idea that the parabolic trajectory of a projectile can be divided into vertical and horizontal components. The second is the idea of measuring motion relative to another moving observer (or ‘frame of reference’). The formula vB (relative to A) = vB – vA (see in2 Physics @ Preliminary, p 8) is used to transform velocities relative to different frames of reference. This formula is sometimes called the Galilean transformation.

Components of a trajectoryThe ideal parabolic trajectory is an approximation that works under two conditions: 1 Air resistance is negligible (gravity is the only external force). 2 The height and range (horizontal displacement) of the motion are both

small enough that we can ignore the curvature of the Earth.

Describe Galileo’s analysis of projectile motion.

Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components.

a b

Figure 1.1.3 Trajectory of the rider’s projectile as seen by (a) the rider and (b) an observer on the ground

Horizontal displacement

Vert

ical

dis

plac

emen

t

Figure 1.1.2 This graph of a parabolic trajectory shows the vertical and horizontal components of displacement separately. The projectile positions are plotted at equal time intervals.

cannonballs, apples, planets and gravity1

6

75° 90°60°

45°

30°

15°

Figure 1.1.4 For a fixed initial speed, maximum range occurs for a 45° angle of launch and maximum height occurs for a 90° angle of launch.

The first condition is true for compact and low-speed projectiles. The second is true in almost all human-scale situations, typically at or near the Earth’s surface. Let’s analyse an example of ideal projectile motion. Recall that the acceleration due to gravity is g = 9.8 m s–2 (see in2 Physics @ Preliminary section 1.3). Here we are going to write it as a vector g. Clearly its direction is downwards.

Consider the trajectory of a ball. We start by separating the horizontal and vertical components of its motion. While the ball is in the air, the only external force on it is gravity acting downwards, so there is a constant vertical acceleration ay = g, illustrated by the changing vertical spacing of projectile positions plotted at equal time intervals in Figure 1.1.2.

The net horizontal force is zero, so, consistent with Newton’s first law, horizontal velocity is constant (ax = 0), which is clear from the equal horizontal spacing of the projectile positions plotted at equal time intervals in Figure 1.1.2.

We can recycle the kinematics (SUVAT) equations from the Preliminary course. (See in2 Physics @ Preliminary section 1.3.)

s = vt (1) s = ut + 12

at 2 (4)

sv= +u

2t

(2)

v 2 = u2 + 2as (5)

v = u + at (3)

Here we need to apply them separately to the vertical (y) and horizontal (x) components of motion. Instead of displacement s, we’ll use ∆x = xf – xi for horizontal displacement and ∆y = yf – yi for vertical displacement. We’ll put subscripts on the initial and final vertical velocities (uy and vy for example). We only need to use SUVAT equations 3, 4 and 5. θi is the launch angle (between the initial velocity u and the horizontal axis). Remember to adjust the sign of g to be consistent with your sign convention. In problems involving gravity, up is normally taken as positive, making the vector g negative (i.e. g = –9.8 m s–2).

In the syllabus, vx2 = ux

2 is included for completeness; but is unnecessary, as it can be derived from vx = ux.

Some properties of ideal parabolic trajectories are:• Atthemaximumheightoftheparabola,verticalvelocityvy = 0.• Thetrajectoryishorizontallysymmetricalaboutthemaximumheightposition.• Theprojectiletakesthesametimetorisetothemaximumheightasittakesto

fall back down to its original height.• Forhorizontalground,initialspeed=finalspeed.• Maximumpossibleheightoccursfora90°launchangle.Themaximum

possiblerange(forhorizontalground)occursfora45°launchangle (Figure 1.1.4).

• Independentoftheirinitialvelocity,allobjectsprojectedhorizontallyfromthesame height have the same time of flight as one dropped from rest from the same height, because they all have a zero initial vertical velocity (Figure 1.1.5).

activity 1.1

pRacTIcaL eXpeRIeNceS


Table 1.1.1 Equations of projectile motion

Horizontal components Vertical components

ux = u cos θi uy = u sin θi

vx = ux vy = uy + gt

∆x = uxt ∆y = uyt + 1—2 gt 2

vx 2 = ux

2 vy 2 = uy

2 + 2g∆y

7

Space

BaLLISTIcS IS a dRag

Air resistance or ‘drag’ introduces deceleration in both the vertical and horizontal directions, distorting the ballistic

trajectory from an ideal parabola. As a projectile becomes less compact, air resistance increases relative to weight. The range decreases, the trajectory becomes less symmetrical, and the final angle becomes steeper. The launch angle for maximum range decreases. In extreme cases (for example, a loosely crushed piece of paper), the trajectory seems to approach Aristotle’s prediction: it moves briefly in a nearly straight line and then drops nearly vertically.

no air resistance

increasing air resistance

Figure 1.1.6 The effect of increasing air resistance

Figure 1.1.5 Multiflash photo of two falling objects. All horizontally projected objects have the same time of flight as an object dropped from rest from the same height.

100 mm

Target practiceYou now have all the equations you need to ‘do some damage’, so let’s launch some projectiles. Safety warning! The following worked example may seem dangerously long because it illustrates several alternative methods of solving projectile problems rolled into one.

Worked examplequestIonYou throw a ball into the air (Figure 1.1.7). You release the ball 1.50 m above the ground, with a speed of 15.0 m s–1, 30.0° above horizontal. The ball eventually hits the ground. Answer the following questions, assuming air resistance is negligible.

a For how long is the ball in the air before it hits the ground (time of flight)?

b What is the ball’s maximum height?

c What is the ball’s horizontal range?

d With what velocity does the ball hit the ground?

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: vx

2 = ux2

v = u + atvy

2 = uy2 + 2ay ∆y

∆x = uxt∆y = uyt + 1

2ayt

2

1.50 m

Figure 1.1.7 Throwing a ball into the air


8

solutIonAlways draw a diagram! Divide the motion into vertical (y) and horizontal (x) components. Choose the origin to be the point of release, so xi = yi = 0. This is not always the most convenient choice of origin.

Use the sign convention + → & +↑

.

Components of initial velocity u (Figure 1.1.8):

ux = +u cos θi = +15.0 cos 30.0° = +13.0 m s–1

uy = +u sin θi = +15.0 sin 30.0° = +7.50 m s–1

The only external force is gravity so vertical acceleration is g = –9.80 m s–2. There is no horizontal force, therefore ax = 0 m s–2 (constant horizontal velocity).

initial ve

locity

u

ux

uy

θi = 30.0°

final velocity v

vx

vy

θf

Figure 1.1.8 Components of initial and final velocities

a The ball hits the ground when vertical displacement ∆y = –1.50 m.

Find final vertical velocity: vy 2 = uy

2 + 2g ∆y = 7.502 + 2 × –9.80 × –1.50 = 85.65

vy = 85.65 = 9.255 m s–1 (must be downwards), so vy = –9.255 m s–1

Find t : vy = uy + gt = –9.255 = +7.50 + (–9.80) × t

Rearrange, solve: t = − −

−9 255 7 50

9 80. .

. = 1.71 s

The ball hits the ground 1.71 s after being thrown.

Alternative method using the quadratic formula ∆y = uyt + 12

gt 2 = –1.50 m

Substitute, rearrange: 1.50 + 7.50 × t + 12

× –9.80 × t 2 = 0

Quadratic, solve for t : t = − ± + × ×

− ×7 5 7 50 4 4 90 1 50

2 4 90

2. . . ..

= –0.179 s or +1.71 s

Two solutions: Reject the physically irrelevant negative solution, so t = 1.71 s.

b At maximum height, vertical velocity vy = 0, so use vy 2 = uy

2 + 2g ∆y.

0 = uy2 + 2g∆ymax = 7.502 + 2 × (–9.80) × ∆ymax

Rearrange, solve: ∆ymax = 7 50

2 9 80

2..×

= +2.87 m above the point of release,

so height above ground = 2.87 m + 1.50 m = 4.37 m above the ground.

Alternative method

Use vy = uy + gt to find the time t when vy = 0, then use ∆y = uyt + 12

gt 2 to find vertical displacement.

c From part a, we know the time of flight t = 1.71 s.

Horizontal displacement in this time is: ∆x = uxt = +13.0 m s–1 × 1.71 s = +22.2 m = 22.2 m (to the right)

9

Space

d x-component of final velocity: vx = +13.0 m s–1

y-component of final velocity: vy = –9.255 m s–1 (down) (from part a)

To find magnitude, use Pythagoras’ theorem (see Figure 1.1.8):

v = v vx y2 2+ = 13 9 2552 2+ . = 15.96 ≈ 16.0 m s–1

Direction: tan θf = v

vy

x= 9 25

13 0..

, so θf = 35.4° down from horizontal

Alternative magnitude calculation

Negligible air resistance, ∴ mechanical energy = kinetic energy + gravitational potential energy and is conserved (see in2 Physics @ Preliminary section 4.2). Near the Earth’s surface, gravitational potential energy U = mgh. Using the ground as h = 0:

Ki + Ui = Kf + Uf

Cancel m: 12

mvi2 + mghi =

12

mvf2 + mghf

Substitute: 12

15.02 + 9.80 × 1.50 = 12

vf2 + 0

Rearrange, solve: v f = + × ×15 0 2 9 80 1 502. . . = 15.94 ≈ 15.9 m s–1

This is the same as for the previous method within the three-figure precision of the calculation, but doesn’t tell us the direction.

In the previous example, time of flight was determined by the vertical component—the flight ended when the ball hit the ground. However, if the projectile hits a vertical barrier such as a wall, then the time of flight is determined by the horizontal component.

Worked examplequestIonSuppose you kick a ball at 22.0 m s–1, 20.0° above the horizontal, towards a wall 21.0 m away (Figure 1.1.9). Ignore air resistance and the ball’s radius.

a What is the ball’s time of flight (before hitting the wall)?

b At what height does the ball hit the wall?

c Is that the greatest height reached by the ball?

solutIon

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: vx

2 = ux2

v = u + atvy

2 = uy2 + 2ay ∆y

∆x = uxt∆y = uyt + 1

2ayt

2

Figure 1.1.9 The ball hits the wall.

Choose the origin to be the initial position, so xi = yi = 0. Use the sign convention +↑

and + →.

ux = 22.0 cos 20.0° (right) = +20.7 m s–1

uy = 22.0 sin 20.0° (up) = +7.52 m s–1


10

1.2 GravityInPtolemy’suniverse,theSun,Moonandplanetseachhadaseparateclockwork-like mechanism to keep it in motion. Copernicus and Kepler greatly improved the picture, but Isaac Newton finally showed there was a single mechanism for them all—the force of gravity.

The calculations of parabolic trajectories in section 1.1 work well close to the Earth’s surface where g is constant. However, if we’re going to venture out into space, we can’t use these simple equations. We need to look at the force of gravity on a larger scale.

Newton’s law of universal gravitationNewton assumed several properties of gravity (see in2 Physics @ Preliminary section 13.5):• All‘massive’objects(thatis,objectswithmass)attracteachother.Thelarger

the masses, the larger the force.

a The ball hits the wall when the horizontal displacement ∆x = +21.0 m.

Substitute: ∆x = uxt = +21.0 m = +20.7 m s–1 × t

Rearrange, solve: t = ++ −

21 0

20 7 1

.

.

m

m s = 1.014 s ≈ 1.01 s

b The ball hits the wall at a height (vertical displacement) of ∆y = uyt + 12

gt 2.

Substitute, solve: ∆y = +7.52 × 1.014 + 0.5 × –9.80 × 1.0142 = +2.587

The ball hits the wall ≈ 2.59 m above ground.

c Check if the ball reaches maximum height of the parabola before hitting the wall.

Time of flight = 1.01 s. vy = 0 at maximum height of parabola.

Find the time taken to reach maximum height.

Substitute: vy = 0 = uy + gt = +7.52 + –9.80 × t

Rearrange, solve: t = 7 529 80..

= 0.767 s which is less than time of flight

The ball would reach the maximum height of the parabola before hitting the wall, therefore the final height is NOT the maximum height for the trajectory.

CheCkPoInt 1.11 Determine the horizontal acceleration of a projectile in flight. Determine its vertical acceleration. (Assume

negligible air resistance.)2 What angle of launch gives maximum horizontal range? What angle of launch gives the maximum possible height?

(Assume negligible air resistance.)3 What is another name for air resistance?4 If you throw a ball horizontally from the roof, and drop another at the same time, which one will hit the ground first?5 Describe the two conditions that must apply so that a trajectory is a parabola.6 List the 8 equations used in calculations of projectile motion. Explain why at least one of them is unnecessary.

11

Space

• Likelightintensity,themagnitudeoftheforcedecreaseswithdistanceaccording to the inverse square law (see in2 Physics @ Preliminary sections 6.1 and 15.1). However, astronomer Ismael Boulliau had suggested this before him.

• Thelawofgravitationisuniversal—itappliesthroughouttheuniverseandisresponsible for the orbits of all the planets and moons.All this is expressed mathematically as the law of universal gravitation:

F Gm m

dG = 1 2

2

where FG is the magnitude of the force of gravitational attraction between two masses m1 and m2 and d is the distance between their centres of mass (see in2 Physics @ Preliminary section 3.6). The universal gravitational constant G (‘big G ’) is 6.67 × 10–11 N m2 kg–2 in SI units. It should not be confused with ‘little g’, 9.8 m s–2.

Morepropertiesofgravitation:• Thedirectionoftheforceactsalongthelinejoiningthe

centres of the two masses and is always attractive.• Theformulaisstrictlycorrectforpointmassesandspheres,

but works well for non-spheres.• Theformulamustbemodifiedifonemasspenetratesthe

surface of the other—gravity would not approach infinity if you were to burrow towards the centre of the Earth.

• Theresultantforceonamassduetothepresenceofothermasses is the vector sum of the individual forces on the first mass due to each of the other individual masses.

Worked examplequestIonCalculate the gravitational force between the Earth and the Moon.

Data: Earth’s mass mE = 5.97 × 1024 kg

Moon’s mass mM = 7.35 × 1022 kg

Average Earth–Moon distance dEM = 3.84 × 108 m

Universal gravitational constant G = 6.67 × 10–11 N m2 kg–2

solutIon

F Gm m

dGE M

EM

=

= × × × × ×

2

11 24 226 67 10 5 97 10 7 35 10

3

. . .

(

–

.. )84 108 2×

= 1.98 × 1020 N

Define Newton’s Law of Universal Gravitation:

F Gm m

d= 1 2

2

TRy ThIS!sligHtly attractiVeYou can see the feeble force of gravity acting between objects in your garage. John Walker’s Fourmilab website describes step by step how you can perform a crude version of the Cavendish experiment in your own garage (see Physics Focus ‘How to weigh the Earth’ at the end of this chapter), using commonly found household items and a video camera. If you’re feeling too lazy to do it yourself, you can just download sped-up videos of the experiment in progress.

Figure 1.2.1 Cavendish apparatus at home


12

PhysICs FeatureDon’t unDerestImate the PoWer oF boreDomboreDom Part 1

Bored? Don’t just write graffiti—try revolutionising physics! In 1665, an outbreak of bubonic plague around London closed Cambridge University,

so Isaac Newton (aged 23) escaped for 2 years to his mother’s farm. He was not a very good farmer, so he fended off his city-boy boredom by inventing calculus and using prisms to show that white light is actually a mixture of colours (the spectrum). To top this off, when he saw an apple fall off his mother’s tree, he wondered if the force accelerating the apple downwards was also responsible for keeping the Moon orbiting the Earth. So he began formulating his theory of gravitation. His mathematics professor was so impressed that a couple of years after Newton returned to Cambridge, he resigned and handed his professorship to Newton.

After this initial investigation, it took Newton another 20 years to fully develop and finally publish his law of universal gravitation.

Now let’s try an example with more than two masses.

Worked examplequestIonA 1000 kg spacecraft is in the vicinity of the Earth–Moon system. The spacecraft is at the origin, the Moon is on the positive y-axis and the Earth is on the positive x-axis (Figure 1.2.2). Given that the Earth–spacecraft and Moon–spacecraft distances are 3.82 × 108 m and 3.91 × 107 m respectively, calculate the resultant gravitational force on the spacecraft.


Moon’s mass mM = 7.35 × 1022 kg


solutIonForce due to Moon: FSM = G

m m

dS M

SM2

= 6 67 10 1000 7 35 10

3 91 10

11 22

7 2

. .

( . )

–× × × ××

= 3.207 N (+y direction)

Force due to Earth: FSE = Gm m

dS E

SE2

= 6 67 10 1000 5 97 10

3 82 10

11 24

8 2

. .

( . )

–× × × ××

= 2.729 N (+x direction)

Magnitude of resultant: Fres N= + =3 207 2 729 4 212 2. . .

Direction: tan..

θ = 3 2072 729

, so θ = +49.6° from the x-axis

Moon

Earth

spacecraft

Figure 1.2.2 A spacecraft in the Earth–Moon system

FSE

FSMFres

θ

θ

spacecraft

Figure 1.2.3 Gravitational force vector diagram. Note; This does not resemble the position vector diagram in Figure 1.2.2.


3. Applications and uses of physics

Figure 1.2.4 Graffiti carved on a stone at the King’s School in Grantham, England, by Isaac Newton, then about 10 years old

13

Space

boreDom Part 2

It is said that, at age 17, Galileo was attending church and, bored, was watching a lantern swing from the ceiling. Using his pulse as a

stopwatch, he observed that the oscillation period of a pendulum barely changed as its amplitude gradually decreased. Back at home he started experiments confirming that the oscillation period depends on pendulum length L, but not at all on mass and only slightly on amplitude. He proposed (correctly) that pendulums could be used to create the first accurate mechanical clocks.

We now know that, consistent with Galileo’s observations, for a simple mass-on-string pendulum the formula for oscillation period T is:

T = 2π Lg

The formula is an approximation, but if the maximum swing angle is less than 15° from vertical, the formula is correct within 0.5%. With this formula and a pendulum, you can measure the value of ‘little g’, which varies slightly between locations around the world.

Figure 1.2.5 Young Galileo watches a swinging lantern in Pisa cathedral.

Weight and gravitational fieldsAs far as we know, the universal gravitational constant G is a fundamental constant, unchanging with position or time. But the acceleration due to gravity g is different on other astronomical bodies, at different heights and even at different positions on the Earth’s surface.

Recall that weight w = mg is defined as the force on an object due to gravity (see in2 Physics @ Preliminary section 3.2); in other words, FG = w = mg. ‘Little g’, the acceleration due to gravity, can also be thought of as the strength of the gravitational field. However, the word weight is usually reserved for the case in which the gravitational field is due to a body of astronomical size, such as a planet.

Any massive object can be described as being surrounded by a gravitational field, a region within which other objects experience an attractive force. Just as for electrical and magnetic fields (see in2 Physics @ Preliminary sections 10.6, 12.3 and 12.4), we can draw diagrams of gravitational field lines (Figure 1.2.6). The arrows on the field lines around a mass, point in the direction of the force acting on another (normally much smaller) test mass. Gravitational field is a vector (g). The density of the field lines at any particular point in space represents g, the magnitude of the field at that point, and the direction of the field lines represents the direction of this vector. Field lines run in radial directions from point masses or spherical masses.

Using a small test mass m, let’s derive g, the magnitude of the gravitational field due to a planet of mass M. The weight w of the test mass is defined as the force on m due to the planet’s gravity; that is:

w = mg = FG = GmM

d 2

activity 1.2



Describe a gravitational field in the region surrounding a massive object in terms of its effects on other masses in it.

Define weight as the force on an object due to a gravitational field.


14

aDivide both sides by test mass m:

gFm

GM

d= =G

2

Newton’s equation for gravitational force is symmetrical—you can choose either mass as the test mass and calculate the field around the other and still get the same magnitude of force when you multiply them together because of Newton’s third law (see in2 Physics @ Preliminary section 3.5)—the two masses are an action–reaction pair. However, if one of the masses is much larger (such as a planet), it is more convenient to calculate the field around it and use the smaller mass as the test mass.

In astronomical situations where one of the bodies (such as a planet or star) is very much larger, the larger body is sometimes called the central body. Because of its large mass, the central body experiences negligible gravitational accelerations compared with a small test mass.

Strictly speaking, the acceleration g is the acceleration of the test mass towards the common centre of mass of the whole system of two masses. However, if the central body is much larger than the test mass, we can ignore its acceleration, so g effectively becomes the acceleration of the test mass towards the central body.

Gravitational field is a vector, so when calculating the resultant field due to several bodies, the approach is identical to calculating the resultant gravitational force due to several bodies—calculate the field due to each individual mass and then find the vector sum of the fields.

Worked examplequestIonCalculate gE the magnitude of the gravitational field at the Earth’s surface.


Earth’s radius rE = 6.37 × 106 m


solutIong

GM

dE

E=2

The test mass is at the Earth’s surface, ∴ d = rE

Substitute: gE = 6 67 10 5 97 10

6 37 10

11 24

26

. .

( . )

–× × ××

= 9.81 m s–2

This should be a very familiar result.

Variations in gravitational fieldNewton’s gravitation equation says that the magnitude of a planet’s gravitational field depends on the mass of the planet and decreases with distance from the planet’s centre. For example, on Earth, the value of g is 0.28% lower at the top ofMtEverestthanatsealevel.Also,becausetheEarthhasaslightlylargerradiusnear the equator than at the poles (the ‘equatorial bulge’), g is slightly lower at the equator. Except at the poles, there is an additional (fictitious) decrease in g

activity 1.3



Figure 1.2.6 Gravitational field lines around the Earth (a) on an astronomical scale and (b) near the surface

b

15

Space

measurements that gets more severe as one approaches the equator. Because of the Earth’s rotation, the (downward) centripetal acceleration (see in2 Physics @ Preliminary section 2.3) of the ground appears to be subtracted from the true value of g. In fact this centripetal effect is responsible for the formation of the equatorial bulge, which was predicted by Newton before it was measured.

TheSunandMoonalsoexertaweakgravitationalforceonobjectsattheEarth’s surface, so the magnitude and direction of g vary slightly, depending on thepositionsoftheSunandMoon.Variationing caused by the positions of the SunandMoonrelativetotheoceansisresponsibleforthepatternoftides.

Strictly speaking, Newton’s gravitation equation written in the form above assumes that the planet is a perfectly uniform sphere. Close to the surface of a planet, local deviations from uniform density can result in small local changes in the magnitude and direction of g. The magnitude will be slightly larger than average when measured on the ground above rock (such as iron ore) of high density ρ (mass per unit volume) and lower above rock containing low-density minerals (such as salt or oil), an effect exploited by geologists in mineral exploration. The Earth’s crust is less dense than the mantle, so variations in thickness of the crust also affect g. Variation in g is measured using a gravimeter, the simplest kind being an accurately known mass suspended from a sensitive spring balance.

Variations in g on larger distance scales around the Earth can be measured using satellites orbiting in low Earth orbit. Deviations in the orbital speed of satellites indicate that, in addition to the equatorial bulge, Earth is also slightly pear-shaped—pointier at the North Pole than the South Pole.

hooke’S LaW

Isaac Newton had enemies, and Robert Hooke (1635–1703) was probably his

greatest. They argued bitterly over (among other things) who first suggested the inverse square law for gravity. Hooke was an accomplished experimental physicist, astronomer, microscopist, biologist, linguist, architect and inventor. He is best remembered for the discovery of (biological) cells and the invention of the spring balance (see in2 Physics @ Preliminary section 3.2), which exploits Hooke’s law F = ‑k x. The force F exerted by a spring is proportional to x, the change in spring length. The ‘spring constant’ k is a measure of the spring’s stiffness. A calibrated spring balance can measure weight, and, if used with an accurately calibrated mass, it can be used as a gravimeter to measure g. Figure 1.2.7 Hooke’s notes on the behaviour of springs

Interactive

Module


16

1.3 Gravitational potential energyWe’ve already mentioned gravitational potential energy (GPE) U = mgh (see in2 Physics @ Preliminary section 4.1) in part d of the worked example accompanying Figure 1.1.7. This formula for GPE is an approximation that only works close to the Earth’s surface, where g is very nearly constant. It’s good enough for projectile motion but, as you now know, g decreases with distance, so we need a more accurate formula to understand energy on an astronomical scale.

Work and GPEFor clarity we’ll use the symbol EP instead of U to denote gravitational potential energy calculated using the more accurate formula, even though the two symbols are really interchangeable. Potential energy is energy stored by doing work against any force (such as gravity) that depends only on position; therefore, gravitational potential energy EP is energy stored by doing work against the force of gravity. It can be shown (using calculus to derive the work done against gravity by changing the separation of two masses) that:

E Gm m

rP = − 1 2

where m1 and m2 are two masses separated by a displacement (or separation) r and G is the universal gravitational constant. Note that EP is always negative and approaches zero as displacement r approaches infinity (Figure 1.3.1). EP for a separation r is the work that would need to be done by a force opposed to gravity in moving the masses together, starting at ‘infinite’ separation where EP = 0 and bringing them to a separation of r (with no net change in speed).

Equivalently, EP is the work done by gravity while the masses are moved apart, starting at a separation of r to a position of ‘infinite’ separation (with no net change in speed). The gravitational potential energy does not depend on the path taken by the masses to get to their final positions; it depends only on the final separation r.

The formula isn’t affected by the choice of which mass to move, although normally we treat a large mass such as the Sun or a planet as an immoveable central body and the smaller mass as a moveable test mass. The formula seems to imply that EP approaches negative infinity as the test mass approaches the centre of a planet. However, this formula no longer applies in this form once one mass penetrates the surface of the other.

Explain that a change in gravitational potential energy is related to work done.

Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field:

E G

m m

rP = − 1 2

CheCkPoInt 1.21 Write down Newton’s law of universal gravitation.2 Define weight.3 What part of Newton’s formula for gravitational force is responsible for the inverse square law behaviour?4 What are two names for the quantity g?5 List three factors responsible for (real) variations in g around the Earth.6 Outline the differences between G and g.

17

Space

Worked examplequestIonA piece of space junk of mass mJ drops from rest from a position of 30 000 km from the Earth’s centre. Calculate the final speed vf it attains when it reaches a height of 1000 km above the Earth’s surface. Assume that above 1000 km, air resistance is negligible.


Earth’s radius rE = 6.37 × 106 m


solutIonAir resistance is negligible, so total mechanical energy (kinetic + potential energy) is conserved. Assume that because of the enormous mass of the Earth, its change in velocity is negligible. Use the Earth as the frame of reference. Don’t forget to convert to SI units.

K i + EPi = Kf + EP

Cancel mJ: 12

mJvi2 – G

m mrJ E

i = 1

2 mJvf

2 – Gm m

rJ E

f

Substitute: 0 – 6.67 10 6.67 10–11f

–× × ××

= − ×5 97 10

30 0 10

12

24

62.

.v 111 × ×

+ ×5 97 10

6 37 1 00 10

24

6

.

( . . )

Rearrange, solve: v f–116.67 10= × × × × × −

− −2 5 97 10

107 37

1030 0

246 6

.. .

= =− −9030 9 031 1m s km s.

Note that this result doesn’t depend on m J.

Figure 1.3.1 Plots of gravitational force (FG) and gravitational potential energy (EP) versus separation between a test mass mt and the Earth mE, starting at one Earth radius rE. The vertical FG and EP axes are not drawn to the same scale.

FG

EP–GmtmE

rE

+GmtmE

rE 2rE0 3rE 4rE 5rEseparation

rE2


18

Escape velocity: what goes up …? Isaac Newton showed that what goes up doesn’t necessarily come down. Normally, if one fires a projectile straight up, the object will decelerate until its velocity changes sign and it falls back down. However, if a projectile’s initial velocity is high enough, the 1/d 2 term in the gravity equation will cause the acceleration g to decrease with height too rapidly to bring the projectile to a stop so it will never turn back—it can ‘escape’ the planet’s gravitational field. The minimum velocity that allows this is called the escape velocity. Strictly speaking, it’s really a speed, because the initial direction of the projectile isn’t critical.

Newton treated the projectile as a cannonball (with no thrust) so that, other than the initial impulse from the cannon, the only force acting on it is gravity. He conceived escape velocity using his force equation, and the escape velocity formula can be derived from it. However, a more modern derivation using energy is easier and similar to the previous worked example.

Let m be the mass of a projectile, M the mass of a planet, ve the initial speed and r the initial position (the planet’s radius if you are on the surface). Assume air resistance is negligible, so total mechanical energy (KE + GPE) is conserved (see in2 Physics @ Preliminary section 4.2).

Ki + EPi = Kf + EPfThe escape velocity represents the minimum limiting case where the projectile

‘just reaches infinite displacement’ with zero speed; in other words, Kf = EPf = 0.

12

0 02mvGmM

re − = +

Rearrange, cancel m:

v

GMre = 2

If the initial speed is greater than this, the projectile will maintain a non-zero speed even as it approaches infinite displacement. Note that the escape velocity depends only on the planet’s mass and the projectile’s starting position r but not on the projectile’s mass.

You may be puzzled that in the above derivation, the total mechanical energy (sum of KE and GPE) was exactly zero. This means that the escaping projectile has just enough (positive) KE to overcome its negative potential energy. When the mechanical energy is less than zero, there is not enough KE to overcome the GPE and the two masses are said to be gravitationally bound. When the total mechanicalenergyME>0,theKEcanovercometheGPEandthetwobodiesare no longer bound together. This concept of binding also applies to the other three fundamental forces (including electromagnetism, which binds electrons to the nucleus of an atom).

The escape velocity from the Earth’s surface is:

2 6.67 10m s

–11× × × ××

= =−5 97 10

6 37 1011 200 11 2

24

61.

.. km s−1

Outline Newton’s concept of escape velocity.

Explain the concept of escape velocity in terms of the: – gravitational constant– mass and radius of the planet.

19

Space

This idealised escape velocity needs to be modified when applied to real spacecraft. First, the derivation ignores air resistance in the atmosphere (hundreds of kilometres thick), which would increase the escape velocity. Second, in a real rocket, engines produce an extra force—thrust—that can accelerate a craft to a higher altitude where the escape velocity is lower. It also ignoresothersourcesofgravitationalfieldssuchastheSun,Moonandplanets.

The escape velocity for a projectile under the gravitational influence of more than one body is given by:

ve total = ve12 + ve2

2 +

where ve total is the escape velocity for the total system and ve1, ve2 … are the escape velocities from the individual bodies within the system, calculated for the projectile using the same starting position in space.

ULTImaTe fRISBee

Was the first artificial object to leave the solar system a giant steel frisbee? In the 1950s, the US started testing nuclear bombs

underground, to minimise atmospheric nuclear fallout. In 1957, during Operation Plumbbob in the Pascal-B test, a nuclear bomb was detonated at the bottom of a 150 m shaft sealed with concrete and a 900 kg, 10 cm thick steel cap. The steel cap fired upwards at enormous speed and was never seen again. Before the test, it was estimated that an extreme upper limit for the speed of the steel cap would be 67 km s–1. This is well above the escape velocity for the whole solar system (43.6 km s–1 from Earth), starting an urban myth that it beat the Voyager probes (launched in 1977) out of the solar system. A later, more realistic, estimate suggested that, at most, the cap had a speed of 1.4 km s–1, reaching an altitude of less than 95 km.

CheCkPoInt 1.31 Define under what circumstances it is suitable to use the simplified formula U = mgh for gravitational potential

energy (GPE).2 Write down the more accurate formula for GPE.3 What limit does GPE approach as the separation of the two masses approaches infinity?4 On what factors does Newton’s idealised escape velocity depend?5 What other factors affect escape velocity in realistic situations?


20

1cannonballs, apples, planets and gravity

chapTeR 1This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

aCtIvIty 1.1: ProjeCtIlesA ball is rolled down a ramp, whose dimensions will be known to you. Predict where the ball will land.Equipment: aluminium track, ball bearing, metre ruler, measuring tape, shoe. Perform a first-hand

investigation, gather information and analyse data to calculate initial and final velocity, maximum height reached, range and time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis.

Discussion questions1 List assumptions you have made in order to make an estimate of the range. 2 Assess how reliable is your method.3 Explain how changing the original angle of the ramp will affect the range

of the ball.

aCtIvIty 1.2: DetermInInG the value oF aCCeleratIon Due to GravItyUse the motion of a pendulum to gather data to determine the acceleration due to gravity. Equipment: pendulum (string and mass), retort stand and clamp, stopwatch, metre ruler, data logger.

Figure 1.4.1 Equipment set-up for this activity

ruler

ball bearing

track

retort stand

string

mass

Perform an investigation and gather information to determine a value for acceleration due to gravity using pendulum motion or computer-assisted technology and identify reason(s) for possible variations from the value 9.8 m s–2.

Figure 1.4.2 Pendulum apparatus set-up

21

Space

Process the information you have gathered using the spreadsheet template. Complete the template to calculate the values of acceleration due to gravity on other planets.

Discussion questions1 Determine which planet has the largest value for acceleration due to gravity

at its surface. (Note that the gas giants Jupiter, Saturn, Uranus and Neptune don’t have

a well-defined boundary between the atmosphere and a solid planet surface. The visible ‘surface’ is fluid, i.e. gas and/or liquid.)

2 Identify the factors that affect the acceleration due to gravity.

Gather secondary information to predict the value of acceleration due to gravity on other planets.

Present information and use available evidence to discuss the factors affecting the strength of the gravitational force.

Analyse information using the expression:

F = mg to determine the weight force

for a body on Earth and for the same body on other planets.

Discussion questions1 Explain what you did in order to make the experiment reliable.2 Galileo originally thought that the period of the pendulum did not depend

at all on the amplitude of the swing. Is this true? Explain how you can take this into account in your experiment.

3 How does your value compare with the accepted value?4 Outline another method that would allow you to achieve the same aim.

aCtIvIty 1.3: GravIty— out oF thIs WorlDUse the spreadsheet template to gather appropriate information to help you predict the acceleration due to gravity at the surface of other planets.

Figure 1.4.3 Spreadsheet template

22


Review questions

chapter summary

PhysICally sPeakInGComplete each definition by using a keyword taken from the list at the beginning of the chapter.

To approach infinite distance from a massive central body, a projectile must start with _________________ .The path of a projectile is known as a _________________ .The formula for converting velocities between frames of reference is the _________________ .A projectile’s maximum horizontal displacement is its _________________ .Universal gravitation and the intensity of light both follow the _________________ .Close to the Earth’s surface and subject only to gravity, a projectile’s path is a _________________ .The acceleration of a _________________ near the central body equals the gravitational field.Close to the Earth’s surface, all objects projected horizontally from the same height have the same_________________ .A _________________ is apparatus used to assist in mineral exploration.If drag is negligible, then a projectile’s range is determined only by initial velocity and _________________ .

• Ifairresistance(drag)isnegligibleandg is very nearly constant (for example near a planet’s surface), then the trajectory of a projectile is a parabola.

• Theformulafortransformingvelocitywithinoneframeof reference into one relative to another frame of reference is called the Galilean transformation:

vB (relative to A) = vB – vA

• Parabolicprojectilemotioncandividedintoverticalandhorizontal components. The vertical component has a downward acceleration of g and the horizontal component has a constant velocity.

• Inparabolicprojectilemotion,theequationsof motion are:

Horizontal components: ux = u cos θi, vx = ux, ∆x = uxt, vx

2 = ux2

Vertical components: uy = u sin θi, vy = uy + gt, ∆y = uyt + 1

2gt2,

vy2 = uy

2 + 2g∆y

• Forhorizontalground,themaximumpossiblerangeoccursfora45°launchangle.Themaximumpossibleheightoccursfora90°launchangle.

• Allobjectsprojectedhorizontallyfromaparticularheight have the same time of flight as one dropped from rest from the same height.

• Ifatrajectoryendswhentheprojectilehitstheground,time of flight is determined by the vertical component. If the projectile hits a vertical barrier, then time of flight is determined by the horizontal component.

• Newton’slawofuniversalgravitation:

F Gm m

dG = 1 2

2

• Gravitationalacceleration(g) towards a central body such as a planet is also called its gravitational field. It depends on the central body mass M and the distance d from its centre:

g GM

d=

2

The force of gravity on an object in that field is called its weight: w = mg.

• Gravitationalfieldg measured near the Earth’s surface varies slightly with distance from the Earth’s centre and density of the surrounding material. The centripetal acceleration of the Earth’s surface also decreases measured values of g (only an apparent effect).

• Gravitationalpotentialenergy(GPE)istheworkdoneby a force opposing gravity in moving masses together starting at ‘infinite’ separation and bringing them to a separation of r (with no net change in speed).

• ThesimpleformulaforGPE(U = mgh) is an approximation that only works at or near the surface of a planet. The more accurate expression is:

E Gm m

rP = − 1 2

• EP approaches zero as separation of the two masses approaches infinity.

• Theminimuminitialvelocitythataprojectileneedstohave in order to escape a planet’s gravitational field is called escape velocity:

vGMre = 2

23

Space

revIeWInG 1 Given that the Earth rotates, account for why when you jump straight up,

you land on the same spot.

2 The high jump and the long jump both involve a run-up and then a jump. Using ideas from projectile motion, briefly compare and contrast the ideal characteristics of the run-up and jump for the two sports.

3 A projectile takes 1.25 s to reach its maximum height. What is its time of flight, assuming the ground is horizontal and drag is negligible?

4 Explain why (assuming negligible air resistance) all objects projected horizontally from the same height have the same time of flight as an object dropped from that height, regardless of their initial speed.

5 Predict what would happen to the magnitude of the gravitational force between two masses:a if one of the masses were doubledb if both masses were doubledc if the distance between the masses were doubled.

6 Describe how (and explain why) g would differ slightly from average at a point on the Earth’s surface above an oil deposit.

7 You’ve seen diagrams of electrical field lines around positive charges in which the arrows point outwards (see in2 Physics @ Preliminary section 10.6). Briefly discuss the possibility of a planet with gravitational field lines that point outwards. Propose how you would expect a test mass to behave there.

8 Without doing a calculation, deduce the speed at which a meteorite would hit the Earth’s surface if it started from rest at a very large distance from the Earth. Justify your answer. Ignore air resistance and gravity of other astronomical bodies. (Hint: The value is one already calculated elsewhere in this chapter.)

9 Read the definition of gravitational potential energy EP in section 1.3 page 16. Explain why it is necessary to specify in the definition that the work is done with no net change in speed. (Hint: What other form of energy is involved?)

solvInG Problems 10 Repeat the calculation in the worked example accompanying Figure 1.1.7,

assuming that the ball lands on the flat roof of a 2.5 m high garage, instead of the ground.

11 Consider the worked example accompanying Figure 1.1.9. Keeping everything unchanged except initial speed:a What would the initial speed of the ball need to be if the ball hit the wall

when it was just at its maximum possible height? What would be its time of flight?

b What would the initial speed of the ball need to be if the ball hit the ground just in front of the wall? What be would its time of flight?

12 By considering the vertical component of velocity and ignoring air resistance, derive an expression (containing initial speed u and launch angle θ) for the time taken for a projectile near the Earth’s surface to reach its maximum height. Then show that the time of flight for a projectile fired over horizontal ground is given by:

tug

= 2sinθ

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using:vx

2 = ux 2

v = u + atvy

2 = uy 2 + 2ay∆y

∆x = uxt; ∆y = uyt + 12

ayt 2


F Gm m

d= 1 2

2


2 = ux 2

v = u + atvy

2 = uy 2 + 2ay∆y

∆x = uxt; ∆y = uyt + 12

ayt 2

24


13 A marble rolls horizontally off the edge of a 1.00 m high table with a speed of 3.00 m s–1. Calculate the speed with which it hits the ground, by:a using the equations of projectile motionb assuming the conservation of mechanical energy (using the simple

version of the equation for GPE).

14 Repeat the calculation in the worked example accompanying Figure 1.2.2, with the positions of the Moon and Earth swapped.

15 Using your own mass, calculate the maximum force of gravity exerted by the planet Mars (m = 6.42 × 1023 kg) on you, given that the closest approach of Mars to Earth is approximately 5.6 × 1010 m. How close would you need to stand to the centre of mass of a 10 tonne truck for the magnitude of the gravitational force it exerts on you to be the same? (1 tonne = 1000 kg)

16 Show that g is 0.28% lower on top of Mt Everest (8848 m) than at sea level. Data: Mean Earth radius r E = 6.367 × 106 m.

17 Calculate the change in GPE in moving a 10 kg object from an initial position 1000 km above the surface of the Earth to a final position at a distance from the Earth equivalent to the mean orbital radius of the Moon (r = 3.84 × 108 m). Assume the Moon is on the opposite side of its orbit at the time and you can ignore its gravitational effect.

18 Using the data and answer from Question 17, calculate the speed at which you would need to project the 10 kg object radially outwards from the initial position so that it would just reach the final position, stop and fall back to Earth. (You can ignore air resistance above an altitude of ~1000 km.)

19 a Calculate the velocity required for a projectile to escape the Sun’s gravitational field (m Sun = 1.99 × 1030 kg) if launched from the orbital radius of the Earth (1.50 × 1011 m), if the Earth and other planets weren’t there.

b Using part a and Earth’s escape velocity (11.2 km s–1), show that the total escape velocity from the solar system for a projectile launched from Earth is 43.6 km s–1. Assume the projectile doesn’t pass near other planets.

extensIon 20 By considering the horizontal component of displacement for a projectile

and your answer for Question 12, derive an expression (containing initial speed u and launch angle θ) for the horizontal range. Either by using calculus or by considering the properties of trigonometric functions, show that the maximum range is attained for a launch angle of 45°.

21 A wildlife reserve ranger needs to hit a monkey in a tree with a tranquiliser dart in order to capture and examine it. The barrel of the dart gun is pointing exactly at the monkey. The angle between the barrel of the dart gun and the horizontal is not 90°. At the instant the ranger fires, the monkey is startled and drops from rest to the ground below.

Show that the dart will hit the monkey. (Hint: Show that by the time the dart reaches the horizontal position of the monkey, both the dart and the monkey have the same vertical position. Assume that air resistance is negligible.)


F Gm m

d= 1 2

2


2 = ux2

v = u + atvy

2 = uy2 + 2ay∆y

∆x = uxt; ∆y = uyt + 12

ayt2

Revie

w Questions

25

Space

PhysICs FoCushoW to WeIGh the earth


F Gm m

d= 1 2

2

Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.

Newton first tested his law of universal gravitation by showing that gravity was responsible for both the acceleration of a falling apple (9.8 m s–2) and the centripetal acceleration (see in2 Physics @ Preliminary section 2.3) of the orbiting Moon. However, he didn’t know the Earth’s mass ME or the value of G, but by using ratios of acceleration and distance squared, he showed that GME = ad 2 is the same for an apple and the Moon, confirming that the same law of gravity applied to both.

In 1798, the Earth’s mass was finally measured. Henry Cavendish (1731–1810) (who discovered hydrogen) measured the average density of the Earth ρE to be 5.448 times denser than water (1000 kg m–3). The experiment was designed by John Michell (1724 –1793) (who first predicted the existence of black holes). Since the Earth’s radius was accurately known, this was equivalent to both ‘weighing the Earth’ and measuring the value of G.

Cavendish used an extremely sensitive ‘torsional balance’ (Figure 1.4.4) to measure the tiny gravitational attraction between two small lead spheres m (attached to a thin 1.86 m rod) and two nearby large lead spheres M. From the angle of twist θ in the calibrated torsion wire, he determined the gravitational force between the spheres. From this, and using Newton’s equation for gravitational force, he calculated the Earth’s average density.

Vibrations, temperature variations and slight air movements would disturb the apparatus, so it was built into a small sealed building, with Cavendish outside, operating the apparatus via cords and pulleys, and making observations through telescopes in the walls.

The shift in position of the smaller masses was about 4 mm.1 Because of Earth’s gravitational field, the Moon

must accelerate towards Earth. Why doesn’t the Moon crash into Earth?

2 Using Cavendish’s value for Earth’s density ρ,

the definition ρ = mass

volume and the mean radius

rE = 6.37 × 106 m, calculate the Earth’s mass and compare it with the modern value.

3 Using Newton’s universal gravitation equation, Cavendish’s value for mE, the modern values for rE and g = 9.8 m s–2, calculate G and compare it with the modern value.

4 Using the modern value for G, calculate the total gravitational force between the spheres measured by Cavendish. (Hint: Calculate the force between a small and a large sphere in a single pair and double it. Ignore the thin rod etc.)

Distance between sphere centres r = 0.225 m, large sphere mass M = 158 kg, small sphere mass m = 0.73 kg

5 Typical laser printer paper weighs 0.080 kg m–2. Calculate the size (in mm) of a square piece of printer paper that would have a weight on Earth equivalent to the force in Question 4.


torsion wire

M

M F

r

F

m

m

θ

Figure 1.4.4 (a) Schematic and (b) cutaway view of the apparatus used by Cavendish to ‘weigh the Earth’

a b

2

26

Explaining and exploring the solar system

2.1 Launching spacecraftIn his book Philosophiae Naturalis Principia Mathematica (Principia for short), Newton used his law of gravity and laws of motion to explain Kepler’s laws of planetary motion, but also predicted the launching of artificial satellites and projectiles capable of escaping Earth’s gravity. Once you understand the physics behind something, it becomes possible to create new technology. In the case of space flight, it took 300 years to release the potential buried within Newton’s equations, via the 2000-year-old Chinese technology of fireworks.

A bite-size history of rocketryFor most of the history of rocketry, starting with the invention by the Chinese of gunpowder (the first rocket fuel or propellant) sometime between 300 bce and 850 ce, the technology was driven mainly by military applications. The Chinese invented the first rockets or ‘fire arrows’ (fireworks tied to arrows). Some of the early milestones of this history are summarised in Table 2.1.1 in the Physics Feature ‘Fire Arrows’ on page 29.

Only in the 20th century were civilian and scientific applications of rocketry (space exploration, Earth monitoring and communications) finally considered to be potentially as important as the military ones.

Getting up thereHow many times have you been told to ‘stop dreaming and be practical’? For scientists and engineers, both dreams and practical know-how were potent tools to turn the understanding of the physics of gravity and motion into the technology of space travel. Most of the important pioneers of rocketry were inspired to pursue dreams of space travel by reading Jules Verne’s (1828–1905) story From the Earth to the Moon, or the stories of HG Wells (1866–1946). But they also had a solid grounding in physics and engineering.

propellant, impulse, exhaust velocity, reaction device, thrust, payload, g-force,

effectively weightless, lift-off, Kepler’s laws, satellite, ellipse, orbital velocity, eccentric,

semimajor axis, periapsis, apoapsis, perihelion, aphelion, perigee, apogee, hyperbola, closed or

stable orbit, geosynchronous, geostationary, medium Earth orbit, semi-synchronous, gravity assist, slingshot effect, re-entry,

orbital decay, drag, lift, supersonic, hypersonic, shock wave, heat shield, ablation

Figure 2.1.1 The Apollo 11 mission: the launch of a Saturn 5 booster—the largest rocket in history—on its way to deliver the first humans to the Moon

27

spacE

Here we’ll concentrate on the important rocket researchers of the 20th century, the period in which the most rapid scientific advances took place. Below is a list their most important contributions.

Konstantin Tsiolkovsky (1857–1935) Tsiolkovsky (also Tsiolkovskii), a Russian mathematics teacher, derived the basic rocketry equations including the ‘Tsiolkovsky rocket equation’ (see Physics Phile ‘This is rocket science’, p 30), used Newton’s definition of escape velocity to calculate it for Earth, and proposed multi-stage rockets and steerable thrusters. He advocated the use of liquid propellants (including liquid hydrogen) because they could be controlled using valves and would give a larger impulse than solids (see in2 Physics @ Preliminary section 4.5). He also wrote science fiction, predicting space stations, and space colonies using biological recycling of food and oxygen and airlocks for moving between a spacecraft and vacuum.

Robert Goddard (1882–1945) Goddard, a US physicist, invented and tested many practical aspects of rockets, launching the first liquid-propellant rockets (liquid oxygen–gasoline) in 1926. He confirmed experimentally that rockets work in vacuum and showed that an hourglass-shaped de Laval steam nozzle greatly increased rocket efficiency. He launched the first scientific payload (camera, thermometer and barometer) that parachuted back to Earth, and steered rockets using vanes to direct exhaust gas and a gimballed (pivoted) nozzle under the automatic control of a gyroscope. He even experimented with very futuristic ion thrusters. Goddard attracted public ridicule by predicting travel to the Moon (see in2 Physics @ Preliminary Physics Phile p 43). He was mostly ignored by the US government, but he strongly influenced Oberth, von Braun and Korolyov (see below).

Robert Esnault-Pelterie or REP (1881–1957) REP, a French aircraft designer, wrote on interplanetary travel, calculated the energies and flight times for trips to the Moon, Venus and Mars and proposed atomic energy to power interplanetary craft. With André Hirsch, he established the REP–Hirsch Prize for aeronautics, the first winner being Oberth (below). In 1931, Esnault-Pelterie conducted early experiments with liquid propellants (petrol–liquid oxygen, benzene–nitrogen peroxide and tetranitromethane) and developed a gimballed nozzle.

Herman Oberth (1894–1989) The German physicist Oberth’s PhD thesis describing space travel was initially rejected as ‘utopian’ (though it was later accepted), so he published it as an influential book By Rocket into Planetary Space. In it he developed equations for space flight, proposed a design for a two-stage rocket using hydrogen–oxygen propellant and described craft for human space exploration. A follow-up book won him the REP–Hirsch Prize, which he used to purchase rocket engines for research assisted by his student Wernher von Braun. He worked (with von Braun) on both the Nazi V-2 rocket program and later the American rocket program. In 1953 he published Man in Space, proposing space stations, space-based telescopes and space suits.

Figure 2.1.2 Konstantin Tsiolkovsky

Figure 2.1.3 Robert Goddard

Figure 2.1.4 Robert Esnault-Pelterie

Figure 2.1.5 Herman Oberth

activity 2.1

pRacTIcaL EXpERIENcEs


Explaining and exploring the solar system2

28

Wernher von Braun (1912–1977) As a student, von Braun (German physicist and aeronautical engineer) tested Oberth’s rocket engines. He was an early amateur researcher in the Spaceflight Society, which was taken over by the Nazis. Under the Nazis von Braun led the team that developed the alcohol–oxygen-fuelled A4 (or V-2) rocket used on Allied cities including London, killing and wounding thousands. After the war, he joined the US army’s nuclear missile program. He dreamed of a civilian space program. In magazines and television, he publicly promoted exploration to the Moon and Mars with permanent colonies and orbiting space stations serviced by re-usable shuttle-type craft.

In 1957 the USSR launched Sputnik, the first artificial satellite, shocking the US and leading to the ‘space race’ of the 60s between the USSR and the US. In response, a civilian space agency, the National Aeronautics and Space Administration (NASA), was formed, and in 1960 von Braun became director of its Marshall Space Flight Center. He became a major figure in the race to the Moon (the Apollo missions) announced in 1961 by President Kennedy. He led the project to construct the largest rocket ever built—the Saturn 5 (Figure 2.1.1).

As is well known, the US won the race to the Moon in 1969, although they spent much of the 60s catching up to many USSR space ‘firsts’. The race also led to rapid development of civilian satellites for communications, Earth surface–atmospheric monitoring and scientific space exploration.

Sergey Korolyov (also Sergei Korolev) (1907–1966) Korolyov, a Ukrainian-born Russian aircraft designer, was known only as the ‘Chief Designer’ of the USSR space program—his name was kept secret until his death. He helped set up the Jet Propulsion Research Group, which launched liquid-fuelled rockets in 1933, and led to the USSR government forming the Jet Propulsion Research Institute, with Korolyov as Deputy Chief. During Stalin’s Great Purge of 1938, Korolyov was imprisoned for 6 years, then released to become a rocket designer in the nuclear missile program, where he quickly improved on the design of captured Nazi V-2 missiles.

Like his US rival von Braun, he dreamed of space travel and tried to convince his government to allow civilian projects. In 1957, he was allowed to launch the first artificial satellite Sputnik into orbit, starting the space race. He oversaw a string of space firsts (and failures): first animal (dog) in orbit, first unmanned Moon landing, first image of the unseen side of the Moon, first man and first woman in orbit, first extra-vehicular activity (space walk), first fly-pasts of Venus and Mars and more. Launch failures of four N1 boosters (rival to von Braun’s Saturn 5) and Korolyov’s death in 1966 helped to lose the race to the Moon for the USSR.

Gerard O’Neill (1927–1992) O’Neill, a US physicist, invented the particle storage ring used in particle accelerators, and an early wireless computer network. He led development of a satellite positioning system—a precursor to the US global positioning system (GPS). Through conferences, papers and books, he was an energetic advocate of space travel. He proposed colonies in cylindrical spacecraft positioned at

Figure 2.1.7 Sergey Korolyov

Figure 2.1.6 Wernher von Braun

Figure 2.1.8 Gerard O’Neill

29

spacE

PHYSICS FEATURE

‘Lagrange points’. (These are five stable locations around pairs of orbiting bodies such as Earth and Moon at which a test mass can remain indefinitely, requiring little or no thrust.) He suggested that colonists would live on the inner surface of these cylinders 3 km in radius and 20 km long. The cylinders would spin, using centripetal force, to simulate gravity, and the inside would be covered with Earth-like geography.

FIRE ARRowS

The following table is a very incomplete summary

of some of the highlights of the 24-century long history of rocketry.


4. Implications for society and the environment


Table 2.1.1 Some milestones in the pre-20th century history of rocketry

300 BCE to 850 At some time between these dates, the Chinese invent gunpowder and fireworks.

1150–1200 The Chinese develop the first rockets, ‘fire arrows’ (fireworks tied to arrows), and projectile weapons including grenades and cannons are used against invading Mongols.

1200–1300 Invading Mongols bring Chinese rocket technology to Europe and the Arabian Peninsula.

1529–1556 Conrad Haas (Austria) proposes the first designs for multi-staged rockets.

1687 Isaac Newton publishes Philosophiae Naturalis Principia Mathematica containing his three laws of motion and the law of universal gravitation. He defines escape velocity and predicts artificial satellites.

~1730 German Colonel von Geissler manufactures rockets (up to 54 kg) for warfare.

1792, 1799 Sultan Tipu (India) uses iron-cased 1 km range rockets against British troops.

1803–1806 Impressed by Tipu, Sir William Congreve (Britain) develops more accurate 3 km range rockets up to 136 kg, which were used successfully against Napoleon’s ships and against the Americans in the war of 1812.

19th century Engineers, scientists, inventors and crackpots experiment with non-military applications of rockets.

1821 Rocket-propelled harpoons are used to hunt whales.

1861–1865 Rockets are used in the American Civil War.

1865 Science fiction writer Jules Verne (France) publishes From the Earth to the Moon.

1903 Konstantin Tsiolkovsky (Russia) publishes reports in which he applies rigorous physics to rocketry and discusses the possibility of space travel.

Figure 2.1.9 The Chinese character for ‘rocket’ translates literally as ‘fire-arrow’.

Identify data sources, gather, analyse and present information on the contribution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun.


30

Forces and rocketsTo understand the forces exerted on rockets and astronauts during take-off, we first need to define some terms.

ThrustTsiolkovsky called a rocket a reaction device. This is because the burning propellant forms hot, high-pressure exhaust gas that is forced through the nozzle at high exhaust speed ve. By Newton’s third law (see in2 Physics @ Preliminary section 3.5), the large force exerted on the exhaust gas results in a reaction (called thrust) back on the rocket, pushing it forward. You can use momentum to calculate the thrust. (See in2 Physics @ Preliminary Worked example, p 70.)

Let’s analyse the [rocket + propellant + exhaust] as our system. The forces ejecting exhaust out of the nozzle are internal forces, so they can’t change the net momentum of the system; therefore, momentum gained by the ejected exhaust ∆(meve) must be cancelled by the [rocket + propellant] gaining ‘equal and opposite’ momentum. An increase in momentum (impulse) of the [rocket + propellant] implies acceleration and, hence, a force (called thrust FT).

Suppose the speed of the exhaust gas ve is constant over a time period ∆t. The impulse J = FT∆t exerted on the [rocket + propellant] and the impulse ∆(meve) exerted on the exhaust gas are equal in magnitude:

J = FT∆t = ∆(meve) = ve∆me

Rearrange: T ee

t=∆∆m

F v

where T e

et=

∆∆m

F v is the mass of exhaust gas lost per unit time. Increasing exhaust

speed ve is important in rocket design because it increases thrust FT . Even if thrust FT = mRa is constant, because the mass mR of the [rocket + propellant] is rapidly reducing, the acceleration a rapidly increases during launch.

Worked exampleQUESTIonThe thrust equation doesn’t only apply to rockets. A fireman holding a hose was not prepared when the water was turned on and was knocked over by the unexpected thrust. Water exited the spout with a speed 39.0 m s–1 and with a flow rate 470 L min–1.

Calculate the force that knocked him over. (Water density is 1000 kg m–3 = 1.00 kg L–1.)

SoLUTIonMass flow rate of water = 470 L min–1 × 1.00 kg L–1/60 s = 7.833 kg s–1

Equation for thrust: FT = vm

ee

t∆∆

= 39.0 m s–1 × 7.833 kg s–1 = 305 N

Rocket engines and stagesThere are two basic kinds of rocket engine, those using solid and those using liquid propellant (Figure 2.1.10). Solid propellant engines are simpler and can achieve maximum thrust faster, but cannot be controlled once they start. Liquid propellant engines are more complicated and slower to start, but can be controlled and produce greater thrust.

A multi-stage rocket can deliver a heavier payload (space cargo) because, when the propellant in each stage is finished, that stage can be jettisoned (falling into the ocean) reducing the rocket’s mass and so allowing for greater payload mass. Rockets from the ‘space-race’ era were typically liquid-propelled stages stacked

Analyse the changing acceleration of a rocket during launch in terms of the: • LawofConservation

of Momentum• forcesexperienced

by astronauts.

ThIs Is RockET scIENcE!

Using calculus and Newton’s second law, Tsiolkovsky derived

his famous ‘delta v’ rocket equation:

∆v vmm

= ei

fln

where ∆v is the magnitude of velocity change during a rocket ‘burn’, ve is the exhaust velocity and mi and mf are the initial and final masses of the rocket (plus remaining propellant). However, a recently discovered pamphlet by William Moore showed he had derived a similar equation in 1813.

31

spacE

vertically. More recent rockets such as the US Space Shuttle and the European Ariane 5 use a combination of liquid- and solid-propelled stages stacked side-by-side. Russian rockets, such as the Proton, use only liquid propellant. Of course, multi-stage rockets are also more complicated and so have more ways to fail. In the US, smaller satellites and military missiles are launched using simpler, solid-fuel rocket engines.

g-force Maybe you’ve heard that astronauts are ‘squashed by g-force’ when a rocket accelerates on take-off. Often the term ‘g-force’ is used to quantify the effects on your body of accelerations experienced in a roller-coaster, car or aeroplane. It’s not an accurate name, because its value is more closely related to acceleration than force, but it is used extensively in aeronautics and astronautics, so we’ll give you a commonly used definition. The ‘g’ refers to acceleration expressed in units of g = 9.8 m s–2. The ‘force’ refers to the fact that a net external force is responsible for that acceleration and it is this force and resulting reaction forces within a body that are responsible for the effects of g-force. Sometimes the term ‘g-load’ is used instead.

Let’s start with the vertical component of motion and g-force. Consider three situations:1 While you are sitting or standing still (not accelerating), there is no net force

on you. Your body is compressed by a pair of balanced forces—weight mg downwards, and the upward normal force N from the seat or floor. (See in2 Physics @ Preliminary p 45.) This compression causes the internal effects of weight. Your body is experiencing the compressive effect of 1 unit of Earth gravity (i.e. g-force = 1).

2 If you are accelerating upwards, such as riding the bottom of a curve on a roller-coaster or in a rocket during take-off, the net force is upwards— the normal force from the seat is larger than your weight. Your body compresses more than usual, as though you are heavier. If your net upward acceleration is 9.8 m s–2 (1g), then your body is compressed as though you are in a gravitational field of 1 + 1 = 2 units of Earth gravity (i.e. g-force = 2).

3 In free-fall (or in orbit), the normal force from the chair disappears and you are no longer compressed. You feel effectively weightless (see in2 Physics @ Preliminary pp 37–38), even though at typical Space Shuttle altitudes you actually have ~90% of your weight on Earth. In this case you are experiencing g-force = 0, the same effect as 0 units of Earth gravity.

The term g-force usually means apparent weight divided by true weight on Earth. Apparent weight is what appears on a set of bathroom scales. Bathroom scales actually measure the magnitude of the normal force, not true weight, so, to calculate g-force, first calculate normal force. Consider vertical motion only. Weight mg is down and normal force N is up. Let av be vertical acceleration and let up be positive: Fnet = mav = N + (–mg)

Apparent weight N: N = mav + (+mg)

The apparent weight increase is caused by the increase in normal force N due to the term mav, which is simply the net force accelerating you.

Vertical g-force = N amg mg

mgm=

+ +v ( )

Vertical g-force = avg +1

Identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch.

Figure 2.1.10 Typical (a) solid-propellant and (b) liquid-propellant rocket engines

exhaust exhaust

nozzle nozzle

combustionchamber

solid fuel–oxidiser mixture

fuel

oxidiser

pumps

combustionchamber

a b


32

The avg term in the g-force formula represents the effects of your acceleration

due to the net force and the +1 represents the background effects of your real weight.

Notice that in free-fall your acceleration is –g, so the g-force is –1 + 1 = 0.If you are holding a rope or are strapped into a harness being pulled upwards

instead of sitting or standing, then the above discussion and formulae still apply except that now your body is being stretched and the normal force is replaced by the tension force.

The horizontal component of g-force, such as when you accelerate or decelerate in a car, is easier to calculate. (See in2 Physics @ Preliminary Physics Phile ‘g-Whiz’ p 11.) As there is no horizontal component of weight mg, the g-force equation simplifies:

Horizontal g-force = ahg

To calculate the resultant g-force, combine the vertical and horizontal components of g-force using vector addition.

Worked exampleQUESTIonYou round the bottom of an upturning curve on a roller-coaster at a speed of 36.0 km h–1. The curve is circular with radius 5.00 m. Calculate the g-force you would experience.

SoLUTIonAt that moment, you experience (upward) centripetal acceleration (see in2 Physics @Preliminary section 2.3). Calculate the vertical g-force.

v = = =− − −36363 6

101 1 1km h m s m s.

avRc m s= = = −

2 2210

5 0020 0

..

Vertical g-force = + = + =avg

19 80

1 3 04.

.+20.0

One effect of g-force is that the apparently increased weight of the blood drains it from the head, affecting vision and consciousness. On average, 4 –5 g causes dimming of vision, 5–6 g visual blackout and above 6 g you experience loss of consciousness (‘g-LOC’). Much larger g-forces can be tolerated for periods of less than about 4 seconds.

To increase g-force tolerance during launch, astronauts face the direction of acceleration. This orientation is called ‘eyeballs in’, because the eyeballs are effectively pushed into their sockets. Also, the seats are oriented with the head and body lying horizontally (Figure 2.1.11). In this way, g-force doesn’t easily force blood into or out of the head. Fighter pilots and astronauts also wear ‘g-suits’ containing inflatable bladders in the trousers which squeeze blood out of the legs and back into the head.

Using a powerful cannon to launch a satellite (see Newton’s thought experiment in section 2.2) would not work because of the enormous g-force from the initial explosion.


ThE FIRsT asTRoNauTs?

According to Chinese legend, in 1500 a senior bureaucrat

(a Mandarin) called Wan Hu tried to launch himself into space by tying 47 gunpowder rockets to a chair. He failed to become the first astronaut by dying in the explosion at launch.

In about 1806 in France, Claude Ruggieri launched a rocket containing a sheep ~300 m into the air, parachuting it back to Earth alive. The police prevented him from turning a small boy into the first astronaut by the same method.

33

spacE

Figure 2.1.11 Gemini 3 astronauts Gus Grissom and John Young strapped into their horizontally oriented seats are being prepared for launch (1965).

Warning! The terms g-force and g-load are not SI quantities. They are informal terms and are sometimes used carelessly. Sometimes g-force and g-load are used to mean the same thing. Sometimes g-load is used to mean only the net acceleration in units of g, not including the effect of gravity. When reading g-force or g-load data, be careful to check which definition is being used.

Force during take-offHere we’ll account for the forces, accelerations and g-force experienced during a typical launch. This example is for a Space Shuttle, but the principles apply to other craft. In the Shuttle, the solid rockets boosters and the main liquid-propelled engines fire-up at the same time, while in a more traditional multi-stage rocket each stage fires sequentially.

Figure 2.1.12 is a graph of g-force experienced by everything within the Shuttle during launch. Just before lift-off (or take-off ), the vector sum of thrust plus the force exerted by the gantry (the crane-like structure holding the rocket) plus the rocket’s weight is zero, so there is no acceleration. Because acceleration is zero, the net force on the astronaut is also zero, so the astronaut’s weight is balanced by the normal force exerted by the seat: g-force = 1 (point A).

After lift-off, thrust is larger in magnitude than rocket weight plus air resistance, so the rocket (and astronaut) accelerate upwards. Now the seat exerts a normal force greater than the astronaut’s weight. The g-force is greater than 1, increasing steadily, along with acceleration as the mass of remaining propellant decreases. Note that throughout the launch process, the craft is ‘pitching over’ from vertical to horizontal motion, so the gravitational contribution to g-force is becoming progressively smaller in the direction of motion.

The ~0.2 drop in g-force between points A and B is due to ‘throttle down’; air resistance-induced pressure on the Shuttle surface reaches a dangerous maximum (‘max Q’), so thrust is deliberately reduced until the atmosphere thins out. Thrust is increased again and g-force increases to between 2 and 3 (point B).

As the mass of remaining propellant decreases, acceleration (and g-force) increases until fuel in the boosters (or lower stage in a traditional rocket) runs out. Acceleration decreases dramatically, but the boosters (or empty stages) are discarded (point C), and the remaining engines provide the thrust. Acceleration increases again as propellant mass decreases. To avoid the astronauts and payload

Analyse the changing acceleration of a rocket during launch in terms of the: • LawofConservation

of Momentum• forcesexperienced

by astronauts.


34

being subjected to a dangerously high g-force, the thrust must be reduced to limit g-force to 3 g (point D).

Once the rocket is in orbit, the rockets stop firing (point E). The only force acting now is weight (providing the centripetal force of orbit), so the rocket and the astronaut are both in free-fall and effectively weightless; the astronaut experiences a zero g-force.

Running startIt takes a lot of fuel to get a spacecraft to a high enough altitude and high enough speed to achieve orbit. You can get higher if (like a pole vaulter) you get a ‘run-up’ before lift-off. Given that Earth rotates rapidly, a rocket already has a large easterly tangential velocity at launch. So, if you launch towards the east, you can use less propellant, carry a larger payload or go into a higher orbit. The closer you are to the equator, the faster your initial speed u (Figure 2.1.13). At the equator u = 465 m s–1.

Worked exampleQUESTIonCompare Earth’s rotational tangential speed vT at the rocket launch facilities at Woomera, Australia (used during the 60s and 70s), and Kourou, French Guiana.

Data: Earth’s radius rE = 6.37 × 106 m

Earth’s rotational period T = 86 164 s

Woomera: Latitude 31.1°S, longitude 136.8°E.

Kourou: Latitude 5.2° N, longitude 52.8°W

SoLUTIonTangential speed: v

RTT = 2π

(See in2 Physics @ Preliminary p 29.)

Radius of rotation R depends on θlat, the latitude angle: R = rE cos θlat


Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket.

0 100 200 300 400 500 600

3.5

3.0

2.5

2.0

1.5

1.0

0.5

E in orbitD accelerating upto orbital speed

C solid rocketboosterseparation

external tankseparation

B reducedair resistance

Alift-off

A

B

C

D

E

Time (s)

g-fo

rce

Figure 2.1.12 g-force during a typical Shuttle launch

Earth’s rotation

v1

v2

tangential velocity

Figure 2.1.13 Rockets are usually launched towards the east, to take advantage of Earth’s rotation. The effect is greatest at the equator.

35

spacE

vr

TTE

lat lat 465= = × × =2 2 6 37 10

86164

6πθ π θ θcos

.cos cos llat

Woomera: 465 cos 31.1° = 398 m s–1

Kourou: 465 cos 5.2° = 463 m s–1

To explore the solar system, you need to reach Earth’s escape velocity. As the Earth also orbits the Sun, you can get an extra boost from Earth’s orbital speed (about 3.0 × 104 m s–1) if you launch at a time of the year when the Earth’s orbital motion points in the desired direction (Figure 2.1.14).

NoT ENough houRs IN a day

The absolute period of Earth’s rotation as determined from the orientation of

distant stars is only 23 h, 56 min and 4 s, or 86 164 s (a sidereal day). The normal 24 hour mean solar day is longer because it includes the extra time needed for the Earth’s rotation to catch up with the extra component of the Sun’s apparent motion in the sky due to the Earth’s orbit.

direction of Sun orbitalmotion

Earth

rotational motion

Figure 2.1.14 Interplanetary missions require higher launch velocities. These launches need to take advantage of the Earth’s orbital velocity.

CHECKPoInT 2.11 Which two people led the space race between the USSR and the US?2 What event triggered the US government to fund a large civilian rocket program?3 What are Lagrange points?4 Name the two basic kinds of rocket engine.5 Explain why a multi-stage rocket allows a heavier payload.6 What is the vertical g-force on someone: a standing stationary? b in free-fall? c in orbit?7 Explain why the acceleration of a firing rocket increases with time.8 Explain one reason why most rockets are launched towards an easterly direction.

2.2 orbits and gravityHow and why do spacecraft stay up there? Newton, using his three laws of motion and his law of universal gravitation, showed not only that gravity provided the centripetal force required to keep the Moon in orbit around the Earth, but also that he could use his laws to explain all of Kepler’s laws of planetary motion.

Newton developed a thought experiment to understand orbit. He imagined standing on a mountain (Figure 2.2.1) and firing a projectile horizontally from a powerful cannon. Gravity would accelerate the projectile towards the ground,

Discuss the importance of Newton’sLawofUniversalGravitation in understanding and calculating the motion of satellites.


36

curving it downwards until impact. If you increased the projectile’s initial velocity, it would travel further around the Earth and its trajectory would be less steeply curved. Eventually you would reach an initial velocity at which the curvature of the trajectory exactly matched the curvature of the Earth itself, so that the projectile would never catch up to the ground—the projectile was now in a circular orbit. The projectile would become an artificial satellite (an object in orbit around a much larger one) in the same way that the Moon is a natural satellite of Earth. Newton also showed that if you increased the velocity further, the orbit would become an ellipse. At high enough velocity (escape velocity), the projectile would never return.

Deriving centripetal forceYou have seen the formula for centripetal acceleration ac, but where does it come from? Look at Figure 2.2.2. Suppose an object moves uniformly in a circle. The magnitudes of the initial and final velocities (vi and vf ) are the same, so call them both v (v = vi = vf ). The distance travelled d = v∆t in going from A to B is R θ (using radians), giving R θ = v∆t, which can be rearranged to:

θ = v tR∆

To find the instantaneous acceleration rather than the average over a long time, use a ∆t (and therefore θ) that approaches zero.

Consider the ∆v vector diagram on the right in Figure 2.2.2. It is an isosceles triangle with two equal sides of length v. For θ approaching zero, the length of ∆v approaches the length of the arc v θ between vf and -vi so:

∆v → v θIf you combine this with the previous equation, eliminating θ, then in

the limit:∆v = v2∆t

R

Divide both sides by ∆t : ∆v

ta

vR∆

= =c

2

Then of course, to get centripetal force, multiply centripetal acceleration by mass (see in2 Physics @ Preliminary p 46). Don’t forget that the magnitude of tangential velocity is v = 2pr ⁄ T. For objects in orbit, the tangential velocity is also called the orbital velocity.

Heaven and EarthNewton showed that the force acting in the ‘heavens’ to keep the Moon in orbit was the same one acting on small projectiles on Earth. He assumed that in both cases the force is given by his law of universal gravitation. At that time, neither the Earth’s mass ME nor the value of G were known, but if you rearrange the formula:

F ma GM m

d= = E

2

their product GME = ad 2 should evidently be the same constant for the orbiting Moon and a falling apple on Earth. Newton showed that this was true.

activity 2.2



Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’sLawofPeriods.

Vi

–Vi

Vf

ΔV

Vf

θ

θ

A B

R R

Figure 2.2.2 This diagram shows the derivation of the centripetal acceleration formula.

Figure 2.2.1 This illustration from Newton’s Principia illustrates the principles behind orbits, and launching an artificial satellite.

37

spacE

Worked exampleQUESTIonShow that GME = ad 2 is the same for a falling apple and the orbiting Moon. Assume that the Moon’s orbit is circular.

Data: Earth’s radius rE = 6.37 × 106 m

Average Earth–Moon distance dEM = 3.84 × 108 m

Moon’s orbital period T = 27.3 days

SoLUTIonApple on Earth: GME = ad 2 = 9.80 m s–2 × (6.37 × 106 m)2 = 3.98 × 1014 m3 s–2

Moon in orbit: Centripetal acceleration avrc =

2

Orbital velocity vr

Ta

r

T= ∴ =2 4 2

2

π π

Orbit radius r = dEM; T = 27.3 × 24 × 3600 = 2.359 × 106 s

GME = ad 2 = 4 4 3 84 102 3

2

2 8π πd

TEM

3

6 2

m)

(2.359 10 s)= 4.0= × ×

×( .

22 10 m s14 3 –2× The two values agree within ~1%.

Extension: (Hard) Can you think of a reason for the ~1% discrepancy? (Hint: See Physics Phile ‘Finding new planets’ page 40.)

Kepler’s laws and satellitesJohannes Kepler (1571–1630), using Tycho Brahe’s data, showed that the known planets and Earth orbit the Sun in ellipses that obey his three laws of planetary motion, but he had no idea why. An ellipse is a circle, stretched along one dimension. The more stretched the ellipse, the more eccentric it is. Most of the orbits of planets of our solar system are very nearly circular, not very eccentric. Comets have highly eccentric orbits.

Figure 2.2.3 defines some properties of elliptical orbits. The semimajor axis is half the length of the ellipse’s longest axis. It can also be thought of as a kind of average radius for the orbit. For a circular orbit, the semimajor axis is the radius. The point of closest approach of the orbit to the central body is called the periapsis. The furthest point is the apoapsis. If the central body is the Sun, then these points can also be called perihelion and aphelion (helios is ‘Sun’ in Greek). If the central body is the Earth, then they are the perigee and apogee (geos is ‘Earth’ in Greek).

Let a be the semimajor axis and let dA and dP be the distances from the central body to the aphelion and perihelion respectively, then look at Figure 2.2.3a and confirm that a is the average of these: a = (dA + dP)/2.

You had a sneak preview of Kepler’s laws of planetary motion in the Preliminary text (see in2 Physics @ Preliminary p 250). Here they are again:1 The orbits of the planets are ellipses, with the Sun at one focus (Figure 2.2.3a).


F G

m m

d= 1 2

2

Solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using:

Fmvr=

2



38

2 A line joining the planet to the Sun sweeps out equal areas in equal times. This means the planet travels faster when it is closer to the Sun (Figure 2.2.3b).

3 For all planets orbiting the Sun, the square of the orbital period T is proportional to the cube of the semimajor axis a. This is the law of periods.

T

a

2

3= constant constant

These laws hold for any orbiting system of two bodies if the mass of the central body is very much larger than the mass of the other. They also apply to circular orbits, since a circle is a special case of an ellipse.

If c is the half the distance between the foci of an ellipse, then eccentricity e is defined as e c

a= . A circle is an ellipse with zero eccentricity (e = 0); both foci are together at the circle centre. For a circle, the semimajor axis becomes the radius, a = r. Orbits of planets in our solar system are very nearly circular, the two most eccentric being Mercury with e = 0.2056 and Mars with e = 0.0934. Pluto has e = 0.2482 but, sadly for Pluto-fans, it was demoted to a dwarf planet in 2006.

Sun

focus focus focus

perihelion(closest tothe Sun)

aphelion(farthest from Sun)

Area = Area

semimajor axis

a b

Figure 2.2.3 (a) A highly eccentric elliptical orbit. (b) Kepler’s law of areas: a line joining the planet to the Sun sweeps out equal areas in equal times.

Discuss the importance of Newton’sLawofUniversalGravitation in understanding and calculating the motion of satellites.

Newton derived Kepler’s laws from his laws of motion and gravitation. He showed that Kepler’s first law (law of elliptical orbits) follows from the inverse square law. By including his three laws of motion, he also proved Kepler’s second law (law of equal areas). These derivations are beyond the syllabus; however, showing Kepler’s third law, the law of periods, for a circular orbit is very easy. Careful! Don’t confuse the semimajor axis a with centripetal acceleration ac.

Suppose a satellite of mass m orbits a central body of mass M and m << M so that the acceleration of M is negligible. In a circular orbit, the satellite’s acceleration is centripetal:

avr

vr

Ta

r

Tc cand= = ∴ =

2 2

2

2 4π π

where v is orbital velocity, r is orbital radius and T is orbital period. The magnitude of gravitational force exerted on m is:

F = mac = GmM ⇒ aGM

r

r

T

T

r GMc = = ∴ =2

2

2

2

3

24 4π π (which is constant)r 2

39

spacE

We derived this for a circular orbit, but it is also true for elliptical orbits if we replace radius r with semimajor axis a.

T

a GM

2

3

24= π

Because here the value of Kepler’s constant is explicit, we will call this the ‘explicit’ form of Kepler’s law of periods.

Worked exampleQUESTIonFrom Earth, you observe (almost edge on) the orbit of Jupiter’s moon Ganymede and determine its orbital period to be T = 7.15 Earth days. You measure the width of the orbit to be w = 2.14 × 109 m. Assuming the orbit is circular, determine Jupiter’s mass.

SoLUTIonAssuming orbit is circular: r

w= = ×2

1 07 109. m

Explicit form of Kepler’s third Law: r

T

GM3

2 24=

π

Rearrange and evaluate: Mr

GT= = × ×

× × × ×−4 4 1 07 10

6 67 10 7 15 24 3

2 3

2

2 9 3

11

π π ( . )

. ( . 66001 90 10

227

).= × kg

An early triumph of the law of universal gravitation occurred when Newton’s friend Edmund Halley used it to show that the trajectory (Figure 2.2.4) of a comet he had observed (now called Halley’s Comet) fitted with the trajectories of two previously observed comets. Halley concluded it was the same comet, and correctly predicted that it would return every 76 years.

The success of Newton’s law of universal gravitation was not simply that it could be used to explain Kepler’s laws, which were already known, but that it could be used to predict other phenomena not yet observed (such as space travel). It could also be used to accurately predict small deviations of planets from Kepler’s ideal orbits around the Sun. For example, when planets pass near each other, local effects of gravity perturb them from perfect Keplerian orbits. The law of universal gravitation can be used to predict these deviations very accurately.

Halley included the effects of perturbations due to planets in his comet calculations. Similar deviations in the orbit of Uranus were attributed to the gravity of a then unknown planet. Neptune, that new planet, was found in 1846 within 1° of the position predicted using the law of universal gravitation. Several astronomers contributed to both the calculations and the observations, resulting in arguments about who deserved credit for discovering Neptune.

The first obvious failure of Newton’s gravitation law was in explaining the observation that the position of the perihelion of Mercury’s orbit was not fixed, but was precessing around the Sun. (See in2 Physics @ Preliminary p 255.) Perturbations due to gravity of other planets and other mechanical effects such as the Sun’s equatorial bulge were able to explain 99.23% of the precession, but the remaining 0.77% required an improved theory of gravitation—Einstein’s theory of general relativity.


r

T

GM3

2 24=

π

EarthMars

Jupiter

Saturn

Halley’sComet

Uranus Neptune

Figure 2.2.4 The eccentric orbit of Halley’s Comet in relation to the nearly circular orbits of Earth and other planets. The angle between the comet’s orbit and the plane of the orbits of the planets is not apparent here.


40

Worked exampleQUESTIonDerive an expression for the magnitude of orbital velocity for a satellite in a circular orbit, in terms of mass of the central body M and orbital radius r. Use this expression to calculate the Moon’s orbital speed, assuming a circular orbit.

Data: Average Earth–Moon distance d = 3.84 × 108 m

Earth’s mass Me = 5.97 × 1024 kg

SoLUTIonGravitational acceleration: a

GM

rg =

2

Resulting centripetal acceleration: a avr

GMrgc = ==

2

2

Rearrange:

v

GMr

= for circular orbits

Moon’s orbital speed: v = × × ×

×=

−−6 67 10 5 97 10

3 84 101020

11 24

81. .

.m s

Note that orbital speed is independent of the mass of the satellite. One of the consequences of this equation is that for circular orbits, the smaller the radius the faster the orbital speed.

Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of theorbitusingKepler’sLawof Periods.

FINdINg NEW pLaNETs

The mass of a large planet is not negligible compared with the mass of its star, so both

the planet and the star orbit the centre of mass of the system with the same period; the planet orbits on a large ellipse and the (more massive) star on a small one. Planets outside our solar system (extra-solar) are normally too far and faint to be seen directly in telescopes, but astronomers can deduce the presence of a very massive planet by detecting the small wobble of the star it orbits, using the Doppler effect. (See in2 Physics @ Preliminary p 259.) As the star moves towards and then away from us periodically, its spectrum shifts towards bluer and then redder wavelengths in succession.

large planet

star

centre of mass

Figure 2.2.5 A large, undetectable distant planet can cause its parent star to wobble detectably.

41

spacE

2.3 Beyond Kepler’s orbits Isaac Newton showed that all orbits consistent with the Law of

Universal Gravitation and his laws of motion fall into one of four possible types: the first two suggested by Kepler’s first law—circles and ellipses—as well as parabolas and hyperbolas. (Remember that a hyperbola is the curve you get when you graph y k

x= ). The central body (if very massive) would be at the focus of all these curves.

Consider Figure 2.3.1 (and Figure 2.2.1). Suppose we increase the speed of the satellite executing the circular orbit. The orbit will become elliptical. If we further increase the speed so that we reach escape velocity, the satellite won’t return. (See ‘Escape velocity: What goes up …?’ p 18). The orbit is a parabola. If we now increase the speed beyond escape velocity, the orbit becomes a hyperbola.

All these orbits are symmetrical in shape and speed—the speed of the satellite at any position is identical to its speed at a mirror image position (although obviously the direction is different).

Circles and ellipses are called closed or stable orbits because the satellite follows a closed curve, repeating its motion periodically and indefinitely if undisturbed. However, an object approaching a much larger mass doesn’t always fall into a stable orbit around it. Assuming that it doesn’t hit the surface of the larger mass, an object moving fast enough will execute an open orbit—a parabola or a hyperbola—flying off after the encounter, never to return.

The kind of orbit depends on the sign of the mechanical energy ME:

ME P= + = −K E mvGmM

r12

2

where m is the mass of the satellite and M is the central mass. Here, r is the instantaneous distance from the central mass. For a circular orbit, r is constant (= the orbital radius), but r varies with time for an elliptical orbit. In the absence

CHECKPoInT 2.21 Who first described the principles behind launching artificial satellites?2 If an object is in a circular orbit and you increase the orbital speed slightly, describe what happens to the shape

of the orbit.3 State the formula for centripetal acceleration.4 State Kepler’s laws. What condition is required in order to apply them to other systems of two bodies?5 Define the terms periapsis, perihelion and perigee.6 Define the terms apoapsis, aphelion and apogee.7 Compare the speed of a satellite at periapsis and apoapsis. Is there an exception?8 A circle is also an ellipse. True or False. Explain.9 Define the semimajor axis of a circle of radius r.0 Name the planet in our solar system with the most eccentric orbit.1 Outline the main reason the planets sometimes deviate from Kepler’s laws. Name the planet that was discovered

because of this and describe (briefly) how.

11

circular

elliptical

parabolic

hyperboliccentral body

Figure 2.3.1 Gravity allows four possible kinds of orbits.


42

of resistive forces such as air resistance, the mechanical energy is constant for any orbit. For stable orbits with semimajor axis a:

ME = − GmMa2

This implies that the lower (more negative) the energy, the smaller the orbit.

There are three possibilities: • Orbitisclosed(circleorellipse),ME<0.Theorbiting

object is gravitationally bound, v < ve (v = orbital velocity, ve = escape velocity).

• Orbitisaparabola,ME=0.Theobjectisborderlinegravitationally unbound, v = ve.

• Orbitisahyperbola,ME>0.Theobjectisgravitationallyunbound, v>ve.

Note that the parabolic orbit described above doesn’t have the same properties as the parabolic trajectory of a projectile near the Earth’s surface.

The projectile motion equations in section 1.1 don’t apply to this parabola. The x- and y-components of the motion don’t separate as easily as for down-to-Earth projectile motion.

A projectile passing by a large massive body (without being captured) follows a hyperbolic orbit (Figure 2.3.3), reaching maximum orbital speed at closest approach. The two arms of a hyperbolic orbit (like all hyperbolas) tend towards asymptotes; that is, the further from the central mass an object is, the closer its path tends towards a straight line and constant speed. Very rarely, if the projectile’s speed coincidentally equals the escape velocity, it would follow a parabolic orbit.

Worked exampleQUESTIonA comet of unknown mass mC passes the Sun (MS = 1.99 × 1030 kg). At perihelion, the centre-to-centre distance is r = 5.08 × 1010 m and the speed of the comet is v = 7.50 × 104 m s–1.

What type of orbit is it? Do you expect the comet to return?

SoLUTIon

ME CC S

CS

C

= − = −

= × ×

12

12

12

7 50 1

2 2m vGm M

rm v

GM

r

m ( . 00 )6 67 10 1 99 10

5 08 10

2

4 211 30

10

. .

.

(

− × × ××

= +

−

mC .. )0 108×

ME > 0, so the orbit is hyperbolic and the comet is not expected to return (unless later interaction with another body reduces its energy sufficiently).

(Careful! When subtracting two numbers of very similar size, check that the answer is not smaller than the smallest significant figure in the calculation.)

asymptotes tohyperbola

hyperbola

Figure 2.3.3 A projectile not captured by a large massive body follows a hyperbolic orbit (or very rarely, a parabolic one).

TRy ThIs!Plasticine orbital mechanicsIsaac Newton showed that all possible orbits are examples of ‘conic sections’, curves resulting from taking planar slices of a cone at different angles. Make an accurate cone out of plasticine or foam and use a sharp knife to cut sections to reveal orbital shapes: Cut horizontally = circle. Cut parallel to one edge of the cone = parabola.Cut less steeply than the parabola = ellipse. Cut more steeply than the parabola = hyperbola.

circle

ellipse

hyperbola

parabola

Figure 2.3.2 All orbits are conic sections.

43

spacE

LEO

GPS

GEO

Altitude (km) Radius (km)

35794 42164

20230 26600

160–2000 6530–8370

Strictly speaking, when calculating gravitational potential energy, we really should include the effect of other astronomical bodies such as nearby stars. However, as long as the other bodies are either too small or too far away, we can get away with calculating the ‘two-body ME’ using only the masses of the central and orbiting body. The rules relating ME to the shape of the orbit only apply to the two-body ME.

Orbits close to homeThe vast majority of artificial satellites (and all Space Shuttle and International Space Station missions) are launched into what are called low Earth orbits or LEOs (160–2000 km above the Earth’s surface, Figure 2.3.4). Because the atmosphere extends into this range, such orbits also experience some air resistance and so are temporary. The LEO band is heavily used and now highly contaminated by orbital debris (space junk), which can be dangerous in a collision. The positions of more than 8500 objects larger than 10 cm are monitored to avoid this.

However, LEOs have several advantages over higher orbits. Because the altitude (height above sea level) is small, it requires less fuel to reach. Some commercial communications satellites use LEOs because, being close to Earth’s surface, transmitters need less power. Military and civilian surveillance and surface monitoring satellites (for example those used by Google Maps) often use LEOs because, being close, photographic resolution of images is high. Also, the orbital period is short (between 1.5 and 2 hours), so there is rapid coverage of the Earth’s surface.

Surrounding the Earth are the Van Allen radiation belts, two belts of energetic electrons and ions trapped from the solar wind by the Earth’s magnetic field (see in2 Physics @ Preliminary p 307). Vehicles in LEO below ~1000 km altitude are protected from this radiation by distance and the atmosphere, so, except for the Apollo Moon missions, human spaceflights do not venture above LEO.

The explicit form of Kepler’s third law allows us to choose a useful orbital period. For example, if the orbit is circular with a radius of 42 164 km, then the orbital period of the satellite equals the rotational period of the Earth (23 h 56 min 4 s; see Physics Phile ‘Not enough hours in a day’, p 35). Such a satellite is called geosynchronous and would appear over the same patch of Earth at the same time each day.

A special case of a geosynchronous orbit is the circular easterly orbit aligned directly over the equator. The satellite follows the Earth’s rotation exactly and so appears stationary in the sky. This is a geostationary or Clarke orbit (Figure 2.3.4). Fixed Earth-bound dish antennas can be pointed permanently at a geostationary satellite. Most communications and broadcast satellites use this orbit.

Another special case of a geosynchronous orbit is the ‘Tundra orbit’, a highly elliptical orbit that is highly inclined to the equator and designed to dwell for extended periods each day over high latitudes.

Medium Earth orbits or MEOs have radii between LEO and geosynchronous orbits. A semi-synchronous satellite has an orbital period exactly half of Earth’s rotation period. Global positioning system (GPS) satellites are in circular semi-synchronous orbits (Figure 2.3.4) at various angles to

Compare qualitatively low Earth and geostationary orbits.


r

T

GM3

2 24=

π

Figure 2.3.4 Ranges of typical satellite orbital radii. GEO, geostationary orbit; GPS, global positioning system (semi-synchronous); and LEO, low Earth orbit. Orbits lying between LEO and GEO are medium Earth orbits (MEO). Below LEO is called ‘sub-orbital’.

aRE you sERIous?

In 1945 in a Wireless World magazine article, science-fiction writer Arthur C Clarke (who wrote

2001: A Space Odyssey) first described how three geostationary satellites could be used to allow continuous, worldwide radio communication, a feat not achieved until the mid-60s.

Aspects of the idea had already been suggested by earlier researchers, although Clarke produced the most complete plan. He didn’t patent his multi-billion dollar idea because a lawyer convinced him it was ‘too far-fetched to be taken seriously’.


44

the equator, so that at any time from anywhere at least six satellites are contactable. Arctic regions are poorly covered by geostationary satellites, so Russia uses communications satellites in Molniya orbits—highly elliptical, semi-synchronous orbits inclined to the equator, designed to dwell over Arctic regions for many hours each day. Russia and the USA both use such orbits for spy satellites.

2.4 Momentum bandits: the slingshot effect

Now you know how to get into orbit, but what are you going to do while you’re up there? To tour the solar system on very little propellant, you will need to learn to hitchhike and steal!

When exploring the outer parts of the solar system with a space probe, you can save fuel and increase payload space by hitching a ride with some of the planets. Planets carry an enormous reserve of orbital kinetic energy (and momentum). By steering a space probe into a temporary hyperbolic orbit around a planet, you can steal some of its momentum using gravity assist (or the slingshot effect).

Consider Figure 2.4.1. A space probe that approaches a planet (for example Jupiter) with an inbound speed of Vin, is steered into a hyperbolic orbit by the Jupiter’s gravity and leaves in a different direction with an outbound speed Vout. Judged from Jupiter’s frame of reference (Figure 2.4.1a), Jupiter is the central mass and, as orbits are symmetrical, the two speeds are equal. Using lower case text for velocities measured in Jupiter’s reference frame, vout = vin.

To calculate velocities as seen by an observer stationary relative to the Sun, we simply add Jupiter’s velocity (relative to the Sun) to each velocity in Jupiter’s reference frame. If we do this, it turns out that Vout>Vin; the trip around Jupiter has increased the speed of the probe (Figure 2.4.1b). Momentum is conserved, so the probe’s gain in momentum is exactly balanced by the planet’s loss of momentum. However, because of the planet’s enormous mass, the decrease in its speed is immeasurably small.

Identify that a slingshot effect can be provided by planets for space probes.

CHECKPoInT 2.31 Name the four shapes of orbits allowed by Newton’s law of universal gravitation.2 Describe the orbit normally followed by a projectile passing (without capture) a massive body. Explain why it is

rarely a parabola.3 List the three different conditions on the sign of two-body orbital mechanical energy, and name the orbits that result.4 Outline why low Earth orbits are temporary.5 Describe the Van Allen radiation belts.6 Define the terms geostationary, geosynchronous and semi-synchronous.

45

spacE

We can also treat gravity assist as a perfectly elastic collision (see in2 Physics @ Preliminary p 66). The planet pulls the probe rather than pushing and the ‘collision’ is gradual, but the conservation of momentum still applies. It is elastic because gravity conserves mechanical energy—there is no ‘friction’. The planet is like an enormously heavy cricket bat and the probe is a small, highly elastic ‘superball’. In the bat’s frame of reference, the bat is a stationary ‘immovable’ object (see in2 Physics @ Preliminary p 66), so in a highly elastic collision the ball’s speed before and after the collision is practically unchanged. In the frame of reference of the batsman, however, the ball has been given an increase in speed by the moving bat.

In 1973, Mariner 10 first used gravity assist (from Venus) to achieve a flyby of Mercury to gather images and other measurements. The Galileo probe was launched from a Space Shuttle in 1989, reached Jupiter in 1995 and studied Jupiter’s moons until 2003. To reduce the explosion hazard to the Shuttle’s astronauts during launch, Galileo’s fuel requirement was decreased by using gravity assist, once from Venus and twice from Earth, to slingshot Galileo to Jupiter. The reduction in Earth’s orbital velocity was of the order of only 10–18 m s–1.

Worked exampleQUESTIonFor gravity assist, the maximum possible speed increase occurs for the (unrealistic) extreme limit at which the spacecraft executes a nearly 180° turn, parallel to the planet’s orbital motion (Figure 2.4.2).

Show that for this special case, the change in speed of the craft in the Sun’s frame of reference is twice the orbital speed of the planet Vp. You can assume that in the Sun’s frame, the probe’s speed is always larger thanVp.

SoLUTIonBe careful! Change in speed Vout – Vin is the change in the magnitude of the velocity, but it is not the same as the magnitude of change in velocity |Vout – Vin|.

All labelled velocities in Figure 2.4.2 lie in one dimension. Upper case variables denote quantities in the Sun’s reference frame, and lower case variables the planet’s frame. As usual, bold = vectors, italic = magnitudes and sign = direction.

Vin, Vout and Vp are respectively the spacecraft’s incoming and outgoing speeds and the planet’s orbital speed in the Sun’s frame of reference. Because of orbital symmetry, in the planet’s frame the spacecraft’s incoming and outgoing speeds vin and vout are equal. Let them both equal v.

Using the Galilean transformation formula vB (rel. to A) = vB ‑ vA to convert velocities in Figure 2.4.2a from the planet’s frame and using the sign convention + → :

vin = (–v) = Vin – Vp = (–Vin) – (+Vp) ∴ v = Vin + Vp (1)

vout = (+v) = Vout – Vp = (+Vout) – (+Vp) ∴ v = Vout – Vp (2)

Equate (1) and (2): v = Vin + Vp = Vout – Vp

Rearrange: ∴ Vout – Vin = ∆V = 2Vp

Jupiter

spacecraft’svelocity outbound

spacecraft’svelocityinbound

spacecraft’strajectory

vout = vin

Jupiter’s velocityrelative to the Sun

Jupiter

resultant Vout

resultant Vin

Vout > Vin

Figure 2.4.1 Gravitational slingshot in (a) Jupiter’s frame of reference and (b) the Sun’s frame of reference

Vin

Vout

Vp

vin

vouta

b

Figure 2.4.2 Hypothetical 180° hyperbolic orbit in (a) the planet’s frame of reference (b) the Sun’s frame


46

hITchhIkER’s guIdE To ThE soLaR sysTEm

Instead of selling burgers, two university students took summer holiday jobs at the NASA Jet Propulsion

Laboratoryandmadethenexthalfcenturyofsolarsystem exploration possible. In 1961, Mike Minovitch proved that gravity assist was possible, and in 1965 Gary Flandro showed (just in the nick of time) that in 1976, 1977 and 1978 the planets would, by luck, be suitably aligned for a space probe to hitchhike across the solar system using gravity assist from Jupiter and Saturn to explore the outer planets Jupiter, Saturn, Uranus and Neptune. This ‘grand tour’ would not be possible again until about 2157. Two of the resulting missions, Voyager 1 and Voyager 2, provided enormous advances in planetary science, including the now-familiar spectacular images of the outer planets and their moons. Voyager 1 is now at the boundary between the solar system and interstellar space.

Voyager 1

Voyager 2

Jupiter9 July 79

Jupiter5 Mar 79

Earth 5 Sept 7720 Aug 77

Saturn12 Nov 80

Saturn26 Aug 81

PlutoAug 89

Uranus27 Jan 86

Neptune01 Sept 89

Figure 2.4.3 Voyager 1 and Voyager 2 take the grand tour of the solar system using gravity assist.

CHECKPoInT 2.41 In a gravity-assist manoeuvre, a probe increases its momentum. Explain how momentum is conserved and why

this might not be obvious to an observer.2 During a slingshot manoeuvre, what is the shape of the probe’s orbit in the planet’s reference frame?3 What can we say about the probe’s speed (far from the planet), before and after the slingshot manoeuvre, when

viewed from the planet’s frame?4 If VP is the orbital speed of the planet, what is the maximum possible speed increase for a probe executing

gravity assist?

2.5 I’m back! Re-entryIt is time to come home. You made it through the launch safely so you’re feeling lucky. However, you still have another dangerous hurdle to jump—coming back to Earth or re-entry.

Orbital decayCraft in low Earth orbits do experience some drag or air resistance, because the atmosphere gradually thins out with altitude in a very roughly exponential trend (Figure 2.5.1). At altitudes above ~1000 km, drag is considered negligible. Most Earth-sensing satellites orbit below this altitude, and so have a limited lifetime.

Drag converts orbital kinetic energy into thermal energy, causing the orbital radius to decrease (orbital decay). The lower the orbit, the greater the air density (and drag) and so the faster the orbital decay. Sustained orbits are not possible at

Account for the orbital decay of satellites in low Earth orbit.

47

spacE

100

10–5

10–10

10–150 100 200 300 400 500 600 700 800 900 1000

Altitude (km)

Den

sity

(kg

m–3

)

Figure 2.5.1 Graph of typical air density in the Earth’s atmosphere versus altitude

altitudes below ~160 km (sub-orbital). Satellites in low orbits can counteract decay by firing rockets from time to time, but only until the propellant runs out.

Eventually, a craft in a decaying orbit spirals down towards the Earth’s surface, to burn up catastrophically like a meteor due to the thermal energy produced by the enormous drag (see in2 Physics @ Preliminary p 57). Drag can be enhanced by solar activity such as temporary increases in the Sun’s output of ultraviolet radiation, which inflates the upper atmosphere, making the orbital decay rate unpredictable. Normally a satellite will burn up completely in the atmosphere, although parts of larger craft (such as the Russian Mir Space Station or the US Skylab) occasionally make it to Earth’s surface.

REsT IN pIEcEs

In the late 70s, NASA’s first space station, Skylab, was crippled with low propellant and damaged gyroscopes.

Enhanced solar activity increased its orbital decay and the first launch of the newly developed Space Shuttle was delayed, preventing repair missions. So, in July 1979, it fell on Australia. Pieces too large to burn up on re-entry landed in a line between Esperance and Rawlinna, Western Australia. Many pieces are in the Esperance Museum, but the largest (Figure 2.5.2) is now in the United States Space & Rocket Center. The Shire of Esperance sent the US government a $400 fine for littering. The fine was finally paid in April 2009 in a Californian radio publicity stunt, using listeners’ donations. Figure 2.5.2 The largest surviving fragment of Skylab

Safe re-entry corridorIf a spacecraft is carrying a human crew, or if the craft needs to be retrieved, then plunging into a fiery re-entry like a meteorite is not an option. A safer return from low Earth orbit normally starts by retro-firing rockets to slow the craft down so it begins to fall into a lower energy, lower altitude orbit. There the higher air density starts to slow the craft further. At the bottom of LEO, orbital speed is nearly 7.8 km s–1. Orbiting spacecraft could not carry enough propellant to ease down to the surface. The crew has no choice but to head bravely towards the Earth at the correct angle, using only high technology and clever physics to protect them.

Safe re-entry is a balance between two forces: drag and lift. Drag is the deceleration force; lift is the force that keeps an aeroplane in the air. Air moving relative to the craft creates pressure differences. If pressure underneath is greater than that above, lift results. The shape and orientation of the craft and the re-entry angle all affect the ratio of these two forces.

Discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface.

Identify that there is an optimum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle.


48

The three main issues behind safe re-entry are: 1 minimising the effects of deceleration (g-force)2 managing the effects of heating3 landing the craft safely in the right place.

The first two issues lead to the existence of a narrow range of safe re-entry angles and speeds (Figure 2.5.3). Drag is both good and bad. Drag provides the spacecraft with brakes, but it also produces the copious amounts of thermal energy that could destroy the craft. If the approach into the atmosphere is at too shallow an angle, drag will be too small, the air flow will provide too much lift and the craft will skip over the atmosphere instead of entering. If the angle is too steep, drag will be too large, producing excessive heat and deceleration g-force, which would destroy the craft and crew.

DecelerationAs at launch, astronauts’ seats during re-entry are oriented perpendicular to, and facing (‘eyeballs in’) the direction of acceleration, but this time acceleration is opposite to velocity, so they look backwards.

Traditional re-entry vehicles (such as were used in the 1960s and 1970s) were teardrop-shaped capsules with the blunt end pointing forward (Figure 2.5.4). They allowed very little or no control once re-entry had begun and provided very little lift. Such a re-entry is called ballistic re-entry and requires larger re-entry angles. This kind of capsule subjected the astronauts to a maximum re-entry g-force of anywhere between 6 and 12. The Apollo re-entry angle was between 5.2° and 7.2°

The Space Shuttle introduced in 1981 has wings that provide lift and flight-control structures (such as elevons, a rudder/speed brake and a body flap) that allow considerable control over the descent, adjusting the vehicle’s aerodynamics to the changing density of the air, and making re-entry more gentle, with a maximum g-force of about 2–3. This degree of control also widens the safe re-entry corridor, allowing a gentle, low-g 1–2° re-entry. This is called glide re-entry. To further decrease descent speed without excessive g-force, the Shuttle performs a series of S-shaped turns by rolling and banking, gently enhancing drag.

The Russian Soyuz capsule, in use continuously (in modified form) since the 1960s, is a more spherical variation of the traditional capsule shape but with attitude control thrusters, which provide some glide control during re-entry. It usually yields a g-force of 4–5, but sometimes up to about 8 for a completely ballistic re-entry.

Ballistic re-entries are high acceleration but quick—between 10 and 15 minutes. A full glide re-entry is low acceleration but slow—Shuttle re-entry, for example, takes about 45 minutes. Soyuz is intermediate and takes about 30 minutes.

HeatingOn re-entry, vehicles travel at well above the speed of sound. The speed of sound is sometimes called Mach 1 (after Ernst Mach (1838–1916) a physicist and philosopher who studied gas dynamics). Twice the speed of sound is called Mach 2 and so on. Supersonic means travelling faster than Mach 1. Hypersonic usually means faster than Mach 5 (oversimplifying somewhat).

Pressure builds up in front of projectiles. Sudden pressure changes normally propagate away as sound (at the speed of sound). In supersonic flight, however,

re-entry corridor

insuff

icient drag

excessive drag

Figure 2.5.3 The safe re-entry corridor

49

spacE

this pressure wave is too slow to move out of the projectile’s way, so the pressure builds up to very high levels, forming a shock wave—the air equivalent of the bow wave in front of a speed boat.

The enormous mechanical energy of orbit must go somewhere. Drag converts it to thermal energy. Contrary to common sense, in hypersonic flight, a blunt projectile with more drag actually gets less hot than a more streamlined one. In the 1950s, Harvey Julian Allen proved this theoretically and explained why the sharp nose cones of intercontinental ballistic missiles were vaporising on re-entry. Hypersonic wind tunnel tests (see Figure 2.5.5) confirmed his theory.

In hypersonic flight, much of the heat generation takes place in the shock wave (the dark line wrapping around the front of both projectiles in Figure 2.5.5). The shock wave does not touch the blunt projectile (Figure 2.5.5a) and so doesn’t transfer the heat efficiently to it, but it does touch the tip of the sharp projectile (Figure 2.5.5b), which gets much hotter. For this reason, re-entry vehicles (including the Space Shuttle) are blunt at the front, and hence the traditional teardrop shape of capsules.

The blunt front of the vehicle is also coated with a suitable heat shield with very high melting and vaporisation temperatures. It is also highly insulating to slow the rate of thermal conduction. Thermal insulating materials in most applications are almost always very porous because tiny pockets of gas are very poor thermal conductors. Some insulator materials are also designed to be highly light-absorbing (black) in the visible and near infra-red parts of the spectrum because such surfaces, when hot, also radiate thermal energy away more efficiently (radiative cooling).

Tiles on the Shuttle surface are made of 90% porous silica fibre, which is an excellent high melting point insulator, but it is brittle. The tiles on the hottest parts (the underside and leading edges) are also coated with a tough black glass to enhance radiation of thermal energy, but also to provide mechanical strength. Broken tiles were believed to be responsible for the destruction during re-entry of the Shuttle Discovery in 2003.

In more traditional space capsule ballistic re-entry, drag is higher, so heat is generated more rapidly, and insulation and radiation alone are not enough. In these cases, the insulating heat shield is also designed to vaporise and erode (ablation). The hot, vaporised and ablated material carries thermal energy away rather than conduct it to the capsule, similar to the way in which evaporating sweat carries away excess heat from your skin. The pressure from this ablation also helps to push away the hot gas convecting from the shock wave. The shield must be thick enough to last the journey and provide sufficient insulation.

Two modern examples of ablating materials are phenolic impregnated carbon ablator (PICA) and silicone impregnated reusable ceramic ablator (SIRCA). In the Chinese space program, one of the ablation materials used is blocks of oak wood. It’s cheap and easy to work. As it chars, it forms charcoal, which is porous and almost pure carbon, making it an extraordinarily good thermal insulator with a very high melting point. Another advantage is that porous carbon is very black and radiates thermal energy efficiently. However, it is mechanically weaker than more ‘high-tech’ ablation materials.

During re-entry, superheated air surrounding the vehicle is ionised. The air becomes a plasma—a conductive soup of free positive and negative charges that, like the Earth’s ionosphere (see in2 Physics @ Preliminary pp 153–4), reflects

Figure 2.5.4 Re-entry vehicles: (a) Gemini 1964–1966 (b) Apollo 1966–1975 (c) Soyuz 1960–present (d) Space Shuttle 1981–present

a

b

d

c


50

radio waves, so the astronauts cannot communicate with the Earth for several minutes during re-entry. This problem has been solved for the Shuttle by communicating via a satellite above it, since only the bottom of the Shuttle has significant ionisation.

LandingDrag depends on the projectile’s cross-sectional area and speed. Drag cannot stop a projectile completely because, during deceleration, drag decreases until it exactly cancels the weight of the projectile and deceleration stops—the projectile has reached terminal speed (see in2 Physics @ Preliminary p 45). The terminal speed of a capsule is too high for it to land safely. To slow the capsule further for the landing, drag is enhanced (and terminal speed decreased) by using parachutes to increase the effective area of the capsule.

The final ‘touchdown’ could be on land (typical of Russian missions) or a ‘splashdown’ in the water (typical of US missions pre-Shuttle). Russian Soyuz also has soft-landing engines that fire just before it touches the ground.

The Space Shuttle lands on a runway, much like an aeroplane (Figure 2.5.6) but it uses parachutes to help it brake. During landing, the Shuttle (which has been described as being ‘like flying a brick with wings’) is controlled entirely by computer.

Another issue is accurate targeting of the landing site. The steeper the re-entry angle, the smaller the horizontal component of motion (range) and so the more accurate the prediction of the final landing site. However, the Shuttle makes up for its shallow re-entry, because its aeroplane-like flight-control structures allow adjustment of the landing path. The shape of the landing path is also designed to be more forgiving. The Shuttle approaches the runway roughly opposite to the landing direction. Four minutes from touchdown, it does a ‘heading-alignment’ loop, to adjust precisely to the direction of the runway (Figure 2.5.6).

Figure 2.5.5 Hypersonic wind tunnel tests. (a) The crescent-shaped shock wave is detached from the blunt projectile, but (b) touches the tip of the sharp projectile.

a b

51

spacE

CHECKPoInT 2.51 Define orbital decay and explain what causes it.2 Because of drag, satellites at altitudes below ~1000 km can do nothing to combat orbital decay. True or False?

Explain.3 What other astronomical body can affect the rate of orbital decay? Explain.4 Discuss how drag is ‘good and bad’ for re-entry.5 Outline what can happen if a spacecraft attempts re-entry with too shallow or too steep an angle.6 Explain why astronauts face backwards during re-entry, unlike at launch.7 Outline why occupants of the Space Shuttle experience lower g-force during re-entry than in the more traditional

re-entry vehicles.8 Define the terms supersonic and hypersonic.9 What is a shock wave?0 Outline why a pointy hypersonic projectile is more likely to melt than a blunt one.1 Explain why a capsule with a parachute slows down more than without one.

11

Mojave

EdwardsAirforce Base

Runway 23

Altitude 25 000 m

Figure 2.5.6 Scale drawing of the relatively gentle descent of the Space Shuttle. The Shuttle is drawn at 1 minute intervals to touchdown. The squares on the ground are 10 nautical miles (18.5 km) wide.


52

2 Explaining and exploring the solar system

chapTER 2This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTIvITY 2.1: DEvELoPMEnT oF SPACE ExPLoRATIonUse the template provided in the activity manual to extract information about your chosen scientist. Process this information to make a short oral presentation to the class.

Discussion questions1 For the scientist that you have researched, list their main contributions

to space exploration.2 Explain how later scientists have benefited from this research.

Extension3 Werner von Braun’s great Russian rival, the ‘Chief Designer’ for the USSR

space program Sergey Korolyov, is not as familiar as some of the names mentioned in the syllabus, despite leading the launch of the first artificial satellite, Sputnik, in 1957. This is probably because his name was kept secret by the communist government of the USSR until after his death in 1966. You may also want to research his contribution to space exploration.

ACTIvITY 2.2: UnIFoRM CIRCULAR MoTIonPerform an experiment that will allow you to determine the relationship between the radius of a satellite’s orbit around the Earth and its gravitational force.Equipment: string, rubber stopper, mass carrier and masses, electronic scales, glass or plastic tube, paperclip, sticky tape, metre ruler, stopwatch.

Discussion questions1 From your experimental data,

determine the mathematical relationship between the orbital radius of a satellite and its tangential velocity for a given centripetal force.

2 Describe the method you would use to determine the centripetal force on a small model satellite.

Identify data sources, gather, analyse and present information on the contribution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun.


F m

vr

=2 tension

mass carrier

string

paperclip

glass orplastic tube

mg

Figure 2.6.1 Force on mass moving in a horizontal circle

53

spacEchapter summary• Rocketthrustisthereactiontotheforceexertedonthe

exhaust exiting the nozzle.• Forafixedmassofexhaustperunittime,thrust

increases as exhaust speed increases.• Asthemassofremainingpropellantdecreases,arocket’s

acceleration (and g-force) increases.• Therearetwokindsofrocketengines:liquidandsolid

propelled.• Byjettisoningusedstages,arocket’smassisdecreased,

allowing more payload to be carried.• g-force is the ‘apparent weight’ experienced during

acceleration, divided by true weight on Earth.

• Verticalg-force=av

g+ 1. Horizontal g-force =

ah

g.

• Toincreaseg-forcetolerance,astronautsareseatedwithbodies horizontal, and looking in the direction of the acceleration (‘eyeballs in’) for both lift-off and re-entry.

• Aspacecraftlaunchedeastwardshastheextrainitialvelocity of Earth’s rotation. This is greatest at the equator (465 m s–1).

• LaunchesinthedirectionofEarth’sorbitalvelocityobtain an initial velocity boost of 3.0 × 104 m s–1.

• Gravityprovidesthecentripetalforceforsatelliteorbits:

Fmv

Rc =2

(for circular orbits)

• Kepler’slaws(applytoanytwobodiesifthecentralbody has a very much larger mass):1 Orbits of planets are ellipses, with the Sun at one

focus.2 Planets sweep out equal areas in equal times.

3 Law of periods: T

a

2

3 = a constant

• ExplicitformofKepler’sthirdlaw: a

T

GM3

2 24=

π• Eccentricityisameasureoftheelongationofanellipse.

A circle is an ellipse of zero eccentricity.• Periapsisisthepositionofclosestapproachtothe

central mass and fastest orbital speed (perihelion for the Sun, perigee for Earth).

• Apoapsisisthepositionoffurthestdistancefromthecentral mass and slowest orbital speed (aphelion for the Sun, apogee for Earth).

• Ifthemassofasatelliteisnotnegligiblewhencomparedto that of the central body, then both masses orbit with the same period around the system’s centre of mass.

• Magnitudeoforbitalspeed:

• Two-bodyorbitalmechanicalenergy

ME = −12

2mvGmM

r:

– ME < 0 (bound), orbit is closed or stable (circle or ellipse); velocity < escape velocity

– ME = 0 (borderline unbound) orbit is parabolic; velocity = escape velocity

– ME>0(unbound),orbitishyperbolic; velocity>escapevelocity

• Forstableorbits,ME = − GmM2a

.

• Orbitsaresymmetricalinshapeandspeed.• LowEarthorbit(LEO):altitudebetween~160and

~2000 km.• Ataltitudesbelow~1000km,dragcausesorbitaldecay.

Upper atmosphere can be inflated by increased solar UV radiation, increasing drag.

• Orbitsbelow~1000kmareprotectedfromtheVanAllen radiation belts by the atmosphere and distance.

• Geosynchronousorbit:T = 1 sidereal day (23 h 56 m 4 s).

• Circulargeosynchronousorbitsovertheequatorarecalled geostationary. These orbits are used extensively for communication satellites (r = 42 164 km).

• Aspaceprobeenteringatemporaryhyperbolicorbitbehind an orbiting planet can gain momentum via gravity assist or the slingshot effect.

• Forgravityassist,themaximumpossiblechangeinspeed of the probe in the Sun’s frame of reference is twice the planet’s orbital speed Vp.

• DragconvertsorbitalKEintothermalenergy.• Safere-entryangle:ifangleistoolow,thecraftwillskip

off atmosphere; if angle is too high, g-force and heating rate are too large.

• Muchheatingtakesplaceinthehypersonicshockwave.Blunt-fronted re-entry vehicles are used because the shock wave is detached from the craft.

• Heatingofaspacecraftonre-entryisreducedbyaninsulating and radiating heat shield.

• Traditionalcapsulesalsouseablationoftheheatshieldto dissipate heat.

• Parachutesdecreaseterminalvelocitybyincreasingtheeffective cross-sectional area.

vGM

r=

54

2 Explaining and exploring the solar system Review questions

PHYSICALLY SPEAKInGFor each type of orbit, fill in the missing information. One has been done already.

name of orbit sign of two-body me oPen or closed v >, =, < vescape bound or unboundGeostationary

Slingshot (in planet’s frame)

Elliptical

Hohmann (see Physics Focus)

Parabolic

Hyperbolic

Halley’s Comet

Circular Negative Closed < Bound

Molniya

REvIEwInG 1 What was the first solid rocket propellant and who invented it?

2 Assuming that propellant is burned at a constant mass per unit time, use the equation for thrust to explain why forcing exhaust gas through a narrow nozzle increases thrust.

3 Listtheadvantagesofbothsolidandliquidpropellants.

4 Discuss why the vertical g-force formula has a ‘+ 1’ term, but the horizontal formula doesn’t.

5 Describe a situation during launch in which astronauts would experience a g-force greater than zero but less than 1.

6 At the bottom of a bungee jump with the cord attached to the ankles, one can easily experience a g-force of 3, the maximum normally allowed for Shuttle launches. Describe three important differences in the way the g-force is experienced in these two situations.

7 Explain why launch facilities are usually built as close to the equator as is practical.

8 Describe the circumstances under which a star would not sit at the focus of a planet’s elliptical orbit.

9 Discuss why we only briefly see Halley’s Comet with an unaided eye every 76 years, even though it is in orbit continuously around the Sun.

10 Outline what happens to the period of a satellite if its semimajor axis is reduced by a factor of 4.

11 A space probe approaches a planet in a hyperbolic orbit. Discuss the condition that must be fulfilled to move it into a stable orbit of the planet and describe how it might be achieved.

12 By re-examining the gravity assist worked example on page 45, show that the magnitude of change in the probe’s velocity in the Sun’s reference frame is twice the probe’s initial speed in the planet’s reference frame (that is, |Vout – Vin| = 2v).

13 A satellite is in a highly elliptical orbit around Earth such that, at perigee, it is briefly at an altitude of less than 1000 km. Over many orbits, the altitude at apogee decreases (the orbit becoming more circular). Explain why this occurs.

14 Listtworeasonswhy(human-crewed)spacestationsarealwaysinlowEarthorbit.


Fmvr=

2


F Gm m

d= 1 2

2


55

spacEReview questionsSoLvInG PRoBLEMS 15 Calculate the two-body gravitational potential energy for a system consisting

of a 1.00 kg test mass sitting on the surface of the Earth. How far would the test mass need to be from the Sun so that the two-body GPE of the test mass–Sun system is the same value? Estimate roughly where that position would be in relation to the orbital radii of the planets.

16 Typically, at launch, the Shuttle’s main engines, with an effective exhaust velocity of 4460 m s–1, produce a thrust of 5.45 × 106 N. The two solid-fuel rocket boosters, with an effective exhaust velocity of 2640 m s–1, produce 1.250 × 107 N each.a Calculate the combined rate (in kg s–1) at which propellant is used

at launch.b Assuming a mass at launch of 2.03 × 106 kg, calculate the Shuttle’s

acceleration at launch and 1 minute later, assuming the above specifications remain constant. (Hint: Don’t forget gravity.)

17 On a roller-coaster, you round the top of a circular hump in the track of 5.00 m radius. You have a g-force meter with you and at the moment you’re at the top it reads a vertical g-force of 0.00.a What is your weight at that moment? b What is the magnitude of the normal force exerted on you by the seat

at that moment?c What is your centripetal acceleration?d Calculate your speed at the top.e Assuming friction and air resistance are negligible, calculate your

horizontal g-force at that moment.

18 Prunella spins a weight (mass m) on a string (length L) in a horizontal circle (Figure 2.6.2) to illustrate the relationship between orbital speed and centripetal force for an orbiting satellite. Renfrew says: ‘Because of the weight, the string isn’t horizontal so the orbital radius is R = L sin θ, and the centripetal force is Fc = T sin θ.’

Prunella then says: ‘Yeah, but as long as the orbital speed v is high enough, θ will be very close to 90° so you can use the approximation that string tension T is the centripetal force and the string length L represents orbital radius R.’

Show that as long as orbital speed v fulfils the condition vR

2

> 7g, then L is no more than 1% larger than the true orbital radius R and T is no more than 1% larger than the true centripetal force Fc.



F G

m m

d= 1 2

2


F

mvr=

2


θL

Tmg

T sinθ

R

θ

Figure 2.6.2 Spinning weight model of a satellite

56

2 Explaining and exploring the solar system

19 In Gerard O’Neill’s proposed colonies in space, people would live on the inner surfaces of rotating cylinders 3 km in radius and 20 km long. Gravity would be simulated via the reaction force to the centripetal force exerted by the cylinder on the occupants inside. Calculate the rotation rate (in revolutions per hour) that would yield artificial gravity, mimicking the magnitude of gravity at the Earth’s surface.

20 Using the mass of Jupiter calculated in the worked example on page 39, predict the period of Jupiter’s moon Callisto, given that its semimajor axis is 1.89 × 106 km.

21 Using the equation for tangential velocity v =2πrT

and the explicit form

of Kepler’s law of periods, re-derive the expression from section 2.2, for the magnitude of orbital velocity of a satellite in circular orbit:

v

GMr

=

22 Using the explicit form of Kepler’s third law, calculate the radii of:a a geostationary orbit b a circular semi-synchronous orbit.

23 Draw a table in which you compare the kinetic energy, the two-body potential energy and the two-body mechanical energy of a geosynchronous and a circular semi-synchronous satellite of mass 2000 kg. Use the data from Question 22 to confirm that the two-body mechanical energy for both orbits is:

ME = −GmMa2

24 A comet passes the Sun (MS = 1.99 × 1030 kg). At perihelion, the centre-to-centre distance is 8.55 × 1010 m and the speed of the comet is 6.69 × 106 m s–1. What kind of orbit is it? Do you expect the comet to return?

25 Gravity assist can also be used as a brake. Show that if the diagram in the planet’s frame in Figure 2.4.1a is unchanged, but the planet’s orbital velocity in Figure 2.4.1b is reversed, then the probe’s speed in the Sun’s frame would decrease.

26 A Soyuz capsule with a crew of three (7460 kg) is in a circular orbit at an altitude of 336 km, having completed a mission to the International Space Station. a Calculate its kinetic energy.b Ignoring the effect of its soft-landing engines, calculate how much

thermal energy is generated by drag during its re-entry. Earth’s mass ME = 5.97 × 1024 kg Earth’s radius at the landing site rE = 6366 km


r

T

GM3

2 24=

π

Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of theorbitusingKepler’sLaw of Periods.

Revie

w Questions

57

spacE

PHYSICS FoCUS L DRIvInG LESSon In oRBIT

Once in a stable orbit, a spacecraft is coasting. One only needs to do work (burn fuel) to swap between orbits. The following are two important examples of manoeuvres between orbits.

CHANGiNG LANES: Circular orbits around Earth (or other body) are like lanes on a highway, the smaller radius orbits being the fast lanes. In 1925, Walter Hohmann devised a fuel-efficient method for ‘changing lanes’. To move from a small orbit to a larger one, an accelerating engine burn of just the right impulse will convert the circular orbit to an elliptical Hohmann transfer orbit (Figure 2.6.3), which joins the initial and final orbits. Once in the elliptical orbit, the engine is shut off again. Once the craft reaches the apoapsis of its elliptical orbit, a second accelerating burn of the right impulse will convert the elliptical orbit into a (now larger radius) circular orbit. This method can also be used in reverse to drop down to a smaller orbit, in which case the two engine burns must be decelerations.

Hohmann transfer orbits around the Sun can also be used for travel between planets. The launch must be timed so the planet is at the periapsis of the Hohmann orbit just as the probe arrives.

OVERTAKiNG: Sometimes spacecraft need to rendezvous (meet) in orbit, such as when the Space Shuttle needs to dock with the International Space Station. If you’re way behind a craft in the same orbit and you need to catch up, you should paradoxically ‘slam on the brakes’—briefly retrofire (fire your engines in reverse) to reduce speed (Figure 2.6.4). This temporarily reduces KE (and total mechanical energy) so you drop down into a lower energy, smaller radius ‘overtaking’ orbit, more than regaining your speed. Smaller radius orbits are faster, so you eventually catch up with the other craft. When you’re about to overtake it in the ‘fast lane’, fire your engines forwards to increase mechanical energy, so you pop back up into the original orbit rendezvous.

1 When you fire a rocket to change your speed, does this violate momentum conservation? Explain.

2 Why doesn’t one just fly from the initial orbit to the final orbit in a straight line instead of following a Hohmann orbit? Does this violate Newton’s first law? Explain.

3 Show that to transfer from an initial circular orbit of radius Ri to a larger final one of Rf, you must use a Hohmann transfer orbit with a semimajor axis a, where Ri + Rf = 2a. (Hint: Draw some diagrams.)

4 Why can’t the Hohmann transfer orbit be a circular orbit? (Hint: Draw another diagram.)

5 Why is the Shuttle in Figure 2.6.4 pointing the ‘wrong way’?

6 When overtaking, you temporarily reduce your KE. Why does your KE increase again in the overtaking orbit? You may use the fact that for a stable orbit,

the two-body ME = ME = −GmMa2

(section 2.3).

ExTEnSIon7 A Space Shuttle releases a communications satellite

into an initial orbit of radius r = 6.66 × 106 m. Using a Hohmann transfer orbit, the satellite moves up to a geosynchronous orbit of rgeo = 4.216 × 107 m. Using the result of Question 3, calculate how long the transfer takes. (Hint: The transfer path is half an orbit.)


initialorbit

final orbit

ellipticaltransfer orbit

satellite

Figure 2.6.3 A Hohmann elliptical transfer orbit

Figure 2.6.4 To overtake, first slow down.

3 Seeing in a weird light: relativity

58

Just some minor problemsYou may have heard it said that some physicists think that a ‘theory of everything’ is just around the corner. This attitude is not new. Many physicists thought this about what is now called ‘classical Newtonian physics’, towards the end of the late 19th century. There were just a few minor problems with understanding the way light travels through space that needed to be fixed, and then the job of physics would be finished. Well, those ‘minor problems with light’ led to the twin pillars of modern physics: quantum mechanics and relativity. Einstein was a key player in both, especially relativity, which comes in two parts. Special relativity replaced Newton’s mechanics and the later general relativity replaced Newton’s universal gravitation. A century later, relativity still defies common sense, bending space, time and the mind, but it has not yet failed any experimental test. In this chapter we will only deal with the theory of special relativity.

‘Common sense is nothing more than a deposit of prejudices laid down by the mind before you reach eighteen.’ A Einstein

special relativity, general relativity, inertial frame of reference, invariant, principle of relativity, fictitious forces, Maxwell’s equations, electromagnetic

wave, mechanical medium, luminiferous aether, interferometry,

beam splitter, aether drag, Michelson–Morley experiment, null result,

postulate, simultaneity, Lorentz factor, time dilation, proper time, twin paradox,

proper length, length contraction, rest mass, proper mass, relativistic mass,

spacetime, mass-energy,

3.1 Frames of reference and classical relativity

Before we talk about relativistic weirdness, recall inertial frames of reference (see in2 Physics @ Preliminary section 3.3). Inertial frames of reference are required for observers using Newton’s laws or the Galilean transformation formula for relative velocity (see section 1.1). An inertial frame is any non-accelerating (including non-rotating) reference frame. Inertial frames can move at a constant velocity relative to each other.

All velocities are relative. There are no absolute velocities and there is no special absolutely stationary inertial frame. The classical Newtonian laws of mechanics and gravity are unchanged (or invariant) when transforming from one inertial frame to another (even though the values of some measurements such as velocity might change). This is called the principle of relativity.

To transform a situation from one inertial frame to another, you simply apply the Galilean transformation formula to all the velocities. You’ve already seen an example of this in the gravity assist example (Figure 2.4.2).

Outline the nature of inertial frames of reference.

Discuss the principle of relativity.

59

Space

Because there is no special inertial frame, no experiment purely within your own frame can detect the velocity of your frame, so absolute velocity is meaningless. You can only compare your frame’s velocity relative to others. An example of this is waiting to depart in a train, looking out the window (Figure 3.1.1) to see that a train next to you is moving slowly away, only to find a few seconds later that, in fact, relative to the station it is your train that is moving. Your acceleration (including vibrations) was negligible—you felt no effect of your uniform velocity.

However, you can feel the acceleration of a non-inertial reference frame, and measure it using an accelerometer. The simplest accelerometer is a pendulum. If a pendulum hangs vertically in a car, your horizontal acceleration is zero. If you are accelerating horizontally, the pendulum will hang obliquely (Figure 3.1.2).

If you are observing from within a non-inertial (accelerating) frame, Newton’s laws appear to be violated. Objects can appear to change velocity without a true net external force; in other words, you experience fictitious forces or pseudo-forces (see in2 Physics @ Preliminary p 39). For example, in a car taking a corner, you experience the sensation of being ‘thrown outwards’ by a fictitious centrifugal force. If viewed from the inertial frame of the footpath, you evidently are pulled inwards by a true centripetal force. (We’ve cheated a bit. The Earth is turning, so the footpath is not strictly an inertial frame. However, Earth’s radius is so large that in most human-scale situations, fictitious forces due to Earth’s rotation are negligible.)

Another view of tests for non-inertial reference frames is that they involve detecting fictitious forces. It’s a two-step process. First, analyse an object within that frame of reference and decide what true external Newtonian forces must act on the object. Then, look for apparently ‘extra’ or ‘missing’ forces—evidence of a non-inertial frame. For example, judged from the inertial frame of the ground, the downward weight mg and the upward normal force N of the seat are the only true forces on an astronaut during launch. Within the accelerating rocket, the sensation of enhanced weight (downwards) associated with g-force has the same magnitude as N but is apparently in the wrong direction and is therefore fictitious.

A pendulum accelerometer hangs obliquely within an accelerating car as though there is a fictitious horizontal component of weight. In free-fall (or orbit), the apparent absence of weight is also fictitious. Your frame accelerates downwards, so true weight becomes undetectable to you, as though your true downward weight is cancelled by a fictitious upward gravity. The effects of neither ‘force’ show up separately on an accelerometer.

Worked examplequestionA Christmas decoration is hanging obliquely inside your car, 5° from vertical and pointing towards the car’s left side. Describe quantitatively the car’s motion (no skidding!).

soLutionOnly two true external forces act on the decoration: tension and weight (Figure 3.1.2). Because there is an angle between them, they aren’t ‘equal and opposite’, so the decoration experiences a net real force and acceleration sideways (in this case centripetal). The net force and acceleration point towards the right side of the car, so the accelerometer (and the car) is steering towards the right.

Figure 3.1.1 Who is really moving?

Try ThiS!Fictitious FunWhile sitting on a playground merry-go-round with a friend, try playing ‘catch’ with a slow moving tennis ball. The fictitious centrifugal and Coriolis forces will ‘cause’ the ball to appear to follow warped trajectories, making it difficult to catch.

TT

mg tan 5°

mg

mg

left right

5° 5°

Figure 3.1.2 Festive season pendulum accelerometer

Interactive

Module

3

60


The ‘centrifugal force’ perceived by the occupants of the car to be pulling the decoration toward the left side of the car is fictitious.

From Figure 3.1.2, the magnitude of the centripetal acceleration is:

aFmc

c= = g tan 5° = 9.80 × 0.0875 = 0.86 m s–2

Note: This test is subjective—it requires personal judgement (hence possible bias). No measurement alone can identify a force as fictitious. For example, no pure measurement can tell the difference between true weightlessness and the fictitious weightlessness of free-fall. You can only tell the difference by looking down and seeing the Earth below; judgement says there should be gravity, but you can’t feel it. The inability of measurement alone to distinguish the effects of true gravity from the effects of g-force is what Einstein used as the starting point for his re-writing of the law of gravitation in his theory of general relativity, but you’ll have to wait until university physics to learn about that!

This approach to distinguishing between inertial and non-inertial frames relies on a classical concept of force. In Einstein’s relativity, the concept of force is more complicated and is used much less.

The term fictitious force doesn’t mean the observed effects are imaginary, as the victims of a cyclone or astronauts who are subject to g-force can attest. It simply means that the apparent force doesn’t fit Newton’s definition of a true force.

It is always possible to re-analyse fictitious forces using an inertial frame and to account for all observed effects using only true Newtonian forces.

activity 3.1

pracTicaL eXperieNceS


Perform an investigation to help distinguish between non-inertial and inertial frames of reference.

FicTiTiouS cycLoNe? yeah righT!

Effects associated with so-called fictitious

forces of Earth’s rotation are not always negligible. The Coriolis force is a fictitious tangential force appearing in rotating frames of reference and is associated with the formation of cyclones.

CheCkpoint 3.11 Define an inertial reference frame.2 Recall the Galilean transformation formula for relative velocities.3 Outline why we usually treat the Earth as an inertial frame, given that it is rotating.4 Discuss whether or not centripetal force is fictitious. 5 In free-fall, you don’t experience any extra apparent forces. Are you in an inertial frame? Explain.6 What apparatus would distinguish true weight from apparent weight due to g-force ?7 The values of some measurements such as velocity might change, but the laws of mechanics are the same in all

frames of reference. True or False? Explain.

Figure 3.1.3 Satellite photo of a cyclone

61

Space

3.2 Light in the Victorian eraThe 19th century was a period of enormous advance in the study of electricity and magnetism. Faraday, Ampere, Oersted, Ohm and others, through theory and experiment, produced a large collection of equations and phenomena. There were hints of connections between electricity and magnetism—an electrical current can produce a magnetic field (see in2 Physics @ Preliminary section 12.3) and a changing magnetic field can induce a changing electric field or current (section 4.1).

The Scottish theoretical physicist, James Clerk Maxwell (1831–1879) collected the existing equations to reduce them down to the minimum number. He reduced them down to eight equations (which expanded to 20 when he included all the x-, y- and z-components). A self-taught electrical engineer called Oliver Heaviside (1850–1925), using the newly developed mathematics of vectors, reduced Maxwell’s equations to four. We now call those four equations Maxwell’s equations.

It puzzled Maxwell that his equations were almost symmetrical in their treatment of electrical and magnetic fields—almost but not quite. So he added a term to his equations, assuming that a changing electrical field can induce a magnetic field (not previously observed). When he did this, he showed that an oscillating magnetic field would induce an oscillating electric field and vice versa, resulting in a self-sustaining electromagnetic wave. From his equations he calculated the speed of that wave to be equal to the speed of light in a vacuum (which is now called c and equals 2.998 792 458 × 108 m s–1).

It was either an astonishing coincidence or strong circumstantial evidence that light is an electromagnetic wave (see in2 Physics @ Preliminary p 84).Heinrich Hertz (1857–1894) experimentally confirmed the predicted speed and properties of these electromagnetic waves.

What is light’s medium?Until then, every existing kind of wave needed a mechanical medium; for example, sound propagates through air, earthquakes through rock, musical vibrations along a violin string, ripples along water and so on (see in2 Physics @ Preliminary section 5.3). To sustain a wave, a medium needs two properties: resilience (or stiffness) and inertia (any density- or mass-related property). The higher the stiffness and the lower the inertia, the higher the wave speed.

It was assumed that light also needs a medium, which was called luminiferous aether or just aether (US spelling: ether). Luminiferous means ‘light-bearing’, and ‘aether’ was the air breathed by the gods of Greek mythology.

So Maxwell developed a model for aether, assigning it bizarre mechanical properties consistent with the behaviour and enormous speed of light. It needed to be far less dense than air but much stiffer than any known material. Despite its stiffness, aether was assumed to penetrate all materials effortlessly. Conversely, it needed to be able to be penetrated without resistance by all objects that move freely through space, including Earth hurtling around the Sun.

If you shout with the wind blowing behind you, then, relative to you, the velocity of sound would be higher than if the air were still. This is because the velocity of sound (and other mechanical waves) is the sum of its velocity relative to the medium and the velocity of the medium itself. In other words, mechanical

Outline the features of the aether model for the transmission of light.

Figure 3.2.1 James Clerk Maxwell

3

62


waves seem to obey the Galilean transformation. It was assumed that light should also obey it, so the speed of light should be affected by the motion of the aether.

However, Maxwell’s equations appeared to allow only one particular value for the speed of light in a vacuum. The Galilean transformation and Newton’s laws imply it is impossible for the speed of light to appear to be the same to all observers with different relative speeds. Perhaps the speed specified by Maxwell’s equations is the speed relative to the aether only. However, this meant that the aether represented a preferred reference frame for Maxwell’s equations, which was inconsistent with the classical principle of relativity.

M and MGiven that the Earth was supposed to be hurtling around the Sun, through the aether at 3 × 104 m s–1, the resulting ‘aether wind’ (or aether drift) relative to Earth should affect measurements of light speed differently according to the time of day and time of year as the Earth rotated and orbited the Sun, changing its orientation relative to the aether.

So in the 1880s, the experimentalist Albert Michelson (1852–1931), joined later by Edward Morley (1838–1923), attempted to measure changes in the speed of light throughout the day due to this shifting aether wind. They used a very sensitive method called interferometry (see section 21.5), which Michelson had used some years earlier to accurately measure the speed of light.

Recall constructive and destructive interference (see in2 Physics @ Preliminary p 102 and p 126). If two light beams are projected onto a screen, then a bright ‘fringe’ occurs at places where the two beams are in phase (constructive interference). Where they are out of phase, destructive interference results in a dark fringe. The pattern of bright and dark fringes is called an interference pattern (Figure 3.2.2).

Interference turns a pair of monochromatic (single wavelength) light beams into an extremely sensitive ruler for which the interference fringes are like magnified ruler markings one light wavelength apart. For visible light, this spacing is less than 8 × 10–7 m and corresponds to time intervals of less than 3 × 10–15 s. If the two light beams travel via different paths, then a very small change in the length of one path will change the relative phase, resulting in a detectable change in the position of fringes in the interference pattern. A change in wave speed along one of those paths should have a similar effect on phase.

Michelson and Morley set up an interferometer in which the light was divided into two perpendicular beams or ‘arms’ by passing it through a half-silvered mirror or beam splitter (Figure 3.2.3). The apparatus was built on a heavy stone optical bench floating in mercury, to allow rotation and damp out vibrations. They assumed that if one interferometer arm was pointing parallel to the aether wind, the speed of light should be slightly different in the two arms. The time of flight of the light in the arm parallel to the aether wind should be slightly longer than that of light along the perpendicular arm. As the Earth (or the apparatus) rotates, this speed difference, as measured by the positions of the interference fringes (Figure 3.2.2), should change with the angle.

Figure 3.2.4 summarises the classically predicted effect of aether wind on the resultant light speed in the two arms of the interferometer. Let’s calculate the expected time difference. Suppose the total distance from beam splitter MS to M1 (or M2) is L, then the round-trip for each arm is 2L.

Describe and evaluate the Michelson–Morley attempt to measure the relative velocity of the Earth through the aether.

Discuss the role of the Michelson–Morley experiments in making determinations about competing theories.

Gather and process information to interpret the results of the Michelson–Morley experiment.

Figure 3.2.2 Interference pattern in a Michelson interferometer illuminated by a mercury vapour lamp. Patterns of different shapes (such as vertical bands) are possible and depend on exactly how the interferometer is aligned.

activity 3.2



63

Space

M2 Ms M1

source eyepiece

source

M1

M2

Ms

l1

l2

aetherwind

Figure 3.2.3 The Michelson–Morley interferometer drawn as (a) a simplified schematic and (b) an actual ray diagram. Multiple reflections were used to make the effective length of the arms very long hence more sensitive to changes in light speed. MS is the half-silvered beam splitter mirror.

In the arm perpendicular to the aether wind (speed v), if c is light speed

relative to the aether, then the resultant light speed is c v2 2− (Figure 3.2.4a) and the time taken for light to do a round-trip is:

tL

c v

Lc v

c

1 2 2 2

2

2 2 1

1

=−

= ×

−

In the arm parallel to the aether wind (speed v), for half the trip against the wind, the speed of light would be c – v, and for the other half with the wind it would be c + v, so the time taken would be longer:

tL

c vL

c vLc v

c

2 2

2

2 1

1

=−

++

= ×−

(Check that you agree that since v is smaller than c, time in the parallel arm t2 is longer than time in perpendicular arm t1.)

Other factors such as thermal expansion or contraction of the apparatus could cause apparent drift in the interference pattern, but the shift due to rotation of the apparatus (or the Earth beneath it) would be a sine wave with a period equal to the rotation period of the apparatus, so any drift not due to rotation could be detected and subtracted. Michelson and Morley graphed the position of interference fringes versus rotation angle at different times of the day, but concluded that the small observed shifts could be explained as drift in the experimental apparatus. Over several years, scientists repeated the measurements, with some reports of possible changes in interference over the day; but eventually the consensus was that any observed effect was well below what was expected by the aether theory and could be explained by drift in the apparatus.

George Fitzgerald and Hendrik Lorentz attempted to squeeze the Galilean transformation into Maxwell’s equations, concluding that charged particles (such as charges in atoms) moving through the aether with speed of v must

shrink in the direction of motion by a factor of 1 2− v 2c/ . The interferometer arm parallel to the aether wind would shrink just enough to compensate for the change in light speed and hence cancel the expected change in the interference pattern.

C

C C

CV

V

V

V

√C2 – V2

C – V C + V

aeth

er w

ind

aeth

er w

ind

Figure 3.2.4 Classical effect of aether wind on light velocity. (a) The resultant velocity perpendicular to the wind and (b) resultant velocity parallel to the wind. Blue = light velocity relative to aether, green = aether velocity and red = resultant light velocity

a

a

b

b

3

64


3.3 special relativity, light and timeAlthough relativity is Einstein’s theory, many of the underlying ideas or mathematical formulae were inspired or anticipated by others including Poincaré, Lorentz and Minkowski. Einstein, being a theoretician, did not conduct laboratory experiments. However, he is famous for making good use of the ‘Gedankenexperiment’ or ‘thought experiment’ to boil abstract ideas down into simple concrete ones. Theory can sometimes be derived by imagining an experiment being done, even if it is impractical. We will mention some of his thought experiments in this section.

Explain qualitatively and quantitatively the consequence of special relativity in relation to: • therelativityofsimultaneity• theequivalencebetween

mass and energy• lengthcontraction• timedilation• massdilation.

Another reason suggested for failure to see the shift was that perhaps aether only penetrates transparent objects, so aether was trapped by large opaque mountains and valleys or buildings and dragged along by the moving Earth, similar to the way in which air is trapped in the fur of a running dog. A flea conducting scientific experiments on the dog’s skin would be unaware that outside the fur, air is whooshing backwards relative to the dog. This idea was called aether drag. If this were true, then at the tops of mountains, closer to outer space, the aether wind might be detectable. Some experimenters repeated the experiment on mountains without success (apart from a controversial partial result).

Failure to detect undeniable effects of aether wind caused some physicists to question if it even existed. Maxwell’s equations only mention electric and magnetic fields. The aether is not required by the equations. Einstein assumed it didn’t exist, but said that relativity was not an attempt to explain Michelson and Morley’s negative result, but rather, he was motivated by the properties of Maxwell’s equations. However, in physics, when experiment and theory say the same thing, you’re probably on the right track. Today almost all physicists agree that there is no aether.

The Michelson–Morley experiment is often called ‘the most famous failed experiment’. It was not exactly a ‘failure’. In 1907, Michelson was awarded the Nobel Prize for Physics for his work. The result of an experiment that fails to find evidence of an expected effect despite careful design and execution is more correctly called a null result. This null result was one of the most important in the history of physics, because it helped bring about a whole new way of seeing the universe.

CheCkpoint 3.21 Describe Maxwell’s circumstantial evidence that light is an electromagnetic wave.2 Discuss why it was assumed that light required a medium or ‘aether’ to propagate in.3 Maxwell’s equations predicted that the speed of light should depend on the speed of the medium, but this was

contradicted by the Michelson–Morley experiment. True or false? Explain.4 In the classical analysis of the Michelson–Morley interferometer, which arm required the longer time of flight? 5 Is it correct to say that the Michelson–Morley experiment didn’t show any change in the interference pattern? Explain.6 Outline how Fitzgerald and Lorentz explained the apparent absence of evidence for aether wind.7 Describe aether drag.8 Discuss which played a greater role in motivating Einstein’s work, the work of Michelson and Morley or that of Maxwell.

65

Space

Speed of lightNewton regarded space and time as absolute. In practical terms, this means that the length of 1 metre, the duration of 1 second and the geometric properties of shapes would be the same to all observers everywhere. Not all physicists agreed, but the success of Newton’s laws silenced any philosophical discussion.

However, Maxwell’s theory (and the Michelson–Morley experiment) pointed to the speed of light in a vacuum being constant to all observers. So Einstein said one of three things must be wrong: the principle of relativity (the invariance of laws of mechanics in all inertial reference frames), Maxwell’s equations or the Galilean transformation (the basis of all of classical mechanics).

The principle of relativity seemed very fundamental to Einstein, so he didn’t reject that. In fact, he extended Galileo and Newton’s principle of relativity to include all laws of physics, not just mechanics. He called it his first postulate.

Following a suggestion by Jules Henri Poincaré (1854–1912), Einstein decided that as the speed of light in a vacuum was invariant in all inertial frames, then that must also be a law of nature, which he called the postulate of the constancy of the speed of light.

Maxwell’s equations accurately described electromagnetic phenomena, so Einstein didn’t want to reject them. So it must be the Galilean transformation (and hence all of classical mechanics) that was wrong. But it is difficult to see how something so simple could possibly be wrong.

Suppose you are on a moving train, shining a torch towards the front of the carriage. To your eyes, the light travels the length of the carriage L. To you, its speed is the length of the carriage divided by the time t it took to get there c = L/t. To an observer at the train station, the light travelled the length of the carriage plus the distance D the carriage travelled in that time: c = (L + D)/t. The arithmetic is so laughably simple. How could both observers possibly get the same value for c? It could only be possible if you and the observer at the train station disagree on the lengths L or D or the time period t. In other words, if the speed of light is constant then length (space) and/or time are not absolute—they must depend on the state of motion of the observer.

So why had no-one noticed until 100 years ago? Classical mechanics had successfully described phenomena for three centuries, but it had never been tested for things moving at close to the speed of light. Classical mechanics and the Galilean transformation are accurate approximations at speeds well below the speed of light. Only when the properties of light itself were examined, did the problems become obvious.

Einstein showed (in several ways), that the speed of light is the ultimate speed limit—no observer can reach the speed of light. As a teenager, he asked ‘What would the world look like if I rode on a light beam?’ He answered as an adult with a thought experiment. A light beam is a wave of oscillating electric and magnetic fields moving at the speed c. If you were in the same reference frame as the light beam, you would observe stationary electric and magnetic fields that vary as sine waves in space, but are constant in time. This is not an allowable solution to Maxwell’s equations, so it is not possible for an observer to travel at the speed of light—it is the ultimate speed limit.

SimultaneityEinstein demonstrated that simultaneity is relative. Events apparently simultaneousto one observer are not necessarily so to all observers. Let’s use Einstein’s own

Describe the significance of Einstein’s assumption of the constancy of the speed of light.

Identify that if c is constant then space and time become relative.

WhaT’S So SpeciaL abouT reLaTiviTy?

Einstein called his 1905 replacement theory for

Newton’s mechanics special relativity. It is a ‘special case’ in the mathematical sense of being restricted to particular conditions—to inertial reference frames. Einstein’s general relativity came 11 years later and was generalised to include non-inertial reference frames. It replaced Newton’s gravity.

3

66


thought experiment. Observer O1 is standing on a train station platform equidistant from points A and B, which appear to O1 to be struck simultaneously by two bolts of lightning (Figure 3.3.1). Because the light travelled the same distance from both points and the light reached O1 at the same time, O1 judges that the lightning bolts struck simultaneously.

Suppose observer O2 is sitting in a high speed train passing the platform without stopping. O1 calculates that at the moment the lightning struck, O2 on the train was also equidistant from A and B and so naively assumes that O2 would also see the events as simultaneous. However, by the time the light reached O1’s eyes, O2 was closer to B and had already seen the light from B but not A, and concluded B was struck first. Only if two events occur simultaneously at the same place, will all observers agree that they were simultaneous.

Similar arguments can show that the order of events is relative. Since an effect must come after its cause, relativity places restrictions on possible chains of cause and effect. Because the speed of light is the universal speed limit, two events separated by distance cannot have any influence over each other over a time scale shorter than the time required for light to travel between them. Because no signal or influence can travel between a cause and its effect faster than light, apparent changes in the simultaneity or order of events based on the passage of light is more than just an optical illusion—it represents a fundamental limitation of reality.

Time dilationThe relativity of simultaneity suggests that time itself is a bit rubbery. Relativity predicts that the rate of passage of time differs, depending on the velocity of the observer.

Consider the following thought experiment (Figure 3.3.2). Suppose you (observer 0) are on a train, moving with speed v relative to the ground, while measuring the speed of light by shining a light pulse vertically towards a mirror on the train ceiling, a distance D from the light source. Outside on the ground is observer v who appears to you to be rushing past with horizontal speed v and watching your experiment. Both you and observer v regard your own reference frame as stationary. However, you both agree on three things: (1) the speed of light, (2) your relative horizontal speed v and (3) the height of the mirror D. You both agree on the height of the mirror because you are both in the same reference frame with respect to vertical components.

Your light source also contains a detector capable of timing the interval t0 between the emission and detection of the light pulse. You do the experiment and use the speed formula c = 2D/t0 to obtain the correct speed of light.

Solve problems and analyse information using: E = mc2

l lv

cv = −0

2

21

tt

v

c

v =

−

0

2

21

mm

v

c

v =

−

0

2

21

Analyse and interpret some of Einstein’s thought experiments involving mirrors and trains and discuss the relationship between thought and reality.

Aplatform

B

O2

O1

v

source

mirror

v

detector

v

vtvsource anddetector

mirror

D

Figure 3.3.1 Lightning strikes at A and B appear simultaneously to observer O1 but not to O2.

Figure 3.3.2 Measuring the speed of light on a train as seen by (a) observer 0 within the train and (b) observer v from the reference frame on the ground

a b

67

Space

Observer v disagrees about what happened. She was carrying an accurate stopwatch and timed the event independently, getting a longer time interval of tv. Using Pythagoras’ theorem, she calculates that the path length of the light (Figure 3.3.2b) was not 2D, but rather:

2 42 12

2 2 2D vt D vt+ = +( ) ( )v v

She agrees on the speed of light c so:

cD vt

t=

+4 2 2( )v

v

Square and rearrange: c 2tv2 = 4D 2 + (vtv)

2 but observer 0 says c = 2D /t0

Eliminate D: c 2tv2 = c 2t0

2 + (vtv)2

Rearrange: tv2(c 2 – v 2) = c 2t0

2

tt

v

c

v =

−

02

21

The factor 1 1 2 2− v c// is called the Lorentz factor (abbreviated as γ). It is always larger than 1, which means that if t0 is the time between ticks on observer 0’s clock then observer v will see observer 0’s clock whiz by at speed v, ticking more slowly with a time tv between ticks. This is time dilation. The word dilation means ‘spreading out’, just like the time between clock ticks.

An observer moving relative to you and observing a clock ticking (or any series of events) stationary in your frame, will judge events to happen more slowly than you observe them. Note that the dilation is only observed in clocks in other frames of reference. You can’t observe time dilation in clocks in your own frame, no matter how fast others think you’re moving.

As there is no preferred inertial frame, the effect is symmetrical; that is, observers can be swapped. You both agree on your relative speed v but both insist the other observer is moving and their clock is running slow. You are both right because time is relative.

A time interval observed on a clock that is stationary relative to the observer is called the proper time for that reference frame. To generalise this idea, t0 can represent the time between any two events (such as the ticks of a clock) that occur in the same place in the frame of the observer. If the events are separated in space, then extra time is required to allow for light to travel between the positions of the two events.

Global positioning system (GPS) receivers estimate your position by measuring how long it takes the GPS signal to travel from the satellites to your receiver. The orbital speed of the satellites is large enough that the calculation needs to take time dilation into account. (It also takes into account a larger effect from general relativity: time runs more slowly in a stronger gravitational field.)

The Lorentz factor γ approaches infinity as speed approaches the speed of light (Figure 3.3.3). This means that time in a frame of reference approaching the speed of light (relative to the observer) will come to a complete stop. In other words, nothing can be seen to happen in such a frame, which is another reason why the speed of light is the ultimate speed limit.

tvt0

Speed VC0

00.2 0.4 0.6 0.8 1.0

7

6

5

4

3

2

1

Figure 3.3.3 Plot of the ratio of a time interval in a moving reference frame to proper time (tv/t0) versus speed in units of c. Note that at speeds below ~0.1c, the ratio ≈ 1, so time behaves nearly classically.

3

68


Worked examplequestionA muon is like a heavy electron, and at low speed it decays with a mean lifetime of 2.2 × 10–6 s. Suppose a beam of muons is accelerated to 80% of the speed of light. What would their mean lifetime be in the laboratory reference frame?

soLutionLifetime in the muon’s frame: t0 = 2.2 × 10–6 s Speed of muon’s frame: v = 0.80c Lifetime in the laboratory frame is tv: t

t

v

c

v s=

−

= ×

−= ×

−−0

2

2

6

2

6

1

2 2 10

1 0 803 7 10

.

..

The twin paradoxWhen two people pass by quickly, observing each other, they both think the other’s clock is running slower. The principle of relativity says you are both right. The twin paradox is a thought experiment in special relativity. Bill goes for an intergalactic cruise travelling at close to the speed of light (in Earth’s frame), while Phil stays on Earth (Figure 3.3.4). During the flight, they both correctly conclude that the other twin’s frame is moving, and so he is ageing more slowly.

But what happens when Bill comes back home? Observations in the same frame should agree. It turns out that Bill is younger than Phil. Does this violate the principle that all inertial frames of reference are equivalent? No. Bill turned around (accelerated) to come home. The situation is no longer symmetrical. Special relativity isn’t enough to explain what Bill saw from his accelerating frame (he needs general relativity). However we have no difficulty talking about what (non-accelerating) Phil saw.

By turning around and coming back, Bill left his original inertial frame and re-entered Phil’s frame, so he should agree with Phil. Phil remained in his inertial frame all along, so his conclusions (that Bill was moving and so is younger) have been consistent with special relativity throughout and, in his frame, correct. If instead Phil had hopped into another craft and caught up with Bill’s inertial frame, then Bill’s original conclusion would have been correct and Phil would have been younger.

This prediction has been confirmed using highly precise, twin atomic clocks and an aeroplane.

Figure 3.3.4 Both twins think the other’s clock is moving slower, but who is older at the end of the journey?

CheCkpoint 3.31 State Einstein’s first postulate and its alternative name.2 State Einstein’s second postulate and its alternative name.3 Outline why Newton’s classical mechanics is so successful despite a fundamental error (the Galilean transformation).4 Explain why the speed of light places restrictions on possible chains of cause and effect.5 Write the formula for time dilation. 6 A clock moving towards you appears to slow down. If the clock were moving in the opposite direction, would it speed up? 7 What is the name given to a time interval measured on a clock that is stationary in your frame of reference?8 In the twin paradox, during a period of constant relative motion, both Bill (astronaut) and Phil (earthling) observe the

other twin’s watch ticking more slowly. Who’s observation is actually correct?

MuoN aNd oN aNd oN

Fast-moving muons are produced in the upper atmosphere by

cosmic ray bombardment. Time dilation extends their normally short lifetimes long enough to allow many of them to make it to Earth, where they are a significant component of Earth’s background radiation.)

69

Space

3.4 Length, mass and energyThe formula for time dilation has already upset our common sense. However, once the clocks start talking to the rulers and the masses, things can only get more bizarre.

Length contractionThere is a grain of truth in Lorentz and Fitzgerald’s suggestion (section 3.2)

that the arm of a Michelson interferometer contracts by a factor of 1 2 2− v c/in the direction of motion. Their formula was correct, but their interpretation that it resulted from motion through the (non-existent) aether was wrong. Also, the contraction doesn’t happen in the frame of reference of the experimenter. Moreover, their hypothesis was ‘ad hoc’; it was designed only to patch a hole in the old theory without resulting in any additional testable predictions. So Einstein re-interpreted their mathematics in light of his theory of relativity.

If an object is moving with speed v relative to the observer, the length of the object in the direction of that motion will be observed to be contracted according to the formula:

l lv

cv = −0

2

21

where l0 is the length judged by an observer who is stationary relative to the object (proper length) and lv is the length judged by an observer in a frame moving with speed v relative to the object. The length contraction only takes place in the dimension parallel to the motion. Just like time dilation:1 the effect is symmetrical, which means the observers can be swapped—both

insist it is the other person’s ruler that is too short 2 you cannot observe a Lorentz contraction within your own frame.

Imagine that observer 1 and observer 2 are trying to measure the length of a rod, but all they have is a stopwatch. They already know accurately (and agree on) their relative speed v. Observer 1 is holding the rod and observer 2 is holding the stopwatch. They whoosh past each other almost touching, both looking at the watch.

Observer 2 is stationary relative to the watch (Figure 3.4.2a), so he knows the reading on his watch is his proper time. As the rod passes by, the watch reads zero at the start of the rod and t2 at the end, so the rod took a time t2 to pass by. Therefore he calculates that the length of the rod in his frame is lv = vt2.

Observer 1 is stationary relative to the rod (Figure 3.4.2b), so she knows that its length for her is the proper length l0. She agrees that the watch says t2, but the moving watch seemed to be ticking too slowly, so the number on the watch must be too small. Using the time dilation formula, she calculates that the time t1 in her frame was longer:

tt

v

c

1 =

−

22

21


l lv

cv = −0

2

21

tt

v

c

v =

−

0

2

21

mm

v

c

v =

−

0

2

21

Explain qualitatively and quantitatively the consequence of special relativity in relation to: • therelativityofsimultaneity• theequivalencebetween

mass and energy• lengthcontraction• timedilation• massdilation.

lvl0

Speed VC

00

0.2 0.4 0.6 0.8 1.0

1.2

1

0.8

0.6

0.4

0.2

Figure 3.4.1 Plot of the ratio of length in a moving reference frame to proper length (lv/l0) versus speed in units of c. Note that as speed approaches c, lv shrinks to zero—another reason why the speed of light is unattainable.

5

10

15

20

2530

35

40

45

50

55

6060

5

10

15

20

2530

35

40

45

50

55

60

vv

Figure 3.4.2 Measuring the length of a rod using a stopwatch as seen by (a) observer 2, holding the watch, and (b) observer 1, holding the rod

a b

3

70


C'C D'

B'B

O

A'

C'

B' A'

B A

v

v

Observer 1 then calculates that the length of the rod is l0 = vt1 or

lvt

v

c

0 =

−

22

21

.

But observer 2 says that lv = vt2, so by substitution and rearrangement

l lv

cv = −0

2

21 .

Worked examplequestionThe distance travelled by light in one year, 9.46 × 1015 m, is called a light-year (ly). The nearest star to our Sun is Proxima Centauri, 4.2 light-years away.

Suppose you are travelling to Proxima Centauri at three-quarters of the speed of light.

a Calculate how long it takes to get there from Earth (measured using your on-board clock).

b Discuss whether this answer is a contradiction.

soLutiona Both you and Earth-bound observers agree on your relative speed 0.75c. In the

spaceship’s frame, the distance to Proxima Centauri is contracted:

l lv

cv ly= − = − =0

2

221 4 2 1 0 75 2 78. . .

tl

c c= = =v ly2 78

0 753 7

..

. years

Earth

Earth

ProximaCentauri

ProximaCentauri

v

vv

Figure 3.4.4 Trip to Proxima Centauri as seen by (a) earthlings and (b) the astronauts

a

b

a

b

c

Figure 3.4.3 A fast-moving vehicle appears contracted horizontally, but also rotated away from the observer. The car is depicted when (a) stationary, (b) moving at high speed and (c) viewed from above. Corner C is normally out of sight, but at high speed, the vehicle moves out of the way fast enough to allow light reflected from C to reach your eyes at O, allowing you to see the car’s back and side at the same time. This is called Terrell–Penrose rotation.

b 3.7 years is less than the 4.2 years that light takes to get there in Earth’s frame. This is not a contradiction because in the spaceship’s frame, light would only take 2.78 years because lv = 2.78 ly.

Note that in the last example, the astronauts thought they experienced a short trip because the distance travelled was contracted, whereas the earthlings thought the astronauts felt their trip was short because their time had slowed.


71

Space

Relativistic mass If you measure the mass m0 of an object at rest in your frame (rest mass or proper mass) and use the classical definition of momentum p = m0v, then in collisions, momentum is not necessarily conserved for all reference frames.

However, momentum is conserved if one instead uses p = mvv where mv is the relativistic mass:

mm

v

c

v =

−

02

21

The relativistic mass of an object increases as its speed relative to the observer increases. As speed approaches c, the mass approaches infinity, so the force required to accelerate an object to the speed of light becomes infinite. This is yet another reason why the speed of light cannot be reached.

When accelerating particles in accelerators, this increase in mass needs to be taken into account, otherwise the machines won’t work.

reLaTiviSTic TraiN craSh

Trains A and B are about to collide head-on, each with

a speed 0.5c relative to the station. So, relative to train B, train A is moving at the speed of light, right? Wrong! The replacement for Galileo’s relative velocity rule in 1-dimension is:

vv v

v(A rel. to B)A B

A vB=

−

−12c

The speed of train A relative to train B is:

0 5 0 5

10 5 0 5

0 8

2

. ( . ). ( . )

.c c

c c

c

c− −

− × −=

7

6

5

4

3

2

1

0

mvm0

Speed VC

0 0.2 0.4 0.6 0.8 1

Figure 3.4.5 Plot of the ratio of relativistic mass mv in a moving reference frame to rest mass m0 versus speed in units of c. As speed approaches c, the relativistic mass approaches infinity.

Worked examplequestionA medical linear accelerator (linac) accelerates a beam of electrons to high kinetic energies. These electrons then bombard a tungsten target, producing an intense X-ray beam that can be used to irradiate cancerous tumours. A typical speed for electrons in the beam is 0.997252 times the speed of light.

Calculate the Lorentz factor and hence the relativistic mass of these electrons, given the rest mass is 9.11 × 10–31 kg.

soLution Lorentz factor γ =

v

c

= =1 113.5

−2

21

1 – 0.997252

m

m

v

c

v =

−

02

21

= 9.11 × 10–31 × 13.5 = 1.23 × 10–29 kg

Note: When calculating Lorentz factors close to the speed of light, use a greater number of significant figures than usual, because you are subtracting two numbers of very similar size.

3

72


phYsiCs FeAtuRe

twisting spACetiMe ... And YouR Mind

There are two more invariants in special relativity. Maxwell’s equations (and hence relativity)

requires that electrical charge is invariant in all frames. Another quantity invariant in all inertial frames is called the spacetime interval.

You may have heard of spacetime but not know what it is. One of Einstein’s mathematics lecturers Hermann Minkowski (1864–1909) showed that the equations of relativity and Maxwell’s equations become simplified if you assume that the three dimensions of space (x, y, z) and time t taken together form a four-dimensional coordinate system called spacetime. Each location in spacetime is not a position, but rather an event—a position and a time.

Using a 4D version of Pythagoras’ theorem, Minkowski then defined a kind of 4D ‘distance’ between events called the spacetime interval s given by:

s 2 = (c × time period)2 – path length2 = c 2t 2 – ((∆x)2 + (∆y)2 + (∆z)2)

Observers in different frames don’t agree on the 3D path length between events, or the time period between events, but all observers in inertial frames agree on the spacetime interval s between events.

In general relativity, Einstein showed that gravity occurs because objects with mass or energy cause this 4D spacetime to become distorted. The paths of objects through this distorted 4D spacetime appear to our 3D eyes to follow the sort of astronomical trajectories you learned about in Chapter 2 ‘Explaining and exploring the solar system’. However, unlike Newton’s gravitation, general relativity is able to handle situations of high gravitational fields, such as Mercury’s precessing orbit around the Sun and black holes. General relativity also predicts another wave that doesn’t require a medium: the ripples in spacetime called ‘gravity waves’.

Figure 3.4.6 One of the four ultra-precise superconducting spherical gyroscopes on NASA’s Gravity Probe B, which orbited Earth in 2004/05 to measure two predictions of general relativity: the bending of spacetime by the Earth’s mass and the slight twisting of spacetime by the Earth’s rotation (frame-dragging)


Mass, energy and the world’s most famous equationThe kinetic energy formula K = 1

2mv 2 doesn’t apply at relativistic speeds,

even if you substitute relativistic mass mv into the formula. Classically, if you apply a net force to accelerate an object, the work done equals the increase in kinetic energy. An increase in speed means an increase in kinetic energy. But in relativity it also means an increase in relativistic mass, so relativistic mass and energy seem to be associated. Superficially, if you multiply relativistic mass by c 2 you get mv c

2, which has the same dimensions and units as energy. But let’s look more closely at it.


E = mc2

l lv

cv = −0

2

21

tt

v

c

v =

−

0

2

21

mm

v

c

v =

−

0

2

21

73

Space

How does this formula behave at low speeds (when v 2/c 2 is small)?

m cm c

v

c

m cv

cv

2 02

2

2

02

2

2

12

1

1=

−

= −

−

Using a well-known approximation formula that you might learn at university, (1 – x )n ≈ 1 – nx for small x:

m cv

c0

22

2

12

1−

−

≈ m cv

c0

22

21

12

+ ×

= m0c 2 + 1

2m0v 2

Rearrange: mvc 2 – m0c 2 = (mv – m0)c 2 ≈ 1

2m0v 2

In other words, at low speeds, the gain in relativistic mass (mv – m0)multiplied by c 2 equals the kinetic energy—a tantalising hint that at low speed mass and energy are equivalent. It can also be shown to be true at all speeds, using more sophisticated mathematics. In general, mass and energy are equivalent in relativity and c 2 is the conversion factor between the energy unit (joules) and the mass unit (kg). In other words:

E = mc 2

where m is any kind of mass. In relativity, mass and energy are regarded as the same thing, apart from the change of units. Sometimes the term mass-energy is used for both. m0 c 2 is called the rest energy, so even a stationary object contains energy due to its rest mass. Relativistic kinetic energy therefore:

m c m cm c

v

c

m cv2

02 0

2

2

2

02

1

− =

−

−

Whenever energy increases, so does mass. Any release of energy is accompanied by a decrease in mass. A book sitting on the top shelf has a slightly higher mass than one on the bottom shelf because of the difference in gravitational potential energy. An object’s mass increases slightly when it is hot because the kinetic energy of the vibrating atoms is higher.

Because c 2 is such a large number, a very tiny mass is equivalent to a large amount of energy. In the early days of nuclear physics, E = mc 2 revealed the enormous energy locked up inside an atom’s nucleus by the strong nuclear force that holds the protons and neutrons together. It was this that alerted nuclear physicists just before World War II to the possibility of a nuclear bomb. The energy released by the nuclear bomb dropped on Hiroshima at the end of that war (smallish by modern standards) resulted from a reduction in relativistic mass of about 0.7 g (slightly less than the mass of a standard wire paperclip).

Worked examplequestionWhen free protons and neutrons become bound together to form a nucleus, the reduction in nuclear potential energy (binding energy) is released, normally in the form of gamma rays. Relativity says this loss in energy is reflected in a decrease in mass of the resulting atom.


eviL TWiNS

The most extreme mass–energy conversion involves antimatter.

For every kind of matter particle there is an equivalent antimatter particle, an ‘evil twin’, bearing properties (such as charge) of opposite sign. Particles and their antiparticles have the same rest mass. When a particle meets its antiparticle, they mutually annihilate—all their opposing properties cancel, leaving only their mass-energy, which is usually released in the form of two gamma-ray photons. Matter–antimatter annihilation has been suggested (speculatively) as a possible propellant for powering future interstellar spacecraft.

3

74


eXpLodiNg a MyTh

It is commonly believed (wrongly) that Einstein was involved in the

US nuclear bomb project. Perhaps this is because, during World War II, the nuclear physicists Leo Szilard, Eugene Wigner and Edward Teller, knowing such a bomb was possible and worried the Nazis might build one, wrote a letter to President Roosevelt suggesting the US beat them to it. They asked their friend Einstein to sign it because, being the most well-known scientist at the time, he would be taken seriously. Apart from that, Einstein did two days’ work on the theory behind uranium enrichment.

CheCkpoint 3.41 Discuss why, if Lorentz and Fitzgerald came up with the correct formula for length contraction, Einstein gets the

credit for explaining relativistic length contraction.2 Write the formula for length contraction. Would a ruler moving lengthwise relative to you appear shorter or longer?3 Define the term proper length.4 To what limit does observed length of a moving object tend as speed approaches c?5 Write the formula for relativistic mass. Would a mass moving relative to you appear larger or smaller?6 Use relativistic mass to justify the statement that the speed of light is the universal speed limit.7 Define all the terms in the equation E = mc 2 and explain what the equation means.8 Explain why an atom weighs less than the sum of its parts.

Calculate how much energy is released when free protons, neutrons and electrons combine to form 4.00 g of helium-4 atoms (2 protons + 2 neutrons + 2 electrons). At room temperature and pressure, each 4 g of helium gas is about 25 L, roughly the volume of an inflatable beach ball.

Data: Mass of proton mp = 1.672622 × 10–27 kg

Mass of neutron mn = 1.674927 × 10–27 kg

Mass of electron me = 9.11 × 10–31 kg

Mass of helium atom mHe = 6.646476 × 10–27 kg

c 2 = 8.9876 × 1016 m2 s–2

soLutionTotal mass of the parts:

mT = 2(mp + mn + me) = 2(1.672622 + 1.674927 + 0.000911) × 10–27 kg

= 6.69692 × 10–27 kg

Reduction in mass:

∆m = mT – mHe = (6.69692 – 6.646476) × 10–27 kg

= 5.0444 × 10–29 kg

Binding energy per He atom:

∆E = ∆mc 2 = 5.0444 × 10–29 kg × 8.9876 × 1016 m2 s–2

= 4.5337 × 10–12 J

Binding energy for 4.00 g (0.004 kg):

4.5337 × 10–12 J × 0.004 kg = 2.73 × 1012 J

mHe

This much energy would be released by the explosion of more than 600 tonnes of TNT.

Some physicists dislike the definition of relativistic mass mv of a moving object and prefer to talk only about the energy of an object (and its rest mass m0). There are problems with the definition, including the fact that relativistic mass doesn’t behave like a scalar, because it can be different along different directions.


75

Space

chapTer 3This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACtiVitY 3.1: FACt oR FiCtion: ineRtiAL And non-ineRtiAL FRAMes oF ReFeRenCePerform an investigation that allows you to distinguish between inertial and non-inertial frames of reference.Equipment: protractor, string, mass (50 g), tape, cardboard, chair on wheels or skateboard.

Discussion questions1 The principle method for detecting a non-inertial frame is measurement

of acceleration. Describe an example of a non-inertial frame in which a typical accelerometer would not appear to measure an acceleration or detect extra fictitious forces.

2 Is there a test that can be performed within a frame of reference to tell if the effect measured by the accelerometer is the result of acceleration of the frame or due to an actual additional force?

ACtiVitY 3.2: inteRpReting the MiCheLson–MoRLeY expeRiMent ResuLtsUse simulations to gather data from the Michelson–Morley experiment. You will gather data as though there is and is not an aether, and then interpret the results.

There are many Michelson–Morley experiment simulations available. Two web-based examples are given on the companion website.

Discussion questions1 Describe what Michelson and Morley were expecting to observe if aether

were present.2 Using the data you have gathered, explain how your observations support

or refute the existence of the aether.3 Recall the interpretation put forward by Michelson and Morley.4 Discuss the importance of this experiment.

Extension1 Research the history of how long the belief in aether persisted in some

physicists after the publication of special relativity in 1905.2 Read the following paper, which contains a thorough review of the history

of the Michelson–Morley experiment, including historical letters to and from several researchers:

Shankland, R S, 1964, ‘Michelson–Morley Experiment’, American Journal of Physics, vol. 32, p 16.

Perform an investigation to help distinguish between non-inertial and inertial frames of reference.

Gather and process information to interpret the results of the Michelson–Morley experiment.

Figure 3.5.1 An accelerometer

010

2030

4050 60 70 80 90

8070

6050

4030

20

100

76


review questions

chapter summary• Inertialreferenceframesarethosethatdonotaccelerate.• Principleofrelativity:Thelawsofmechanicsarethe

same in all inertial reference frames. Einstein extended it to all laws of physics (first postulate of relativity).

• Whenjudgedwithinanon-inertialframe,fictitiousforces are perceived.

• Maxwell’sequationsforelectromagnetismpredictedonly a single possible speed for light, which was assumed to be relative to a hypothetical medium called aether.

• MichelsonandMorleyfailedtodetectchangesinspeeddue to aether wind, using an interferometer. Fitzgerald and Lorentz made the ad hoc suggestion that things contract when moving relative to the aether, hiding the effect of the changing relative speed of light.

• Einsteinandothersarguedthataetherwasnotrequiredby Maxwell’s equations and was inconsistent with the principle of relativity.

• Secondpostulateofrelativity:Thespeedoflightisconstant to all observers.

• Thespeedoflightisthefastestpossiblespeed.• Thefinitespeedoflightmeansdifferentobservers

disagree on the simultaneity and order of events. Only events at the same time and place are agreed by all observers to be simultaneous.

• Lorentzfactor:γ =

−

1

12

2

v

c

• Propertime t0 is a time interval measured on a clock stationary in the observer’s frame.

• Properlength l0 is the length of an object stationary in the observer’s frame.

• Properorrestmassm0 is the mass of an object stationary in the observer’s frame.

• Timedilation: tt

v

c

v =

−

02

21

Clocks (and all time-dependent phenomena) evolve in time more slowly if they are moving relative to the observer’s frame.

• Lengthcontraction: l lv

cv = −0

2

21

Length lv of an object moving relative to the observer’s frame contracts in the direction of motion.

• Relativisticmass: mm

v

c

v =

−

02

21

Mass of an object mv moving relative to the observer’s frame increases.

• Twoobserversinseparateinertialframeswillagreeontheir relative speed v.

• However,bothobserverswilljudgetheother observer to be moving and, hence, subject to time dilation, length contraction and relativistic mass increase. They disagree, but both are correct because these three quantities are relative. Only when two observers are in the same frame will they agree on these.

• Massandenergyareequivalent:E = mc 2. A small mass is equivalent to a large energy.

phYsiCALLY speAkingInertial _______________ have _______________ status in _______________ mechanics.

_______________’s laws apply in these frames. If one performs measurements

in _______________ , then _______________ forces might be perceived. Classical

mechanics and _______________ relativity both agree that physical laws are

_______________ in _______________ frames. However, they disagree on the

_______________ of the speed of light. According to _______________’s equations,

the _______________ of the speed of light does not _______________ between frames,

so light doesn’t obey the transformation formula of _______________ . Because of

this, measurements of _______________ , _______________ and _______________ within

a reference frame moving relative to the _______________ , will depend on the

_______________ of that frame.

Use the words below to complete the following paragraph:

Galileo, Newton, Einstein’s, Maxwell,

constancy, fictitious, change,

non-inertial frames, length, observer,

classical, value, invariant, mass, time,

frames, speed, inertial, special

77

Space

ReViewing 1 You have a priceless Elvis Presley doll hanging from

your rear-vision mirror at a constant angle from vertical. Elvis’s feet lean towards the front of the car. Are you driving:A forwards at uniform speed? B backwards at uniform speed? C forwards but accelerating?D forwards but decelerating?

2 In a car that is cornering, is the centripetal force exerted on you by the seat belt fictitious? Centrifugal force normally refers to the fictitious force you feel pushing you outwards when you steer a car. Some people have suggested re-defining centrifugal force as the outward reaction force you exert on the seat belt in response to the centripetal force it exerts on you. Re-defined in this way, is centrifugal force still fictitious? Justify your answers.

3 At the end of the 19th century, no-one was able to travel at close to the speed of light, and clocks, rulers and mass balances weren’t sensitive enough to measure relativistic changes. So why did the problems with classical physics start to become obvious then?

4 Explain why interferometry is an extremely sensitive method for measuring short differences in time or length.

5 Explain why Michelson and Morley performed their experiment at different times of the day and year.

6 If we were an entire civilisation of blind people relying on sound instead of light to decide the simultaneity of events, would our equations for relativistic length, time and mass contain c = 340 m s–1 (the speed of sound in air) instead? What’s so special about the speed of light? Discuss.

7 In Figure 3.3.2b, the dimensions of the light path have been drawn correctly. However, for simplicity, two aspects of the train’s appearance to observer v have been left out. Describe two changes that would need to be made to Figure 3.3.2b to represent these effects more correctly.

8 Suppose our relativistic twins Bill and Phil both got into spacecraft, went off in opposite directions and took journeys at relativistic speeds that were mirror images (judged from Earth). Predict and explain:a how their apparent ages will compare when they

come back homeb how their apparent ages will be judged by stay-at-

home earthlings.

9 Prunella and Renfrew, two observers in inertial frames moving relative to each other, will always agree on their relative speed v. A third observer, Thor, standing between them, sees them both coming towards him from opposite directions, at equal speeds. Is it correct to say that relative to Thor, Prunella and Renfrew are both moving at a speed of v|2?

10 A stretch-limo drove into a small garage at near light speed. The garage attendant slammed the garage door behind the car. For a brief time the attendant saw that the relativistically shortened limo was completely contained between the closed garage door and the rear garage wall. A short time later, the still-moving car smashed through the back wall. As far as the driver was concerned, the garage was shortened and the limo was too long for the garage so the limo was never contained between a closed door and the intact back wall. Reconcile the two differing accounts of what happened. (Hint: See section 3.3.)

11 Show that mc2 has the units and dimensions of energy.

12 In a perfectly inelastic collision, two colliding objects stick together. In a symmetrical inelastic collision between two identical objects, the final speed is zero in the frame of their centre of mass. Given that mass-energy is conserved in an inertial frame, is the mass of the system the same as before the collision? Explain. (Hint: What happens to kinetic energy in an inelastic collision?)


l lv

cv = −0

2

21

tt

v

c

v =

−

0

2

21

mm

v

c

v =

−

0

2

21

78


soLVing pRobLeMs 13 Depending on your answer to Question 1, calculate

the magnitude of your speed or acceleration if the Elvis Presley doll hangs at a constant angle of 10° from vertical.

14 The caption for Figure 3.2.3b states that increasing the length of the arms would increase sensitivity to changes in the speed of light. Justify this, using the equations given in that section.

15 Supposing the aether hypothesis were correct, show that (in agreement with Fitzgerald and Lorentz’s suggestion) if the length L of the interferometer arm parallel to the aether wind shrinks to

′ = −L L v c1 2 2

then the difference t2 – t1 between the times of flight for the two arms would be zero. Use the equations given for t2 and t1 in section 3.2.

16 In the worked example of your trip to Proxima Centauri (Figure 3.4.4), one member of the crew had a mass 80 kg at launch. Assuming his normal diet and physiology were maintained, what would you expect his mass to be during the trip:a as measured on the spaceship? b as judged from Earth?

17 Your rival in the space race plans a trip to Alpha Centauri, which is slightly further away (4.37 ly). She wants to do the trip in 3.5 years (one-way) as judged by her own on-board clock. a What speed (as a fraction of c) does she need to

maintain? b How long does the trip take as judged from Earth?

18 Calculate the total energy in the two gamma ray photons produced when an electron meets a positron (an anti-electron) (me = 9.11 × 10–31 kg).

19 For subatomic particles, a more conveniently sized (non-SI) unit of energy is the electron volt (eV). The conversion is E(eV) = E(J)/e where e = 1.60 × 10–19 C, the charge on an electron. A mega-electron volt (MeV) is 106 eV.

For the worked example on page 71, show that the kinetic energy of the electron in the medical linac beam is 6.4 MeV (me = 9.11 × 10–31 kg). What is the total energy of that electron?

20 Estimate the total energy (in joules) released by the Hiroshima bomb (∆m0 = 0.7 g).


l lv

cv = −0

2

21

tt

v

c

v =

−

0

2

21

mm

v

c

v =

−

0

2

21

Revie

w Questions

21 In their rest frame, muons have a mean lifetime of 2.2 × 10–6 s. However, measurements (at various altitudes) of muons produced by cosmic rays indicate that, on average, they travel 6.00 × 103 m from where they are produced in the upper atmosphere before decaying. Calculate their average speed (as a fraction of c).

22 Show that if the speed of light were infinite, the following equations would revert to their classical form.

l lv = −0 1 2c/2v

tt

v = 0

−1 2c/2v

mm

v = 0

−1 2c/2v

vv v

v(A rel. to B)A B

A vB

=−

−1 2c/

23 Research the history of relativity and list up to five historically important experimental confirmations of its predictions. Make a timeline of the events. Note that some experiments may pre-date relativity. For example, in 1901 W Kaufmann measured the increase in an electron’s mass as its speed increased. If possible, identify whether such examples came to Einstein’s attention before he formulated his theory.

Analyse information to discuss the relationship between theory and the evidence supporting it, using Einstein’s predictions based on relativity that were made many years before evidence was available to support it.

a

b

c

d

79

Space

phYsiCs FoCusCAn’t MeAsuRe the speed oF Light

The French metric system, which evolved into the Système International d’Unités or SI units, was

originally based on ‘artefact’ standards. The standard metre bar and kilogram were real objects (or artefacts) in Paris. Artefacts can degrade or be damaged, and making copies for standards labs is expensive, slow and unreliable. Artefact standards are now being replaced by fundamental physical property standards. One second is now defined as a certain number of periods of oscillation of a very stable light frequency in the spectrum of cesium-133 (in atomic clocks).

Measurement standards often involve sensitive interferometry. The metre was changed in 1960 from the original bar to a certain number of wavelengths (measured interferometrically) of a colour from the krypton-86 spectrum.

Being invariant, the speed of light is very useful for standards. Interferometric measurements of the speed of light became so precise that the weakest link was the experimental difficulty in reproducing the krypton-86 standard metre. So in 1983 the speed of light was fixed by definition at exactly 299 792 458 m s–1 and the standard metre was redefined as the distance travelled by light in 1|299 792 458 of a second. Now, any lab with an interferometer and an atomic clock can produce its own standard metre.

Since 1983, by definition, the speed of light can no longer be measured. Traditional procedures for measuring the speed of light should now be called ‘measuring the length of a metre’.

The last artefact standard, the platinum–iridium kilogram in Paris, appears to be changing mass slightly. The Avogadro project at Australia’s CSIRO is trying to develop a replacement for it with a procedure for making and testing (almost) perfect spheres of silicon that could be made in standards labs around the world without the need to copy the original sphere directly. The spheres are measured using interferometry, with the best result so far being an overall distortion from sphericity of 30 nm and an average smoothness of 0.3 nm.

1 Explain why standards based on fundamental physics properties are preferable to artefacts.

2 Justify (in light of relativity) the statement that the speed of light is an especially good property on which to base a measurement standard.

3 The 1960 metre standard was based on light from krypton-86. Explain why it needed to specify the light

source and why the new metre standard doesn’t.4 Given that the value of the speed of light is now

arbitrarily fixed, discuss why they didn’t just make the speed of light a nice round number such as 3.000 000 00 × 108 m s–1.

5 A single atomic layer of silicon is approximately 5.4 × 10–10 m thick. For the best silicon sphere in the Avogradro project, to approximately how many atomic layers does the reported distortion from sphericity and average smoothness correspond?

2. The nature and practice of physics



Discuss the concept that length standards are defined in terms of time in contrast to the original metre standard.

Figure 3.5.2 One of CSIRO’s accurate silicon spheres

80

1Multiple choice(1 mark each) 1 Ignoring air resistance, all projectiles fired

horizontally from the same height above horizontal ground will have the same:A horizontal velocity. B time of flight.C range. D final speed.

2 Which of the following orbits has a two-body mechanical energy greater than zero?A Geostationary B Elliptical C ParabolicD Non-returning comet

3 You have just rounded the top of a curve on a roller-coaster. The g-force meter you are carrying reads exactly zero. Which one of the following is true?A Your weight is the centripetal force.B Your weight is zero.C Your weight is equal and opposite to the normal

force exerted on you by the seat.D Your weight is equal and opposite to the tension

in your body.

4 The Michelson–Morley experiment demonstrated that:A the aether wind was undetectable.B waves do not require a medium.C one arm of the interferometer contracted in

response to the aether wind.D aether is trapped by mountains and valleys and

dragged along with the Earth.

5 Observer A on the ground, watches a train (containing observer B) rush past at speed v. Both make measurements of things in each other’s frame of reference. From the following list of statements, choose the statement they disagree on.A The other observer’s frame of reference is moving

with speed v.B The apparent length of my own metre ruler is

longer than the apparent length of other observer’s metre ruler.

C Observer B’s watch appears to run slower than observer A’s watch.

D The height of the train carriage ceiling is 2.2 m above the carriage floor.

Short response 6 The escape velocity from the Earth’s surface, based

on Newton’s original concept, is 11.2 km s–1. Briefly explain two ways in which this number is not quite applicable to real Earth-surface launches. (2 marks)

7 Calculate the potential energy of a 2500 kg satellite in a geostationary orbit around the Earth. Assume a sidereal day is 23 h 56 min 4 s. (3 marks)

8 In their rest frame, charged pions have a mean lifetime of 2.60 × 10–8 s. A particular beam of charged pions travel an average distance of 30 m before decaying. Calculate their speed (as a fraction of the speed of light). (4 marks)

9 Explain why if you are in a circular orbit and you briefly retro-fire your engines to slow down, you move to a faster orbit. (3 marks)

The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 30 marks. It should take you approximately 54 minutes to complete this review.

81

Space

10 A 9000 kg helicopter is parked at the equator and then later near the North Pole. Assuming the two locations are chosen so that the gravitational potential energy is the same at the two spots, estimate the difference in relativistic mass at the two locations. At which location will the mass be larger? (Hint: Velocity is low, so use the classical expression for kinetic energy.) (3 marks)

extended response 11 Critically discuss the following proposition: ‘The

Michelson–Morley experiment was an embarrassment for physics because, despite a large effort, it failed to find what it was looking for and so it should be relegated to the dustbin of physics history.’ (5 marks)

12 The following formula relates the length of a pendulum (L) to the period of its swing (T).

TLg

= 2π

During your studies in physics you carried out an experiment to determine acceleration due to gravity. Describe and explain a method you would use to perform this measurement. (5 marks)

2 Motors and Generators

Figure 4.0.1 A generator produces electricity in each of these wind turbines.

82

The first recorded observations of the relationship between electricity and magnetism date back more than 400 years. Many unimagined discoveries followed, but progress never waits. Before we understood their nature, inventions utilising electricity and magnetism had changed our world forever.

Today our lives revolve around these forms of energy. The lights you use to read this book rely on them and the CD inside it would be nothing but a shiny coaster for your cup. We use magnetism to generate the electricity that drives industry, discovery and invention. Electricity and magnetism are a foundation for modern technology, deeply seated in the global economy, and our use impacts heavily on the environment.

The greatest challenge that faces future generations is the supply of energy. As fossil fuels dry up, electricity and magnetism will become even more important. New and improved technologies will be needed. Whether it’s a hybrid car, a wind turbine or a nuclear fusion power plant, they all rely on applications of electricity and magnetism.

Context

83

InQUIRY ACtIVItY

BUIld YoUR own eleCtRIC motoR

Many of the devices you use every day have electric motors. They spin your DVDs, wash your clothes and even help cook your food. Could you live without them, and how much do you know about how they work?

The essential ingredients for a motor are a power source, a magnetic field and things to connect these together in the right way. It’s not as hard as you think. All you need is a battery, a wood screw, a piece of wire and a cylindrical or spherical magnet. Put these things together as shown in Figure 4.0.2 and see if you can get your motor to spin. Be patient and keep trying. Then try the following activities.1 Test the effects of changing the voltage you use. You could add another

battery in series or try a battery with a higher voltage.2 Try changing the strength of the magnet by using a different magnet or

adding another. What does this affect?3 Try changing the length of the screw, how sharp its point is or the material

it is made from. Does it have to be made of iron?

Figure 4.0.2 A simple homopolar motor

84

electrodynamics: moving charges and magnetic fields

4electric current, conventional current, magnetic field, force,

direct current (dC), alternating current (AC), voltage, potential difference,

right-hand grip rule, right-hand palm rule, motor effect,

current-carrying conductor

Strange but trueSome of the most surprising discoveries in science have come from the relationship between electricity and magnetism. Who would have thought that a current-carrying wire can act as a magnet? Could anyone have guessed that moving a magnet could create electricity, or that moving charges get a push from magnetic fields?

All these amazing facts were discovered by people who were curious, creative and dedicated. They were interested enough to find out how things work, and now it’s your turn. Dust off your imagination and get ready to picture the concepts behind many applications of electricity and magnetism.

4.1 Review of essential conceptsIn this module you will develop and apply an understanding of the relationship between electricity and magnetism. First we need to review some essential concepts you encountered in the Preliminary course (see in2 Physics @ Preliminary Chapters 10 and 12). You will need to recall these concepts throughout this module and apply them to new situations.

Electric currentIn the presence of an electric field, charged particles and free ions will move towards a region with an opposite electric charge. In this module we will be considering the movement of electrons within metal conductors such as copper. If you connect a wire to the terminals of a battery, as in Figure 4.1.1, an electric field is set up along the length of the wire between the two ends. The free electrons in the metal wire are then attracted towards the positive terminal. The movement of these electrons is called an electric current. Protons are bound tightly inside the nuclei of atoms and the nuclei are essentially fixed in position. This means that when a wire is connected to a battery the only charges free to move inside the wire are the free electrons.

85

Motors andGenerators

An electric current is defined as the rate of flow of net charge through a region. In the wire in Figure 4.1.1, a number of electrons will flow through the area A each second, and each electron carries a charge of –1.6 × 10–19 coulombs. If the current is 1 ampere (or 1 amp), it carries 1 coulomb of charge per second through area A, which is more than 6 billion billion electrons each second.

We now know that the nature of an electric current is actually a flow of electrons. However, in electric circuits we often consider currents as if they were a flow of positive particles. This type of current is called conventional current and it is in the opposite direction to the flow of the negatively charged electrons. This confusing situation has arisen because current was first thought to be a flow of positive particles. Conventional methods for determining the direction of other physical quantities, such as magnetic fields and forces, have been developed using the conventional direction of current. So we will stick with these historical conventions.

It is important throughout this module to consider the direction of an electric current as the direction in which a positive charge would flow through a conductor. In Figure 4.1.1, conventional current flows from the positive terminal towards the negative terminal, as indicated by the arrow along the wire.

Direct current and alternating current An electrical current in which the charges only flow in one direction is

called direct current (DC). This current is commonly used in small portable electronic devices and is supplied by a battery. One way of illustrating this type of current graphically is shown in Figure 4.1.2.

The red line on the graph is a direct current measured by a digital ammeter. The sign of the current (+ or –) represents the direction in which the current is travelling. You can see that in this example the current has a constant value and direction over time.

In contrast to direct current, an alternating current (AC) is continually changing direction. The sign of the terminal at each end of an AC circuit alternates between positive and negative over time. Each time this occurs, the electric field within the wire changes direction. This reverses the direction of the force on the charges within the wire and the current changes direction accordingly. This type of current is good for transporting electrical energy over large distances and is commonly used in larger appliances. In Figure 4.1.2, an AC current measured by a digital ammeter is shown as a blue line. As the current changes direction the blue line moves above or below the horizontal axis. The corresponding change in sign of the current indicates a change in the current’s direction.

Potential difference, emf and voltageThe work done by any electrical device can be traced back to the creation of a difference in electrical potential energy. In Figure 4.1.1, a chemical reaction separates the charges inside the battery. This causes a difference in electrical potential energy between the positive and negative terminals of the battery. The battery energy given to each coulomb of charge within the battery is measured in volts and is commonly called an emf (ε). In an ideal battery it is equal to the voltage measured at the terminals when the battery is not connected to a circuit. For this type of battery this is usually about 1.5 V.

–

––

–areaA

copper wire

+

–

Figure 4.1.1 A battery creates an electric field within the wire and a current flows.

10 20 30 40 50

Cur

rent

(A

)

Time(ms)

DC

AC

+

0

–

Figure 4.1.2 A graph of AC and DC over time

Flick oF a switch

The charges in a current-carrying wire travel along the wire much more slowly than the speed at

which you normally walk. Why then does a light come on instantly when you flick the switch? All the free electrons in the wire start moving at the same time under the influence of the electric field in the wire. When you flick the switch, this field travels along the wire at close to the speed of light. Almost instantly all the free electrons are moving and losing energy in the light bulb.

electrodynamics: moving charges and magnetic fields4

86

When a wire is connected to the battery’s terminals (see Figure 4.1.1), an electric field is set up within the wire. The electric field drives electrons from the negative terminal, through the wire to the positive terminal. As the electrons move through a circuit element, such as a light bulb, they collide with the ions in the metal lattice. During these collisions they lose kinetic energy to the metal lattice. The metal lattice then loses this energy as heat; in the light bulb it is also lost as visible light.

The energy lost in the circuit element corresponds to a loss of electrical potential energy by the charges in the current. The difference in potential energy per unit of charge between the two points either side of a circuit element is known as the potential difference, potential drop or voltage (V). We can measure this potential difference by connecting a voltmeter to the circuit in parallel with the circuit element.

Resistance and Ohm’s lawRecall that the structure of a metallic conductor is essentially a lattice of metal atoms (or ions) surrounded by a ‘sea’ of free electrons. If a potential difference V is established within the metal, these free electrons will flow as a current I.

The amount of current that flows due to this potential difference is determined by the electrical resistance of the material, which is defined by R R

VI

= and measured in ohms (Ω). A material that has a relatively high resistance will only conduct a relatively small current for a given potential difference.

The resistance in a conductor is a result of the collisions of the moving charges with the ions of the metal lattice. Basically, the more collisions the free electrons have with the lattice the higher the resistance. Recall that the length, cross-sectional area, temperature and type of material within the conductor influence resistance.

In many circuit components, the ratio of voltage divided by current is a constant. This relationship is known as Ohm’s law and describes the relationship in which V and I are proportional for a circuit, or circuit component, with a fixed resistance R.

Resistance and powerThe wire filament of the light bulb in Figure 4.1.1 has a very high resistance and, therefore, a large amount of the available energy within the circuit is lost within the bulb. This energy heats the filament in the blub to such a high temperature that it emits visible light.

Whether energy in a circuit is lost as heat or in turning an electric motor, the rate at which energy is converted into another form is called electric power P :

=energy transferred

time taken for transferPower

Recall that: P = IVSubstituting Ohm’s law V = IR: P = I 2R or P

VR

=2

where P is power in watts (W).

Watts are equivalent to joules per second (J s–1), so we can determine the energy lost:

Energy = Pt

where energy is in joules (J) and time is in seconds (s).

Birds on a wire

Have you ever wondered why birds can happily sit on

power lines and not get electrocuted? No, those wires are not insulated. For a current to flow through a bird on a wire, there would have to be a potential difference between its feet. If the bird could stand on the wire and touch any other object such as the ground or another wire then it would get the shock of its life.

Figure 4.1.3 There is no potential difference between the bird’s feet.

87


Magnetic fields produced by electric currentsA magnetic field exists in a region of space if a magnet at some point in that space would experience a magnetic force. From in2 Physics @ Preliminary section 12.3, you know that a current-carrying conductor produces a magnetic field. You can observe this circular magnetic field around a current-carrying wire if you place a magnetic compass needle at various locations around the wire (see Figure 4.1.4).

We represent magnetic fields by solid lines, called magnetic field lines, and label these lines with the symbol B. These lines represent the places where the magnetic field has the same strength. We also place arrow heads along these magnetic field lines to indicate the direction of the field. By convention, the arrow heads indicate the direction in which the north pole of a magnet would point within the magnetic field (see the compass needles in Figure 4.1.4).

You can work out the direction of the magnetic field in Figure 4.1.5 by using the right-hand grip rule. Grip the wire with your right hand and point the thumb in the direction of the conventional current along the wire. Remember, this is the direction in which positive particles would flow, from the positive to the negative terminal. Your curled fingers will now point in the direction of the magnetic field around the wire.

There are several conventions we need to recall when we draw two-dimensional diagrams showing currents and magnetic fields. For example, when we view the situation shown in Figure 4.1.5 from above, we represent this as shown in Figure 4.1.6. The symbol in the centre indicates that the current is coming towards you (out of the page). When viewing the situation from below, we represent this as shown in Figure 4.1.7. The symbol in the centre indicates the current is flowing away from you (into the page). Notice that the magnetic field lines shown in Figures 4.1.6 and 4.1.7 are drawn more closely near the wire, where the field becomes stronger.

The and symbols indicate that the current is flowing out of and into the page respectively. You can remember this convention if you imagine that is the head of an archer’s arrow coming out of the page at you. The crossed feathers in the back of the arrow are represented by , indicating that the arrow is pointing away from you.

We also represent magnetic fields in two-dimensional diagrams as shown in Figure 4.1.8. To show that the magnetic field points into or out of the page, we use or • respectively. Use the right-hand grip rule to determine the

B

I

Figure 4.1.4 A straight wire carrying a current deflects the compasses around it in a circular pattern.

magnetic field B

electric current I

Figure 4.1.5 The curled fingers point in the direction of the magnetic field when the thumb points in the direction of the conventional current along the wire.

B

Figure 4.1.6 Magnetic field lines around a conventional current going out of the page

B

Figure 4.1.7 Magnetic field lines around a conventional current going into the page


88

direction of the magnetic field around the wire in Figure 4.1.8. You should see that the magnetic field lines would go into the page on the right-hand side of the wire (represented by ) and come out of the page on the left (represented by •).

An extension of the situations you have reviewed above is the magnetic field around a current-carrying wire loop. Again we use the right-hand grip rule to determine the direction of the magnetic field around the current-carrying wire (Figure 4.1.9). Notice that the magnetic field in the centre of the loop always points in the same direction, no matter where your hand is around the loop. In the following chapters there are many situations that involve loops of wire carrying currents; therefore, it is important you are familiar with the magnetic field that is produced around them.

Most applications of magnetic fields in current-carrying wire loops actually involve more than one loop. Each loop is called a turn, and many turns together are known as a solenoid (Figure 4.1.10). A solenoid is simply a long coil of wire, and the magnetic field produced is similar to that of a bar magnet, with a north and south pole at each end. The direction of the magnetic field through the centre of the solenoid is determined by using a special version of the right-hand grip rule (Figure 4.1.10). In this situation, you must curl your fingers in the direction of the conventional current around the solenoid and your thumb will point in the direction of the magnetic field. Your thumb will point to the end of the solenoid that forms a north pole. A coil such as this is used to make an electromagnet or simply to produce a magnetic field.

S N

BB

conventional current I

Figure 4.1.10 The right-hand grip rule is used to find the direction of the magnetic field inside the solenoid.

Figure 4.1.8 Magnetic field lines into () and out of (•) the page for a wire carrying a conventional current upwards in the plane of the page

B

(a) (b)

+–

Figure 4.1.9 (a) The right-hand grip rule can be used for a current loop. (b) Magnetic field lines around a single wire loop

the north Pole?

Did you know that the Earth’s north geographic

pole is actually its south magnetic pole? The north pole of a magnet is attracted towards the north geographic pole and therefore it must be a south magnetic pole. Did you also know that the Earth’s magnetic field changes direction? At irregular intervals of about 250 000 years the polarity of the Earth’s magnetic field flips and points in the opposite direction. Scientists are not sure of the effects of this flip or for how long the field disappears during each flip.

GeographicNorth Pole

Geographic South Pole

MagneticNorth Pole

Magnetic South Pole

S

N

Figure 4.1.11 The Earth acts as though it has a south magnetic pole near the geographic north pole! The ‘magnetic north pole’ is the place to which the north end of a compass appears to point.

89


4.2 Forces on charged particles in magnetic fields

We have already seen that free charged particles move when placed in an electric field, because they experience a force (Figure 4.1.1). This force is caused by the attraction and repulsion that charged particles experience within the field. Knowing why this occurs, it might seem strange to hear that a charged particle also experiences a force due to a magnetic field. There is, however, one important difference. A charged particle only experiences a force in a magnetic field when the particle is moving relative to the magnetic field, or if the strength of the magnetic field is changing. It is important to know that stationary charges or charges moving parallel to the magnetic field do not experience a force.

In Figure 4.2.1a, a positively charged particle, let’s say a proton of charge q, is travelling upwards with a velocity v within a horizontal magnetic field B. The proton experiences a force F in a direction perpendicular to both the magnetic field and the direction in which it is moving. The force is given by F = qvB. The way we can tell the direction in which the force is acting is to introduce another right-hand rule. This rule is commonly called the right-hand palm rule (or the right-hand push rule) and is illustrated in Figure 4.2.1b.

To find the direction of the force that a positive particle will experience when moving through a perpendicular magnetic field:1 Place your open right hand with the fingers pointing in the

direction of the magnetic field (north to south).2 Place your thumb at right angles to your fingers and in the

direction in which the particle is travelling.3 The force on the positive particle will be directed out of your

palm and at right angles to your hand.

We can conclude that the proton in Figure 4.2.1a will experience a force out of the page at right angles to both the magnetic field and the direction in which the proton is travelling. Note that if we know the direction of any two of the three quantities represented in the right-hand palm rule, we can use this rule to determine the direction of the third quantity.

CheCkpoInt 4.11 Define the nature and direction of conventional direct current.2 Construct two-dimensional diagrams illustrating the magnetic field around a current-carrying wire from two

different perspectives (end-on and side-on).3 Sketch and compare two diagrams illustrating the magnetic field around a bar magnet and a current-carrying

solenoid.

direction in which theparticle is travellingwith velocity v

NS

direction of the force Fon the positive particle

direction of themagnetic field

B

v

F

B

+

a

b

Figure 4.2.1 The right-hand palm rule is used to find the direction of a force acting on a positively charged particle in a magnetic field.


90

4.3 the motor effectIn the previous section we saw that charges moving in a magnetic field experience a force. If a current-carrying conductor is placed in an external magnetic field, the wire also experiences a force and this is called the motor effect. This effect occurs because the charges within the wire are travelling through the magnetic field and experience a force, just as they would if they were free charged particles (see Figure 4.2.1a). Remember, we are considering current to be a flow of positive particles. It is actually the negative electrons that experience the force from the magnetic field. It works just fine to use conventional current and consider the force is acting on positive particles to keep things simple.

We can now use the right-hand palm rule we saw in section 4.2 to work out the direction of the force on a current-carrying wire in an external magnetic field (see Figure 4.3.1). To find the direction of the force that acts on a current-carrying wire that is perpendicular to an external magnetic field:1 Place your open right hand with the fingers pointing in the direction of the

magnetic field (north to south).2 Place your thumb at right angles to your fingers and in the direction in which

the conventional current is flowing (from the positive to negative terminals in the circuit).

3 The force experienced by the wire will be directed out of your palm at right angles to your hand.

Identify that the motor effect is due to the force acting on a current-carrying conductor in a magnetic field.

try this!Bending Beams of particlesAsk your teacher if they can show you a beam of electrons in a Crookes magnetic deflection tube (see Figure 4.2.2). With teacher supervision, bring the north pole of a magnet close to the front of the tube and observe the effect on the beam of electrons. By the right-hand palm rule, negative particles experience a force out of the back of your hand. Try to predict the direction in which electrons will be deflected when you place the south pole of the magnet close to the front of the tube. Now try it and then explain it to a friend. Do they agree with your explanation?

Figure 4.2.2 A Crookes magnetic deflection tube produces a beam of electrons.

CheCkpoInt 4.21 Explain why a stationary charged particle experiences a force when you move a magnet past it.2 Identify the direction in which the proton in Figure 4.2.1a would be moving for it to experience a force into the page.

activity 4.1

Practical eXPeriences


91


When we allow current to flow through a wire within the magnetic field in Figure 4.3.1, we see that the wire moves out of the page, at right angles to both the magnetic field and the direction of the conventional current. Each positive particle of the conventional current within the wire in Figure 4.3.1 experiences a force due to its motion within the external magnetic field. Since these positive particles are within the wire, the force acts on the wire.

phYSICS FeAtUReloUdSpeAkeRS

An excellent example of an application of the motor effect is a loudspeaker. This device is a key part

of telephones, televisions and any other appliance in which an electrical signal needs to be converted into sound for us to hear. Figure 4.3.2a shows the most fundamental parts of a typical loudspeaker labelled A to D. Figure 4.3.2b illustrates its operation via the motor effect.

A loudspeaker contains a current-carrying coil (C), which is commonly called the voice coil. This coil is wound around a hollow cardboard tube (B) and the tube is fixed to the cone of the speaker (A). The voice coil is suspended inside a cylindrical permanent magnet (D) that provides a uniform magnetic field at right angles to the coil. An alternating current is passed through the voice coil, causing the cone to rise and fall due to the motor effect. Each time the speaker cone pushes outwards on the air, it creates a wave of pressure that travels away from the speaker. These waves of air pressure are sound waves. By varying the frequency of the alternating current in the voice coil, the frequency (or pitch) of the sound can be varied. This means that speakers can generate a variety of sounds and reproduce sounds recorded elsewhere.

At the moment in time shown in Figure 4.3.2b, the current is travelling out of the page on the left side of the voice coil. Using the right-hand palm rule

you can see that a force will be exerted upwards on the voice coil on this side. This force is exerted in the same direction around the circumference of the voice coil and causes the speaker cone to move upwards. So we can see that now we know about the motor effect and the right-hand palm rule we can explain how some everyday things work.

Identify data sources, gather and process information to qualitatively describe the application of the motor effect in:– the galvanometer– the loudspeaker.

A cardboard cone

B cardboard tube fixed to cardboard cone and wrapped in voice coil

C wire coil (called the voice coil)

D permanent cylindrical magnet

N NS

direction of forceon the voice coil

Figure 4.3.2 A loudspeaker converts electrical energy into sound. (a) A cut-away section showing the parts of the loudspeaker and (b) a simplified cross section showing the direction of the magnetic field

A

BC

D

NS

F

I

+direction in which the

current is flowing

F direction of the force

on the positive particle

direction of themagnetic field

I

B

a

b

Figure 4.3.1 The right-hand palm rule for a current


92

Quantifying the motor effectIf we place a current-carrying wire within a magnetic field (see Figure 4.3.3) the force on the conductor is given by:

F = BIl sin θ

where F is magnitude of the force on the wire in newtons (N), l represents the length of the wire inside the magnetic field in metres (m), I is the size of the current in amps (A), B is the strength of the magnetic field in tesla (T) and θ is the acute angle between the magnetic field and the wire.

Note that you can rearrange this equation and make any of the variables the subject to find their values.

worked example QUeStIonIf the wire in Figure 4.3.3 has a length of 5 cm within a 0.2 T magnetic field, the wire is at 30º to the field and it contains a current of 0.5 A, what is the force exerted on the wire?

SolUtIonFirst we convert the length of the wire into the appropriate units: 5 cm = 5/100 = 0.05 m

Then substitute: F = BIl sin θ = 0.2 × 0.5 × 0.05 × sin 30

= 2.5 × 10–3 N out of the page

Qualitative analysis of factors affecting the motor effect If you have an equation that describes a relationship, the easiest way to see

how the variables affect the subject of the equation is to place some numbers in the equation and see what happens.

Remembering that the motor effect is the force F that a current-carrying conductor experiences in a magnetic field, let’s look at how the other variables in the equation affect the magnitude of F. As you read through this section keep referring back to the equation and check that you come to the same conclusions.

Magnetic field strength, BIf the value of B was very small, then the right side of the equation would be multiplied by a very small number. Conversely, if the value of B was very large, the right side of the equation would be multiplied by a very large number. Since the magnitude of the force F is equal to the right-hand side of the equation, F is clearly directly related to B.

To take our analysis one step further, consider multiplying the value of B by 2. Looking at the equation, we see that if we do this, the effect on the value of F is the same as multiplying F by 2. We can now say the force F is directly proportional to magnetic field strength B; that is, as B increases by some factor (say 2 times) F also increases by that same factor.

Having understood the analysis above, it should be easy now to see the following relationships.

Discuss the effect on the magnitude of the force on a current-carrying conductor of variations in:• thestrengthofthemagnetic

field in which it is located• themagnitudeofthecurrent

in the conductor• thelengthoftheconductor

in the external magnetic field• theanglebetweenthe

direction of the external magnetic field and the direction of the length of the conductor.

I

NS

Bl a

θ

a = l sin θ

sin θ = a l

Solve problems and analyse information about the force on current-carrying conductors in magnetic fields using: F = BIl sin θ

Figure 4.3.3

93


Current, I As for magnetic field strength, by inspecting the formula we can see

that force F will be directly proportional to current I.

Length, l Similarly, F is directly proportional to the length l of the current-

carrying conductor within the magnetic field. Be particularly careful to remember that l is the total length of the wire within the magnetic field. It is noteworthy that l is regarded as a vector, but current I is not.

Angle, θWhen the wire is parallel to the magnetic field, the angle θ is zero. Inspecting Figure 4.3.4 you can see that if θ is zero degrees then sin θ is also zero. When you substitute zero for sin θ in the motor effect equation, you see that the force must be zero. This shows us the interesting situation that the force on a current-carrying conductor in a magnetic field is zero when the conductor is parallel to the magnetic field lines.

When the current-carrying conductor is perpendicular to the magnetic field, θ is 90°. From Figure 4.3.4 you can see that sin 90 is 1. Since 1 is the maximum value for sin θ, when we substitute 1 into the equation the force F will be the maximum value it can be for each set of the other variables.

This shows that the force F is a maximum value when a current-carrying conductor is perpendicular to the magnetic field B.

Inspecting Figure 4.3.4 you can see that as θ increases from 0° to 90° the value of sin θ increases towards a value of 1. The rate of this increase is not constant (i.e. the graph is not a straight line). So we can only say that force F depends on θ, as it is not directly proportional to θ.

sinθ

θ

–1

1

90 180 270 360

Figure 4.3.4 Graph of θ versus sin θ

nanotuBe loudsPeakers: no MaGnets

A group of Chinese researchers has developed a loudspeaker that

consists only of a thin film of carbon nanotubes driven by an AC input signal. The sound generation is attributed to a thermoacoustic effect. Changes in the current flowing through the film are reflected in the film’s temperature. Those temperature changes excite pressure waves in the surrounding air and these are sound waves. The film is flexible and can be stretched and still operate unimpeded. Perhaps loudspeakers won’t have magnets in the near future!

CheCkpoInt 4.31 Describe the relative directions of the force, the current and the

magnetic field when a current-carrying wire experiences the maximum possible force due to the motor effect.

2 Compare the relationships of B, I, l and θ to F in the equation F = BIl sin θ.

3 Explain the motor effect.

4.4 Forces between parallel wiresIn many applications of electric circuits there are wires bundled tightly together and running parallel to each other. If we want to explore the interaction of these wires, we need to bring together two of the facts we have learned so far.

The first is that current-carrying conductors produce a magnetic field. The second is that a current-carrying conductor experiences a force when inside a magnetic field. We will also need to apply two of the right-hand rules we have learned to determine the direction of the magnetic fields and the forces.


94

wire 1 wire 2 wire 1 wire 2

F1

magnetic fieldsaround parallel wires

magnetic fielddue to I1


I2I1 I1 I2

wire 1 wire 2

F2

a b c

Figure 4.4.1 Determining the forces on two parallel wires with currents flowing in the same direction

try this!the motor effectTake a piece of insulated wire about 5–10 metres long. Stretch it out between two retort stands so that there are two pieces of wire running parallel within a few centimetres of each other. Connect the ends to a 12 V battery and insert a tapping key switch at one end of the circuit. When connected briefly, the currents will run antiparallel to each other. Caution: Connect these wires for a very short time only, as they carry a large current. Predict what will happen when you press the switch. Now observe! How did you go?

Qualitative analysisLet us first consider the situation in which we have two parallel current-carrying wires with currents that are travelling in the same direction.

In Figure 4.4.1a we use the right-hand grip rule to determine the direction of the magnetic fields around the wires. Now, to understand what is happening to each wire, we will consider what is happening to one wire at a time.

In Figure 4.4.1b we are looking at what is happening to wire 2. The magnetic field generated by the current in wire 1 travels into the page around wire 2. Using the right-hand palm rule, we can see that wire 2 experiences a force towards wire 1.

In Figure 4.4.1c we now see what is happening to wire 1. The magnetic field of wire 2 comes out of the page around wire 1. Therefore the right-hand palm rule shows that wire 1 experiences a force towards wire 2.

The conclusion we can come to is that when two parallel current-carrying conductors have currents travelling in the same direction, the two conductors are forced towards each other.

Describe qualitatively and quantitatively the force between long parallel current-carrying conductors: Fl

kI I

d= 1 2

Now let’s consider the two parallel current-carrying wires with currents that are travelling in opposite directions.

In Figure 4.4.2a the right-hand grip rule shows us the direction of the magnetic fields around the wires. To understand what is happening to each wire, we will again consider each wire in turn.

Let’s look at what is happening to wire 2 first (Figure 4.4.2b). The right-hand grip rule shows the magnetic field of wire 1. This field travels into the page around wire 2. The right-hand palm rule shows that wire 2 experiences a force away from wire 1.

Figure 4.4.2c shows what is happening to wire 1. The magnetic field of wire 2 goes into the page around wire 1. The right-hand palm rule then shows that wire 1 experiences a force away from wire 2.

The conclusion we can now come to is that when two parallel current-carrying conductors have currents travelling in the opposite direction, the two conductors are forced away from each other.

It may be easy for you to remember the two conclusions above about the direction of forces on parallel wires, although remembering the result is generally less important than knowing how you got there. If you forget the conclusions

95


above and you know how to work them out yourself you can never get them wrong. You will apply similar methods in other problems later in this module. So if you are comfortable with these methods now, it will be easier later. If at any time you have trouble using your right-hand rules, come back to the relevant part of this chapter and revise. You will meet the skills you have used here several more times yet and each occasion is a chance to test your knowledge.

Quantifying the relationshipUsing our right-hand rules, we have determined that parallel wires exert forces on each other. To quantify these forces, let’s start with an equation we have seen already. Recall the equation for the force on a current-carrying conductor:

F = BIl sin θFor parallel wires, each wire is at right angles to the magnetic field of the other

wire. The sin θ term in the above equation is therefore equal to 1 (see Figure 4.3.4) so the equation becomes:

F = BIlInspecting this formula we can see that the current I and length l can be

measured relatively easily. To work out F, the size of the force, we now need to calculate B, the strength of the magnetic field around the wire.

The strength of the magnetic field around a current-carrying conductor can be determined using the equation:

BI= k

dwhere the proportionality constant k is 2 × 10–7 N A–2, I is the current in amps, and d is the distance away from the wire in metres.

Let’s consider the situation in Figure 4.4.1b. The magnetic field is being produced by wire 1, so the current I1 will be used in calculating the magnetic field strength. The force we are calculating is acting on wire 2, so the current we should use in this part of the calculation is I2. Combining the two previous equations and inserting the correct currents in each we get:

FI

I= kd

l12

Rearranging this gives:F I Il

kd

= 1 2

where F is the force on each wire in newtons and l is the length the wires are parallel in metres, so F/l is the force on each metre of wire. k is the proportionality constant 2 × 10–7 N A–2, I1 and I2 are the currents in the two wires in amps, and d is the distance between the two wires in metres.

wire 1 wire 2 wire 1 wire 2magnetic fields

around antiparallel wiresmagnetic field

due to I1


I2I1 I1 I2

wire 1 wire 2

a b c

Figure 4.4.2 Determining the forces on two current-carrying wires with currents in opposite directions


96

worked exampleQUeStIonFor the situation shown in Figure 4.4.3 calculate the magnitude of the force acting on each wire.

SolUtIonFrom Figure 4.4.3, l = 0.5 m, l1 = 1.5 A, l2 = 1.0 A and d = 2/100 = 0.02 m.

Use: F I Il

kd

= 1 2

Rearrange to make F the subject: FI I

= kd

l1 2

Substitute: F = × × × ×−2 10 1 5 1 00 02

0 57 . ..

.

= 7.5 × 10–6 N

Note that as this question asked for only the magnitude of the force, you do not have to include a description of the direction. If asked, you should add that the force on each wire is directed towards the other wire.

More qualitative analysisOur last stop in our look at parallel wires is to analyse the relationships expressed in the equation:

F I Il

kd

= 1 2

If we follow the process we used in section 4.3, we can see the following relationships:• F is directly proportional to the length l. To see this easily we rearrange the

formula to make F the subject. As l increases F increases.• F is also proportional to both I1 and I2. • F is inversely proportional to the distance d between the two wires. This

means that as the distance increases F decreases, or as the distance decreases F increases.

FeelinG the Pinch

The piece of copper pipe shown in

Figure 4.4.4 was crushed by lightning. Just like two parallel wires carrying currents in the same direction, the sides of this pipe were pulled together when a current of more than 100 000 amps was present.

Figure 4.4.4 Dramatic evidence of parallel conductors experiencing a force

CheCkpoInt 4.41 Identify the two key facts that explain the interactions of two parallel current-carrying conductors.2 Describe the interactions of two parallel current-carrying conductors.

2 cm

0.5 m

1.0 A1.5 A

Figure 4.4.3 Two parallel current-carrying wires

Practical eXPeriences

97


chaPter 4This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACtIVItY 4.1: the motoR eFFeCtObserve the effect of a current-carrying wire that is placed in an external magnetic field and relate it to the mathematical formula:

F = BIl sin θEquipment: 2 strong horseshoe magnets or ceramic magnets on an iron yoke, long wire, power supply, retort stand, clamp.

Perform a first-hand investigation to demonstrate the motor effect.

Solve problems and analyse information about the force on current-carrying conductors in magnetic fields using: F = BIl sin θ

ceramic magnetson an iron yoke

low voltage DCpower supply

flexible wire

hanging from a retort stand

Discussion questions1 Describe the motor effect.2 Discuss what happened when the current direction was changed.

Figure 4.5.1 Experiment set-up

98

4 electrodynamics: moving charges and magnetic fields chapter summary

review questions

• Currentistherateofflowofchargethrougharegion,and in circuits the direction of a current is that of a positive charge, called conventional current.

• Acurrentinwhichthechargesonlyflowinonedirection is called direct current (DC). A current in which the charges move back and forth is an alternating current (AC).

• Resistanceisameasureofhoweasilyacurrentflowsandit is defined as the ratio of voltage over current.

• Electricalpoweristherateatwhichenergyistransferredwithin a circuit component.

• Electriccurrentsproduceamagneticfieldaroundawire. The direction of this field can be determined by the right-hand grip rule.

• Amagneticfieldisdepictedbylineswithanarrowindicating the direction in which the north pole of a magnet points within the field.

• Thesymbols×and• are used to show the direction of a magnetic field into and out of the page respectively.

• Thesymbols and are used to show the direction of a current into and out of the page respectively.

• Chargedparticlesmovinginamagneticfieldexperiencea force. When these charges are moving in a wire, the wire experiences a force called the motor effect.

• Theright-handpalmrulerelatestheperpendiculardirections of force, magnetic field and motion in the motor effect.

• Aloudspeakerisagreatexampleofanapplicationofthemotor effect.

• Themotoreffectisquantifiedbytheequation F = BIl sin θ where the force is proportional to the magnetic field strength (B), current (I), length of the wire within the magnetic field (l ) and the angle between the wire and the magnetic field (θ).

• Parallelcurrent-carryingwiresexperiencethemotoreffect due to each other’s magnetic fields and this

phenomenon is quantified by the equation F I Il

kd

= 1 2.

phYSICAllY SpeAkIngAcross

4 Equation to determine the force between two current-carrying wires (7, 3)

7 emf stands for this (13, 5)

9 Application of the motor effect that converts electrical energy to sound

10 Quantity related to the energy given to electrons in a circuit

11 Force experienced by current-carrying wire in an external magnetic field (5, 6)

Down

1 Electrons moving in one direction (6, 7)

2 Unit of power

3 The branch of electricity that deals with moving charges and magnetic fields

5 Rate of use of electrical energy

6 Opposition to the flow of electrons

8 Device that converts electrical energy to kinetic energy using the motor effect

4

1

2 3

5

6

7 8

9

10

11

99


ReVIewIng 1 Describe the difference between DC and AC.

2 Outline the journey of an electron through the circuit in Figure 4.1.1, noting the energy transformation.

3 Explain why a bird can sit on an electrical power cable and not get electrocuted.

4 Describe what is meant by metals having a ‘sea of electrons’.

5 a Recall the factors that affect the resistance of a wire.

b State how they affect it.

6 Given that the definition of power is P = W /t, show that the equation for electrical power is P = VI.

7 Describe what happens to a charged particle in a magnetic field.

8 Compare the paths of two charged particles entering magnetic fields. The first is in a constant magnetic field and the second is in a magnetic field that is gradually increased.

9 State the motor effect in words.

10 Sketch a graph to show how changing the angle of the wire in a magnetic field changes the force experienced.

11 Explain how current can create sound in a loudspeaker.

12 Draw a labelled energy transformation diagram of a loudspeaker.

SolVIng pRoBlemS 13 Calculate the resistance in a circuit that has a battery

supplying 10 V and current flow of 2.3 A.

14 Determine what happens to current in a circuit when the thickness of the wire is doubled and the voltage is increased to four times the original.

15 Using the right-hand palm rule, determine the direction of the unknown quantity B, I or F.a

b

c

16 Determine the direction the particle will move when it enters the magnetic field that is shown.a

b

17 Determine the force on a current-carrying wire (I = 2 A) of length 0.5 m that is placed in a magnetic field of 3 T.

18 A physics student did an experiment to measure the force on a wire placed in an external magnetic field. The field is altered and the results are recorded below.

force (n) magnetic field (t)0.05 0.1

0.08 0.2

0.11 0.3

0.17 0.4

a Graph the results.b Determine the value of the gradient of the graph

and what the gradient represents.c Given that the length of the wire is l = 0.2 m and

the current is I = 2 A, comment on the accuracy of the student’s results.

+

+

Revie

w Questions

5

100

Induction: the influence of changing magnetism

electromagnetic induction, Faraday’s law, magnetic flux,

magnetic field strength B, magnetic flux density, emf, perpendicular area, Lenz’s law, law of conservation of energy, eddy

currents, induction cooktops, resistive heating, eddy current braking

Electromagnetic induction The discovery of electromagnetic induction was a giant step on the path to modern technology. Our understanding of this phenomenon required a great deal of new physics and involved the work of many individuals. One man, Michael Faraday, led the way with his experimental genius and intuitive diagrammatic reasoning.

Faraday lacked the mathematical skills to numerically describe his discoveries, but James Clerk Maxwell took Faraday’s understanding and eventually quantified all electromagnetism. Faraday’s law and Lenz’s law provide us with the tools to explain and predict eddy currents and to understand their applications. Later they will help us in our quest to uncover the secrets of motors, generators and transformers.

He noticed that this current was only induced while the magnet was moving. He moved the magnet out of the coil, and this time he measured a current in the opposite direction within the coil (Figure 5.1.1b). Following many other experiments, he put forward his general principle of electromagnetic induction that a changing magnetic field can cause a current to be generated in a wire. This change can be caused by either the relative motion of the field and the coil or by a change in the strength of the magnetic field.

In light of Faraday’s conclusion, let’s offer a simple way to understand his observations. In Figure 5.1.1 we can consider the change in the magnetic field through the coil to be represented by the number of magnetic field lines passing within the coil. In Figure 5.1.1a, notice that the magnetic field around the bar magnet is not uniform. You see that as you get closer to the poles of the magnet the magnetic field lines get closer together, indicating that the field is stronger. Therefore, as the magnet gets closer to the coil more field lines pass within the coil, indicating that the magnetic field within the coil gets stronger. So, as the magnet gets closer to the coil the strength of the field within the coil is changing and this induces the current. This gives us a general explanation for electromagnetic induction. Now let’s take things a bit further.

Faraday’s law: explaining electromagnetic inductionElectromagnetic induction can be summarised by Faraday’s law.

The induced emf in a coil is proportional to the product of the number of turns and the rate at which the magnetic field changes within the turns.

Faraday’s law is quantified by an equation, and it is very useful to analyse the equation to understand the relationships involved. This equation is:

Let us spend a little time now looking at what each of the variables in this equation means and then we can understand the relationships.

Magnetic flux ΦB Magnetic flux is a measure of the ‘amount’ of magnetic field passing

through a given area. There are two variables that determine the value of magnetic flux: the strength of the magnetic field B and the area the field is passing through A. Magnetic flux ΦB is measured in weber (Wb) and can be expressed by the equation:

ΦB = BA⊥

Magnetic field strength B, a quantity we are already familiar with, is also called magnetic flux density. This quantity is measured in tesla (T), or equivalently in webers per square metre (Wb m–2). This is a measure of the field strength per square metre.

A ⊥ is an area that is perpendicular to the magnetic field lines. If a magnetic field passes through a circular wire loop of area A (Figure 5.1.2a) and the loop is at an angle to the field (see Figure 5.1.2b), then the field passes through an ‘effective’ area A⊥ that is perpendicular to the field and is smaller than area A (see Figure 5.1.2c). Note that flux could also be calculated using the perpendicular component of the magnetic field strength, B⊥, and the total area A.

Outline Michael Faraday’s discovery of the generation of an electric current by a moving magnet.

5.1 Michael Faraday discovers electromagnetic induction

Michael Faraday (1791–1867) was born into a working-class family in London in 1791. He received little formal schooling and started work at the age of 12. At 14 he became a bookbinder’s apprentice and set about educating himself with the books he was able to access. Over this time he developed a keen interest in science and began attending scientific lectures. In 1813 he became a research assistant for the prominent scientist Sir Humphry Davy (1788–1829). In the following years Faraday became renowned as one of the greatest experimental scientists.

One of his numerous experimental discoveries was the phenomenon of electromagnetic induction in 1831. Electromagnetic induction is the generation of an electric current by a changing magnetic field. Faraday showed that when he moved a magnet near a wire coil, a current flowed within the coil. He first moved a magnet into one end of a wire coil (Figure 5.1.1a). As he did this he measured a current in the coil on a galvanometer (a type of ammeter).

101

motors andgenerators

He noticed that this current was only induced while the magnet was moving. He moved the magnet out of the coil, and this time he measured a current in the opposite direction within the coil (Figure 5.1.1b). Following many other experiments, he put forward his general principle of electromagnetic induction that a changing magnetic field can cause a current to be generated in a wire. This change can be caused by either the relative motion of the field and the coil or by a change in the strength of the magnetic field.

In light of Faraday’s conclusion, let’s offer a simple way to understand his observations. In Figure 5.1.1 we can consider the change in the magnetic field through the coil to be represented by the number of magnetic field lines passing within the coil. In Figure 5.1.1a, notice that the magnetic field around the bar magnet is not uniform. You see that as you get closer to the poles of the magnet the magnetic field lines get closer together, indicating that the field is stronger. Therefore, as the magnet gets closer to the coil more field lines pass within the coil, indicating that the magnetic field within the coil gets stronger. So, as the magnet gets closer to the coil the strength of the field within the coil is changing and this induces the current. This gives us a general explanation for electromagnetic induction. Now let’s take things a bit further.

Faraday’s law: explaining electromagnetic inductionElectromagnetic induction can be summarised by Faraday’s law.

The induced emf in a coil is proportional to the product of the number of turns and the rate at which the magnetic field changes within the turns.

Faraday’s law is quantified by an equation, and it is very useful to analyse the equation to understand the relationships involved. This equation is:

ε = n(∆ΦB/∆t)

Let us spend a little time now looking at what each of the variables in this equation means and then we can understand the relationships.

Magnetic flux ΦB Magnetic flux is a measure of the ‘amount’ of magnetic field passing

through a given area. There are two variables that determine the value of magnetic flux: the strength of the magnetic field B and the area the field is passing through A. Magnetic flux ΦB is measured in weber (Wb) and can be expressed by the equation:

ΦB = BA⊥

Magnetic field strength B, a quantity we are already familiar with, is also called magnetic flux density. This quantity is measured in tesla (T), or equivalently in webers per square metre (Wb m–2). This is a measure of the field strength per square metre.

A ⊥ is an area that is perpendicular to the magnetic field lines. If a magnetic field passes through a circular wire loop of area A (Figure 5.1.2a) and the loop is at an angle to the field (see Figure 5.1.2b), then the field passes through an ‘effective’ area A⊥ that is perpendicular to the field and is smaller than area A (see Figure 5.1.2c). Note that flux could also be calculated using the perpendicular component of the magnetic field strength, B⊥, and the total area A.

Perform an investigation to model the generation of an electric current by moving a magnet in a coil or a coil near a magnet.

Define magnetic field strength B as magnetic flux density.

Describe the concept of magnetic flux in terms of magnetic flux density and surface area.

activity 5.1

PraCtICaL eXPerIenCes


N

A

A

N

Figure 5.1.1 The changing magnetic field of a moving magnet can induce a current in a coil of wire.

our sun’s magnetIC InfLuenCe

Large outbursts from the Sun cause changes in the strength of the

magnetic field at the Earth’s surface. This changing magnetic field induces currents in long metal pipelines and the wires of power grids, especially at high latitudes. Currents of 1000 amps have been measured in pipelines in Alaska, causing accelerated corrosion. Large currents in power grids have overloaded circuits and left millions of people in the dark for hours.

a

b

Induction: the influence of changing magnetism5

102

A A⊥

B

P

a b c

This line is the perpendicularheight of area A.

Figure 5.1.2 (a) A circular coil of area A. (b) A side view of the coil at an angle to the magnetic field. (c) The ‘effective’ area of the coil perpendicular to the magnetic field viewed from point P (b)

The product of the magnetic flux density B and the area A ⊥ gives a measure of the total ‘amount’ of magnetic field passing through that area. This is the magnetic flux ΦB.

The delta (∆) symbols in Faraday’s law mean a change in some quantity. So the two terms with delta symbols attached are differences between an initial value and a final value. The term ∆ΦB stands for the change in the magnetic flux and is given by:

∆ΦB = ∆ΦB final – ∆ΦB initial

The ∆t term represents a period of time. This is the period of time over which the change in flux ∆ΦB is measured.

We can see that the ∆ΦB/∆t term in Faraday’s law is actually the rate of change of magnetic flux, just as acceleration is the rate of change of velocity (aav = ∆v/∆t). The term ∆ΦB/∆t tells us how fast the flux is changing.

emf εThe symbol ε stands for emf, measured in volts. It is the difference in electrical potential between the two ends of a coil (X and Y in Figure 5.1.3). This emf creates an electric field within the wire of the coil and a current is established, provided the circuit is complete (X and Y are not connected in Figure 5.1.3). A current will flow as long as there is a change in the magnetic field within the coil.

Qualitative analysis of Faraday’s lawLet’s now look at the relationship between emf and the other terms in the equation. Recall that the equation for Faraday’s law is:

ε = n(∆ΦB/∆t)

The emf is proportional to the rate of change of the magnetic flux (∆ΦB/∆t). If there is a large change in flux in a small amount of time, then ∆ΦB/∆t is large; therefore, the emf produced will be large. This emf is responsible for the induced current in a closed loop of wire. If the emf is large, then so is the induced current. Another way of saying this is that if we wanted to induce a large current, we would change the magnetic field within the coil as much as possible in the shortest time possible. To do this in the example we have seen, we could move the magnet more quickly, use a stronger magnet or make the perpendicular area as large as possible for the coil.

The emf is also proportional to n, the number of turns in the coil. Again, if we wanted to create a large current we would want to have as many turns in the coil as possible.

N S

X Y

B

Figure 5.1.3 A moveable magnet passes through a stationary coil with terminals X and Y.

103


By inspecting all the variables in Faraday’s law we can conclude that an induced potential difference, and therefore an induced current in a coil, is proportional to the rate of change of magnetic flux (∆ΦB/∆t) and the number of turns (n) within the coil. Further, we can conclude that induced currents are produced by changing the magnetic field strength B, the relative motion between B and the area A, or changing the perpendicular area A ⊥ of the coil.

Our next challenge is to find the direction of an induced current, and for that we need Lenz’s law. We will cover this in section 5.2.

Induction without relative motionNow that we have a basic understanding of electromagnetic induction, we will look at another example.

In one of his early experiments Faraday wound two coils of wire around an iron ring (Figure 5.1.4). He noticed that a current was induced in coil B for a short time after the current in coil A was switched on or off. To explain this induced current, recall that a current travelling through a wire produces a magnetic field around the wire. We can then follow similar reasoning to that in the different situation described earlier. The list of events below explains why Faraday observed induced currents for a short time after he connected or disconnected coil A. The steps below are indicated in Figure 5.1.5, which shows the currents in both coils and the magnetic field produced in coil A.1 Coil A is connected by closing the switch and a current begins to flow.2 The current in coil A produces a magnetic field around the iron ring

(see Figure 4.1.9). This field gets stronger as the current increases to its maximum value. The changing magnetic flux from coil A causes a changing magnetic flux in coil B, which induces an emf and therefore a current in coil B. The rate of change of magnetic flux in coil B is positive, rapid at first but then slows down (Figure 5.1.5), so the induced current is positive, high at first but decreases rapidly.

3 The current and magnetic field in coil A both reach their maximum value. The rate of change of magnetic flux in coil B is now zero, so the induced emf and current are also zero.

4 Coil A is disconnected by opening the switch. The current in coil A decreases rapidly, but does not stop immediately because there is normally a brief spark in the switch that allows current to continue flowing briefly.

5 The current in coil A, and therefore the magnetic field it produces, rapidly decreases. The changing magnetic flux from coil A causes a changing magnetic flux in coil B that, in turn, induces an emf and therefore a current. The rate of change of magnetic flux in coil B is negative, initially high but rapidly decreases. Therefore, the induced current is negative, high at first but decreases rapidly to zero.

The example above illustrates our previous conclusions that it is the change in the magnetic field (more precisely, the changing magnetic flux)passing through a wire coil that induces a current, and the induced current is proportional to the rate of change of magnetic flux. Now we can see that the direction of the induced current is determined by whether the magnetic flux is increasing or decreasing.

Describe generated potential difference as the rate of change of magnetic flux through a circuit.

try thIs!skipping currents

Take a length of wire about 30–40 metres long and

connect the ends to a sensitive ammeter. Lay the wire out in a large open space and swing it like a skipping rope. Can you induce a current by changing the size of the loop and using the Earth’s magnetic field? Can you explain why this should work?

powersupply

switchsoftironring

layers of copper coilinterwound withcotton and calico

coil A

coil B

+

–G

Figure 5.1.4 Basic set-up of Faraday’s experiment

12 3

45

IB

BA

IA

Figure 5.1.5 The behaviour of currents and magnetic fields in Faraday’s iron ring experiment


104

N

S

NN

S

S

I

B

direction ofmovement


5.2 Lenz’s lawHeinrich Lenz (1804–1865) independently made many of the same discoveries as Faraday. He also devised a way to predict the direction of an induced current in a closed conducting loop due to a changing magnetic field. This method is called Lenz’s law and it states that an induced current in a closed conducting loop will appear in such a direction that it opposes the change that produced it. This means that the induced magnetic field from a wire loop will oppose the change in magnetic flux that causes the induced current. Let us look at the example shown in Figure 5.2.1 to understand this better.

Figure 5.2.1 (a) Bar magnet moves towards a wire loop. (b) The magnetic field due to the induced current

As the north pole of the magnet gets closer to the wire loop in Figure 5.2.1a, the magnetic flux passing through the coil increases. This change in flux induces a current in the wire loop. Now let’s find the direction of the current to agree with Lenz’s law.

The north pole of the magnet is coming towards the coil, so the magnetic flux pointing downwards through the coil is increasing. Lenz’s law says that the magnetic field produced by the induced current should oppose this change in flux, so the induced flux should point upwards through the coil. Using the right-hand grip rule for a wire coil or solenoid, we see that the current would need to flow in the direction shown in Figure 5.2.1b, to produce an upwards-pointing magnetic field within the loop. This field is pointing in the opposite direction to the changing magnetic flux, so it reduces the changing magnetic flux from the approaching magnet. Simply, you can think of the interaction of these two magnetic fields as if the north poles of two bar magnets are facing each other. The two north poles repel each other, and in this way the induced field acts to minimise the increasing magnetic flux within the coil.

a b

ChECkpoint 5.11 Outline Faraday’s discovery of induction by a moving magnet and summarise his conclusion.2 Define magnetic flux in terms of magnetic flux density.3 Using the term magnetic flux, explain why removing a magnet quickly from a coil induces a relatively large current.

105


Now, let’s look at what would happen when you move a magnet away from the loop with its north pole facing the loop (Figure 5.2.2).

In this case the magnetic flux in the coil is decreasing and pointing downwards. The induced magnetic field should therefore be also pointing downwards to add to the reducing field and try to minimise the change. Using the right-hand grip rule, we see that we need a current as shown to produce a downwards-pointing induced magnetic field. Again, you can think of the interaction of these two magnetic fields as the interaction of two magnets in which a north pole is facing a south pole. The two ends of the magnets are attracted to one another and in this way the induced field acts to add to the decreasing flux within the coil. These examples leave us with Lenz’s law as a tool to determine the direction of an induced current in a wire coil due to a changing magnetic flux.

The law behind Lenz’s law: the law of conservation of energyWhile we were learning about Lenz’s law you may have wondered why the current needs to produce a magnetic field to oppose the change in flux? Well the answer lies in the law of conservation of energy. This law states that energy cannot be created or destroyed, only converted from one form to another. In the case of Lenz’s law, it is the fact that energy cannot be created that is important.

If the induced current in Figure 5.2.1 was in a direction that added to the changing flux through the coil there would be an attractive force on the magnet. This would mean that the magnet’s motion would cause it to be pulled through the coil. The amount of energy you could get from the induced current (heat from electrical resistance) and the induced magnetic field would be much more than that put in initially to move the magnet and change the flux through the coil. This would mean you would be getting something (energy) for free (without doing any work), which is not possible.

Energy must be ultimately converted from the work done to move the magnet into heat energy from the electrical resistance within the wire coil.

There must be a balance between the energy that goes into a system and the energy that comes out. So, whether you push the magnet towards the loop or pull it away from the loop, you will always experience a force that resists the motion. This force is an attraction or a repulsion between the magnetic fields of the magnet and the wire loop.

So far in this chapter we have explained the cause of magnetic induction using Faraday’s law, used Lenz’s law to find the direction of an induced current, and the right-hand grip rule to find the direction of this current’s magnetic field. Keep these ideas in mind, as we will apply them in the next two chapters.

Account for Lenz’s law in terms of conservation of energy and relate it to the production of back emf in motors.

ChECkpoint 5.21 Define Lenz’s law.2 Describe the induced current and magnetic field in Figure 5.2.2 if the south pole of the permanent magnet is

pointing towards the wire coil.3 Justify the current and magnetic field shown in Figure 5.2.1b in terms of the law of conservation of energy.


N

S

Figure 5.2.2 Induced current due to a decreasing magnetic field in accordance with Lenz’s law


106

5.3 Eddy currentsWe have seen that charges moving in a magnetic field experience a force in accordance with the right-hand palm rule (sections 4.2 and 4.3). This effect occurs for free charges and charges within conductors. When a current-carrying wire experiences a force within an external magnetic field, we call this the motor effect. When charges within a wire experience a force within a changing magnetic field, inducing a current, we call this electromagnetic induction. Many conductors that experience a changing magnetic field and produce an induced current are much larger than a wire. We call these induced currents eddy currents.

Eddy currents can be produced by the relative motion of a conductor and a magnetic field. These eddy currents are small loops of current within the conductor. They are the same as induced currents in wires subjected to changing magnetic flux, except that the currents are not confined to a loop of wire. These currents are set up in accordance with Lenz’s law and produce magnetic fields that act to minimise the change in magnetic flux within the path of the current.

Figure 5.3.1 shows a square piece of copper sheet swinging like a pendulum through a magnetic field. When this piece of metal moves through the magnetic field, we notice that there is a braking effect, slowing its swing. After a few swings it comes to rest. It stops much more quickly than it does when we remove the magnetic field. To explain this situation we can use the right-hand rules or approach the problem in terms of Lenz’s law. When we use Lenz’s law, we use the right-hand grip rule for solenoids or coils to predict the direction of magnetic fields and eddy currents.

As the copper square swings into the magnetic field on the left of Figure 5.3.1 let’s consider what happens to a positive charge (as shown) on the leading edge of the square. If this positive charge was within a piece of wire it would experience a force F1 upwards as shown. If this was a square loop of wire, this force would generate a conventional current moving anticlockwise around the loop. As this charge is not confined to a wire, the charge moves upwards initially and then loops around to form a complete circuit (an eddy current). Using the right-hand grip rule, we can see that the eddy current (I1) shown would produce

Explain the production of eddy currents in terms of Lenz’s Law.

++N S

F1 I2I1

F2 causingbraking effect

F4 causingbraking effect

F3

direction of swing

Figure 5.3.1 A square metal sheet is swung through a uniform magnetic field between two bar magnets. A braking effect is observed due to induced currents and their magnetic fields, in accordance with Lenz’s law.

107


a magnetic field out of the page (indicated by the N for the north pole of the current’s magnetic field). The flow of positive charges in the direction of F1 is a current. This current experiences a force due to the uniform magnetic field. This force F2 opposes the motion of the copper, acting as a braking effect.

As the copper square leaves the magnetic field (on the right in Figure 5.3.1) Lenz’s law tells us that the eddy current should produce a force to slow the square’s departure from the field. The positive charge shown experiences a force F3 upwards, as shown. This causes a flow of positive charges in the direction of F3. This current experiences a force due to the uniform magnetic field. This force F4 opposes the motion of the metal sheet, again acting as a braking effect.

If you have access to the equipment that demonstrates the situation in Figure 5.3.1, then try observing it for yourself. You may also be able to observe the effect of cutting slots through the piece of metal swinging in the field. The slots limit the size of the eddy currents that can be produced and therefore the size of the induced magnetic fields, and the braking effect is considerably less. We will meet this idea of reducing the size of eddy currents again in chapters 6 and 7.

try thIs!racing magnetsFind two identical magnets. Get a piece of copper or aluminium sheet and a sheet of a non-metal, such as glass, with a surface similar in smoothness to the surface of the metal. Place the two sheets at the same angle (say 60º) to the table surface and place the magnets at the same height on each sheet (see Figure 5.3.2). Now predict which magnet is going to win the race and why. Now race! Did everyone agree? Explain your observations to a friend.

glassaluminium orcopper

60°

Figure 5.3.2 Which magnet will win?


108

phYSiCS FEAtURE

indUCtion Cooking

Induction cooktops are a great example of a growing application of eddy currents. The main appeal of

induction cooking is its efficiency and fast heating. Heat is not transferred to the pan from a hot plate or flame in induction cooking. The heat is generated within the pan itself and then flows into the food being cooked. This means that minimal heat is lost to the air before reaching the food, making this much more efficient than other cooking methods.

The operation of an induction cooktop is illustrated in Figure 5.3.3. A rapidly changing strong magnetic field is generated in a large wire coil, using an alternating current. Both the intensity and direction of this field change continuously over very short periods of time. The resulting rapidly changing magnetic flux within the base of the frying pan induces strong eddy currents, causing resistive heating.

Resistive heating by eddy currents occurs when the charges flowing within the metal collide with the ions in the metal lattice. Kinetic energy is transferred to the metal ions as vibrations, and this increases the temperature of the metal. The amount of heat Q produced by resistive heating is proportional to the resistance of the material R and the square of the current I , as shown by Joule’s law:

Q = Pt = I 2Rtwhere Q is in joules, power P is in watts or joules per second (J s–1), I is in amps A, R is in ohms (Ω) and t is in seconds (s).

From Joule’s law we can see that a large current and relatively high resistance would result in a large amount of resistive heating. This explains why specialised cookware is required to gain maximum efficiency from induction cooking.

Another method of heat production within induction cookware is a process called magnetic hysteresis losses. When a magnetic field is applied to, and then removed from, a magnetic material such as iron, a permanent magnetic field remains within the material. If a magnetic field is then introduced in the opposite direction to this remnant field, some energy is expended reducing the remnant field to zero before the field can build in the other direction. Energy is dissipated in this process as heat in the material and therefore also raises the temperature.

AC

B

eddy currentsproduced inbase of frypan

coil supplied withhigh frequency AC

rapidly changingmagnetic field

ceramic surface

Figure 5.3.3 The frypan on an induction cooktop up heats due to eddy currents.

ChECkpoint 5.31 Explain the formation of eddy currents in a small, flat, square metal sheet that falls between the poles of a magnet.2 Describe how Lenz’s law can be used to predict the formation of eddy currents.

activity 5.2


Activity Manual, Page 39 Gather, analyse and present information to explain

how induction is used in cooktops in electric ranges.


109


ChaPter 5This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACtivitY 5.1: gEnERAting ELECtRiC CURREntUsing the equipment listed, write up an investigation that will allow you to generate alternating current. Once you have produced alternating current, investigate how changing the distance between a coil and a magnet, the strength of the magnet and the relative motion between the coil and the magnet will affect the electric current produced.Equipment: coil of wire (transformer coil), galvanometer, magnet (either electromagnet or a series of permanent magnets).

Discussion questions1 Describe the relationship between the distance between the coil and the

magnet and the electric current produced.2 Determine how the strength of the magnet affects the current produced.3 What effect does making the magnet move instead of the coil and vice

versa have on the current produced?

ACtivitY 5.2: MAking USE oF EddY CURREntSResearch induction cooktops and check to see if advertised claims about their efficiency are true. Look at the use of eddy currents in braking. What forms of transport use it and where could it be applied?

Discussion questions1 Explain why AC and not DC must be used for an induction cooktop to

work.2 Discuss the efficiency claims of induction cooktops in comparison to

traditional cooktops.3 List the advantages and disadvantages of eddy current braking. (Hint: See

Physics Focus on page 112.)

Perform an investigation to model the generation of an electric current by moving a magnet in a coil or a coil near a magnet.

Plan, choose equipment or resources for, and perform a first-hand investigation to predict and verify the effect on a generated electric current when:• thedistancebetweenthecoil

and magnet is varied•thestrengthofthemagnet

is varied•therelativemotionbetween

the coil and the magnet is varied.

Plan, choose equipment or resources for, and perform a first-hand investigation to demonstrate the production of an alternating current.

Gather, analyse and present information to explain how induction is used in cooktops in electric ranges.

Gather secondary information to identify how eddy currents have been utilised in electromagnetic braking.

110

5 Induction: the influence of changing magnetism Chapter summary

• MichaelFaradaydiscoveredthatacurrentcanbegenerated by a changing magnetic field when moving a magnet within a wire coil.

• MagneticfieldstrengthB in tesla (T) is equivalent to magnetic flux density in webers per square metre (Wb m–2).

• MagneticfluxΦB is a measure of the magnetic field passing through a certain area. This is equal to the magnetic flux density B multiplied by the perpendicular area A⊥ through which the field is passing.

• Anemfproducedinacoilisproportionaltotherate of change of magnetic flux and the number of turns in the coil.

• Inducedcurrentsareproducedbychangingthemagnetic flux due to relative motion, changing the flux density or changing the perpendicular area.

• Lenz’slawandtheright-handgriprulecanbeusedtopredict the direction of an induced current.

• Lenz’slawstatesthataninducedcurrentinaclosedconducting loop will appear in such a direction that it opposes the change that produced it.

• Lenz’slawcanbeexplainedintermsofthelawofconservation of energy.

• Theproductionofeddycurrentscanbeexplainedinterms of Lenz’s law.

• Applicationsofeddycurrentsincludeinductioncooktops and eddy current braking.

review questionsphYSiCALLY SpEAkingThe theme of this word search is induction. There is a twist; there is no list provided, so you have to work out the words that have been included. Find the 10 hidden words that have to do with induction, list them and write their definitions.

Y Q R A B R Y M I U D T D T E W F Q V Z

U J E Q Z V L N H V H L L F F U C U F Y

B R O D S Q D A O G E C O N A B P I K V

A K Z M Q U C B E I A S Z H L P O E O J

W N P V C A S X F D P R E A B F T Y T C

W R H T U Z L C C X I X F V D P K C K V

P A I R G E I Q B K W S H N G Q O T T B

V O L L B T D G N I K A R B M E O H F Z

N Q S S E X T D S P Y R I Z Q L C S A A

C H L N Z D B U Y G E T Q F A H A C R L

F B G Z D N D I G C E V C P X N G O A I

J A Z B I F E G H V U O Y X W A C K D C

M P B S Q O N L G O O R I T M R Q A A G

C N G D V M E C S O C M R S K D R K Y X

E G T S M K D G B K O A T E K H X H S Q

L Y T R F H B A Y K C V S N N P F F L H

V K E M F L U X D E N S I T Y T F V A T

H T T Q M Z Z K C Q Z T F C C B P F W B

F O V H Q N R X D Y V E J T H K D G R Y

O X B Y J L W L K B B V N Z Q M S D G Q

111


REviEwing 1 Outline Faraday’s experiment that led to the discovery

of electromagnetic induction.

2 Recall the factors that affect the size of the induced emf that is created by induction.

3 Magnetic flux is a measure of the magnetic field passing through a certain area. Use this statement to explain how magnetic flux can be altered.

4 Outline how emf can be used to produce current.

5 Describe the place of relative motion in inducing emf.

6 Explain how Lenz’s law supports the law of conservation of energy.

7 Compare induced current in a wire and eddy currents.

8 Explain why a solid square piece of copper swinging through a magnetic field will slow more quickly than one with slits in it.

9 Compare and contrast the use of eddy currents in induction cooktops and electromagnetic braking. (You may need to refer to Physics Focus on page 112.)

SoLving pRoBLEMS 10 Using Lenz’s law, predict the direction of current

in the following situations. Sketch these diagrams, showing the induced currents.a c

b d

11 A conductive wire is placed on rails of an electrical circuit and forced to move to the right, as shown in the diagram below.

a Determine which is the positive and which is the negative end of the wire.

b Determine the direction of the current in the circuit.

12 Explain how eddy currents can be a problem.

13 Give examples of how eddy currents can be of use.

14 Predict the direction of the eddy currents in the following examples.a b

15 Justify the claim that induction cooktops need

special cookware.

Revie

w Questions

NS

A

NS

A

X

Y

expanding wire loop

B

wire

rails

bulb

rotating metal discsquare metal sheet

112

5 Induction: the influence of changing magnetism

phYSiCS FoCUSEddY CURREntS Stop ME in MY tRACkS

Brakes are used in vehicles when it is necessary to reduce speed quickly. When a vehicle brakes,

friction is usually used to convert kinetic energy into heat. Typically this involves a brake pad made of a heat-resistant material being forced against a metal disc attached to a wheel. This leads to wear on brake components and requires the ongoing cost of replacement. Some high-speed trains and modern roller-coasters now use eddy current braking. This method requires no physical contact between brake components and therefore no wear and tear.

Some roller-coasters use stationary magnets to induce eddy currents in the moving roller-coaster.

High speed trains in Europe have electromagnets fixed to the train, to induce eddy currents in the rails beneath, as shown in Figure 5.4.1.

A series of current-carrying coils are suspended a few millimetres above the top of the rail. Their magnetic fields are produced with a polarity that is in the opposite direction to the neighbouring coil (Figure 5.4.1) by alternating the direction of the direct current within the coil.

The operation of these brakes is outlined in Figure 5.4.2. A direct current is passed through the coil, producing a magnetic field and the soft iron core intensifies the field. These fields extend into the iron rail and, as the train moves, the top of the rail experiences a changing magnetic flux. In accordance with Lenz’s law, this changing flux is opposed by the production of eddy currents and their associated magnetic fields (see Figure 5.4.2b). The braking effect is derived from the force that each current experiences due to the magnetic field of the other. For example, the eddy current in the rail experiences a force due to the magnetic field of the coil and an equal and opposite force is experienced by the coil due to the magnetic field of the eddy current.

The kinetic energy of the train is converted to heat through resistive heating and magnetic hysteresis losses. A graph of typical rail heating is shown in Figure 5.4.3. The adoption of this technology is not yet widespread, as there are environmental and structural problems associated with excessive rail heating. Use is limited to rail lines with rails that can withstand the temperature changes without affecting their performance.

B

coil woundaround softiron core

top ofstationaryrail

directionof train’smotion

locationof coil

rail

+

–

a b

I

Figure 5.4.2 (a) Simplified diagram of eddy current braking system. (b) Induced eddy currents in rail seen from above

Figure 5.4.1 An eddy current brake in action

N N N NS SSSS S S SN NNN

coils top of rail

113


1 Identify and describe the function of eddy currents in train eddy current brake systems.

2 Calculate the amount of kinetic energy Ek that must be dissipated as heat to stop a train with a mass m of 5 × 104 kg travelling at a velocity v of 300 km h–1, using Ek = ½mv 2. (Remember to convert velocity to the appropriate units.)

3 If typical braking forces of 100 kN are applied, identify the associated change in track temperature.

4 Assess the environmental impact of an excessive rise in rail temperature if eddy current braking was used on a high speed train in Australia.

5 Identify further research that could be conducted to allow the widespread adoption of train eddy current brakes.

ExtEnSion6 Research other methods of applying eddy currents

to braking and compare these methods to the one outlined above.

7 The kinetic energy lost when braking can be recycled by regenerative braking systems. Research these systems and outline their operation.

H4. Assesses the impacts of applications of physics on society and the environment

H5. Identifies possible future directions of physics research

Gather secondary information to identify how eddy currents have been utilised in electromagnetic braking.

30

25

20

15

10

5

00 20 40 60 80 100 120 140 160

Braking force (kN)

Cha

nge

in r

ail t

empe

ratu

re (

°C)

Figure 5.4.3 Graph of rail heating

6

114

Motors: magnetic fields make the world go around

DC motors, coils, armature, rotor, stator, electromagnets, torque,

commutator split‑ring, commutator brushes, current‑carrying loop,

galvanometer, back emf, supply emf, single‑phase, three‑phase,

AC induction motor, ‘squirrel cage’ rotor, shaded‑pole induction motor

The magic of motorsMuch of the activity and infrastructure that makes our lives of convenience possible is hidden from our view. Electric motors contribute significantly to our modern existence but remain a mystery. To most they are magical devices that convert electricity into motion with the flick of a switch. What better way to appreciate our comfortable lives than to understand one of the things that drives it. This chapter will equip you to contemplate how one invention, the electric motor, contributes to your quality of life in the age of technology.

6.1 Direct current electric motorsDirect current (DC) motors transform electrical potential energy into rotational kinetic energy. They can be powered by relatively lightweight batteries and are therefore easily portable. The most common type of DC motor found in

battery-operated toys is shown in Figure 6.1.1. These motors typically draw a current from a number of batteries. This current flows through a wire coil within an external magnetic field. The coil experiences a force we know as the motor effect, which causes the motor to rotate.

The essential components that allow a DC motor to operate are summarised in Table 6.1.1.

Figure 6.1.1 Parts of a simple DC motor

armature or rotor commutator brushes

commutator contactsstator (curved magnets) coils

115

Motors andgenerators

Table 6.1.1 Parts of a DC motor

Part DescriPtion anD roleCoils Many loops of wire that carry a direct current. These wires experience

a force (due to the motor effect) that causes the motor to turn when current is flowing.

Armature This is the part of a motor that contains the main current-carrying coils or windings. For DC motors this is the rotor, but for AC motors it is usually the stator.

Rotor In a DC motor, the rotor consists of coils of wire wound around a laminated iron frame. The frame is attached to an axle or shaft that allows it to rotate. The iron frame is laminated to reduce heating and losses due to eddy currents. The iron itself acts to intensify the magnetic field running through the coils.

Stator Stationary permanent magnets (electromagnets in large DC motors) that provide an external magnetic field around the coils. Permanent magnets are curved to maximise the amount of time the sides of the coil are travelling perpendicular to the magnetic field, to maintain maximum torque.

Commutator split-ring A device with metal semicircular contacts that reverse the direction of the current flowing in each coil at every half rotation. This reversal of the current makes continuing rotation possible in a DC motor.

Commutator brushes Conducting contacts (generally graphite or metal) that connect the commutator to the DC power source.

Describe the main features of a DC electric motor and the role of each feature.

Identify that the required magnetic fields in DC motors can be produced either by current-carrying coils or permanent magnets.

Brushless dC Motors

The designs of many modern DC motors are more complex than the simple example we study in this chapter. The

brushless DC motor has numerous significant advantages, and is used commonly in computer cooling fans (Figure 6.1.2). The cylindrical permanent magnets in the rotor have alternating poles around the circumference. Small electronics switch the direction of the current in the stator coils. The magnetic field from the stator coils rotates and this causes the rotor to rotate.

Now that we have seen the basic structure of a DC motor, we need to turn our attention to understanding and explaining its operation. In section 4.3 we encountered the motor effect, but now we will apply it to rotating a current-carrying coil. Figure 6.1.1 shows that motors contain many coils of wire. For simplicity we will begin by looking at a single coil and how it would act as a motor.

Let’s torque about rotating coils A torque is the turning effect (or turning moment) of a force. The force

that causes an electric motor to turn is the motor effect, so we must see how to calculate a torque before we can begin to perform calculations for motors.

The idea of torque is illustrated in Figure 6.1.3a, which shows a force F due to a person sitting on a see-saw. In this case the force is acting perpendicular to the see-saw. The magnitude (τ) of the torque involved is calculated using the equation:

τ = Fd

Define torque as the turning moment of a force using: τ = Fd

Figure 6.1.2 The brushless DC motor of a computer cooling fan

N

SS

N

Motors: magnetic fields make the world go around6

116

where torque in newton metres (Nm) is given by the distance d from the pivot in metres multiplied by the force F in newtons. If the force is not acting at right angles to the see-saw (Figure 6.1.3b), it is the perpendicular component F⊥ of the force that is used. As F⊥ = F cos θ, the formula for torque becomes:

τ = F⊥d = F cos θd

Notice that the angle θ between F⊥ and F is the same as the angle θ the see-saw makes with the horizontal. We will use this fact again soon. Note that you could also use τ = Fd⊥.

Worked exampleQUESTIONCalculate the torque exerted on the see-saw in Figure 6.1.3b if a force of 980 N was acting at a distance of 4.0 m from the pivot at an angle θ of 30°.

SOLUTIONτ = F⊥d = F cos θd

= 980 × cos 30 × 4.0

= 3390 Nm

Quantifying torque on a coilWhen a current-carrying loop (or coil) is placed in a magnetic field, it is the force we know as the motor effect that applies a torque on the coil. Figure 6.1.4a shows a current-carrying loop within a magnetic field. If we know the direction of the current in the coil, we can use the right-hand palm rule to give us the direction of the forces on each side of the coil. Recall from our previous study of the motor effect that charges moving parallel to the magnetic field do not experience a force. This explains why only the forces FWX and FYZ act on the coil between points W and X, and Y and Z respectively, when it is in the position shown in Figure 6.1.4a. During the coil’s rotation, forces are experienced by sides XY and WZ, but they essentially cancel each other out and the coil is not free to move in these directions. Now let’s apply our new understanding of torque to this coil.

The forces FWX and FYZ apply a torque to the coil, and at the moment shown in Figure 6.1.4a the magnitude of each torque (τ) is given by:

τ = F⊥d

We will use the wire between points W and X as our example and consider the situation in Figure 6.1.4b in which the coil has rotated. We must now account for the fact that force FWX is not acting perpendicular to the coil, so we use the formula seen previously:

τ = F⊥d = FWX cos θd

As the force FWX is the motor effect, we recall the equation:

FWX = BIl sin θ

Note that θ in this equation is the angle we met in section 4.3. It is the angle of a current-carrying wire to the magnetic field through which it passes. This is not the same angle as in the cos θ term of the previous equation. Since the

Identify that the motor effect is due to the force acting on a current-carrying conductor in a magnetic field.

d

d

F

FF⊥

pivota

b

θ

θ

cosθ =F⊥F

A

axis of rotation

axis of rotation

rotation

θ

θ

B

FWX

FWX

FYZ

FYZ

F⊥

W

X

WX

YZ

Y

Z

a

b

d

d

l

Figure 6.1.3 A force acting on the end of a see-saw exerts a torque. (a) Force at right-angles to the see-saw and (b) force angled to the see-saw.

Figure 6.1.4 (a) The current-carrying coil is able to rotate about the axis of rotation (dashed line). (b) The same coil a short time later has rotated slightly, as seen from point A along the axis of rotation.

117


two sides of the coil that experience a force (WX and YZ in Figure 6.1.4) are always perpendicular to the field, θ = 90°. This makes sin θ = 1 and therefore:

FWX = BIl

Combining this equation with τ = FWX cos θd (using the other θ!) gives:

τ = BIl cos θd

where l is the length W to X (WX) and d is half the length of X to Y (XY/2).

Now we consider the whole coil. The two forces FWX and FYZ are equal and both make the coil rotate in the same direction. The total torque on the coil is:

τtotal = 2BIl cos θd

The product 2ld is the area of the coil A. This gives:τ = BIA cos θ

Motors have more than one coil, so we make one final adjustment to this expression, giving:

τ = nBIA cos θ

where n is the number of turns in the coil. This formula quantifies the torque (turning effect) on a rotating coil in newton metres (Nm). It is important to note that the angle θ in this equation is the angle of the coil relative to the magnetic field, as shown in Figure 6.1.4b.

Worked exampleQUESTIONThe coil in Figure 6.1.5 contains 50 turns and is carrying a 6.0 × 10–2 A current in a 5.0 × 10–3

T magnetic field. Calculate the torque on the coil if the coil has rotated 30° relative to the field.

SOLUTIONUsing τ = n BIA cos θ and converting the lengths to metres:

τ = 50 × 5.0 × 10–3 × 6.0 x 10–2 × (0.06 × 0.05) × cos 30

= 4.5 × 10–5 Nm

Operation of a simple DC motorLet’s follow the rotation of a coil within a simple DC motor to understand how motors work. We will analyse the situation shown in Figure 6.1.7. In Figure 6.1.7a the coil of a DC motor is connected to a battery via a split-ring commutator at A. At the moment shown in Figure 6.1.7a, the two curved metal contacts of the commutator split ring direct the conventional current around the coil through points W, X, Y and Z in turn. Figure 6.1.7b shows the coil in Figure 6.1.7a as seen from the point A along the axis of rotation. The coil is seen end-on, and we can see the direction in which the current flows along sides WX and YZ. These sides experience a force F that imposes a torque τ on the coil, causing it to turn. The magnitude of the torque on the coil at positions 1–5 is shown on the graph in Figure 6.1.7c. Let’s now follow this coil through half a rotation and see what happens to the current and torque on the coil.

Recall that the torque τ on the coil in Figure 6.1.7 is given by the equation:

τ = nBIA cos θ

Solve problems and analyse information about simple motors using: τ = n BIA cos θ

Describe the forces experienced by a current-carrying loop in a magnetic field and describe the net result of the forces.

axis ofrotation

θB

F

F

6 cm

5 cm

IS N

Figure 6.1.5

activity 6.2

PraCtICal eXPerIenCes



118

In this example the values of all the variables on the right-hand side of the equation remain constant, except for the angle θ. The variation in the cos θ term accounts for the variation in the magnitude of the perpendicular force F⊥ experienced by the sides WX and YZ. This force determines the amount of torque acting on the coil, and therefore the torque varies accordingly as the coil rotates. • Atposition1inFigure6.1.7b,theperpendicularforceF⊥ on the coil is at a

maximum and therefore the torque is at a maximum. This occurs when θ is zero and cos θ is 1.

• Atposition2,thecoilhasrotated45°withrespecttothemagneticfield.Themagnitude of the perpendicular force (F⊥) acting on the coil is less than at position 1. This means that the torque acting on the coil has reduced. At this point, cos θ is approximately 0.7. Note that real motors have curved magnets in the stator to ensure that each coil maintains θ = 0 as long as possible. This provides maximum torque for the longest time possible during each rotation. In the example in Figure 6.1.7, the magnets in the stator are not curved, so only at positions 1 and 5 is θ = 0.

• Atposition3,theperpendicularcomponentoftheforceF⊥ has dropped to zero. Consequently, there is no torque acting on the coil at this instant. Here θ is 90° and cos θ is 0, so torque is zero. At this point the brushes and split ring within the commutator have broken contact momentarily and no current flows in the coil. The current needs to be reversed in the coil at this point, so that the motor continues to turn for the next half rotation.

• Betweenposition3and4,thecommutatorhasreconnectedthecoilandreversed the direction of the current flowing in the coil (now flowing through points Z, Y, X and W in turn). The opposite sides of the split-ring commutator are now connected to the terminals of the battery. If the current had remained in the original direction once the motor rotated past point 3, the forces on the coil would have been in the opposite direction to the initial rotation. This would stop further rotation.

• Asthemotormovesthroughposition4,theperpendicularcomponentoftheforce F⊥ is increasing. This means the torque is increasing and will reach a

axis ofrotation

B

F

FW

X Y

Z

A

+ –

θ = 0°

θ = 45°

θ = 90°

θ = 45°

θ = 90°Time

0 max.Position

1

Torque

I

2

3

4

5

θ

F

F

F⊥

F⊥

F⊥F⊥

FF

F

F

F

F

F

WX

YZ

F

WX

YZ

YZ

YZ

WX

WX

WX

YZ

S N

a

b c

Figure 6.1.7 (a) A simplified model of a DC motor. (b) The coil as seen from point A in part a, showing forces and current over a half rotation. (c) Torque on the coil over a half rotation

try thIs!MoDel a siMPle Motor

To get a better understanding of the forces on a coil as it rotates,

make a model of a simple motor. You need some coat-hanger wire, a few bamboo skewers, Blu-Tack, two small magnets, a cork and paper. Construct a coil as shown in Figure 6.1.6. Here the green skewer indicates the magnetic field. The red skewer points on the magnets represent the direction of the force due to the motor effect. Work with a friend and model the forces on a coil as it rotates.

Figure 6.1.6 A simple model of an electric motor

119


maximum again at position 5. At position 5 the coil has performed half a rotation and the pattern we have seen from positions 1 to 5 will repeat until the coil is back to its original orientation at position 1.

PHYSICS FEATUREGALvANOmETERS

The galvanometer was developed in the 1800s to measure the relative strength and direction of electrical currents. Its

descendant is the analogue ammeter (Figure 6.1.8), which is calibrated in units of amperes and its basic structure is shown in Figure 6.1.9. In an ammeter, the coil is connected in parallel to a low-resistance wire within the ammeter. When a current is passed into the ammeter, only a small amount of the total current flows through the coil and this is proportional to the total current.

When a current passes through the coil shown in Figure 6.1.9, it experiences a force due to the motor effect. This force exerts a torque on the coil, causing it to rotate around the pivot. A spring provides resistance to the torque, and the needle comes to rest when the torque on the coil equals the torque from the spring. The torque on the coil is proportional to the current passing through the coil. Therefore the amount the coil and needle move indicates the size of the current.

To ensure that the needle moves (or deflects) by the same amount with each ampere of current being measured, the magnets are curved around the coil and iron core. This ensures that the magnetic field is always perpendicular to the flow of current along the sides of the coil. A uniform maximum torque will then be experienced by the coil through its whole range of movement. This means that the scale you read on the ammeter can be uniform (i.e. the scale has the same-sized divisions throughout).

Identify data sources, gather and process information to qualitatively describe the application of the motor effect in:

• thegalvanometer• theloudspeaker.

CHECkPOINT 6.11 Explain how each part of a DC motor contributes to its operation.2 Define torque and explain how torque varies during the rotation of a DC motor.3 Compare the features of a DC motor and a galvanometer.

activity 6.1



N S

pointer

force permanent magnet

magnetic field

spring

coil

pivot

movingcoil

soft ironcore

Figure 6.1.8 A ‘moving coil’ ammeter

Figure 6.1.9 Cross-section of a ‘moving coil’ galvanometer


120

6.2 Back emf and DC electric motorsWe now know that when a current is applied to the coil of a motor, the coil experiences a force. This force exerts a torque on the coil and the rotor begins to rotate. If we continue to apply the same current to the coil, the net force on the coil continues increasing the motor’s speed (recall Newton’s second law). This may make you wonder why a motor doesn’t just keep speeding up. We know from experience that they don’t because our toys with motors don’t accelerate forever.

The main reason DC motors reach a maximum operating speed is that a back emf is generated in the coil as the motor rotates. Backemfisapotential difference across the terminals of the motor created by the changing magnetic flux passing through the wire coils within the motor (see section 5.1).

Look back at Figure 6.1.7 for a moment. At position 1 the magnetic flux ΦB through the coil is zero. At position 2 in Figure 6.1.7b, the magnetic field is pointing right to left through the coil and the magnetic flux is increasing. This tells us that there should be an emf induced that would produce a magnetic field pointing from left to right to reduce the increasing magnetic flux. This field would be produced by a current flowing through Z, Y, X and W in turn. We can see from Figure 6.1.7a that this would be a current that opposes the one generated by the supply emf (through W, X, Y and Z in turn). The result of this would be that the net current in the coil would be less than the supply emf could generate. Another way to think about it is to recall the equation for Faraday’s law:

ε = n(∆ΦB/∆t)

We can see that this changing flux ∆ΦB over time ∆t would generate a potential difference or emf ε.

This potential difference (back emf ) is in the opposite direction to the applied potential difference (supply emf) that causes the rotor to turn. As the speed of the motor increases the back emf increases, as ∆ΦB/∆t increases. Eventually this potential difference cancels out most of the applied potential difference and virtually no current flows through the coil. At this point there is no net torque acting on the rotor and it turns at a constant speed.

Let us analyse this situation using Figure 6.2.1. At time A on the graph, the motor is connected to a DC power source. The applied potential difference causes a current (blue line) to flow in the motor’s coil and this builds quickly to its maximum value. Once the coil starts turning, a back emf is generated due to the changing flux within the coil. This back emf is in the opposite direction to the applied potential difference and therefore reduces the net potential difference, which in turn reduces the current flowing in the coil as the speed of rotation increases.

AttimeB,themotorhasreacheditsmaximumrotationalspeed.Heremost of the supply emf has been cancelled out by the back emf. If the motor has no load attached to it, only a small current continues to flow. This residual current is required to overcome any friction within the motor and any voltage drop due to losses such as resistive heating. There is no net torque on the coil betweentimesBandCandthemotoroperatesataconstantspeed.

Account for Lenz’s Law in terms of conservation of energy and relate it to the production of back emf in motors.

Explain that, in electric motors, back emf opposes the supply emf.

motor current

Cur

rent

(A

)

Time (ms)00

A B C D

Figure 6.2.1 Graph of net current for a DC motor averaged over many cycles

try thIs!a back eMfConnect a DC motor to a battery and place an ammeter in the circuit to measure the current flowing through the motor. Predict how the amount of current flowing will change if you apply a significant load on the motor. Try it! How did you go?

121


At time C, a large load is applied to the motor, such as the motor turning a wheel to move a toy car. The motor slows down quickly under this load and the amount of back emf is reduced. This means that the applied potential difference is greater and therefore a larger current flows through the coil.

At time D, a larger current continues to flow through the coils. If the motor is not designed to handle the resistive heating produced by this larger current,themotormay‘burnout’.Burnoutoccurswheninsulationmeltsathigh temperature, and may cause other components to melt. The motor will then cease to operate efficiently.

BaCk eMf Measures Motor sPeed

In some motors back emf is used to measure their speed.

The input voltage is turned off for a short amount of time and the back emf is measured. Since the back emf is proportional to the speed of the motor, a calibration can be applied and the speed can be calculated.

CHECkPOINT 6.2 1 Define the term back emf.2 Describe the relationship between back emf and supply emf during

the operation of a motor.3 Analyse the production of back emf in terms of Lenz’s law.

6.3 Alternating current electric motors

Many large appliances in your home contain motors that use single-phase 240 V alternating current (AC). Single-phase AC was illustrated in Figure 4.1.2 as a blue curve, showing a current changing direction many times a second. To connect an appliance to the mains power, you insert a plug (Figure 6.3.1) into the wall socket. This allows single-phase AC to be supplied by the active pin and the circuit to be completed by the neutral pin.

Some industrial motors require more power and torque than can be supplied by single-phase AC. These motors are typically supplied with three-phase 415 V AC by a plug with four or five pins. Figure 6.3.2 shows the AC signal available for three-phase equipment. This involves three alternating currents that can be applied to different coils within a motor at any one time.

neutralactive

earth

Figure 6.3.1 A typical plug for an electrical appliance in Australia

+

0

–

Cur

rent

Time

Figure 6.3.2 Each of the three AC in three-phase power is 120° out of phase with the others.


122

unIversal Motors

The most common type of motor in appliances around the house is the universal motor. This

type of motor is capable of using both AC and DC. Figure 6.3.4 shows that coils act as electromagnets in the stator of these motors. These provide stronger fields and are lighter than permanent magnets. The coils are connected in series with the rotor coils and use a single-phase alternating current. These motors are found in many household appliances in which a variable speed is required, such as drills and blenders.

armature coils

stator coils

carbon brushesshaft

segmented commutator

spring-loadedbrush holders

Figure 6.3.4 A typical universal electric motor, showing the main components. Some motors would have additional stator coils. The commutator feeds current to the armature coils in the position where most torque will be experienced.

Figure 6.3.3 In this cutaway image of a three-phase AC induction motor, you can see the stator, consisting of electromagnets arranged to form a hollow cylinder. Within the stator sits the rotor, which is mounted on the motor’s shaft.

rotor

electromagnets

activity 6.3



AC motors have the same basic components as DC motors but lack the need for others. They all contain a rotor and stator. Their magnetic fields can be generated by current-carrying coils. They utilise the motor effect to transform electrical potential energy into rotational kinetic energy. Let’s explore examples of the AC induction motor that exist both in industry and in the home.

Three-phase AC induction motorsThe simplest induction motor is the three-phase AC induction motor (Figure 6.3.3). These motors are used in industry for their efficiency and reliability. Three-phase AC is fed to the coils in the stator. In Figure 6.3.5a, each pair of magnetic poles in the stator is fed one phase of the AC signal. The peak current of each phase is reached sequentially around the stator (Figure 6.3.5b), creating a magnetic field (the stator field) that rotates.

123


Time Cur

rent

+

–

0

2

2

3

3

rotor

stator pole

1

1

conducting bars

end ringi ii iii

a b

Figure 6.3.5 (a) In a three-phase motor, as the current in each pair of opposite coils peaks, the field appears to rotate, dragging the rotor around with it. (b) A ‘squirrel cage’ rotor. The rotor is made of iron laminations to cut down undesirable eddy currents. The induced currents flow lengthwise in copper or aluminium rods which are joined at the ends (as in a squirrel cage).

These motors contain a ‘squirrel cage’ rotor (Figure 6.3.5b) that does not require the input of an external current. Around the circumference of these rotors are a number of parallel conducting bars. These bars are joined at the ends by an end ring that allows current to flow from one bar to another. The rotating stator field induces a current in these bars and a magnetic field is induced in accordance with Lenz’s law. This induced magnetic field interacts with the rotating field from the stator and the resulting forces cause a torque on the rotor. This causes the rotor to spin without the need for a commutator as in a DC motor. As these motors do not need brushes and therefore have fewer moving parts, they are more efficient and more reliable.

Another way to understand the operation of an AC induction motor is to consider a positive particle within one of the conducting bars in the stator. Let’s consider the bar marked A in Figure 6.3.6a. As the rotating magnetic field moves upwards past bar A this is equivalent to the bar moving downwards in a stationary magnetic field. Figure 6.3.6b shows this equivalent situation in which bar A moves relative to the magnetic field. Using the right-hand palm rule (see section 4.2) we see that a positive particle in bar A would experience a force into the page. This is equivalent to a current being induced into the page in bar A. Now we must use the right-hand palm rule (see section 4.3) to deduce the direction of the motor effect on a current-carrying conductor. This indicates that a force is exerted upwards on bar A and this is in the same direction as the rotating field in Figure 6.3.6a. This force on bar A is the same as the force experienced by each bar as the magnetic field rotates. These forces, in the same direction as the rotating stator field, exert a torque on the rotor and are responsible for its rotation.

‘squirrel cage’conducting bars

2

2

stator

Arotatingmagneticfield

B

force on bar A due to motor effect

motion of bar Arelative to the magnetic field

B

a b

IFigure 6.3.6 (a) The magnetic field due to

one phase of an AC inductor motor. (b) The force acting on squirrel cage rotor bar A in part (a) due to the rotating magnetic fields


124

Single phase AC induction motorsA more complicated, but widely used, AC induction motor is the ‘shaded-pole’ AC induction motor shown in Figure 6.3.7. These motors require only a single phase of AC and can be found in most household electric fans. In these motors the alternating current is passed through a coil wrapped around the soft iron casing. The stator field induced by the alternating current passes through the casing and through the squirrel-cage rotor. Figure 6.3.7a shows the rotor removed and leaning on the motor. A cross-section of a squirrel cage rotor is also shown (Figure 6.3.7b) and clearly shows the conducting bars within the rotor.

Four small copper shading rings can be seen within the stator in Figure 6.3.7a. These are inserted into the stator on each side of the rotor on opposite poles. The currents induced in these shading rings in accordance with Lenz’s law act to delay the magnetic flux passing through the rotor. This produces an asymmetric magnetic field passing through the rotor shown in Figure 6.3.7c. This leads to a changing magnetic field in each cycle of the alternating current that sweeps across each pole of the stator. The sweeping change in magnetic field strength across the rotor is essentially the same as the rotating magnetic field we studied in the three-phase AC induction motor. This rotating magnetic field causes a torque on the rotor in the same way as outlined for the three-phase induction motor.

try thIs!inDuction exerts forcesSuspend a small piece of a lightweight conductor (e.g. aluminium foil) from fishing line. Use your right-hand palm rule to predict the force acting on the foil as you move a strong magnet vertically past the foil. Can you observe any movement? If not, use the scientific method to discover why it’s not working and try again.

Figure 6.3.7 (a) A shaded-pole AC induction motor taken from a small household fan, with (b) a cross-section of the rotor. The conducting bars can be seen clearly within the laminated rotor. (c) The principle of a simple single-phase, ‘shaded pole’ induction motor. The distorted (or ‘shaded’) field causes the rotor to turn in one direction in preference to the other.

a

shading rings

bconducting bar

squirrel cage rotor

Describe the main features of an AC electric motor.

Gather, process and analyse information to identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry.

thick, coppershading ring

shading rings delaythe phase of part ofthe field to producea rotating field

rotor

c

125


Table 6.3.1 Characteristics of motors

tyPe efficiency (%) aDvantages DisaDvantages coMMon aPPlicationsSimple (brushed) DC motor

40–90 Low cost, battery powered, speed easily controlled

Short working life, high maintenance (brushes), sparking and ozone production

Toys, power tools, treadmill exercisers, automotive starters

Brushless DC motor 30–90 Long working life, low maintenance, high efficiency

High cost of some designs, requires a controller

CD and DVD players, computer hard drives

AC universal motor 40–60 High starting torque, compact design, high running speeds

Less efficient than equivalent DC motor

Blenders, vacuum cleaners, hair dryers, portable power tools, sewing machines

Three-phase AC induction motor

70–90 High starting torque, high power, high efficiency, good power to weight ratio

Requires three-phase power Industrial machinery, pumps and compressors

Single-phase (shaded-pole) AC induction motor

20–35 Inexpensive, long working life, high power, multi-speed

Inefficient, low starting torque

Fans

CHECkPOINT 6.31 Compare the features of an AC induction motor and a simple DC motor.2 Construct a flow chart to account for the operation of a three-phase AC induction motor.3 Justify the choice of a three-phase AC induction motor for use in industrial machinery.

The right motor for the jobNow that we have seen a few examples of some common electric motors, let’s consider for a moment why they are chosen for their common applications. The initial price and operating cost are key factors in making a decision. Operating costs depend on factors such as energy efficiency and replacement of parts due to wear and tear. The amount of torque required and how often a motor will be put under load are also critical factors. These will determine how much current is needed, the strength of the magnetic field required and numerous other parameters. Of course there are other constraints including portability, size and weight to consider. All these factors lead to a lot of homework for an engineer who is trying to design a machine or appliance driven by a motor.

A summary of the characteristics of the different types of motors discussed in this chapter is provided in Table 6.3.1. Can you see why they are used in their common applications?


126

6 Motors: magnetic fields make the world go around


ACTIvITY 6.1: APPLICATIONS OF THE mOTOR EFFECTPart A: Make a loudspeaker and determine how the motor effect is used to make it work. Equipment: two horseshoe magnets, strong sticky tape, thin insulated wire, cardboard, power supply, alligator clips and wires. Part B: Make a working galvanometer and determine the differences in this application of the motor effect.Equipment: PVC-covered copper wire (150 cm) with bare ends, wooden base board, armature block, magnets, split pins, knitting needle, rivets, wire strippers, drinking straw, rheostat (10–15 ohms, rated at 5 A or more).

Discussion questions1 Outline how the motor effect is used to make music in a loudspeaker.2 Explain how a loudspeaker differs from a motor in its use of the

motor effect.3 Determine the difference between the way in which the motor effect

is used in a loudspeaker and in a galvanometer.

ACTIvITY 6.2: mOTORS AND TORQUEMake a motor like the one shown and note what factors change its performance. Calculate the torque of your motor. Equipment: insulated wire, magnets, magnetic field sensor and data logger (if available), paperclips, Blu-Tack, connecting wires with alligator clips, power supply.

Discussion questions1 Investigate the factors that determine the effectiveness of the motor.2 Calculate the amount of torque in your motor and list ways in which

torque can be increased.

ACTIvITY 6.3: AC INDUCTION mOTORS Using the equipment supplied, make a model of an AC induction motor and relate each part to the parts in a real AC motor.Equipment: aluminium foil, fishing line, retort stand and clamp, ceramic magnet.

Discussion questions1 Outline how the metal is made to move.2 Explain why the AC induction motor is so efficient.

Identify data sources, gather and process information to qualitatively describe the application of the motor effect in:• thegalvanometer• theloudspeaker.

Solve problems and analyse information about simple motors using: τ = nBIA cos θ

Perform an investigation to demonstrate the principle of an AC induction motor.

armature

brush

commutator

N S

Figure 6.4.1 A simple motor

127

Motors andgeneratorsChapter summary

review questionsPHYSICALLY SPEAkINGThe terms in the following list belong to two distinct groups. Group these terms into the two groups and add the definition of each term. Create a diagram to display the relationship between them.

• Split-ringcommutator • Fan• Squirrelcage • Bearings• Brushes • Armature• Stator • Magnets

REvIEwING 1 a Identify the type of motor in Figure 6.4.2.

b Identify each of the labelled features.c Construct a table to list the parts you have

identified and the role each plays.

2 Explain why radial (curved) magnets in a motor allow for greatest efficiency.

3 Describe the differences and similarities in the way permanent and current-carrying coils produce magnetic fields within a DC motor.

4 Define a galvanometer and outline what it is used for.

5 A galvanometer has a spring attached to the centre of it, distinguishing it from a simple DC motor. Give reasons for its presence.

6 Recall Lenz’s law and explain how Lenz’s law accounts for the conservation of energy.

7 Describe what is meant by back emf.

8 Determine how back emf is produced in a motor.

9 Explain what a manufacturer does to a motor to account for back emf.

• ThemainfeaturesofaDCmotoraretherotor(coils,shaft and frame), stator (permanent or electromagnets), commutator split ring and commutator brushes. The roles of these components are summarised in Table 6.1.1.

• Torqueτ is the turning effect of a force F. τ = Fd where d is the distance from the pivot to the point where the force is applied.

• Acurrent-carryingloopinanexternalmagneticfieldexperiences forces due to the motor effect that generate a torque.

• Thetorqueτ on a current-carrying coil can be quantified by the relation τ = nBIA cos θ where n is the number of turns in the coil, B is the strength of the magnetic field, I is the strength of the current in the coil, A is the area of the coil and θ is the angle between the plane of the coil and the magnetic field lines.

• Thetorqueonacoilinanexternalmagneticfieldvariesas the coil rotates. The torque is at a maximum when the plane of the coil is parallel to the magnetic field

and zero when the coil is perpendicular to the magnetic field.

• Thecurrent-carryingcoilofagalvanometerexperiencesa torque due to the motor effect. This torque is balanced by a spring and this causes the needle to deflect by an amount proportional to the current flowing. This allows the determination of the magnitude and direction of the current being measured.

• WhenaDCmotorrotates,itscoilsexperienceaninduced emf (back emf ) set up in accordance with Lenz’s law. This back emf opposes supply emf and reduces the current flowing through the motor’s coils.

• ACinductionmotorscontainasquirrel-cagerotor(conducting bars, shaft and frame) and stator electromagnets.

• ACinductionmotorsgeneratearotatingmagneticfieldthat induces currents in the squirrel-cage rotor. These current-carrying conductors in the rotor experience a force due to the motor effect, which exerts a torque on the rotor.

Figure 6.4.2

D

E

CBA

128

6 Motors: magnetic fields make the world go around

10 a Label the parts of the AC motor in Figure 6.4.3.b Explain what you could do to this motor to make it

into a DC motor.

11 Give examples of where an AC motor would be used.

12 Compare and contrast DC and AC motors.

13 Explain how an AC induction motor works.

14 Explain how a single-phase induction motor gets started.

15 Distinguish between situations in which AC universal motors and AC induction motors would be best suited.

SOLvING PROBLEmS 16 Calculate the maximum torque that is generated by a

force of 460 N applied to an object at a distance of 3 m from its axis of rotation.

17 Calculate the torque in a square coil with sides of length 3 cm. The current in the coil is 2 A and it is placed in a magnetic field of 0.3 T.

18 Students undertook to measure the torque produced by a simple DC motor.

The motor contained 100 turns and the square armature was 0.03 m in length. The motor is attached to a piece of string holding a mass (Figure 6.4.4).

The motor was turned on and allowed to wind up the mass until it stalled and stopped. At this point the radius of the windings r was recorded. The current supplied to the motor was gradually increased and the process repeated.

The table of results is shown below.

Mass = 0.5 kg τ = F × d = mg × d

torque (nm) current a raDius (m)0.27 0.1 0.055

0.54 0.2 0.11

0.81 0.3 0.165

1.08 0.4 0.22

1.35 0.5 0.276

1.62 0.6 0.331

a Draw a graph of torque versus current.b Determine the gradient of the line.c What quantity does this value represent?d From this value, determine the magnetic field

in which the armature is spinning.

motor

motor shaft

string

mass

edge of table

r

Figure 6.4.4

Revie

w Questions

Figure 6.4.3

A

B C

D

E

F

129


PHYSICS FOCUS

LINEAR mOTORS

Figure 6.4.5 is of a maglev train—a train that floats on its rail and moves at very high speeds. You will

study the floating mechanism in detail in Module 3 ‘From Ideas to Implementation’, but let’s have a look at how the train actually moves forward.

The maglev train operates solely on electric power. The propulsion method is via an electric motor—but one with a difference—it is a linear motor. This is, in principle, an electric motor that has been unwrapped and flattened. Magnetic fields on the train and rail are continually created to attract and repel each other. It is these interactions that apply the forces to propel the train forward.

Linear motors produce motion by moving in a straight line rather than the traditional rotational motion. There are two main parts: the stator unrolled (the primary) and laid flat on the rail, and the secondary, which is the glider on the train that floats over the rail.

Why use this type of motor? Reasons include:• Ithasnomovingpartssothereisnowearandtear.• Thetrainridesonanaircushion,solessenergyis

lost due to friction.• Electromagnetsareusedforbraking,sothetrain

is a lot quieter.

1 Outline how a DC motor works.2 Outline how an AC induction motor works.3 Compare the uses for an AC induction motor with

that of an AC universal motor.4 Draw a diagram of a DC motor and explain what

can be done to it to make it an AC motor.5 Determine how torque is calculated in a motor.6 Justify the use of linear motors in such applications

as the maglev train.7 Evaluate the cost associated with maglev trains and

standard trains.

RESEARCH8 Find out exactly how a linear motor works.9 Compare the torque produced in a standard motor

with the linear force in a linear motor.

Figure 6.4.5 A linear motor propels this maglev train.


7

130

Generators and electricity supply: power for the people

generator, transformer, step-down transformer, step-up transformer, flux leakage, magnetic hysteresis,

power stations, substation, transmission towers, insulators,

lightning protector

Technology that changed our livesWidespread access to affordable electricity has arguably been the single greatest catalyst for change in modern society. It has had an impact on every aspect of our lives, from our health to our wealth and even what we do in our leisure time. Some key developments that enabled this revolution were the invention of the AC generator and the transformer. Their significance is far beyond the reaches of the power lines they service and, as we shall see, they involve more than meets the eye.

7.1 AC and DC generatorsIn chapter 6, we saw that the real value of electric motors was that they convert electrical potential energy into rotational kinetic or mechanical energy. A logical question that follows is ‘how do we get the electrical potential energy that turns these motors?’. For many of our household applications, the answer is ‘we use a generator’. When you switch on an electrical appliance at home, you may not realise where the energy comes from. In some distant power plant there is a huge rotating machine as big as a building that looks like a huge electric motor. The difference is that this generator is producing electricity instead of using it.

The simplest AC motor design is the synchronous AC motor (Figure 7.1.1). If this motor was supplied with 50 Hz AC from the power point, it would spin at 50 revolutions per second in synchronisation with the electrical signal. This design has no practical application, but more complicated designs are used in clocks and tape drives, due to their constant speed.

AC supply

Slip-ring commutator continuouslyconnects the rotating coil to the AC supply.

SN

Permanent magnetsor electromagnets

provide an externalmagnetic field.

Figure 7.1.1 A synchronous motor uses a slip-ring commutator to feed AC to the motor.

131

motors andGenerators

Although it lacks applications as a motor, this design would produce an electric current if you turned the coil with your hands. In this way it is acting as a generator. When we turn the motor ourselves, its coil experiences a changing magnetic field and an emf is induced (recall Faraday’s law). If the loop is connected by a circuit, a current flows, and we have turned the motor into a generator.

The parts of a generator are essentially the same as those of a motor (see Table 7.1.1). The difference is that we physically turn a generator and it produces electrical potential energy. So the operation of a generator is the opposite of that of a motor. In this section we will only consider simple generators; we will see a more complicated version in section 7.3.

Table 7.1.1 The parts of a simple generator and their function

Part DescriPtion anD roleArmature The armature is the part of a generator that contains current-carrying coils. These carry an induced

current caused by a changing magnetic field. For simple generators these are the coils in the rotor, but in other generators the armature is in the stator.

Coils These are the many loops of wire that carry electrical current. In many generators there are two sets of coils. One set is the electromagnets that provide the magnetic field (in simple generators this field is provided by permanent magnets). The other set is in the armature and these electromagnets carry the current produced by the generator.

Rotor The rotor generally consists of coils of wire wound around a laminated iron frame. The frame is attached to an axle or shaft that allows it to rotate. The iron frame is laminated to reduce heating and losses due to eddy currents. The iron itself acts to intensify the magnetic field passing through the coils.

Stator In simple generators the stator is the stationary permanent magnets or electromagnets that provide an external magnetic field around the rotor. These magnets are curved to maximise the amount of time the sides of the rotor coil are travelling perpendicular to the magnetic field. In some generators the stator contains the armature coils and a magnet turns as part of the rotor to produce a changing magnetic field.

Commutator split ring

A simple DC generator contains semicircular metal contacts (a split ring) that reverse the direction of the current flowing out of the rotor coil every half rotation. This reversal of the current ensures that the current being produced is DC.

Commutator slip ring

A simple AC generator has two circular metal contacts. Each slip ring is connected to one end of the coils in the rotor. These provide an alternating current that changes direction every half rotation. In more complicated examples these provide a current to an electromagnet in the rotor and a current is produced in the stator.

Commutator brushes

Brushes are conducting contacts (generally of graphite or metal) that connect the commutator to the external circuit.

Figure 7.1.2a shows a typical hand-operated generator found in schools. This generator has the features of both an AC and a DC generator. With the flick of a switch you can connect to the components for either type of generator and produce a current by winding the handle. If you have access to one of these make sure you try it out and try to identify all its parts.

A simple AC generator Figure 7.1.3a shows a simple model of an AC generator. This generator

therefore contains a slip-ring commutator that simply connects each end of the rotating coil to an external circuit. The rotor of the generator shown is being turned by hand. Let’s follow a full rotation of this generator as shown in Figure 7.1.3b and use the graph (Figure 7.1.3c) to gain an understanding of its operation. Figure 7.1.3c shows the amount of flux ΦB passing within the coil WXYZ. It also shows the induced emf ε caused by the rate of change of flux ∆ΦB/∆t in accordance with Faraday’s law (see section 5.1). This emf would result in a current flowing within the coil, if the output terminals were connected.

Describe the main components of a generator.

activity 7.1



Generators and electricity supply: power for the people7

132

Figure 7.1.2 (a) A typical AC and DC generator found in high schools and (b) a close-up showing the main parts of this generator (see Table 7.1.1 for details)

a b

• Atposition1(Figure7.1.3b),theplaneofthecoilWXYZliesparalleltothemagnetic field. This means that no magnetic field lines are passing within the coil, so the amount of magnetic flux ΦB within the coil is zero. Recall Faraday’s law:

ε = n(∆ΦB/∆t)

which shows that the emf ε induced in the coil is proportional to the rate of change in magnetic flux ∆ΦB/∆t passing within the coil—not the flux value itself. At position 1, the emf is at a maximum because the rate of change of the magnetic flux is at its maximum. This can be seen in Figure 7.1.3c in which the tangents to the curve (illustrating the slope of the curve) for the magnetic flux ΦB represent the rate of change of magnetic flux ∆ΦB/∆t. The tangent at position 1 is at the maximum positive value and, therefore, so is the induced emf, according to Faraday’s law.

• Iftheoutputterminalsofthegeneratorareconnected,theinducedemf in the coil will cause a current to flow. At position 1, this occurs in the direction marked (i.e. through Z, Y, X and W in turn). To determine the direction of this current, let’s consider a positive charge in the wire between points W and X at position A. The right-hand palm rule shows that a charge moving upwards at position A, as the coil rotates within the magnetic field, will experience a force towards point W. You could also determine the direction of the current using Lenz’s law. Review Lenz’s law and then give it a try.

• Atposition2thecoilhasrotated45º.Theamountoffluxpassingthroughthecoil has increased from zero and will continue to increase until position 3. The rate of change of magnetic flux ∆ΦB/∆t at position 2 has decreased and will continue to decrease until the coil reaches position 3. Around position 3 the flux is changing relatively slowly. At position 3 the flux reaches a maximum and then starts to decrease. Since the rate of change of flux ∆ΦB/∆t is zero (as shown by the slope of the tangent) at position 3, the emf must also be zero according to Faraday’s law. This means there will be no current flowing through the coil at this point.

try thIs!MoDelling generatorsUse the model of a simple motor suggested in section 6.1 to model the operation of a generator. Add slip-ring and split-ring commutators by sticking pieces of aluminium foil around the cork. Ensure you can describe and explain the current that would be produced during a full rotation.

output terminals

coils or windings

stator magnets

slip-ring commutator to produce AC

rotor and armatureN

S

slip-ring commutator to produce DC

Interactive

Module

133


a LInear Generator

With energy shaping up to be the single biggest issue for

humans in the future, there are many new devices available to help us save energy. A torch without batteries is just one example. These torches contain a permanent magnet that can slide back and forth inside a coil of wire. The current generated in the coil is stored in a capacitor and is then ready for use in the torch.

• Oncethecoilrotatespastpoint3,thefluxΦB begins to decrease and the rate of change of flux ∆ΦB/∆t is increasing. We can see this as the slope of the curve for ΦB gets bigger (as shown by a tangent to the curve). When the flux ΦB is decreasing, ∆ΦB = ∆ΦB final – ∆ΦB initial is a negative value and so is ∆ΦB/∆t. This means that according to Faraday’s law the induced emf now has the opposite sign to the sign it had before position 3. This means that the current produced by this emf is in the opposite direction to the one that would have flowed before position 3. This explains why we get an AC current, because at positions 3 and 7 in Figure 7.1.3b the current will change direction.

• Betweenpositions3and5thefluxdecreasesandiszeroatposition5.Therateof change of flux increases to a maximum at position 5. This means that the emf increases to a maximum at point 5 and, therefore, so does the induced current. Notice that in Figure 7.1.3b the current flowing along the side of the coil WX is now flowing into the page. You can use the right-hand palm rule to show that the current is now flowing through points W, X, Y and Z in turn. This further illustrates that the current has reversed direction at point 3 and we are producing an AC current at the output terminals.

• Followtherotationthroughtoposition9andyouwillseethatthecurrentchanges direction again at position 7 and the full cycle is completed at position9.Eachrotationinducesanemfinthecoilthatisinonedirection for half a rotation and in the opposite direction for the other half of a rotation. This alternating emf can produce an alternating current and our explanation of the operation of AC generators is complete.

A simple DC generatorNow that we have seen how an AC generator works, let’s look at a DC generator.

Figure 7.1.5a shows a simple model of a DC generator. The obvious difference is the split-ring commutator connecting the coil to the external circuit. We have seen this situation in a simple DC motor (see Figure 6.1.7) but now it is acting as a DC generator. Note that the graph for emf ε is the emf that would be measured at the output terminals. This is therefore affected by the inclusion of a split-ring commutator that is fundamental to the functioning of a DC generator. Let’s focus on what is different about the operation of a DC generator.

rotation

axis ofrotation

B

W

P

X Y

Z

A

0 +max.–max.Position

output terminals

ε and ΦB

ΔΦB

Δt

tangents showrate of change

1

2

3

4

5

6

7

8

9

F εΦB

II

a

b c

WX

WX

WX

WX

WX

WX

WX

WX

WX

YZ

YZ

YZ

YZ

YZ

YZ

YZ

YZ

YZ

v

FB

S

Time

N

Figure 7.1.3 (a) A simplified model of an AC generator and (b) the coil as seen from point P in part (a), showing the direction of the current flowing over one rotation. (c) A graph of magnetic flux and induced emf over one rotation

Figure 7.1.4 A torch without batteries

magnet

motion of magnet

coil


134

• Asbefore,weseethatthechangeintherateofchangeofmagneticflux∆ΦB/∆t in the coil causes a changing induced emf (Figure 7.1.5c). The operation of this generator is the same as the AC example until we get to position 3. At position 3 the brushes of the commutator reverse the contact between the coil and the external circuit.

• ThecurrentflowinginthecoilWXYZisinthesamedirectionasforanACgenerator (see Figure 7.1.3b). The difference is that now this current only flows in one direction in the external circuit. As the current in the coil changes direction at position 3, the split-ring commutator reverses the connection. This causes the current to flow in the same direction in the external circuit throughout the entire rotation.

• FromthegraphofemfinFigure7.1.5cyoucanseethattheemfinducedinthe external circuit rises and falls but is always in the same direction. This produces a current that also rises and falls and is always in the same direction. This is DC. We usually see a DC (or emf ) represented on a graph as a straightlinewithonevalue(e.g.theredlineinFigure4.1.2).Thismightbethe case with a current from a battery, but not for the output direct from a DC generator.

• Atposition7weseethatagainthesplit-ringcommutatorreversesthecontactto the external circuit. A current in this external circuit would continue to flow in the same direction, even though the current in the coil again changes direction.

axis ofrotation

B

W

P

X Y

Z

A

0 +max.–max.Position

output terminals

ε and ΦB

ε

ΦB

∆ΦB∆t

tangents showrate of change

Time

1

2

3

4

5

6

7

8

9

F

WX

WX

WX

WX

WX

WX

WX

WX

WX

YZ

YZ

YZ

YZ

YZ

YZ

YZ

YZ

YZ

S NI I

Figure 7.1.5 (a) A simplified model of an DC generator and (b) of the coil as seen from point P in part (a), showing one complete rotation. (c) A graph of magnetic flux and induced emf over this rotation

reGeneratIve brakInG

An energy-efficient way to slow a moving vehicle is to

convert the kinetic energy of the vehicle into electricity. This energy can then be stored and reused to get the vehicle moving again. Traditionally braking was achieved by using friction and the energy was lost as heat. Modern trains and electric hybrid automobiles now use these systems to increase efficiency. The electric motors on the wheels of the vehicle are used as generators when the brakes are applied, and energy is stored in batteries or a capacitor.

a

b c

135


Comparing AC and DC generatorsThe differences in the structure and operation of simple AC and DC generators should now be clear. The two styles of commutator are essential to the production of either AC or DC and are the main difference. As DC generators have more complicated commutators, they are generally less reliable. The continual electrical reconnection and the extra mechanical wear in the split-ring commutator reduces the service life of a DC generator and increases running costs.Electricalarcingbetweenthesplitringandthebrushesalsoproduceselectromagnetic radiation. This can cause interference in other electronics and to radio communication.

When we consider the operation of simple generators, we see that both styles of commutator connect the rotating coils to an external circuit. The commutators in a DC generator perform the additional function of reversing the direction of the connection to the outside circuit each half revolution. This acts to maintain the current in the external circuit flowing in one direction, even though the current within the rotating coil is changing direction. We will see in section 7.3 that more complicated generators produce their current in the stator. We will learn that this is the case in a generator used in a large-scale AC power plant. This configuration is best for generating high currents and is designed to produce three-phase electricity.

Describe the differences between AC and DC generators.

Gather secondary information to discuss advantages/disadvantages of AC and DC generators and relate these to their use.

PHYSICS FEATURE

ComPARIng moToRS AnD gEnERAToRS

A comparison between motors and generators is now relatively easy. First, take a look at Tables 6.1.1

and 7.1.1 to identify the similarities and differences.You may have noticed that we had already seen the

principles behind the operation of a generator when we studied back emf in motors (section 6.2). The induced emf that opposes supply emf in a motor would produce a current if there was no supply emf—and you would have a generator.

Figure 7.1.6 shows the combination of a motor and a generator. This illustrates the conversion of mechanical energy to electrical energy in the generator and back again in the motor. It also shows the similarity in construction of both an AC motor and an AC generator. This combination of a motor and a generator is commonly known as a motor-generator set or

dynamotor. These can be used to convert DC electricity to different voltages. This could occur if the set contained a DC motor and DC generator. As we will see in the next chapter, there is a handy device that can easily convert AC to different voltages, but it is more difficult with DC. Motor-generators can also be used to convert AC to DC, DC to AC and AC to AC at another frequency. Modern electronics have now allowed all these conversions to be made without the use of a motor-generator, but there are still some applications where electronics are not practical.

Compare the structure and function of a generator to an electric motor.

back emf generatedby turning of motor

voltage applied to motor

AC voltage output is throughslip rings and brushes

mechanicalenergy output

motor

generator

mechanical energyinput is convertedinto electrical energy

AC voltage is applied to themotor coil through slip rings

S

N

N

S

Figure 7.1.6 A motor-generator set or dynamotor


136

Transformers are devices that can easily convert alternating currents to higher or lower voltages. They consist of two coils of insulated wire wound around an iron core (see Figure 7.2.1). You may notice that they have the same basic construction as Faraday’s iron ring we studied earlier(seeFigure5.1.4).AswesawinChapter5, Faraday observed an induced current in a secondary coil only when there was a changing current in the primary coil, and this is illustrated in Figure 5.1.5. The changing current in the primary coil created a changing magnetic flux that passed through the secondary coil. This changing flux induced an emf in the secondary coil and an induced current was measured. It is important to note that when a direct current was applied to the primary coil it did not produce a continually changing magnetic field, so transformers are only practical for AC.

To quantify the voltage conversions performed by transformers, we recall Faraday’s law from Chapter 5:

ε = n(∆ΦB/∆t)

7.2 TransformersOureverydayuseofelectricityrequiresalargevarietyofelectricalcurrentsandvoltages. With DC, this usually means that we have the right battery for the job. With AC, it’s another story altogether. When you plug in your laptop computer, it typically requires an AC input of about 18.5 V, but the average value of the voltagefromthepowerpointisabout240V.(Note:theAustralianstandardisnow 230 V.) The conversion of AC to the correct voltage occurs, with a range of +10% to –6%, in a small box on the laptop’s power cable, which contains a device called a transformer.

Discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer.

Describe the purpose of transformers in electrical circuits.

CHECkPoInT 7.11 Use labelled diagrams to outline the structure of simple AC and DC motors.2 Compare the structure and function of motors and generators.3 Describe and contrast the structure and function of AC and DC generators.

changing magnetic flux

AC supply

primary coilwith np turns secondary coil

with ns turns

loadvoltage Vp voltage Vs

Figure 7.2.1 In an ideal transformer, the iron core ensures that all the flux generated in the primary coil also passes through the secondary coil.

try thIs!exPloring electroMagnetisMTo observe the magnetic flux produced by a changing current, place a wire with many loops near or wrapped around a magnetic compass. Connect a battery to your coil and observe the compass needle as you connect and disconnect the circuit.

260240

220

200

S16

0

140

120100 E

W

8060

4020

N340

320

300280

Figure 7.2.2 A current through a coil induces a magnetic flux.

137


Let’s consider applying a 50 Hz input voltage Vp across the primary coil of the transformer in Figure 7.2.1, which has np turns (or loops) of wire. If we assume there are no losses, this will produce the same changing magnetic flux ∆ΦB/∆t within both coils. This changing magnetic flux will produce a 50 Hz alternating voltage Vs across the secondary coil with ns turns. Substituting into Faraday’s law gives:

V nt

V ntp p

Bs s

Band=

=

∆Φ∆

∆Φ∆

Dividing one expression by the other gives:

V

V

n

np

s

p

s=

This expression shows us the relationship between the number of turns and the voltages across each coil of a transformer.

If we analyse the expression above, we can see that if the primary coil has more turns (loops) than the secondary coil (i.e. np/ns > 1), the voltage Vp across the primary coil will be greater than the voltage Vs across the secondary coil.

This is called a step-down transformer because it produces AC with a lower voltage. When the secondary coil has more turns than the primary coil (i.e. np/ns < 1), the voltage Vs measured across the secondary coil is higher than the input voltage Vp. This is called a step-up transformer because it produces an AC output with higher voltage. Figure 7.2.3 is a simplified diagram showing the change in magnitude of the output voltage Vs compared to the input voltage Vp for these two types of transformers.

Worked exampleQUESTIonCalculate the number of turns required in the secondary coil of a transformer to produce 18.5 V AC from 230 V AC if there are 100 turns in the primary coil.

SoLUTIonUsing:

V

V

n

np

s

p

s=

and rearranging to make ns the subject gives:

n nV

Vs ps

p=

Substituting Vp = 230 V, Vs = 18.5 V and np = 100 gives: ns = ×10018 5230

.

= 8 turns

Transformer efficiency and design The law of conservation of energy states that energy cannot be created or

destroyed, but can be converted from one form to another. This means that the amount of energy you put into a transformer must equal the amount you get out. This is true for all transformers, but some of the energy that went into a transformer is always converted to heat. It is for this reason that transformers are never 100% efficient.

Identify the relationship between the ratio of the number of turns in the primary and secondary coils and the ratio of primary to secondary voltage.

Compare step-up and step-down transformers.

Solve problems and analyse information about transformers using:

V

V

n

np

s

p

s=

Vp

Vs

t

t

Figure 7.2.3 The simplified graphs show the AC output of a step-up transformer (blue line) and a step-down transformer (red line) in the secondary coil of a transformer. The input voltage in the primary coil is shown for comparison (black line).


138

Efficiencyisoftenexpressedasapercentageandcanbedeterminedusingthefollowing equation:

Efficiency (%) = ×100useful power output

total power input

The losses of energy that occur within a transformer cause the useful electrical poweroutputtobelessthanthepowerinput.WesawinChapter4thatpoweristhe rate at which energy is converted and can be determined using the following relationships:

P = IV, P = I 2R and PVR

=2

where P is electrical power in watts (W), I is current in amps (A), V is potential difference (or voltage) in volts (V) and R is resistance in ohms (Ω).

In formulating the equation: V

V

n

np

s

p

s=

we assume that there are no losses; that is, that transformers are ideal. We will continue this assumption whenever we do calculations for transformers, but it is important to consider transformers in a more realistic way, to understand their losses and design considerations.

Using the equation above, we can determine another relationship for a transformer that illustrates the changes in current. In an ideal transformer the power input in the primary coil (Pp) and the power output in the secondary coil (Ps) would be:

Pp = IpVp and Ps = IsVs

Rearranging and substituting into the relationship above gives:I P

I P

n

ns p

p s

p

s=

Pp = Ps for an ideal transformer, so:

II

n

ns

p

p

s=

This illustrates that both the voltage and the current output of a transformer change with respect to the input. We have therefore gained another insight into the workings of transformers. By inspecting both equations:

I

I

n

n

V

V

n

ns

p

p

s

p

s

p

s

and= =

we can see that for a step-up transformer (ns > np), the transformer increases the voltage but decreases the current of AC electricity. Alternatively, for a step-down transformer (ns < np) the voltage decreases but the current increases. Let’s see how this situation satisfies the law of conservation of energy.

Since energy (in joules) must be conserved and power (in joules per second) is the rate at which energy is converted, then power must also be conserved if no energy is being stored in a transformer. As power is equal to the product of the current and voltage:

Pp = Ps and IpVp = IsVs

neon LIGhts

Neon lights are one very visible application of

transformers. Typically about 25 000 volts are required to excite the gas in a neon tube, so a step-up transformer is used to convert the voltage from 230 V AC. In Figure 7.2.4, the red light is produced by neon gas and the violet-blue light is produced by argon.

Figure 7.2.4 Neon and argon discharge tubes

139


As we already know, the voltage is changed in a transformer and, for the above relationship to hold, the change in current must be inversely proportional to the change in voltage. We have already shown that this is the case by deriving the two expressions: I

I

n

n

V

V

n

ns

p

p

s

p

s

p

s

and= =

Combining these expressions gives:

II

V

Vs

p

p

s=

This expression shows that current and voltage are inversely proportional. For example, in a step-up transformer Vs is larger than Vp so Vp/Vs must be less than 1. In this case, Is/Ip is also less than 1. This means that Is is less than Ip.

We now see that transformers obey the law of conservation of energy when they alter the current and voltage of AC electricity. For an ideal transformer, the input power is equal to the output power and this is maintained by the inverse relationships between input and output current and voltage.

These findings explain the design features evident in Figure 7.2.5, which shows a step-down transformer. The increase in current in the secondary coil is the reason a larger diameter wire is used for the secondary coil.

As discussed in Chapter 5, larger currents produce greater losses due to resistive heating. The amount of heat Q produced by resistive heating is proportional to the resistance of the material R and the square of the current I as shown by Joule’s law:

Q = Pt = I2Rt

where Q is in joules, power P is in watts or joules per second (J s–1), I is in amps, R is in ohms and t is in seconds (s). So we see that larger currents produce more heat.

The resistance of a metallic conductor is described by the following equation:

R =ρlA

where ρ is resistivity, l is length and A is cross-sectional area. This equation shows that resistance in a wire is proportional to its length and inversely proportional to its cross-sectional area. So, by increasing the diameter of the wire in the secondary coil of the transformer (Figure 7.2.5), the resistance is decreased and this minimises resistive heating.

We have now identified that resistive heating is one of the mechanisms responsible for energy losses in all transformers. The transformer shown in Figure 7.2.5 is a very common style of transformer in household appliances because it minimises another loss mechanism—flux leakage. The transformer in Figure 7.2.5 has a central iron core around which the primary and secondary coils are wound. The overall shape of the iron core acts to contain and direct the magnetic flux from the primary coil through the secondary coil. This ensures that the maximum amount of flux passes through the secondary coil and therefore maximises induction in the secondary coil. If a significant amount of flux did not pass through the secondary coil, then energy would not be transferred and could be lost by other mechanisms.

Explain why voltage transformations are related to conservation of energy.

Gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome.

Figure 7.2.5 A step-down transformer in which the coils are wound around a central laminated iron core

primary coil

secondary coil


140

The remaining mechanisms for losses in transformers are sometimes referred to as core losses. The first of these is loss due to magnetic hysteresis as the magnetic field continually changes direction. We were introduced to this effect in our discussion of induction cooktops (see Chapter 5 Physics Feature p 108). In induction cookware hysteresis served a useful purpose in heating food, but in a transformer we wish to minimise the effect. This can be done by making the core of a transformer from a magnetically ‘soft’ iron that does not stay magnetised once an external magnetic field is removed. Magnetically ‘soft’ iron is therefore used in transformer cores to reduce heat loss through magnetic hysteresis. This soft iron is also used in the rotors of motors and generators and in the stators of AC motors to reduce losses due to magnetic hysteresis.

The other type of core loss in transformers is resistive heating due to eddy currents. To reduce this effect, the soft iron cores of transformers are made from thin laminated sheets that are electrically insulated from each other. Figure 7.2.6a shows a simple set-up of two coils that would act as a transformer. This diagram shows a moment in time in which the current in the primary coil is producing an increasing magnetic flux in the direction indicated. In accordance with Lenz’s law, currents would be induced in both the secondary coil and as eddy currents in the iron core (Figure 7.2.6b). To minimise the energy lost as heat due to these eddy currents, the core is made from thin laminations (Figure 7.2.6c) that are electrically insulated from one another. This minimises the magnitude of the eddy currents and therefore the amount of heat lost through resistive heating.

We have now seen that transformers come in many shapes and sizes, but they all have the same basic structure. They contain a magnetically soft, laminated ironcoreandtwocoilsofinsulatedwire.Largetransformerscanbeupto99%efficient, but there are always some losses as heat. These losses, although small in proportion, can affect the performance of a transformer, if allowed to build up. In small transformers, air removes heat by convection, but in large transformers oil is used (Figure 7.2.7). The oil is passed through cooling tubes and heat is dissipated to the air. This type of transformer is used in electrical substations and at power plants as part of the distribution of electricity to the community. We will explore this more in the next section.

Figure 7.2.6 (a) A simple model of a transformer. (b) Large eddy currents are set up in the core. (c) Smaller eddy currents are set up in a laminated core.

Figure 7.2.7 An industrial transformer used in the transmission of electricity

coolingtubes

oil tank

high voltageterminals

low voltageterminals

low voltage coils

high voltage coils

laminated core

the sounds of aC

If you ever get a chance to see a large industrial transformer

you will probably hear it hum. This is the sound of the laminations in the core vibrating. As the direction of 50 Hz AC output alternates back and forth, the magnetic field within the core also changes direction. The alternating magnetic flux exerts forces on the core laminations and causes them to vibrate.

primary coil

secondarycoilI

I

increasingmagneticflux Φ

increasingmagneticflux Φ

crosssection

cross section

cross section

eddy currents

eddy current

solid core

laminatedcore

soft iron core

B

a

b

c

141


CHECkPoInT 7.21 Identify a device that uses a transformer and outline the role the transformer plays in the operation of this device.2 Define step-up and step-down transformers.3 Explain why transformers do not work with DC electricity.4 Calculate the voltage produced by a transformer for which np/ns is 0.01 and the input voltage is 12 V.5 Recall the law of conservation of energy and analyse its application to the operation of a transformer.6 Identify the losses that occur in transformers and outline the design features that minimise these losses.

7.3 Electricity generation and transmission

PHYSICS FEATUREEDISon AnD WESTIngHoUSE

In the late 1800s a rivalry developed in the US over the type of electricity that should be generated.

Thomas Edison (1847–1931) had established a system based on DC electricity. It was primarily used for household lighting and Edison had patents on much of the technology involved. One of the major drawbacks of this system was that it had to be distributed at the voltage used in households, as transformers could not be used with DC to easily alter the voltage and current. Transmission therefore required large currents, and this limited the distance DC could be transmitted because of the huge power losses as a result of resistive heating. This meant that power stations had to be dotted all over a large city and distribution was not practical in rural areas.

When Edison was establishing his system there was no viable alternative to DC. Nicola Tesla (1856–1943), an employee of Edison at one stage, eventually developed AC motors and generators. The patents for these were promptly purchased by George Westinghouse (1846–1914). Alternating current had the advantage that voltage and current could be changed using transformers. It could therefore be generated at low voltages and converted to high voltages for transmission using low currents.

This allowed transmission over long distances

with acceptable energy losses, and made AC more economical.

Edison embarked on a campaign to discredit the use of AC and promote it as unsafe. To do this he used an AC generator to publicly electrocute animals, including an elephant. The showdown between Edison’s DC and Westinghouse’s AC came when a proposal was put forward to generate hydroelectricity at Niagara Falls. Several designs were considered, including those from Westinghouse and Edison. Westinghouse was awarded the contract and, due to its many advantages, AC electricity was eventually adopted as the standard worldwide.

Gather secondary information on the competition between Westinghouse and Edison to supply electricity to cities.

activity 7.2



Figure 7.3.1 (a) Thomas Edison and (b) George Westinghouse

a b


142

stator

neutral

three phaseoutput

Time(ms)

Cur

rent

or

EM

F +

–

0 V

A

A A

C

C

C

B

B

B

1020

30 40

N

S

a

b

AC power generation and deliveryToday,morethan99%oftheworld’selectricityisgeneratedasAC.Most of the electricity generated in New South Wales is produced by burning coal in a power plant like the one shown in Figure 7.3.2.

Heat from the burning of coal in these facilities is used to produce steam. This steam is used to turn turbines, which are connected to large generators that produce electricity. These generators typically produce three-phase 23 kV AC at 50 Hz. Figure 7.3.3 shows the main features of athree-phasegeneratorusedinapowerstation.Oncegenerated,electricity is transmitted to consumers through a vast network, such as theonesummarisedinFigure7.3.4.

The most common use of this energy is in our homes. It arrives there in wires above your street or below your feet in underground cables. In order to get there efficiently, electricity goes through the following steps on the way to your home.

The 23 kV 50 Hz output voltage from a power station’s generators is passed through step-up transformers to produce 330 kV or even 500 kV AC. High voltages mean low currents and this reduces power losses during transmission due to the resistance of the lines. Large transmission lines are used to transport this electrical energy over long distances to a terminal substation. Fuel sources for power stations are typically a long way from major population centres, so this long-distance transmission is essential.

At the end of the transmission line, step-down transformers in a terminal substation convert the voltage to 132 kV or 66 kV for transmission to zone substations (see Figure 7.3.5). Higher voltages are less important for these shorter transmission distances. As the voltages decrease, the effort required to insulate each wire from the others decreases and so does the cost involved.

At zone substations this power is again stepped-down using transformers. Most commonly it is converted to 11 kV AC and distributed into communities.

Typically, transformers on power poles finally step-down the AC to 400Vand230Vforindustrialandhouseholduserespectively.Thesevoltages are low enough so that discharges cannot occur through air within appliances, and the insulation of wires from each other is easy and economical.

A typical power pole in a suburban street resembles the one shown in Figure 7.3.6. Three of the four main wires shown carry the three phases produced in the power plant’s generator (see Figure 7.3.3). The other wire is the neutral that completes the circuit. The potential difference between neutral and a single phase is 230 V. The difference betweenanytwoofthethreephasesis400V.Asinglephaseandtheneutral are connected to each household and all four wires deliver three-phase powertoindustry(seeFigure7.3.4).

Explain the role of transformers in electricity substations.

Figure 7.3.2 A coal-fired power station

Figure 7.3.3 (a) A three-phase generator. The DC output is fed to the rotating magnet via slip rings (not shown). The ends of each of the three stator coils are connected together—this becomes the ‘neutral’. (b) The other ends carry the high voltages, which are one-third of a cycle apart.

143


POWER STATION23 kV50 Hz

step-up transformers

step-down transformers132 kV & 66 kV AC

step-down transformers11 kV AC

HIGH VOLTAGETRANSMISSION LINES

TERMINALSUBSTATION

ZONESUBSTATION

330 kV or 500 kVAC transmission

400 V

230 V

123N

factorythree-phase 400 V AC

householdsingle-phase 230 V AC

400 V and 230 VAC distribution

11 kVAC distribution

132 kV or 66 kVAC transmission

pole step-downtransformer

GENERATION

TRANSMISSION

SUB-TRANSMISSION

DISTRIBUTION

CONSUMER

Figure 7.3.4 A typical electricity supply network in New South Wales


144

Power losses during transmission and distributionIn order to deliver electrical power efficiently across NSW the power loss between the generator and the customer must be as small as possible. The main losses during transmission result from the resistance in the transmission wires and the induction of eddy currents. First, let’s look at losses due to resistance.

Using a relationship we have seen before:

P = I 2R

we see that power loss P in transmission lines is proportional to the resistance R and the current I squared. Therefore, if we double the resistance in transmission wires we would double the power loss. But if we double the current we increase the power loss by a factor of four. So obviously, we want the resistance in transmission wires as low as possible but, even more importantly, we want to use a relatively small current. This is achieved using transformers to step-up AC to as

Discuss the energy losses that occur as energy is fed through transmission lines from the generator to the consumer.

Figure 7.3.5 An electricity substation Figure 7.3.6 A power pole in Marrickville, Sydney. Note that the four main wires are not insulated. They are spaced far enough apart to stop discharges occurring. They are insulated from the pole by small ceramic insulators.

PHYSICS FEATURETRAnSFoRmERS In THE HomE

The use of transformers does not end when electricity is fed into your home. The

components of many household appliances require higher or lower voltages than the 230 V available from the wall socket. A good example is the microwave oven. The component within a microwave oven that produces the microwaves typically requires thousands

of volts, while the control circuits and control panel on the front only use small voltages. This means that a microwave oven would require both step-up and step-down transformers to supply the power for its components.

Discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer.

insulator

four wires carrying three-phase AC and the neutral wire

145


high a voltage as practical. During this process the current is reduced to a minimum.

Another way to illustrate why we want to reduce transmission current comes from another relationship we saw earlier:

R =ρlA

where ρ is resistivity, l is length and A is cross-sectional area. This equation shows that the resistance R of a wire is proportional to its

length l . This means that, as the distance of transmission increases, the resistance also increases. A typical resistance for one type of high-voltage transmission line isabout0.4ohmsperkilometre.Wecanseethatiflargecurrentsareused,transmission will quickly become uneconomical over the hundreds of kilometres typically required. A large proportion of the energy generated would be lost as heat due to resistive losses. The best solution is again to minimise the current being transmitted.

The two previous paragraphs highlight the importance of transformers in the distribution of electricity. Transformers can readily step-up voltages and in doing so they minimise the magnitude of the current involved. They can also step-down the voltage when required to a value suitable for the consumer. This ability of transformers has changed modern society enormously, and we will take a closer look at this soon.

The equation R =ρlA

also shows us that the resistance of a wire is inversely

proportional to its cross-sectional area A. So making the diameter of the wire larger will reduce its resistance. Unfortunately, thick copper wires are very heavy and require larger structures to support them. Larger structures cost more money, so a compromise is required. Aluminium is used as a conductor in high-voltage transmission lines. Although it has a higher resistance than copper, it is much lighter, so the wires have a larger diameter to reduce resistance without being too heavy.

Losses during AC transmission also occur due to the formation of eddy currents. The constantly alternating current in transmission lines produces a constantly changing magnetic flux. This can induce eddy currents in nearby conductors and energy will be lost. To minimise this, transmission lines are held at a distance from metal transmission towers by insulators. Losses due to eddy currents also occur in the cores of transformers. Recall that these losses are minimised by constructing these cores from thin laminations of magnetically soft iron.

Transmission structuresA typical high-voltage AC transmission tower is shown in Figure 7.3.7. The three sets of wires carry the three phases generated in the power station. As with lower voltage transmission (Figure 7.3.6), high-voltage transmission lines are not coated with insulation.

These three sets of wires must be kept at large distances from each other and from the metal tower. This not only minimises losses due to eddy currents, it also minimises the chance of current flowing through the air (electrical discharge) between conductors or to the ground through the tower. In ideal conditions, 500 kV

Gather evidence and analyse secondary information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use.

activity 7.3



lightning protector

insulators

high voltage transmission lines

Figure 7.3.7 A high-voltage transmission tower


146

transmission lines must be approximately half a metre away from the tower that supports them. In conditions in which the insulators are wet and covered in pollution deposits, this distance must be greater. Typically, therefore, these wires are held more than a metre from their supporting tower.

Figure 7.3.8 shows that insulators used for lines are typically long chains of ceramic components. These components are disc shaped and have deep corrugations on their underside. These features increase the distance any current would have to travel in a discharge. The corrugations also minimise the chance of pollution deposits settling on the surface and keeps some of the insulator dry in wet weather.

Gather and analyse information to identify how transmission lines are: • insulatedfromsupporting

structures • protectedfromlightning

strikes.

hIGh-voLtaGe dC transmIssIon

Modern electronics have made the task of converting high-voltage AC to high-voltage DC (HVDC) relatively simple, and HVDC transmission is now

economically viable for distances of more than several hundred kilometres. These systems lose less power due to eddy currents induced in metal support structures and require only two transmission lines. For short distances, these savings do not offset the cost of the extra electronics, but if the transmission distance is long enough, the savings are significant. HVDC is the method of choice for transmitting electricity by underwater cables due to the excessive losses involved. The Victorian and Tasmanian electricity grids are connected across Bass Strait via a 400 kV DC link called Basslink. This provides a two-way link along which electricity flows according to the demand at either end. In this way, either state can make up for a shortfall of electricity in the other.

insulator

cement

cup

porcelain

ball

arcing horn

conductor

a

b

Figure 7.3.8 (a) A technician services insulators on a power line. (b) Insulators used are typically long chains of ceramic components.

The impact of AC generators and transformers on societyThe development of AC generators and transformers in the 20th century enabled the widespread availability of electricity. AC generators efficiently convert the energy from sources such as coal into electricity. Transformers enabled the efficient long-distance transmission of this electricity to the households of whole nations.

These developments have arguably been the single greatest catalysts for change in modern society.

Before the availability of household electricity, a significant part of the daily ritual was the lighting of fires for cooking and heating. Refrigeration was generally not available, so many perishable foods could only be eaten fresh or when in season. None of the electrical appliances that make our lives so convenient had been developed, so much more time was devoted to chores and manual tasks. As you can imagine, these differences would have resulted in a very different life for the average citizen in a society without electricity.

The availability of electricity has also had many positive and negative effects on society.• Theabilitytotransmitelectricityefficientlyoverlongdistances

has allowed more people in large cities to live further from the

147


city centre. It is not uncommon in modern society for people to commute large distances to work.

• Peopleinmostruralareasenjoythesameaccesstoelectricityascitydwellers.This has improved their quality of life and allowed many people to remain in country areas, who might have otherwise decided to move to the city.

• Thequalityoflifeimprovedwhenelectricitywasaffordableineveryhome.This changed the way we live through the many appliances available. Many of us find it hard to conceive a life without a refrigerator, television, DVD player or computer.

• Tasksperformedbyelectricalmachinerydecreasedtheamountofunskilledlabour required and increased unemployment in certain parts of society.

The widespread generation of electricity has also had a significant effect on the environment.• Whenindustrywasabletomoveawayfromthepowerplantsincitycentres

this took pollution away from many people’s homes. This improved the quality of their environment and improved their health.

• Theavailabilityofelectricpowerforheatinghasreducedtheneedtoburnwood or coal in houses, significantly improving air quality in large cities.

• Theuseofhydro-electricitygenerationrequiredtheconstruction of large dams. These destroyed the habitat of many plants and animals. It also displaced many people from their homes, and this continues today as our demand for electricity grows.

• Theconstructionoflong-distancepowerlinesalsorequiredthe destruction of habitat and increased rates of erosion where vegetation has been removed.

• Radioactivewastefromnuclearpowerplantsrequireslong-term storage and poses a threat to the environment if not contained.

• Theburningoffossilfuelsfortheproductionofelectricityhas led to a significant increase in atmospheric carbon dioxide levels. Scientists believe this will have significant consequences in the near future through global climate change.

• Theburningoffossilsfuelshasalsoproducedpollutants such as sulfur dioxide. This contributes to the formation of acid rain, which can damage forests, aquatic life and man-made structures.

Assess the effects of the development of AC generators on society and the environment.

Discuss the impact of the development of transformers on society.

PHYSICS FEATURELIgHTnIng PRoTECTIon

Figure 7.3.7 shows a typical high-voltage transmission tower. The two wires at the very

top of the tower are called lightning protectors or shield conductors. These wires are placed above the tower and transmission lines to significantly reduce the chance of lightning striking the transmission lines. At regular intervals these wires are connected directly to the ground. These connections act to transfer huge currents from lightning strikes directly to earth. This protects the transmission system from dangerous spikes in current and voltage, which can damage substation transformers and cause blackouts.

CHECkPoInT 7.31 Explain how transformers allow the transmission of AC power over long distances.2 Identify the role of transformers in household appliances.3 Outline the energy losses in high-voltage transmission lines and the steps taken to minimise them.4 Describe how transmission lines are insulated from supporting structures and protected from lightning strikes.5 Assess the impact of transformers and AC generators on society and the environment. Justify your answer.


148

7 Generators and electricity supply: power for the people


ACTIvITY 7.1: AC AnD DC gEnERAToRSResearch how AC and DC generators work then place the information that you have found into a table outlining the advantages and disadvantages of each. Use this information to create a list of applications that would benefit from using each.

Discussion questions1 Compare the structure of AC and DC motors.2 List applications for which AC and DC generators are preferred and

explain your choice.

ACTIvITY 7.2: EDISon AnD WESTIngHoUSE ResearchthecompetitionbetweenWestinghouseandEdisontodominatesupply of electricity. Use this information to present a 5 minute speech on the topic.

Discussion questions1 Outline the benefits of both DC and AC electricity supplies.2 StatereasonsthatEdisonusedtoshowthatACwasmoredangerous.

Gather secondary information to discuss advantages/disadvantages of AC and DC generators and relate these to their use.

Analyse secondary information on the competition between Westinghouse and Edison to supply electricity to cities.

Gather and analyse information to identify how transmission lines are:– insulated from supporting structures– protected from lightning strikes.

Perform an investigation to model the structure of a transformer to demonstrate how secondary voltage is produced.

Gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome.

Gather and analyse secondary information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use.


ACTIvITY 7.3: PoWER To THE PEoPLEIn this investigation, you will gather first-hand data to see how a transformer works. You will then gather information from secondary sources in order to look into how heating due to eddy currents are overcome in a transformer.Equipment: transformer with removable coils, power supply, ammeter, voltmeter, connecting wires, light bulb.

Discussion questions1 Outline the difference between a step-up and step-down

transformer.2 Explain why the first-hand data does not fit the equation for

a transformer.3 Outline methods used to overcome heating in transformers due

to eddy currents.

149

Chapter summary motors andGenerators

review questions

• Thedifferencebetweenmotorsandgeneratorsisthatmotors use electricity to produce mechanical energy and generators produce electricity from mechanical energy.

• Themaincomponentsofageneratorarethearmature,coils, rotor, stator, split-ring or slip-ring commutator and commutator brushes.

• Acurrentisproducedintherotatingcoilofageneratordue to the changing magnetic flux within the coil.

• AnACgeneratorcontainsaslip-ringcommutator;aDC generator contains a split-ring commutator.

• Theoperationofatransformerreliesonthechangingmagnetic flux from the primary coil passing through the secondary coil. This changing flux then induces an emf and produces a current in the secondary coil.

• Thelawofconservationofenergystatesthatenergycannot be created or destroyed, only converted from one form to another.

• TransformersalterthevoltageandcurrentofACelectricity. There is an inverse relationship between the voltage and current produced by a transformer. If the voltage is increased the current is decreased and vice versa. This is in agreement with the law of conservation of energy.

• Foranidealtransformer,therelationshipbetween the number of turns in the coils of a transformer and the voltages in either coil is:

V

V

n

np

s

p

s=

• AtransformerthatincreasesthevoltageofACelectricityis called a step-up transformer; one that decreases the voltage is called a step-down transformer.

• Lossmechanismsintransformersincludefluxleakage,resistive heating by eddy currents and hysteresis loses. These are minimised by using a suitably shaped and magnetically ‘soft’ laminated iron core.

• TheincreaseinACvoltageandsubsequentdecreaseincurrent in transformers allows the efficient transmission of AC power over long distances.

• Householdelectricalappliancescontaincomponentsthat require various voltages and it is transformers within these appliances that provide these voltages.

• Themainlossesinthetransmissionofelectricityaredueto resistive losses and eddy currents. These losses are minimised through appropriate isolation of conductors and minimising resistance.

• Transmissionlinesareisolatedandinsulatedfromtheirsupporting structures by insulating components called insulators. The higher the voltage, the larger the length of the insulator required to minimise the occurrence of electrical discharge.

• TransformersandACgeneratorshaveenabledthewidespread distribution of AC electricity. This has had an enormous impact on society and the environment. These impacts include changes in where many people live, their quality of life, the nature of their employment, their health and that of our environment.

PHYSICALLY SPEAkIngCopy and complete the table below by adding a definition for each key term.

terM Definition

Step-up transformer

Generator armature

Generator

Substation

High-voltage insulator

150

7 Generators and electricity supply: power for the people

REvIEWIng 1 Look at Figure 7.4.1. Identify each of the parts that have been labelled.

2 Create a table to compare and contrast the parts and function of an electric motor and a generator.

3 Explain how you could tell the difference between an AC and DC motor.

4 a Identify where energy is lost during the distribution of AC from the time the electricity is generated to when it reaches the household consumer.

b Describe the energy that is lost.

5 Outline methods of reducing energy lost along the path from generation to consumer.

6 List the applications that would be best suited to using an AC generator and those that would be more appropriate using DC generators.

7 State the features of AC and DC generators that make them appropriate to their use.

8 a List the effects that generators have had on society and the environment.b Assess these effects.

9 Give reasons why transmission lines need to be insulated from supporting structures and describe how it is done.

10 Explain how transmission lines are protected from lightning strikes.

11 a Identify parts A–E of the transformer in Figure 7.4.2.b Identify the function of each of these parts.c If np = 250 and ns = 750, identify if this is a step-up or a step-down

transformer.

12 Explain how current is produced in the secondary coil of a transformer.

13 Discuss how the number of turns in the coils of a transformer determines the voltage in the secondary coil.

14 Show that V

V

n

np

s

p

s=

is true for a transformer that is 100% efficient.

15 Outline the energy transfers for a transformer that is not 100% efficient.

16 Determine how currents in primary and secondary coils of a transformer are related.

17 Explain why transformers are needed in order to bring electricity effectively from a power station to the consumer.

18 Many appliances that are used in the home have a transformer attached to their power cord. Explain the presence and need for these.

A

B

W

P

X Y

II

Z

D

F

B

C

Figure 7.4.1 An AC generator

changing magnetic flux

E

D

C

B

A

voltage Vp voltage Vs

Figure 7.4.2 Parts of a transformer

151


19 Give examples of how transformers have impacted on the way in which we live.

20 Discuss the production of eddy currents in transformers and the problems that arise from them.

21 Outline methods of reducing the effects of eddy currents in transformers.

SoLvIng PRobLEmS 22 Calculate the voltage in the secondary coil of a transformer that has 20 turns

in the primary coil and 300 turns in the secondary coil if the input is 240 V.

23 Determine the efficiency of a transformer that has a turns ratio of 10:500, an input voltage of 50 V and output of 200 V.

24 Complete the following table.

coils in PriMary PriMary voltage (v) coils in seconDary seconDary voltage (v) steP-uP or steP-Down100 6 200

320 240 50

50 000 30 500 240

25 Complete the following table by calculating the unknown quantities.

current in PriMary coil (a)

voltage in PriMary coil (v)

current in seconDary coil (a)

voltage in seconDary coil (v)

turns ratio

5 6 240

240 2 12

0.5 0.1 1000

Revie

w Questions

152

The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 30 marks. It should take you approximately 54 minutes to complete this review.

2multiple choice(1 mark each) 1 Two current-carrying wires of different lengths are

placed side by side as shown in Figure 7.5.1.

The length of the 2 m wire and the separation between the two wires are doubled. Identify the magnitude of the new force acting on each wire.A F B 2F C 4FD F/2

2 Identify the answer that has the correct part matched to the correct role.

Part roleA Armature Provides magnetic field

B Brushes Allow current direction to change within the motor

C Commutator Maintains electrical contact without tangling wires

D Magnet Part of motor or generator that contains current-carrying coils or windings

3 Determine the output voltage from a household transformer for which the ratio of windings is 40:500.A 3000 V B 19 V C 0.05 VD 240 V

4 Look at Figure 7.5.2.

Identify the answer that lists its parts correctly.

i ii iii iv

A Axis of rotation

Force Output terminals

Current

B Force Current Axis of rotation

Output terminals

C Axis of rotation

Output terminals

Force Current

D Output terminals

Axis of rotation

Current Force

5 A series of substations can be found in suburban streets. The role of these substations is mainly to:A step up the voltage.B step down the voltage.C boost the power being transmitted.D add extra branches of wires to serve more houses.

2 m

0.8 m

0.5 cm

S N

III

IV

III

Figure 7.5.2 A generator

Figure 7.5.1

153


extended questions 6 Describe the main features of an AC induction motor

and their roles. (3 marks)

7 Outline the procedure you followed in order to investigate three factors that affect the generation of an electrical current. (3 marks)

8 Discuss the how competition between Edison and Westinghouse led to the efficient supply of electricity to cities. (2 marks)

9 Explain how eddy currents are used in an induction cooktop to cook a steak. (2 marks)

10 A wire is placed in an external magnetic field as shown in Figure 7.5.3.a Determine the direction of the force on the wire.b Calculate the size of the force experience by

the wire. (3 marks)

0.1 T magnetic fieldwithin the shadedarea

0.5 A

1 c

m

1 cm

B

axis of rotation

40°4 cm

4 cm

SN

I

12 Compare and contrast how the motor effect is used in a galvanometer and a loudspeaker. (3 marks)

13 From the four chapters in this module, identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry. (4 marks)


Figure 7.5.3 A wire in a magnetic field

Figure 7.5.4 A simple electric motor

11 A simple electric motor has dimensions as shown in Figure 7.5.4. The single coil carries a 0.1 A current and sits within a 0.05 T magnetic field at 40º to the vertical.a Explain why the magnets in a real motor would

be curved.b Determine the force generated by the motor

effect on each side of the motor.c Calculate the torque in the motor. (5 marks)

3 from ideas toimplementation

154

Figure 8.0.1 Semiconductor integrated circuits have produced laptop computers that have much more computing power than even the most powerful early military computers that used thermionic devices.

In just over 100 years we have gone from not knowing what constitutes matter to understanding its subatomic structure. From the discovery of the mysterious cathode rays came an understanding that these are fundamental particles that hold solid matter together. Technology, such as X-rays, sprang from this discovery and showed that their use in medical imaging enabled better patient diagnosis.

An understanding of the way electrons move in space and solids gave rise to the plethora of electronic devices that have transformed our technology and way of life. The control of electron motion in a vacuum resulted in early thermionic devices (‘valves’), such as the triode, to control the flow of current in electronic circuitry. From these came television, radio, radar and the start of the modern electronic age. The study of electron motion in solids resulted in miniature semiconductor devices that replaced thermionic devices and enabled faster, more efficient and much more sophisticated electronics, such as integrated circuits, to be made. These devices are the backbone of all modern-day electronics, and include mobile phones and computers, as well as electronics and instrumentation used in homes, hospitals, industry, and in space.

The accidental discovery of superconductors (solid material through which electrons travel unimpeded) may one day transform our technology yet again to produce much faster computers and more efficient transportation, energy production and transmission.

Context

155

Figure 8.0.2 CDs and DVDs can be used as spectrometers.

InquIry aCtIvIty

CDs: wInDows to vIewIng photons

Neon signs and most street lights use discharge tubes to create the many different-coloured lights. In all of these, an electrical current is passed through a gas at reduced pressure. The light from the glowing gas is characteristic of the type of gas in the tube. A spectrometer is a device that separates the colours of the spectrum of the gas by using the same principle of interference as early research into X-ray diffraction. You can make your own spectrometer of visible light at home simply by looking at the light reflected from the back of a CD or DVD.

Ordinary light (from the Sun and an incandescent bulb) will show you the continuous spectrum of white light that makes up the colours of the visible spectrum. Wait until it’s dark and find lighting that does not use incandescent lamps. Fluorescent tubes or bulbs are commonly available lighting that uses electrical discharges. These are filled with mercury vapour and an inert gas. 1 Compare the spectrum of colours from a fluorescent lamp with that of an

incandescent lamp. Does it come in discrete colours or is it continuous like the light from the incandescent lamp?

2 Use your spectrometer under the street lighting. Can you tell if this lighting has the same gas as your fluorescent lamp or is it different? Or does it use a mixture of gases?

8

156

from cathode rays to television

fluorescence, geissler tubes, cathode rays, cathode ray tubes,

discharge tubes, aston dark space, cathode glow, Crookes dark space, negative glow, Faraday dark space,

positive column, anode glow, cyclotron motion, oscilloscope, Cro,

timebase, sawtooth, blanking, shadow mask

Mysterious raysIn today’s terminology, ‘cathode rays’ are collimated beams of electrons in an evacuated vessel. They were given this label at a time when not even the structure of the atom was known. The discovery of these ‘mysterious’ rays led to the discovery of the electron, and paved the way not only for understanding the atom but has also led to technology such as television, radar, electronics, the oscilloscope and lighting.

8.1 Cathode ray tubesPhysicists began passing electrical current through air at lower pressure as soon as the vacuum pump was invented by Otto von Guericke in 1650. Among the most notable physicists to experiment with this was Michael Faraday, who, in 1838, reduced the pressure of air in a glass vessel and applied a high voltage to electrodes imbedded at each end of the tube. He noted that a narrow channel of light was produced between the negative (cathode) and the positive (anode) electrodes. Today we use this phenomenon to produce lighting in all shapes and sizes (such as neon signs) and are generally known as discharge tubes.

In 1855 Heinrich Geissler (1815–1879), a master glassblower and skilled instrument maker, improved the vacuum pump. Geissler, in collaboration with the physicist Julius Plücker (1801–1868), reduced the pressure in the tubes to the point where the colours and patterns of the gaseous discharge disappeared and were replaced by a green glow (also known as fluorescence) that came from the glass. Moreover, Geissler was able to seal the tubes and maintain the very low pressure such that a vacuum pump was no longer needed. Julius Plücker called these tubes Geissler tubes.

try this!Human-powered ligHtingYou can make a fluorescent light tube temporarily give out flashes of light by rubbing it quickly with a piece of fur or wool in a darkened room.

Figure 8.1.1 The high voltage produced in static electricity will temporarily light up a fluorescent tube.

157


We now know that this green glow is caused by electrons leaving the cathode and striking the glass at high speeds, causing it to glow, but during Geissler and Plücker’s time this was a complete mystery. The name cathode rays was first used by Eugene Goldstein, a German physicist, who coined the name because it appeared that unknown beams were being emitted by the cathode. The tubes displaying this property came to be known as cathode ray tubes. The next great challenge for physicists at the time was to find the composition of these cathode rays from their behaviour.

The cathode ray debate: charged particles or electromagnetic waves?A variety of cathode ray tubes were made by a number of physicists in order to study the behaviour of cathode rays and hopefully determine their nature.

In about 1875, the most prominent of these physicists Sir William Crookes placed a piece of metal in the shape of a Maltese cross in the path of the cathode rays as shown in Figure 8.1.3. This produced a sharp shadow on the glass behind the cross, indicating that cathode rays travel in a straight line.

A magnetic field at right angles to the direction of the beam caused it to deflect in the same way that negatively charged particles would deflect (see Figure 8.1.4). Crookes concluded that cathode rays consisted of negatively charged particles.

A moveable paddle wheel struck by the cathode rays started to rotate and roll along the tube (see Figure 8.1.8). This implied that cathode rays carried energy and momentum. Crookes also went on to show that the properties of cathode rays did not depend on the type of cathode material.

From all of these experiments, the British physicists (such as Crookes) concluded that cathode rays were beams of negatively charged sub-microscopic particles. However, German physicists such as Heinrich Hertz (1857–1894) carried out experiments that seemed to contradict the British viewpoint.

Hertz was unable to deflect the cathode rays with an electric field between two parallel plates placed inside the tube. This suggested that cathode rays did not have a charge. He also found that cathode rays could pass through very thin gold foil without damaging it. Thus Hertz and other German physicists concluded that cathode rays had no mass or electric charge, and were possibly a form of electromagnetic radiation.

There were also objections to the paddle wheel experiment because it was thought that the cathode ray heated one side of a paddle, causing the gas in contact with it to heat up and expand thus causing the paddle, to move.

The German physicists knew that electrical current can cause a magnetic field, but they were unable to detect such a field around the cathode rays. These contradictions strengthened their view of the wave nature of the cathode rays.

From a modern perspective, we can easily resolve the debate between the British and German physicists. Cathode rays pass through a thin gold foil because the space between atoms is much larger than the size of an electron, and so a few atomic layers of gold allow the electrons to pass without much chance of a collision with a gold atom. However, the atomic structure of matter was unknown at the time. It was difficult to measure a magnetic field due to the cathode rays because the current was extremely small—a very sensitive magnetic field probe is needed to make such a measurement.

Explain that cathode ray tubes allowed the manipulation of a stream of charged particles.

Explain why the apparent inconsistent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves.

A

– +

near vacuum

glass tube

anodecathode

switch high voltage(variable)

Crookes tube(cathode ray tube)

mask holderanode

cathode

Figure 8.1.2 A high voltage across electrodes at each end of an evacuated glass tube causes the glass near the positive electrode (anode) to glow.

Figure 8.1.3 A Maltese cross placed in the path of cathode rays produces a sharp shadow on the glass indicating that cathode rays travel in a straight line.

+collimator anode

cathodeS N

magnet

–

Figure 8.1.4 Cathode rays are deflected by a magnetic field in the same way that negatively charged particles are deflected.

from cathode rays to television8

158

physICs FeatureanatoMy oF a DIsCharge tube

Energy-efficient lighting, such as fluorescent tubes and neon signs, are known as discharge

tubes. These are evacuated glass vessels filled with gas at approximately 1% of atmospheric pressure and two electrodes at opposite ends of the tube. A large potential difference between these electrodes causes an electrical current to flow through the gas in the tube, resulting in colours that depend on the type of gas and its pressure.

Figure 8.1.5 illustrates an operating discharge tube with the characteristic bright and dark spaces. The Aston dark space next to the cathode is very thin and may go unnoticed. This is followed by the first luminous region called the cathode glow. Next is the Crookes dark space and the negative glow (a luminous region). The Faraday dark space follows this and is the largest of all the dark spaces. The largest luminous region, known as the positive column, follows and is the most prominent feature of the discharge. The positive column may display periodic regions of bright and dark spaces known as striations. Finally, there is the anode dark space and then the anode glow adjacent to the anode.

An operating discharge tube consists of a mixture of ions, electrons and neutral gas atoms. The discharge starts because energetic particles (such as electrons and protons) continually stream from outside the Earth and strike the surface and atmosphere. This so called ‘cosmic radiation’ can strike gas atoms and produce free ions and electrons in the discharge tube. An electric field causes the free electrons to gain sufficient energy to ionise gas atoms and produce further free ions and electrons, which in turn will accelerate to produce more ionisation. This avalanche of ionisation is the way an electrical discharge is started.

To maintain the current through the discharge, electrons (known as secondary electrons) are continually ejected from the cathode surface by the bombardment of ions that are attracted to it. These secondary electrons cause further ionisation, which results in further ion bombardment of the cathode surface, and the process repeats.

Bright regions in the discharge are areas where the electrons have sufficient energy to ionise the gas. Some of the ions recombine with electrons, which results in the emission of light. Not all atoms will be ionised. Some will simply have their electrons gain energy while remaining bound to the atom or molecule; this is known as excitation. All excited electrons will fall back to their previous energy state and, in doing so, give out light. A minimum energy is required to excite or ionise an atom or molecule.

Many discharge characteristics are related to the average distance that electrons travel between colliding with the gas atoms or molecules. An increase in gas pressure leads to an increase in the density of gas particles, and results in an electron travelling a shorter distance between collisions, and vice versa.

At low pressure, electrons have sufficient distance to accelerate and reach the required ionisation energy; however, they may strike the sides of the vessel more often than they strike the gas. A minimum number of ionisations are required to sustain the avalanche of electrons that maintain the discharge. Below a minimum pressure, it becomes difficult to start an electrical discharge along the tube.

+

+

e–e–

–

Figure 8.1.6 Ions that strike the cathode liberate secondary electrons, which help sustain the discharge.

cathodeglow

negativeglow anode

glowpositivecolumn

Aston CrookesFaraday

anode darkspace

– +

Figure 8.1.5 A discharge tube showing the typical bright and dark spaces

activity 8.1

praCtiCal eXperienCes


159


anatoMy oF a DIsCharge tube (continued)

At higher pressures there is a large density of gas particles, so there is a high probability that electrons will strike them. However, too high a pressure results in very frequent electron collisions such that they do not reach sufficient energy to ionise gas particles. Thus, the electric field strength must be increased for them to reach this energy. There is an optimum pressure for starting a discharge at a relatively low electric field.

Electrons in dark regions have not reached sufficient energy to ionise or excite the gas. Figure 8.1.7 shows the potential difference and the magnitude of the electric field strength between the cathode and anode during a discharge. The largest potential difference occurs between the negative glow and the cathode. This is due to the ions being attracted to the cathode and thus crowding around it. Any electron outside this region will see an almost equal amount of positive charge (from the ions) and negative charge (from the cathode), resulting in a much lower electric field between the negative glow and the cathode.

Let us now trace the path of electrons from cathode to anode. Electrons ejected from the cathode are accelerated by the high electric field until they have sufficient energy to cause the cathode glow. This depletes the electron energy gained, so they must accelerate once more through the cathode dark space until they have sufficient energy to cause the negative glow. Once again this process depletes their energy, and they must accelerate through the Faraday dark space until they gain sufficient energy to cause the ionisation and excitation in the positive column.

Striations may result in the positive column for the same reasons that other bright regions occur. However, the electric field is more uniform in this region, so the bright and dark regions (the striations) are uniformly spaced. Again, electrons gain sufficient energy to ionise and excite the gas, which causes a luminous region and depletes the energy of the electrons. The electrons then accelerate through a region (a dark space) until they gain sufficient energy to excite and ionise the gas, and so on.

Clear luminous and dark regions will be produced if electrons that leave the cathode remain in step. This is not possible in practice. Some electrons will strike gas particles before others and have their energy depleted sooner. This results in a spread of energies as electrons move along the tube. Thus, the luminous regions are diffuse and do not have well-defined boundaries. If electrons become too out-of-step with each other, then the striations will start to merge into each other and will not be distinguishable from the dark spaces between them. This may happen at too high a pressure.

As the pressure is lowered, striations become more widely spaced because there is a lower gas particle density and the electrons will travel longer, on average, before striking a gas particle. The converse happens at higher pressures. Electrons strike gas particles more frequently, so the striations will be more closely spaced.

Finally, electrons are collected by the anode, which also repels any ions, thus forming the anode dark space. This is a region of high electric field that causes the electrons to accelerate to the anode and to excite or ionise the gas around it, causing the anode glow.

As a rule, the positive column is less bright than the negative glow and differently coloured. Helium results in a red cathode glow, a green negative glow, and a reddish-purple positive column. Neon produces yellow, orange and red in the respective regions. Nitrogen emits pink, blue and red. Each gas has a characteristic set of colours, depending on the degree of excitation that a bound electron experiences.

V

E

– +Distance along the tube

Figure 8.1.7 The potential difference V and the magnitude of the electric field strength E along the axis of a discharge tube


160

Finally, the apparent lack of deflection of the cathode rays by an electric field arose because the cathode ray tubes were not in a total vacuum—there was residual air in the tube, albeit at a low pressure. A fraction of the electrons in the beams (cathode rays) struck air molecules and ionised them. As shown in Figure 8.1.9, the positively charged ions are attracted to the negatively charged plate and the electrons are attracted to the positively charged plate. This essentially neutralises the charge on the plates and results in zero electric field between them. Therefore, there can be no deflection of the cathode ray.

A resolution to the debate started with the French physicist Jean Perrin (1870–1942), who showed that a metal plate acquired a negative charge when it was struck by cathode rays. British physicist, Joseph John Thomson (1856–1940) was able to definitively show that cathode rays can be deflected by an electric field simply by producing an even lower pressure in the tube—thus reducing the number of ions. Moreover, he measured the charge to mass ratio of electrons. To quantitatively examine his work, we must first revise the motion of electric charges in electric and magnetic fields.

+collimator anode

cathode

–

paddle wheel

+++++–

–

+

+cathode rays are not deflected,and behave in the same way as light

– ––––

Figure 8.1.8 Cathode rays caused a paddle wheel to turn, implying they carried both momentum and energy.

Figure 8.1.9 The cathode ray is not deflected by oppositely charged plates because ions and electrons neutralise the charge on the plates.

CheCkpoInt 8.11 Describe Faraday’s work with cathode ray tubes.2 Outline the contribution of Geissler and Plücker to the development of discharge tubes.3 Recall the different types of cathode ray tubes that were developed to understand cathode rays.4 Create a table that lists the observations and inferences made from each of the tubes.5 List the evidence that led Hertz to believe that cathode rays were not particles.6 Give a reason for not being able to detect the magnetic field from cathode rays.7 Explain why Hertz could not detect a deflection when he applied an electric field across the cathode rays.8 Describe how the debate about whether cathode rays were particles or waves was finally resolved.

8.2 Charges in electric fields Any region in space in which there is an electric force on a charged object is

said to contain an electric field. The field points in the direction of the force on a very small positive charge, known as a test charge. The magnitude of the field on the test charge—the electric field strength—increases in proportion to the force.

A way of visualising the electric field is to draw lines that indicate the direction of the force on the test charge. Closely spaced lines mean higher electric field strength and vice versa. Equally spaced lines indicate it is a uniform field.

Electric field lines are not real lines. They are used to give a qualitative description of the field. We can only draw a finite number of lines. The field is actually continuous—it exists everywhere in space.

The electric field lines around a point positive or negative charge are shown in Figure 8.2.1. Although this is a two-dimensional drawing, the lines actually radiate outwards for a positive charge and inward for a negative charge in three

Discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates.

a b

–+

Figure 8.2.1 The electric field lines for (a) a positive point charge radiate outwards, and (b) a negative point charge radiate inwards.

activity 8.2



161


dimensions. The greatest repulsive or attractive force is in the region of the lines with the closest spacing—near the charge in this case.

There are some guidelines for drawing electric field lines for two or more charges:• Thelinesmustbeginonapositivechargeand/orendonanegativecharge.• Largerchargeshavemorelinesstartingorendingonthem.• Linescannotcross.• Linesarealwaysatrightanglestoaconductingsurface.

The electric field lines for a pair of opposite and equal point charges are shown in Figure 8.2.2. The number of field lines leaving the positive charge is equal to the number of lines ending on the negative charge. The lines are radial at very close distances to the charges. These lines are more closely spaced near the charges to indicate higher electric field strength in that region.

The electric field lines between two equal charges q of the same sign are shown in Figure 8.2.3. The region at the centre, between the charges, is where the electric field strength is zero, because the electric fields of the two charges cancel each other.

The field lines between two charged parallel metal plates of opposite sign but equal magnitude are shown in Figure 8.2.4. Equally spaced and parallel lines indicate that the field is equal in magnitude and direction and is said to be uniform. The field lines start to curve near the edges and become unevenly spaced, indicating a non-uniform field (called the edge-effect).

Identify that charged plates produce an electric field.

+ –

positive

negativea b

+ + – –

Figure 8.2.2 The electric field lines for two point charges of equal magnitude but opposite sign

Figure 8.2.4 The electric field lines between two oppositely charged parallel metal plates

Figure 8.2.3 Electric field lines around (a) two equal positive charges, and (b) two equal negative charges

positive

negative

+

+ + + + + + + + + + +

– – – – – – – – – – –

d

E

Figure 8.2.5 Parallel plates with a uniform electric field between them

Electric field strength between parallel platesA side-view of a uniform field between two plates is shown in Figure 8.2.5 (we have ignored edge effects).

The electric field strength E was defined as the ratio of the force F on a small positive charge q given by:

EFq

=

(see in2 Physics @ Preliminary section 10.6).It has units of force per unit charge, which in SI units is newtons per

coulomborN/CorNC–1. However, in practice neither the force between the plates nor the charge on them is easily measured. We need a more convenient expression in terms potential difference V between the plates.

The equal and opposite charges on the plates were produced by applying a potential difference V between them, using a power supply or a battery. Energy from the power supply moved the electrons from one plate to the other, resulting in equal and opposite charges on the plates. Recall that it was shown that the

Describe quantitatively the electric field due to oppositely charged parallel plates.


162

energy required to move an object in the direction of the force is known as work and is defined as:

Work = force × distance = W = F × d(see in2 Physics @ Preliminary section 4.3).

In this case, the displacement d is the separation of the parallel plates. Rearranging this formula as follows:

FWd

=

and substituting this expression for force in the expression for electric field given previously, we obtain:

EW

q d=

×

Recall also that the work done on charges is related to the potential difference V by the following expression:

W = q × V

(see in2 Physics @ Preliminary section 10.8).Substituting this into the previous expression for electric field strength,

we obtain:E

q Vq d

=××

EVd

=

Thus, the electric field strength E is easily calculated from the plate separation d and the potential difference between them V. From this equation, you can see that an alternative unit for the electric field strength is volts per metre (V m–1).

Worked examplequestIonTwo parallel plates are separated by d = 1.0 cm and have a side length of L = 2.0 cm (Figure 8.2.6). A potential difference of V = 10 V is applied between them. An electron enters halfway between the plates with a velocity of v0 = 2.7 × 106 m s–1 parallel to the plates.

Calculate:

a the electric field strength E between the plates

b the vertical force F experienced by the electron as it travels between the plates

c the acceleration of the electron

d the time it takes for the electron to travel the full length of the plates

e the vertical displacement of the electron just as it exits the plates

f the change in kinetic energy of the electron between its entry to and exit from the plates.

Note that the electron charge e = – 1.6 × 10–19 C, and its mass me = 9.11 × 10–31 kg.

solutIona The electric field strength E is given by:

EVd

= = = − −100 010

1000 1 1

.)V m (or N C

Solve problem and analyse information using: F = qvB sin θ F = qE and

EVd

=

+ + + + + + + + + + +

– – – – – – – – – – –

e– d

E

L

Figure 8.2.6 Two parallel plates with a uniform electric field

163


b The vertical force is given by:

F = e × E = 1.6 × 10–19 × 1000 = 1.6 × 10–16 N upwards

c The acceleration can be obtained from Newton’s second law (see section 3.4 of in2 Physics @ Preliminary):

a

Fm

= = ××

= 1.76 × 1014 m s–2 = 1.8 × 1014 m s–2 downwards−

−e

1 6 10

9 11 10

16

31

.

.

d The motion of the electron is similar to projectile motion (see Module 1 ‘Space’). There is only acceleration in the vertical direction. We are concerned with the horizontal component of the velocity, which remains constant. Therefore, the time t to travel the length L of the plates is given by:

tL

v= =

×= × −

06

90 020

2 7 107.4 10 s

.

.

e The electron only accelerates vertically, so we use the acceleration from part c and one of the SUVAT equations (see in2 Physics @ Preliminary section 1.3) to work out the vertical displacement s given by:

s ut at= + 12

2

where u is the initial vertical velocity = 0 m s–1, t is the time to travel the length of the plates (from part d), and a is the vertical acceleration = 1.8 × 1014 m s–2 (from part c). Therefore, the displacement is given by:

s = + × × × × = =−0

12

1 8 10 7 4 10 0 0049 4 914 9 2. ( . ) . .m mm

f The change in kinetic energy is due to the vertical deflection of the electron by the electric field. This is work done by the field on the electron, and is given by:

W e V= ×∆

where ΔV is the change in potential that the electron experiences due to its vertical deflection s. The relationship between potential difference and the displacement s is given by:

∆V E s= ×

Combining this with the above expression for work gives:

W = e × E × s = 1.6 × 10–19 × 1000 × 0.0049 = 7.8 × 10–19 J

CheCkpoInt 8.21 Explain the meaning of the arrows and spacing of lines when drawing electric field lines.2 Calculate the electric field strength at the location of a charge of 1.28 × 10–18 C that experiences a force of

1.1 × 10–18 N.3 Sketch the electric field lines around:

a a point positive chargeb two oppositely charged parallel metal plates.


164

8.3 Charges moving in a magnetic field

A charged particle moving in a magnetic field will experience a force, as discussed in Module 2 ‘Motors and generators’. Consider a magnetic field directed into the pages (see Figure 8.3.1) and a positively charged particle that enters the field from the left. The particle will experience a force as given by the right-hand palm rule (refer to Module 2 ‘Motors and generators’). The force F on the particle changes direction every time the particle changes direction, such that the resulting motion is circular. This circular motion is known as cyclotron motion. A negatively charged particle will undergo circular motion in the opposite sense.

The magnitude of the force F on a charge q that moves with a speed v at right angles to a magnetic field B is given by:

F = qvB

The speed of the charged object remains constant even though its direction changes. This means that the magnetic force simply changes the direction of the particle without adding or subtracting energy, provided that the magnetic field is constant. In general, the magnitude of the force on a charged particle with a velocity at an angle θ with respect to the magnetic field is given by:

F = qvB sin θ

Note that a particle that is parallel to the field (θ = 0°) does not experience a force. A charged particle that enters a magnetic field at an angle other than θ = 0° or 90° travels in a spiral along the magnetic field (also known as helical motion), as shown in Figure 8.3.2. However, the component of the velocity at right angles to the field is still circular.

A particle of mass m with charge q enters a uniform magnetic field B at an angle θ and a speed v. Recall from the Module 1 ‘Space’ that the force directed towards the centre of circular motion is known as the centripetal force. In this case, the centripetal force is the magnetic force; that is: centripetal force = magnetic force

mvr

qvB2

= sin θ

where r is the radius of curvature of the particle.Rearranging this expression, we obtain the following expression for the radius:

rmv

qB=

sin θ

Identify that moving charged particles in a magnetic field experience a force.

Describe quantitatively the force acting on a charge moving through a magnetic field:

F = qvB sin θ

Solve problem and analyse information using: F = qvB sin θ F = qE and

EVd

=

+

B

F

+qv

r

Figure 8.3.1 A positively charged particle enters into a magnetic field directed into the page and undergoes anticlockwise circular motion.

+ v

B

θv

B

a b

F

+

v sinθ

v cosθ B

c

+

Figure 8.3.2 (a) A proton enters a magnetic field at an angle θ. (b) The component of the velocity at right angles to the field undergoes circular motion. (c) The total motion is helical along the magnetic field.

165


Worked examplequestIonProtons are occasionally ejected from the Sun at high speeds towards the Earth. These can be caught in the Earth’s magnetic field and are trapped along magnetic field lines in a region known as the Van Allen belt. A proton travelling at a speed of v = 1.0 × 107 m s–1 strikes the Earth’s magnetic field at an angle of 30° at a distance of 3000 km above the surface where the magnetic field B = 3.5 × 10–5 T. Assume that the Earth’s magnetic field at this distance is uniform. The charge on a proton is 1.6 × 10–19 C and it has a mass of 1.67 × 10–27 kg.

a Calculate the magnitude of the force on the proton.

b Calculate the radius of curvature of the proton.

c Determine if the cyclotron motion of the proton will cause it to come in contact with the Earth.

solutIona The magnitude of the force is given by:

F qvB= = × × × × × °− −sin . . . sinθ 1 6 10 1 0 10 3 5 1019 7 5 –17=30 2.8 × 10 N

b The radius of curvature r is given by:

r

mvqB

= = × × ×× × ×

−

− −sin. .

. .θ1 67 10 1 0 10

1 6 10 3 5 10

27 7

19 5 ssin.

306 0 103

°= × m

c No, the proton’s cyclotron motion will not cause it to come in contact with the Earth’s surface, since the cyclotron radius is 6 km and the proton is 3000 km above the surface.

CheCkpoInt 8.31 Outline what will happen to a negative charge moving to the left at right angles to a magnetic field that is

directed out of the page.2 Calculate the magnitude of the force experienced by an electron that travels at right angles to a magnetic field

of 2 T at a speed of 3 m s–1.

8.4 thomson’s experiment In 1897 Joseph John (JJ) Thomson not only provided a definitive resolution

to the debate about whether cathode rays were particles or electromagnetic waves, but he also measured the charge to mass ratio of the main constituent of cathode rays—the electron.

Thomson’s cathode ray tube is shown schematically in Figure 8.4.1. Cathodes rays (electrons) were accelerated from the cathode to the anode, which consisted of two anodes aligned along the axis of the tube and separated by a small distance. Each anode had a horizontal slit cut into it so that cathode rays could pass through. The separation of the anodes produced a flat beam that passed between two parallel plates and struck the end of the tube, which had a fluorescent screen painted on the inside surface. The beam produced a well-defined narrow horizontal line on the screen.

Outline Thomson’s experiment to measure the charge/mass ratio of an electron.


166

The parallel plates deflected the beam vertically upwards by placing a positive charge on the top plate. A magnetic field (using an electromagnet outside the tube) was applied at right angles to the electric field and the direction of the beam. The direction of the magnetic field was such that it deflected the beam downwards.

Adjusting the magnetic and electric forces such that they were equal and opposite resulted in the beam passing through undeflected.

We now quantitatively describe this experiment and how it leads to the charge to mass ratio of the electron.

Equating the electric to the magnetic force on a particle with a charge q and speed v in a magnetic field B and an electric field E, we obtain the following expression: Magnetic force = electric force

qvB = qE

From this we obtain an expression for the speed:

vEB

=

Recall that the relationship between the magnetic force and centripetal force (see section 8.3) is given by: mv

rqvB

2=

This enables an expression for charge to mass ratio to be obtained:qm

vrB

=

Substituting the expression for v into this equation:

qm

E

rB=

2

Thomson was able to calculate the radius of curvature r from the deflection of the beam on the fluorescent screen when the electric field was switched off and magnetic field switched on. He could calculate the magnitude of the electric field E because he knew the spacing d of the plates and the potential difference V between them, and made use of the relationship E=V/d. Finally, knowing the number of turns in the wires of the electromagnet and the current flowing through it, he was able to calculate the magnitude of the magnetic field B.

Thomson found that the charge to mass ratio always came to:qm

= × −1 76 1011 1. C kg

regardless of the cathode material, indicating that a fundamental particle was being emitted. This, in essence, marks the discovery of the electron. In 1891, the Irish physicist George Johnstone Stoney (1826–1911) suggested that the fundamental unit of electricity be called an electron—6 years before Thomson’s publication of his now famous experiment.

cathode rays electromagnet

electromagnet charged plates

fluorescentscreen

largevoltage

cathode

anodes

magnetic field

Figure 8.4.1 Thomson’s cathode ray tube apparatus for measuring the charge to mass ratio of electrons

167


CheCkpoInt 8.41 Explain the purpose of Thomson’s experiment.2 Describe how Thomson’s experiment obtained the charge to mass ratio of the electron.

8.5 applications of cathode raysA cathode ray oscilloscope (or simply oscilloscope or CRO) is a device used to measure the variation of voltage in time across an electrical component. Both the CRO and TV have elements in common with Thomson’s original cathode ray tube apparatus. Here we are not referring to the new plasma orLCDtelevisionsets,buttheolderstylescanningelectronbeam sets.

Cathode ray oscilloscope (CRO)The display screen of a cathode ray oscilloscope, as seen by theuser,isshowninFigure8.5.1.Despitethemanyknobs and switches on its front control panel, the CRO simply displays a real-time graph of voltage (y-axis) versus time (x-axis).

The oscilloscope has had many uses by scientists and technicians in the design and operation of electronic and electrical equipment. The way this real-time graph of voltage versus time was achieved can be understood by looking at the main components inside the CRO. A schematic diagram of a CRO (see Figure 8.5.2) consists of an electron gun, horizontal and vertical deflection plates and a fluorescent screen.

The electron gun (see Figure 8.5.3) produces and accelerates the electron beam. It consists of a heater, cathode and anode. The heater is a wire filament with a large enough current flowing through it so that it reaches a high temperature. Electrons in the wire then acquire enough energy to escape the wire and enter the region between cathode and anode. They accelerate toward the anode and pass through a hole at its centre. The shape of the anode and its separation from the cathode enables a narrow and focused beam of electrons to travel to the fluorescent screen. Some electron guns can become more sophisticated than the description given here, to achieve better focusing.

The horizontal deflection plates cause the beam to sweep horizontally across the fluorescent screen (from left to right) by periodically changing electric field between the plates—this is the time axis. The speed at which the beam sweeps across the screen is controlled by the timebase dial on the front control panel of the CRO. This changes the frequency at which the beam sweeps across the screen. The waveform of the potential difference across the horizontal plates is shown in Figure 8.5.4 and is commonly known as a sawtooth waveform. Electronics in the CRO blocks the beam on its way back to the left of the screen so that it doesn’t retrace itself (known as blanking).

Outline the role of:• electrodesintheelectrongun• thedeflectionplatesorcoils• thefluorescentscreen

in the cathode ray tube of conventional TV displays and oscilloscopes.

Figure 8.5.1 The display screen and the control panel of a cathode ray oscilloscope

anodecathode

heater

horizontallydeflecting plates

vertically deflecting plates

path of electrons

bright spoton screenwhereelectrons hit

fluorescentscreen

electron gun

Figure 8.5.2 A cathode ray oscilloscope deflects an electron beam with vertical and horizontal electric fields between parallel plates.


168

The vertical deflection plates cause the beam to move up or down in synchronisation with an input voltage. For example, a sinusoidal voltage will display a sinusoidal waveform (known as a trace) on the screen.

TelevisionCathode ray tube (CRT) television sets used the principles of the cathode ray tube for most of the 20th century. These are now being superseded by plasma and liquid crystal display television sets, which use different operating principles and allow a larger display area with a sharper image. However, the CRT television holds quite a significant historical place in this form of communication.

A schematic diagram of a colour CRT television set is shown in Figure 8.5.5. Its basic elements are similar to those of the CRO. The main difference is the method of deflecting the electrons. Magnetic field coils placed outside the tube produce horizontal and vertical magnetic fields inside it. The magnitude and direction of the current determine the degree and direction of electron beam deflection. Recall your right-hand palm rule for the force on charged particles in a magnetic field. The vertical magnetic field will deflect the electrons horizontally; the horizontal field will deflect them vertically.

The picture on the screen is formed by scanning the beam from left to right and top to bottom. The electronics in the television switches the beam on and off at the appropriate spots on the screen in order to reproduce the transmitted picture. However, to reproduce colour images, colour television sets need to control the intensity of red, blue and green phosphors on the screen. Three separate electron guns are used, each one aimed at one particular colour. The coloured dots on the screen are clustered in groups of red, blue and green dots that are very close to each other and generally cannot be distinguished by eye without the aid of a magnifying glass. For this reason a method of guiding the different electron beams to their respective coloured dots was devised. A metal sheet, known as a shadow mask (Figure 8.5.6) and consisting of an array of holes, is placed behind the phosphor screen. Each hole guides the three beams to their respective coloured phosphor as the beams move horizontally and vertically. Black and white television sets did not need the shadow mask since they had only one beam.

heater

cathode (negative)

electrons'boil' offthe heatedcathode

anode (positive)

electronbeam

electrons attractedto the positive anode

collimator

Figure 8.5.3 The components of an electron gun used in both cathode ray oscilloscopes and CRT televisions

VTime

VTime

sawtooth voltage for timebase sinusoidal vertical voltage

Figure 8.5.4 A sawtooth voltage waveform on the horizontal deflection plates of a CRO sweeps the electron beam across the screen to display the sinusoidal waveform on the vertical deflection plates.

electron gun

magneticcoils

electronbeam

fluorescent screen

Figure 8.5.5 A television picture tube showing the electron gun, deflection coils and fluorescent screen

169


electronguns

deflectingcoils

focusingcoils

glass

fluorescent screen

vacuum

mask

phosphor dots on screen

fluorescentscreen

mask

holes inmask

bluebeam

redbeam green

beam

RGB

electronbeams

Figure 8.5.6 A colour CRT television set has three electron guns that will only strike their respective coloured phosphor dots with the aid of a shadow mask.

try this!do not adjust your Horizontal!If you have access to an old black and white TV set or an old style monochrome computer monitor, try holding a bar magnet near the front of the screen and watch how the image distorts. This occurs because the magnetic field deflects the electrons that strike the screen. DO NOT do this with a colour TV set. This can magnetise the shadow mask and cause permanent distortion of the image and its colour. You can move a bar magnet near the back of a colour TV set to deflect the electrons from the electron gun and therefore distort or shift the image without causing permanent damage to the TV set.

Can an osCillosCope be used as a television set?

The similarity between the cathode ray oscilloscope (CRO) and CRT television suggests that a CRO can be used as a television set. In

fact, there have been some devices that have made use of the CRO as you would a computer monitor. So, in principle, it can be used as a television. One is then forced to ask ‘why did they need to deflect the beam in a television set with magnetic fields rather than with electric fields as in the CRO?’

In principle all television sets could be made in the same design as a CRO; however, it is much easier and cheaper to deflect the beam with a magnetic field on the outside of the tube rather than embed electrodes in the glass and inside the vacuum—this is a little trickier. So now another question arises: ‘why not deflect the beam of the CRO using magnetic fields, wouldn’t it result in cheaper CROs?’

Cathode ray oscilloscopes are precision instruments. The horizontal sweep rate must be able to be increased to very high frequencies in order to detect signals that change very quickly. Electric fields can be made to change very quickly without significant extra power requirements. However, a magnetically deflected system requires higher and higher voltages with increasing horizontal and vertical deflection frequencies in order to maintain the same current in the coils, and therefore, the same angle of beam deflection – thus having a significantly greater power requirement. Cathode ray tube television sets, however, only operate at fixed and relatively low scanning horizontal and vertical frequencies. Thus it is simpler and cheaper for the mass market to deflect with a magnetic field.

CheCkpoInt 8.51 Outline the purpose of a CRO.2 List the main parts of a CRO.3 Describe the role of each of these parts in the CRO.4 State the similarities and differences between the cathode ray tube CRO and CRT TV.


170

8 from cathode rays to television

Chapter 8This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

aCtIvIty 8.1: ChangIng pressure oF DIsCharge tubesConnect a set of discharge tubes that are at different pressures to an induction coil or high voltage power supply and observe the different striation patterns. The patterns are hard to see unless the room is very dark.Equipment: induction coil, connecting wires, discharge tubes at different pressures, DC power supply.

Discussion questions1 Drawalabelledsetofdiagramsshowingthedistinctpatternsthatoccur

during the evacuation of the tube.2 Describe the change in the striation pattern with changing pressure.

aCtIvIty 8.2: CathoDe ray tubes evolveConnect the induction coil or high voltage power supply to each of the cathode ray tubes. Observe the behaviour of the electron beam under the influence of a magnetic or an electric field. Note the result of placing a Maltese cross or a paddle wheel in the path of the electron beam. Equipment: induction coil, connecting wires, cathode ray tubes in the following set-ups: Maltese cross, magnetic field (you can use a permanent magnet), electric field, paddle wheel, DC power supply.

Discussion questions1 Listwhichexperimentsupportedtheideathatcathoderayswerewavesand

which supported the particle theory. State the observation that led to the particle or wave conclusion.

2 Identify the properties of cathode rays that were determined from these experiments.

Perform an investigation and gather first-hand information to observe the occurrence of different striation patterns for different pressures in discharge tubes.

Perform an investigation to demonstrate and identify properties of cathode rays using discharge tubes:•containingaMaltesecross•containingelectricplates•withafluorescentdisplay

screen•containingaglasswheel.

Analyse the information gathered to determine the sign of the charge on cathode rays.

171

Chapter summary from ideas toimplementation

• Cathoderaysareelectronbeamsinanevacuatedvessel.• Cathoderaytubesaretheevacuatedglassvesselsin

which cathode rays travel.• Adischargetubeconsistsofananodeandacathodeat

the ends of a glass vessel filled with a low-pressure gas that glows when a high voltage is applied between the anode and the cathode.

• Geisslertubesaresealeddischargetubesthatdonotrequire a vacuum pump.

• Cathoderayswerethoughttobenegativeparticlesdueto their direction of deflection when subjected to a magnetic field.

• Cathoderaysmovedapaddlewheel,implyingtheycarried momentum.

• Initialinconclusiveexperimentsledtothedebatebetween British and German scientists about whether cathode rays consisted of particles or were a form of electromagnetic wave.

• Thompson’sexperimentprovideddefinitiveproofoftheparticle nature of cathode rays from their motion in electric and magnetic fields. Moreover, he measured the charge to mass ratio of the electron.

• Cathoderays(orelectronbeams)areusedinthecathode ray oscilloscope (CRO) and cathode ray tube (CRT) television.

• TheCROisusedtodisplayachangingvoltageintime.• TheCROdeflectstheelectronbeamwithelectricfields,

while the CRT television deflects them with magnetic fields.

• ColourCRTTVsetshavethreeelectronsbeams— one for each of the coloured phosphors (red, green, blue) on the screen.

review questionsphysICally speakIngThe items in the columns are not in their correct order. Copy the table and match each of the definitions to the apparatus.

apparatus definition

Oscilloscope Shows that electrons travel in a straight line by the shadow cast on the tube

Discharge tube Uses magnetic fields to deflect electron beams

CRT TV An instrument used to measure a time-varying voltage

Maltese cross A glass vessel containing gas at low pressure that can be ionised by high voltage

revIewIng 1 State the origin of the name cathode rays.

2 State one inconsistency of the behaviour of cathode rays and how it was solved.

3 Explain the significance of Jean Perrin’s experiment.

4 Describe the different components that constitute a discharge tube.

5 Construct a table that names the features in a discharge tube, their characteristics and positions.

6 Briefly explain how a discharge is started and maintained in a discharge tube.

7 Describe how light is produced inside a discharge tube, in terms of the electrons in an atom.

8 Explain what would happen when a positive charge is placed in an electric field.

9 Estimate the difference in the force experienced by a charge placed in an electric field when the charge is tripled.

10 Describe how the force on a charge in a magnetic field varies as the angle of entry changes.

11 Compare and contrast the method of producing an image in a CRO and a CRT TV.

172

8 from cathode rays to television

solvIng probleMs 12 Calculate the electric field needed for an electron to

experience a force of 3.2 × 10–16 N.

13 Calculate the work done in order to move an object through 10 cm with a force of 3 N.

14 Consider Figure 8.6.1.

Figure 8.6.1 A positive charge between two plates

a Calculate the work needed to move a positive charge from the bottom plate to the top plate.

b Calculate the electric field between these two plates.

c Determine the direction of the force experienced by the charge.

d Calculate the magnitude of the force experienced if the charge is a proton.

15 As a charge moves through an electric field it gains kinetic energy. Obtain an expression for the kinetic energy gained in terms of the electric field strength E and the distance s travelled by the charge q.

16 Calculate the acceleration of the electron while in an electric field strength of 1000 V. The mass of an electron is 9.11 × 10–31 kg and its charge is 1.60 × 10–19 C.

17 Using the information from Question 16, calculate the final velocity of an electron initially at rest, when it had travelled a distance of 1.00 cm.

18 Calculate the force on an electron that enters a magnetic field of 0.1 T at a speed of 3.2 × 106 m s–1.

19 a Calculate the radius of curvature of a proton that enters a magnetic field 0.1 T at 2.1 × 106 m s–1.

b Determine the magnitude and direction of the force that is experienced by the proton.

20 Using the value that Thomson determined for charge to mass ratio 1.76 × 1011 C kg–1 and the electron charge of 1.6 × 10–19 C, determine the mass of an electron.

10 mm

100 V

0 V

+

CRT SED

deflectingyoke

electrongun

phosphorphosphorelectron

emitter

spacer

Revie

w Questions

173


physICs FoCuswhere to FroM here?

Science has come a long way since the first experiments with discharge tubes. The applications

range from a CRO in scientific fields to TVs for the general public. But development has continued.

TVs have changed; more and more households now have either an LCD or plasma screen TV. There are also surface-conduction electron-emitter display TVs (SED-TVs), which are still at the planning stage. These use the traditional cathode ray tube method but instead of having one electron gun for each colour (red, blue, green), these sets have a surface-conducting electron emitter (SCE) for each coloured phosphor dot on the screen—there may be up to a million of them!

Each SCE is made of two carbon layers with a gap between them; one with a negative electrode, the other with a positive electrode. A voltage of only 10 volts is needed to make an electron appear at one side of the gap. These cross the vacuum to strike the different phosphor dots lining the glass, resulting in a glow.

The big advantage with this system is that the matrix that controls the SCEs allows each pixel to be activated simultaneously rather than in the traditional method of one row at a time.

But development is not only on the entertainment front. Lighting has benefited from discharge tube technology. Commercial fluorescent lights have been around since 1938. They consist of a vacuum tube containing mercury vapour, filled with an inert gas such as argon and coated with phosphor powder.

Neon lights are similar but are filled with inert gases such as neon, argon and krypton. These glow with different gas-dependent colours, when an electrical current is passed through them.

The most recent and economical is the compact fluorescent light. Set to eventually replace all incandescent lights in Australia, these lights are basically a twisted fluorescent tube.

1 List the parts that are needed to make a discharge tube.

2 Outline the necessity for having low pressure inside the tube.

3 Compare and contrast the Geissler tube and the fluorescent light.

4 Determine why fluorescent lights are more efficient than incandescent lights.

5 List advantages and disadvantages of SED TVs. Suggest why a design that seems to encompass all the features wanted by consumers is not yet being made commercially.

6 Explain the need for magnetic coils in CRT TV sets.7 Give reasons why magnetic coils are not needed in

the SED TV.

H3. Assesses the impact of particular advances in physics on the development of technologies

H4. Assesses the impacts of applications of physics on society and the environment


luminescence

glass substrate

glass substrate

phosphor

electron beamselectrode

Vf Va

Va

electron emittermetal back film

colour filterblackmatrix

nanogap several nm

scatteringfieldemission

Figure 8.6.2 A diagram of a surface-conduction electron-emitter display TV

9

174

Electromagnetic radiation: particles or waves?

electromagnetic wave, transmitter, receiver, standing waves, nodes,

anti-nodes, refraction, polarisation, hertz, black body, classical theory,

ultraviolet catastrophe, quanta, photons, spectrometer, Planck’s constant,

photoelectric effect, photocathode, photoelectrons, cut-off frequency, work

function, stopping potential, photocell, dynodes, photomultiplier tube,

positron emission tomography (PET), gamma photons

The wave nature of lightThere was considerable debate about whether light was a wave or a stream of particles, long before there was a similar debate about cathode rays. However, by the 19th century there was growing experimental evidence that light had wave properties, but the nature of these waves was not known.

9.1 Hertz’s experiments on radio waves

A suggestion about the nature of light waves came from the Scottish physicist James Clerk Maxwell (1831–1879). From his studies of Michael Faraday’s experiments with electricity and magnetism, he derived four fundamental equations that linked electricity and magnetism. Remarkably, these equations predicted that oscillating electric charges should produce a wave that travels through space at the speed of light. This wave consisted of oscillating electric and magnetic fields at right angles to each other, and was called an electromagnetic wave.

Although Maxwell’s equations did not conclusively prove that light was an electromagnetic wave, he strongly suspected that it was, because the predicted speed of these electromagnetic waves was the same as the speed of light. Moreover, his equations predicted that all electromagnetic waves travelled at the speed of light, regardless of their frequency. Sadly, he died at age 48 before his theory could be tested experimentally.

In 1887 Heinrich Hertz (1857–1894) was the first to experimentally verify Maxwell’s electromagnetic wave theory. Hertz’s apparatus was essentially a radio wave transmitter and receiver. Today, almost all of us carry out a more

Activity 9.1



175

fRom IdEAS ToImPLEmENTATIoN

sophisticated version of Hertz’s experiment when we tune in to a radio station. More importantly, we are able to select different radio stations by their transmitting frequency. Hertz needed to make an antenna that transmitted at a specific frequency and a receiver that was tuned to that frequency. This was not such an easy task when electronics were not available.

Hertz used a number of different transmitters and receivers during the course of his experiments. Each successive apparatus was refined to give more accurate results. The basic principle of Hertz’s apparatus is shown in Figure 9.1.1. The transmitter consisted of a pair of metal rods placed end to end with a small gap between them. He used an induction coil to place charges of opposite signs on these rods at very large potential difference, causing a spark to jump across the gap. This caused electrical current to oscillate back and forth across the gap and along the rods, thus producing an electromagnetic wave. The frequency of the oscillation was determined by the dimensions of the rods. His first apparatus produced a frequency of 50 million cycles per second, but he had no way of measuring that. The mathematics for calculating the frequency was known during Hertz’s time and so he was able to calculate this oscillation frequency.

The electromagnetic wave emitted by the transmitter was detected by one of several receivers. The receiver shown in Figure 9.1.1 is a loop of wire with a gap.

The natural oscillating frequency of this loop had to match the frequency of the transmitter. The dimensions of the loop and gap determined this frequency and were accurately calculated by Hertz. The receiver showed that a spark jumped across the gap even if it was placed many metres away from the transmitter, thus indicating that an electromagnetic wave has been transmitted in air.

Hertz measures the speed of radio wavesHertz was able to obtain a fairly accurate measurement of the speed of his radio waves by a number of methods. His most famous paper for measuring their speed in air relied on the phenomenon of standing waves. Recall that two waves of the same frequency, wavelength and amplitude can form standing waves if they overlap while travelling in opposite directions (see in2 Physics @ Preliminary section 7.4). Figure 9.1.2 shows a standing wave produced by a rope tied to a wall. The nodes are points of zero amplitude and no motion (thus the term standing wave). The points of maximum amplitude are known as anti-nodes.

Outline qualitatively Hertz’s experiments in measuring the speed of radio waves and how they relate to light waves.

plate

rod

spark gap

a

b

metal loop

to induction coil

Figure 9.1.1 General schematic representation of all of Hertz’s apparatuses, which consisted of (a) a transmitting dipole antenna and (b) a receiver of various designs

TRY THIS!Any experiment in A thunderstormYou can reproduce Hertz’s experiment at home. The next time there is a thunderstorm, turn on your radio to the AM band (FM does not work for this experiment). Turn the tuning dial to a frequency at which no radio station is transmitting and listen. You will hear crackles and hisses of thunder near and far. Incredibly, you can hear more crackling than the number of lightning flashes that you see, because not all lightning flashes are large enough to be visible. This is the essence of Hertz’s experiment—electromagnetic waves produced by the motion of charge in lightning travel to your radio receiver at the speed of light and are detected.

Figure 9.1.2 A standing wave on a rope (A = anti-node, N = node)

A A AN N N

Hertz formed a standing wave of electromagnetic radiation by reflecting the radio waves from a large flat zinc plate, as shown in Figure 9.1.6, in which only a standing wave of the electric field is shown. Hertz then moved his receiver coil along this wave. A spark in the gap was produced at the anti-nodes but not at the nodes. The distance travelled between nodes or anti-nodes is half a wavelength. Doubling this

Electromagnetic radiation: particles or waves?9

176

PHYSICS FEATURE

THE EvolUTIon oF HERTz’S ExPERImEnTHertz’s first apparatus produced an electromagnetic wave with a wavelength of 6 m. There are difficulties with precision experiments when using such a long wavelength. You need to be far enough away from the transmitter so that you are not simply detecting a spark in your receiving antenna due to purely electrostatic effects as a result of being too close to the high-voltage induction coil. Moreover, the spark in the detecting antenna becomes very weak if the distance between it and the transmitter is too great. You have to contend with reflections from the walls of the room and the objects within it. You must also face the problem of tuning the transmitter and receiver to the same frequency when there is no instrument that can help you. Heinrich Hertz overcame all of these issues.

Figure 9.1.4 shows the evolution of the different types of transmitters and receivers used by Hertz. The first transmitter consisted of zinc spheres at the ends of the two rods (see Figure 9.1.4a). The antenna could be tuned by sliding the spheres along the rods. The receiver was a square loop of wire with a gap. The second type of antenna, was tuned by changing the area of the flat plates at the end of the rods (see Figure 9.1.4b). The receiver was a circular loop with a gap. The first two antennas produced a wavelength of 6 m. The next generation of transmitter and detector (Figure 9.1.4c) was a smaller antenna and a similar receiver (no loop) for a wavelength of 0.66 m.

Because these antennas were smaller, they were placed in parabolic reflectors (Figure 9.1.5) to direct and detect the beam without much loss. This configuration produced the most accurate results.

Figure 9.1.3 Heinrich Hertz was a skilled experimental physicist.

λ

6 m

6 m

0.66 m

transmitters

a

b

c

receivers

Figure 9.1.4 The different types of antennas and receivers used by Hertz include (a) a dipole transmitting antenna with spheres and a square loop detector for 6 m wavelength, (b) a transmitting antenna with square plates and a circular loop receiver for 6 m wavelength and (c) a small dipole antenna and similar dipole receiver for 0.66 m wavelength.

to induction coil

transmittingantenna

receivingantenna

parabolicreflector

Figure 9.1.5 Hertz’s transmitting and receiving antennas used parabolic reflectors to direct the electromagnetic wave without much loss of energy with distance.

177


distance gave the wavelength λ. Hertz was able to calculate the frequency f of the transmitted radio waves. The speed of these waves c was then determined using the following well-known wave speed formula:

c = f λ Hertz used a different frequency and found that the

speed of the waves remained the same. Although this didn’t prove that the speed of these radio waves was the speed of light, it was strong supporting evidence of Maxwell’s theory that the speed of all electromagnetic radiation was the same. Moreover, the speed of these waves was exactly the measured speed of light.

Hertz also showed that the path of these waves could be bent in the same way as light (refraction) by passing them through a large asphalt prism. (See in2 Physics @ Preliminary section 8.3.)

Hertz showed that the electric and magnetic fields of the radio waves had a unique direction in space, known as polarisation. The electric field in Figure 9.1.6 points vertically. When the detecting loop was at an anti-node and the direction of the gap aligned with the electric field (vertically), then a spark jumped across the gap. When the loop was rotated so that the gap was at right angles to the electric field, there was no spark.

It took great skill for Hertz to show that polarised electromagnetic waves exist and have a finite speed equivalent to the speed of light. Moreover, these waves shared other properties with light such as reflection and refraction. This set the scene for the emergence of radio communications and hence the modern field of telecommunications—mobile phone technology is a sophisticated version of Hertz’s experiment. Hertz did all of this before his life was cut short at the age of 36. One of the greatest honours that can be bestowed upon a scientist is to name a unit of measurement after them. The international unit of frequency is no longer called cycles per second—it is known as the hertz.

HEINRICH HERTz: ‘ooPS’

Hertz initially published his estimate of the speed of electromagnetic waves as

2 × 108 m s–1. This relied on the accuracy of his calculation of the oscillation frequency of the transmitter. It was an embarrassing moment when Poincaré, a great French mathematician, wrote him a letter pointing out that he had not included a factor of the square root of 2 in his frequency calculation. As a result, his actual measured speed was 2.8 × 108 m s–1, which is only 7% off the actual value of 3 × 108 m s–1. So don’t feel so bad next time you make a numerical mistake—even great scientists do it.

2 m1 m 3 mDistance from the wall

4 m 5 m 6 m 7 m 8 m

Figure 9.1.6 Standing electromagnetic waves produced by reflection from a large zinc plate enabled Hertz to measure the speed of light.

CHECkPoInT 9.11 Describe Maxwell’s contribution to understanding the connection between electricity and magnetism.2 Describe an electromagnetic wave.3 Recall the two predictions that Maxwell made about electromagnetic waves.4 Describe how Hertz produced and controlled the frequency of electromagnetic waves.5 Outline the requirements needed of the receiver in order for it to detect any given frequency.6 Recall the definition of a standing wave.7 Sketch a diagram of a standing wave in a rope, labelling a node and anti-node.8 Outline how Hertz manipulated the equipment in order to determine the wavelength of the electromagnetic wave

produced.9 Define polarisation and explain how Hertz showed that electromagnetic waves were polarised.


178

9.2 Black body radiation and Planck’s hypothesis

The success of Maxwell’s theory of electromagnetic waves led other physicists to apply it to the long-standing problem of radiation from hot objects. As an object such as a piece of metal is heated, its temperature rises and it emits colours from a dull red to orange and then to yellow, eventually becoming a bluish-white (if the object has not melted). This is shown in Figure 9.2.1 for metal parts at increasing temperatures.

The brightness or intensity (power radiated per unit area) of radiation depends on both wavelength and temperature, as shown by the plot in Figure 9.2.2. The maximum intensity of the curve shifts to smaller wavelengths as the temperature is increased. This simply means that as an object becomes hotter, its colour changes from red towards blue. The challenge for physicists was to mathematically predict the exact form of this curve using electromagnetic wave theory.

An object that absorbs all the wavelengths of the spectrum is referred to as a black body absorber. In a strange twist of terminology, an object that emits all the wavelengths of the spectrum is called a black body emitter. The curves shown in Figure 9.2.2 are known as black body radiation curves. The Sun is close to being a black body emitter and so its temperature can be determined from this plot of intensity versus wavelength.

An object that is painted black is not usually an ideal black body, as some radiation is always reflected from its surface. An ideal black body is generally pictured as a cavity with a small hole (see Figure 9.2.3). Any radiation that enters the hole becomes trapped inside the cavity and undergoes many reflections before eventually being absorbed. Similarly, a heated cavity with a small hole acts as a black body emitter. A small fraction of the radiation from within the cavity exits through the hole, so the intensity of this radiation can be measured (see Figure 9.2.2).

A number of physicists attempted to calculate the mathematical relationship for the black body radiation curve on the basis of electromagnetic wave theory and thermodynamics (the physics of the movement of heat). We will call this

Figure 9.2.1 Increasing temperatures cause these metals to radiate different colours from dull red to yellow.

Figure 9.2.2 The intensity of light from a hot object is dependent on wavelength.

10

8

6

4

2

00 1.0 2.0 3.0

Inte

nsit

y I

(arb

.uni

ts)

ultraviolet visible infra-red

Wavelength λ (μm)

λmax T = 6000 K

5000 K

4000 K

3000 K

λmax

TRY THIS!sun power Use a magnifying glass to focus the Sun’s rays to a point on a wad of tissue paper—it should catch fire. Now colour another tissue paper with black ink, say from a felt-tip pen or by dipping it in black ink. Let it dry. Focus the Sun’s rays onto this blackened tissue. You should find that the paper catches fire much more quickly. This is because black absorbs more of the incoming radiation and therefore heats up more quickly.

Activity 9.2



179


Figure 9.2.3 A spherical cavity with a hole acts as a black body absorber of radiation.

the classical theory approach. They assumed that the walls of the cavity were made from tiny oscillators that emit electromagnetic waves—just as Hertz had assumed for his transmitting antenna. Although this theory was able to reproduce the shape of the graph for large wavelengths (see Figure 9.2.4), it failed badly at the shorter wavelengths. Here the calculated intensity increased towards infinity, which violated the law of conservation of energy. This was called the ultraviolet catastrophe, because this started to occur at the ultraviolet end of the radiation spectrum.

Max Planck (1858–1947) solved this problem in 1900 and was able to mathematically reproduce the black body radiation curve by making the following radical assumptions:• TheemittersinthewallsofthecavitycanonlyhaveenergiesE given by

E = nhf where f is the emitted frequency in hertz, h is a constant (now called the

Planck constant = 6.63 × 10–34 J s) and n is an integer. The energy E is measured in joules.

• Theemitterscanabsorborradiateenergyin‘jumps’,orquanta. Two consecutive energy states of an emitter differ by hf.

These assumptions were radical ones since classical physics said that objects can have any energy. Max Planck’s theory defied this assumption. Incredibly, Planck himself resisted this theory wholeheartedly for many years and thought of it as nothing more than a mathematical convenience. However, this idea of the quantised nature of energy heralded the start of quantum physics, which, together with relativity, became the foundations of the modern physics we use today.

Einstein’s contribution to quantum theory: the photon Planck originally restricted his concept of energy quantisation to emitters in the walls of a black body cavity. He still believed that the electromagnetic energy radiated as a wave. However, in 1905 Einstein called into question this view of electromagnetic radiation and of light in particular. He proposed that radiant energy is made of concentrated bundles of energy that later came to be called photons. That is, Einstein had proposed that light is made of particles and is not a wave, as shown in Figure 9.2.5.

Einstein assumed that such an energy bundle is initially localised in a small volume of space, and that it remains localised as it moves away from the source with velocity c (the speed of light). Ironically, we still express the speed of photons in terms of the wave properties of frequency f and wavelength λ, as follows:

c = f λ

Identify Planck’s hypothesis that radiation emitted and absorbed by the walls of a black body cavity is quantised.

Figure 9.2.4 The classical theory of black body radiation could only explain the long wavelengths, Planck’s theory could explain all wavelengths.

Rad

ianc

e

classical theory

Wavelength (μm)

experiment andPlanck theory

0 1 2 3 4

Identify Einstein’s contribution to quantum theory and its relation to black body radiation.

Explain the particle model of light in terms of photons with particular energy and frequency.

Identify the relationships between photon energy, frequency, speed of light and wavelength: E = hf

and c = f λ

Figure 9.2.5 Light can be shown as (a) an electromagnetic wave and(b) a collection of particles called photons.

photon withenergy hf

x

x

y

y

z

z

c

c

E

B

ba


180

PHYSICS FEATUREmAx AvoIdS A CATASTRoPHE

Detailed understanding of Max Planck’s mathematical manipulations is beyond the scope

of this book. Even Planck, at first, looked on his solution as nothing more than a mathematical act of desperation. But we can gain some qualitative insights into his theory by adopting a view that is slightly different from that of his original work.

Imagine that the electromagnetic wave emitters in the walls of the cavity produce standing waves, as shown in the simplified one-dimensional representation in Figure 9.2.7 in which the walls on either side of the waves represent the walls of the cavity. The intensity at a particular wavelength of the radiation from the cavity is related to the energy in the wave at that particular wavelength.

Both the classical theory and Planck’s theory assumed that the energy in the wave is equal to the energy from the emitter. However, according to classical theory, the average energy for each emitter in the walls is kT, where k is a known constant (the Boltzmann constant) and T is the temperature of the walls in kelvin. Therefore, the energy per standing wave is also kT. An understanding of the way intensity is measured is important to understanding the difference between the ultraviolet catastrophe and Planck’s radiation law.

An instrument known as a spectrometer is used to measure the intensity of the black body radiation. All spectrometers take in a small wavelength range simultaneously—they never measure just a single wavelength. For example, a typically good spectrometer cannot distinguish between wavelengths of 6000.0 × 10–10 m and 6000.1 × 10–10 m. Note from Figure 9.2.7 that the difference in wavelength between two consecutive standing waves becomes progressively smaller with increasing frequency. This implies that at higher frequencies more wavelengths can enter the spectrometer. Since each wavelength contributes an equal amount of energy, and more of them are entering the spectrometer with increasing frequency, this will register as an increasing intensity since the total energy entering the spectrometer increases. As the wavelength becomes shorter, and

therefore the frequency increases, the intensity will increase towards infinity. This is the origin of the ultraviolet catastrophe, and which, of course, does not happen.

Planck restricted the energy of the emitters in the cavity to integer multiples of hf, which is the same as only allowing integer multiples of a certain frequency f. This restricted the number of standing waves that can occur. He also used a well-established law of thermal physics that states that a particle is more likely to have a lower than a higher energy. That means there is a dramatic decrease in the number of emitters with higher energies. As the frequency becomes very high, and therefore the wavelength very short, these two restrictions ensure that the number of emitters with these higher frequencies becomes very small and approaches zero. This registers as a diminishing intensity in the spectrometer and, thus, reproduces the actual black body radiation curve.

Figure 9.2.6 Max Planck’s radical hypothesis avoided the ultraviolet catastrophe and started a new field of physics called quantum physics.

Figure 9.2.7 A simplified representation of the standing waves inside a metal cavity

181


Einstein assumed that the energy E of the photon is related to its frequency f by the equation:

E = hf

where h is Planck’s constant = 6.63 × 10–34 J s. In classical theory, the intensity was determined by the electric field of the electromagnetic wave. In the new quantum theory, it is the number of photons that determines the intensity.

He then used this idea to explain peculiar properties of metals when they are irradiated with visible and ultraviolet light, known as the photoelectric effect.

Worked exampleqUESTIonA laser pointer emits light with a wavelength of 6.50 × 10–7 m. The power of the laser is 1.00 × 10–3 W.

Assume that Planck’s constant h = 6.63 × 10–34 J s, and the speed of light c = 3.00 × 108 m s–1.

a Calculate the energy in each photon of the laser beam.

b Calculate the number of photons emitted each second.

SolUTIona The energy E in a photon is given by

E = hf

where h is Planck’s constant and f is the frequency of the light. We can calculate the frequency from the following:

c = f λ

Therefore:

E hc= = × × ×

×= ×−

−λ6 63 10

3 00 10

6 50 103 06 1034

8

9–17.

.

.. J

The energy per photon is 3.06 × 10–17 J.

b The laser has a power of 1.00 × 10–3 W. This means it emits 1.00 × 10–3 joules per second. The number of photons per second is given by:

number of photons per second

photons per second

powerenergy per photon

= = ××

= ×

−

−1 00 10

3 06 10

3.27 10

3

17

13

.

.

Solve problems and analyse information using: E = hf and c = f λ

CHECkPoInT 9.21 Describe what happens to the wavelength of light emitted as the temperature of an object is increased.2 Define a black body.3 Outline how classical theory described black body radiation.4 Explain the ultraviolet catastrophe.5 Outline the assumptions Planck used to solve the problem.6 Explain Einstein’s revolutionary thoughts on light.


182

Figure 9.3.1 Electrons ejected from the cathode register a current on the ammeter, when the cathode is struck by ultraviolet light.

Figure 9.3.2 The energy of photoelectrons increases with increasing frequency. No electrons are ejected below the cut-off frequency f0.

evacuated quartz tube

electrons

V

A

– +

light

R

Max

imum

kin

etic

ene

rgy

(eV)

3

2

1

0

–W

0.5 1.0 1.5

Frequency ( 1015 Hz)

f0

f

pota

ssiu

mso

dium

zinc

tung

sten

plat

inum

9.3 The photoelectric effect In his experiments on radio waves, Hertz noticed that the sparks were easier

to produce in the gap of the detector loop whenever it was directly exposed to the ultraviolet light from the spark of the transmitting antenna. Placing an ordinary glass plate between the transmitter and the detector reduced the brightness of the spark, because ordinary glass blocks ultraviolet light. Replacing the glass plate with a quartz glass plate caused the spark to be more easily produced, because quartz glass allows ultraviolet light to be transmitted. He concluded that ultraviolet light has an effect on the receiving loop, but he was not in a position to explain it. However, he realised that it was an extremely important phenomenon and stopped his wave research in order to study it. Hertz had discovered the photoelectric effect.

The photoelectric effect is the ejection of electrons from the surface of a polished metal when light is shone on it. The metal on either side of the gap of Hertz’s detector loop emitted electrons when struck by ultraviolet light, which assisted in producing the spark. Ironically, the results of this experiment were used by Einstein to prove the particle nature of light, in contrast to Hertz who wanted to prove its wave nature.

The photoelectric effect was studied by other physicists who used a cathode ray tube, as shown in Figure 9.3.1. The tube either had a quartz window or was made of quartz, which allowed ultraviolet light to pass through and strike the cathode, known as a photocathode. This caused electrons, known as photoelectrons, to leave the cathode and be collected by the anode, thus registering a current on the ammeter.

The photoelectric effect was considered to be strange because:• Noelectronsareejectedfromthemetalbelowacertainfrequencynomatter

how intense the light. This is called the cut-off frequency f0. This is in contrast to classical theory, which says that all electromagnetic waves have energy and, if you wait long enough, all electrons can gain energy and leave the surface.

• Thekineticenergyoftheelectronsincreasesasthefrequencyoftheincidentlight increases (that is, going from red to blue to ultraviolet and beyond) as shown in Figure 9.3.2. However, there is no change in the electron energy if the frequency of the incident light is constant but the intensity is increased. Increasing the intensity simply increases the number of electrons but their energy remains the same. From classical wave theory, an increase in intensity should result in an increase in the energy of the electrons.

• Thereisnodelaybetweenthetimethelightisshoneonthesurfaceandthetime the electrons are emitted, no matter how dim the light source. From classical theory, electrons would require a length of time to gain enough energy from a low-intensity light source in order to escape the surface of the metal.

Einstein explained the photoelectric effect by assuming that an electron is ejected when it absorbs a photon that has energy E = hf. Although electrons involved in electrical conduction in a metal are free to move around, they are still bound to the metal as a whole. Energy is needed to separate them from the metal. This energy is called the work function W. The value of W depends on the type of metal. The photoelectric effect is mainly a surface phenomenon, so the surface must be free of oxide films, grease or other surface contaminants. Einstein then simply conserved energy by stating that:

Describe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but failed to investigate.

183


Photon energy = energy of ejected electron + energy needed by the electron to leave the metal.

This is encapsulated in the equation:hf = Kmax + W

where Kmax is the maximum kinetic energy of the ejected electrons. Rearranging this equation, we get:

Kmax = hf – W

From this we can see that the kinetic energy is zero (that is, no ejected electron) when hf = W. This accounts for the cut-off frequency. In addition, the kinetic energy increases as the frequency increases. Increasing the light intensity merely increases the number of photoelectrons and therefore increases the current emerging from the surface, but does not change the kinetic energy of the electrons. Einstein’s theory also accounts for the absence of a time delay, because all of the energy is supplied in a concentrated bundle; it is not spread uniformly over a large area,asinwavetheory.In1921EinsteinwontheNobelPrizeinPhysicsforhisexplanation of the photoelectric effect.

The graph in Figure 9.3.3 shows the dependence of the photoelectron current on the applied potential difference between photocathode and anode. There are two interesting features of this graph. The first is that there is a current even when there is no potential difference. At this point, all the energy to produce the current comes from the incoming photons. The second, and more interesting, feature is that the current is stopped by a certain negative potential difference placed on the photocathode.Notsurprisingly,thispotentialiscalledthestopping potential Vstop and is used to determine the maximum kinetic energy of the electrons. The ammeter will read zero current when eVstop equals maximum electron kinetic energy.

Worked exampleqUESTIonA zinc photocathode is irradiated with ultraviolet light. A voltage of –2.0 V is required to stop the photoelectrons.

a Calculate the maximum kinetic energy of the photoelectrons. Use the electron charge e = –1.6 × 10–19 C.

b Calculate the wavelength of the ultraviolet light striking the photocathode, given the work function of zinc W = 6.88 × 10–19 J and Planck’s constant h = 6.63 × 10–34 J s.

SolUTIona The maximum photocathode kinetic energy Kmax is given by:

Kmax = eVstop = 1.6 × 10–19 × 2.0 = 3.2 × 10–19 J

b Use Einstein’s photoelectric effect formula: Kmax = hf – W

where Kmax is obtained from part a, h is Planck’s constant and the zinc work function W = 6.88 × 10–19 J. In order to calculate the wavelength we first require the frequency, which is obtained by manipulating Einstein’s formula:

fK W

h.=

+= × + ×

×= ×

− −

−max . .

.

3 2 10 6 88 10

6 63 101 52

19 19

3411015 Hz

We can now use the relationship between frequency and wavelength:

λ = = ××

= × −cf

3 00 10

1 52 101 97 10

8

157.

.. m

Figure 9.3.3 Current of photoelectrons versus applied potential difference between cathode and anode

–Vstop 0 Potential difference

Cur

rent

Activity 9.3



Activity 9.4



Interactive

Module


184

Figure 9.4.1 The photomultiplier tube uses the photoelectric effect and can multiply the number of electrons produced by millions of times.

anode

photomultiplier outputs an electrical impulse to electronic circuits

incominglight photon

photo cathode electron

dynode

CHECkPoInT 9.31 Define the photoelectric effect.2 Outline Hertz’s observations regarding the photoelectric effect.3 Describe the significance of the cut-off frequency.4 Complete the table below by listing the observation of the photoelectric effect that classical theory predicted

incorrectly.

observAtion ClAssiCAl prediCtionElectrons are not emitted from the surface of a metal no matter how intense the light if the frequency of the light is below the cut-off frequency.

5 Explain how the work function helps explain the measured energy of photoelectron.6 Outline the significance of the stopping voltage.

9.4 Applications of the photoelectric effect

The photoelectric effect presents a convenient way of producing an electrical signal from light. The apparatus used to produce the traditional photoelectric effect (a vacuum tube with a photocathode and anode), called a photocell (photoelectric cell), was not of practical use outside the classroom for demonstrating the photoelectric effect. The current that was produced was far too small for most applications.

A modification to this tube was made by placing a series of specially coated surfaces, called dynodes, between the photocathode and the anode (see Figure 9.4.1). This is known as a photomultiplier tube. The dynodes are at successively increasing positive voltage with respect to the cathode, so they attract the electrons emitted by the photocathode. Each incident electron causes many other electrons to be emitted from the dynode surface. This effect is multiplied at the next dynode, and so on until a large number of electrons reach the anode and register a relatively large current. This makes the photomultiplier tube extremely sensitive to low levels of light, such that it is possible to detect individual photons. The material from which the photocathode is made has been improved and it can eject electrons from incident light ranging from the ultraviolet to the infra-red range of the electromagnetic spectrum.

This high sensitivity of the photomultiplier tube makes it suitable for the detection of small changes in light intensity. This has opened up many applications in astronomy, nuclear physics, blood testing and medical imaging. One method of medical imaging is called positron emission tomography (PET). One use of this method is the imaging of tumour cells within the body. The patient is injected with a short-lived radioactive chemical that contains molecules that attach themselves to particular tissue types such as tumour cells. This chemical emits two very high energy photons (known as gamma photons) that exit the body in opposite directions and strike a ring of photomultipliers around the patient (see Figure 9.4.2). Each photomultiplier has a material in

Identify data sources, gather, process and present information to summarise the use of the photoelectric effect in photocells.


185

front of it, known as a scintillator, that gives out flashes of light when the gamma photons strike it. The flashes of light are converted to electrical signals by the photomultiplier tubes. Computers calculate the time of arrival of the individual gamma photons and are able to locate their point of origin in the body. In this way, an image of the tumour cells can be generated from the different pairs of gamma photons arriving from different locations.

Another use of the photoelectric effect is in night-vision devices used by the military to see in almost total darkness by using the few photons coming from stars.

If we broaden our definition of the photoelectric effect to mean any process by which light is converted to electricity, without the need to eject an electron from a surface, then we can include many devices made from semiconductor materials. Although the electron stays within the solid, it becomes free from the atom to which it was bound, when struck by a photon. However, one important use of this type of photoelectric effect is the conversion of sunlight into electrical energy in solar cells. More detail will be given on this topic in Chapter 10 ‘Semiconductorsandtheelectronicrevolution’.

Figure 9.4.2 (a) A positron emission tomography scanner uses photomultiplier tubes to obtain (b) an image of tumour cells.

detectors

gamma rayscreated

positron-electroncollision

CHECkPoInT 9.41 Explain why the classic photocell is not used in commercial

applications.2 Outline how a photomultiplier tube works.3 Give examples of applications in which the photoelectric effect

is used.

a

b

PRACTICAL EXPERIENCESCHAPTER 9This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTIvITY 9.1: RECEIvIng RAdIo wAvESProduce sparks with the induction coil. Tune the portable radio to a position on the AM band at which there is no radio station. Move around the classroom and listen to the clicking sounds on the radio that are in step with the sparking from the induction coil.Equipment: induction coil, power supply, radio.

Discussion questions1 Explain how your experiment differs from the one carried out by Hertz.2 Describe what you notice as you move further away from the coil.

Perform an investigation to demonstrate the production and reception of radio waves.


PRACTICAL EXPERIENCES9 Electromagnetic radiation: particles or waves?

ACTIvITY 9.2: BlACk BodY RAdIATIonGather information to assess Einstein’s contribution to quantum theory and black body radiation. Present your information in the form of a newspaper article relating what you have found to people without scientific backgrounds.

Discussion questions1 Define the UV catastrophe.2 Outline how Planck explained the difference between observations and theory.3 State Einstein’s contribution to black body radiation.

ACTIvITY 9.3: PHoToCEllSPart A: Gather information about how the photoelectric effect is used in photocells. Part B: Set up an electrical circuit to investigate the energy that a photon of light gives to an electron. You can carry out the experiment first hand or use a computer simulation.

Discussion questions1 List devices that can be classified as photocells and explain how their

working principle is related to the photoelectric effect.2 Use the graph of electron energy (y-axis) and photon frequency (x-axis)

to calculate the Planck constant.

ACTIvITY 9.4: EInSTEIn vERSUS PlAnCkGather information about the differing views held by Planck and Einstein and use it to write an informed paragraph on your views.

Discussion questions1 Outline Einstein’s and Planck’s view of scientific research.2 List political pressures on Planck and Einstein during the times of their

main discoveries, and assess the influence of these pressures on Planck’s and Einstein’s research.

Identify data sources, gather, process and analyse information and use available evidence to assess Einstein’s contribution to quantum theory and its relation to black body radiation.

Identify data sources, gather, process and present information to summarise the use of the photoelectric effect in photocells.

Solve problems and analyse information using: E = hf and c = f λ

Process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces.

186

Chapter summary• Maxwell’stheorypredictedthatallelectromagnetic

radiation travels at the same speed.• Maxwellstronglysuggestedthatlightisan

electromagnetic wave.• Hertzperformedanumberofexperimentsto

demonstrate the propagation of electromagnetic waves.• Hertzalsomeasuredthespeedofelectromagneticwaves

by causing them to become standing waves, then measuring the wavelength, calculating the frequency and using c = f λ.

• Ablackbodyisanobjectthatiscapableofabsorbingoremitting all wavelengths of the electromagnetic spectrum.

• Classicalphysicscouldnotreproducethemeasuredcurve of intensity versus wavelength of radiation emitted from a black body.

• Theultravioletcatastropheisapredictionbytheclassical electromagnetic wave theory that the radiation intensity from a black body becomes infinite at very short wavelengths, which is false.

• MaxPlanckwasabletopredicttheblackbodyradiationcurve by assuming that the energy of the oscillators in the black body walls is quantised in integer steps of hf, where h is Planck’s constant and f is the frequency.

• Planckthoughtthatthequantisationofenergywasonlya mathematical trick and not related to reality.

187


Review questions

• Einsteinrealisedthatthisquantisationwasrealandthatlight is made of particles, each with a bundle of energy given by hf.

• Theparticleoflightiscalledaphoton.• Thephotoelectriceffectistheemissionofelectrons

from the surface of a metal that has been struck by light.

• Einsteinexplainedthephotoelectriceffectintermsofthe photon model of light.

• Thephotoelectriceffecthasapplicationsinallformsoflight detection, medical applications and the conversion of light into electrical energy.

PHYSICAllY SPEAkIngCreate a visual summary of the concepts in this chapter by constructing a mind map incorporating the following terms and equation:

hertz, radio waves, quanta, photon, Maxwell, photoelectric effect, black body, UV catastrophe, photomultiplier, work function, Einstein, kinetic energy, threshold, frequency, polarisation, E = hf

REvIEwIng 1 State the significance of Maxwell’s equations.

2 Outline the purpose of Hertz’s experiments.

3 Outline the procedure followed by Hertz to determine the speed of radio waves.

4 Describe how a standing wave is set up.

5 Explain how Hertz used standing waves to determine the wavelengths of radio waves.

6 Outline how Hertz’s results support Maxwell’s predictions.

7 Describe the results that Hertz obtained to support the idea of polarisation.

8 Recall the relationship between wavelength of radiation and its intensity.

9 Outline how classical theory predicted the ultraviolet catastrophe.

10 Explain the idea of quanta.

11 Explain how Einstein expanded on Planck’s idea of quantisation of oscillators in a black body cavity.

12 Discuss Einstein’s contribution to the understanding of the photoelectric effect.

13 Justify the use of quartz glass for the photoelectric effect experiment.

14 Relate the frequency of light on a photocathode to the kinetic energy of the emerging photoelectrons.

15 Discuss the significance of the work function in understanding the behaviour of photoelectrons.

16 What determines the maximum kinetic energy of a photoelectron?

17 What is a photomultiplier tube and how does it work?

18 Describe how the photoelectric effect can be used in PET scans to detect tumours.

SolvIng PRoBlEmS 19 Calculate the energy of a photon of red light with

a wavelength of 656 nm.

20 Figure 9.5.1 is a graph of voltage versus frequency of light.a Convert the voltage scale to energy of

photoelectrons.b Determine the cut-off frequency.c Use information from the graph to determine the

work function of the material.d Explain what will happen to the graph if the

intensity of light is increased.e Explain changes in the graph if the cathode was

made of a different material.f Calculate the Planck constant from the graph.

Revie

w Questions

V0 (V)

f (1015 Hz)0.25 0.50 0.75 1.0

3

2

1

0

–1

Figure 9.5.1 Frequency versus voltage of light

10

188

Semiconductors and the electronic revolution

valence electron, valence level, conduction level, valence, conduction

bands, energy gap, band gap, forbidden energy gap, solid state physics,

electron volt (eV), hole, doping, n-type semiconductor, donor, p-type

semiconductor, acceptor, donor energy level, acceptor energy level, extrinsic,

intrinsic, microprocessor, diode, diffusion, depletion region, forward

bias, reverse bias, photovoltaic cells, thermionic devices, transistors, plate, triode, electronics, bipolar transistor,

emitter, collector, base, field-effect transistors (FET), source, drain, gate,

MOSFET, integrated circuits

Electronics transforms the worldElectronics has transformed almost every aspects of modern life, from television, computers and mobile phones to medical diagnostics and treatment. The discovery of cathode rays led to devices such as radios, televisions and computers. The earliest computers, which used cathode ray tube devices, weighed several tonnes and filled very large rooms. Today, a basic mobile phone has much more computing power than the most powerful early computer. This reduction in size (and power consumption) came with the discovery of semiconductors. The working parts of semiconductor devices are so small that a microscope is needed to see their structure. An example of semiconductor circuitry is the microprocessor, the brains of your computer. It is no larger than a coin but contains millions of circuits.

Figure 10.0.1 An old valve computer weighed several tonnes but had much less computing power than your mobile phone.

189


10.1 Conduction and energy bands Electrical conduction occurs as a result of the flow of charge. Gases, which

are normally electrical insulators, are made to conduct large currents during lightning strikes, due to the movement of ions and electrons created by the ionisation of air atoms. Ionisation is the removal of the outer electron of an atom, which is known as a valence electron. The energy required to remove this electron is usually supplied by the electric field during lightning. In general, we are not interested in the exact motion of the charges in a current, we just want to know whether there will be electrical conduction. A simple and abstract way of visualising whether there will be conduction in gas atoms is shown in Figure 10.1.1. The vertical axis is energy and the horizontal lines indicate the position an electron can have. The lower line (or level), known as the valence level, is the energy it has while being bound to the atom. The upper level, known as the conduction level, is the energy it must acquire to become free and contribute to electrical conduction.

Placing two individual atoms together will result in two pairs of these levels (see Figure 10.1.2a). Note that the levels are not at exactly the same energy but become shifted slightly vertically with respect to each other. This odd behaviour can be explained by the laws of quantum physics, which are beyond the scope of this section. Basically, energy levels of electrons in close proximity to each other are forbidden from being the same, so placing five atoms close together results in five pairs of these lines, as shown in Figure 10.1.2b.

To form a solid we must place many atoms, say 1023, close together, and so we have 1023 pairs of energy levels (see Figure 10.1.3). It is impossible to draw these individual levels because they are very close to each other and too numerous. We simply represent them by two shaded areas called energy bands. The lower and upper bands are labelled as valence and conduction bands respectively. The energy gap between them is referred to as the band gap or forbidden energy gap.

Electrical conduction in a solid occurs because some electrons have gained enough energy to be in the conduction band and therefore become free electrons. They are no longer localised to a particular ion but are shared by all ions in the lattice, since they are free to move throughout the solid. Unlike conduction in a gas, the ions in solids are not free to move, so the only charge carriers are the electrons.

Quantum physics determines that no electron can have an energy in the energy gap. Moreover, only a fixed number of electrons are allowed per energy level in the energy bands. When an energy level is occupied by the maximum number of electrons allowed by quantum physics, we say that the ‘level is filled’. Any further electrons will start to occupy the level above it until this new level is filled, and so on until a whole energy band is filled.

Ordinary experience tells you that not all solids conduct electricity. All metals are good conductors; however, materials such as glass and plastic are insulators. There is also a third type of solid known as a semiconductor, such as silicon and germanium, which conduct electricity but much less than metals. The energy band diagram, shown in Figure 10.1.4, can be used to illustrate the difference in the conductivities of these three types of materials.

Identify that some electrons in solids are shared between atoms and move freely.

Describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance.

Figure 10.1.1 An abstract energy level diagram helps visualise the ionisation energy of an atom, which would produce free charges for an electrical current.

Figure 10.1.2 Energy level diagrams for (a) two atoms and (b) five atoms in close proximity

Figure 10.1.3 The energy band structure of solids

Ene

rgy

Ene

rgy

Ene

rgy

a

b

energy gap

conduction band

valence band

Semiconductors and the electronic revolution10

190

The type of atom, the bonding between atoms and the number of valence and inner electrons of an atom are some factors that determine the type of energy band diagram for the resulting solid. This is the area of physics known as solid state physics, which relies heavily on quantum physics. Figure 10.1.4a shows an energy band diagram for conductors. There is no clear energy gap and therefore no distinct conduction or valence band. We can say that the conduction band is partially filled and therefore the solid contains free electrons.

The energy band diagrams for a semiconductor and an insulator are similar (Figure 10.1.4b, c). The much smaller energy gap of the semiconductor distinguishes it from an insulator. The valence band of an insulator is full and, due to the relatively large size of its energy gap, none of the electrons in the valence band have enough energy to reach the conduction band. However, some electrons in the valence band of the semiconductor can acquire enough thermal energy to jump across the small gap into the conduction band. The number of conduction electrons that result is much less than for a conductor, so a semiconductor is a much poorer conductor than a metal but a better conductor than an insulator such as plastic or glass.

Compare qualitatively the relative number of free electrons that can drift from atom to atom in conductors, semiconductors and insulators.

ChECkpOinT 10.11 Describe how gases, which usually have insulating properties, can be made to conduct electricity.2 Describe why the energy levels of two neighbouring atoms are not exactly the same.3 State why only electrons are charge carriers in solids.4 Define the terms electrical conductor and insulator.5 Describe how it is possible for electrons in the valence band of semiconductors to reach the conduction band.

Figure 10.1.4 Energy band diagrams for (a) a conductor, (b) a semiconductor and (c) an insulator

energy gapoverlap

a b cconduction band

valence band

10.2 Semiconductors The size of the energy gap determines whether a solid is a semiconductor

or an insulator. A convenient unit for measuring the energy gap is the electron volt (eV). This is a unit of energy and is related to the joule:

1 eV = 1.6 × 10–19 J

Most semiconductors have an energy gap less than 5 eV. Insulators have greater energy gaps. Examples of energy gaps for semiconductors and insulators are given in Table 10.2.1. The most common semiconductors are silicon (Si) and germanium (Ge).

191


Some electrons in the valence band of a semiconductor will gain enough energy from the ambient thermal (heat) energy in the solid to jump up to the conduction band. This occurs readily at normal room temperature. Heating a semiconductor further increases the number of electrons jumping from the valence band to the conduction band, thereby making the solid more conducting. For example, heating a piece of silicon will lower its resistance. At absolute zero (T = 0 K), there is no thermal energy available for the electrons, all of which remain in the valence band; therefore, at absolute zero the semiconductor behaves as an insulator.

Figure 10.2.1 Holes are created in the valence band of a semiconductor when electrons are excited up to the conduction band.

applied electric field

electrons holes

conductionband

narrowenergy gap

valenceband

v

v

v

v

Table 10.2.1 Energy gaps in electron volts of common semiconductors at room temperature

Semiconductor cryStal energy gap (eV)Si 1.14

Ge 0.67

InP 1.35

GaP 2.26

GaAs 1.43

CdS 2.42

CdTe 1.45

ZnO 3.2

ZnS 3.6

Diamond 5.4

trY tHiS!Semiconducting pencilGraphite is a form of carbon that is used in ‘lead’ pencils (there is no lead in pencils, just graphite and clay, but we will refer to it as the lead of the pencil). Graphite also behaves as a semiconductor. Obtain a length of pencil lead (about 10 cm) and attach each end to the probes of a digital ohm meter. It should measure units of ohms. Now heat it (with, say, a match) and observe that the resistance decreases. The resistance may decrease by up to 0.5 Ω and it will be more for thinner leads.

This decrease in resistance is due to an increase in the number of electrons jumping from the valence band to the conduction band. The resistance will increase once more when the lead cools.

Electrons and holesWhen an electron is excited into the conduction band, it leaves a vacancy in the valence band. This vacancy is known as a hole (see Figure 10.2.1). The hole behaves like a positively charged particle and moves in the opposite direction to the electron. In reality, the other electrons in the valence band move to fill the vacancy, but in doing so they leave behind another vacancy, resulting in the apparent motion of the hole. It is much simpler to treat the hole as a positively charged particle than to track the motion of all the electrons that move in the opposite direction. The magnitude of the charge on an electron and a hole are the same at 1.6 × 10–19 C, but they are opposite in sign. Therefore, the motion of both electrons and holes contribute to the motion of electrical current.

DopingAt normal room temperature (300 K) the average kinetic energy of an electron is about 0.026 eV. However, the energy gap of silicon is 1.14 eV. It would appear that the electrons don’t have enough energy to jump to the conduction band, so why is silicon, or any of the other semiconductors listed in Table 10.2.1, slightly conducting at room temperature? The answer is that 0.026 eV is an average energy. This means there are some electrons that have a much lower energy and

Identify absences of electrons in a nearly full band as holes, and recognise that both electrons and holes help to carry current.


192

some that have a much higher energy. It is these higher energy electrons that make it across the gap, but there are not many of them.

Physicists have found that the conductivity of semiconductors can be improved by introducing impurities into the crystal lattice, a process that is called doping. In order to understand how an impurity atom can improve the conductivity, we must examine what happens at the crystal lattice level.

We will use silicon as an example, but similar explanations can be applied to all other semiconductors. Silicon belongs to group 4 of the periodic table, which means that the silicon atom has 4 electrons available for bonding with other silicon atoms (see Figure 10.2.2a). Silicon can be doped with phosphorus P by replacing a silicon atom with a phosphorus atom. Phosphorus is in group 5 of the periodic table and has 5 electrons available for bonding, but it can only bond with 4 atoms in the silicon crystal lattice. This leaves one electron unbonded (see Figure 10.2.2b). This semiconductor is called an n-type semiconductor. (n represents the negative charge of the unbonded electron.) Impurities that produce unbonded electrons are called donor impurities. These unbonded electrons are weakly attached to their atoms and can easily be moved into the conduction band by ambient heat, and thus can contribute to an electrical current.

Similarly, silicon can be doped with boron, which belongs to group 3 and therefore has 3 electrons available for bonding with 3 atoms of silicon. There is no electron available for the fourth silicon atom to bond with and, as a result, a vacancy in charge has been created, which we call a hole. An electron from a neighbouring silicon atom can now jump across to fill this vacancy (see Figure 10.2.2c). This, in turn, creates a vacancy for the atom it left behind, and so can be viewed as the movement of the hole. As already mentioned, a hole behaves like a positive charge since it moves in the opposite direction to the electron. The semiconductor behaves as though it has positive charge carriers and is therefore called a p-type semiconductor. Impurities that produce a charge vacancy are called acceptor impurities.

Note that both n- and p-type semiconductors remain electrically neutral even though there is an unbonded electron or a vacancy is created. The neutral impurity atoms were added to an already neutral solid, so the net charge remains zero.

Both n- and p-type doping improve the conductivity of a semiconductor by creating an excess of negative and positive charge carriers respectively. The unbonded electron in n-type semiconductors needs a small amount of energy to move to a different location. Similarly, very little energy is required to cause a neighbouring electron to occupy the vacancy of a p-type semiconductor.

The energy-band diagram of an n-type semiconductor is shown in Figure 10.2.3. A new energy level has been created in the energy gap just below the conduction band. This level is occupied by the unbonded electrons of the impurities and is called the donor energy level. This level is very close to the conduction band. The difference in energy between the two levels is usually 0.026 eV or less, and so most of the electrons in the donor level are able to jump up to the conduction band at room temperature, making the semiconductor much more conducting than the undoped crystal.

a Hole in a bottle

A way of picturing the concept of a hole in a semiconductor

is to consider a bottle that is almost full of water with a small air bubble underneath its closed cap. Turning the bottle upside down will cause the bubble to move upwards. The question is: ‘did the bubble move up or did the water move down?’ The answer is that both are true, but it is much simpler to follow the motion of the bubble than the more complex motion of the water. In the same way, we describe the motion of the hole in a semiconductor rather than the more complex motion of all the valence electrons.

Figure 10.2.2 The silicon lattice (a) has no doping, (b) has been doped with phosphorus and (c) has been doped with boron.

a

b

c

+4 +4 +4

+4 +4 +4

+4 +4 +4

+4 +4 +4

+4 +3 +4

+4

electron hole

+4 +4

+4 +4 +4

+4 +5 +4

+4 +4 +4

193


The energy-band diagram for a p-type semiconductor in Figure 10.2.4 shows that there is an extra energy level, called an acceptor energy level, near the valence band. This level is full of vacancies (holes), allowing the valence electrons to jump up and occupy these vacancies and therefore contribute to an electrical current. The gap between the acceptor level and the donor (valence) level must be equal to or less than 0.026 eV for conduction to take place at room temperature.

A semiconductor is labelled as extrinsic if conduction is dominated by donor or acceptor impurities. Otherwise, it is known as an intrinsic semiconductor.

Identify differences in p- and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes.

Describe how ‘doping’ a semiconductor can change its electrical properties.

Figure 10.2.3 Energy-band diagram of an n-type semiconductor

donor impurity energy level

conduction band

valence band

Figure 10.2.4 Energy-band diagram of a p-type semiconductor

acceptor impurityenergy level

holes

conduction band

valence band

ChECkpOinT 10.21 Distinguish between a semiconductor and an insulator in terms of the size of the energy gap.2 Describe what is meant by a hole.3 Describe doping in semiconductors.4 Explain the difference between n-type and p-type semiconductors.5 Identify elements that can be used to make n-type and p-type semiconductors.6 Distinguish between acceptor and donor energy levels.

10.3 Semiconductor devicesCombinations of p- and n-type semiconductors can be used to make all modern electronic devices, ranging from the simple remote control to the sophisticated computer microprocessor. The basis of this technology is a result of the properties of the junction between p- and n-type materials.

The p–n junctionA p–n junction is a p-type material joined onto an n-type material. In reality, the p–n junction is made from a continuous single crystal in which the concentration of impurities has been made to change abruptly from p-type to n-type as we cross the junction. However, to clarify the physics of the junction region, we will assume that the p-type and n-type crystals are initially separated and then brought together.

The p–n junction is used as the most basic electronic device, the diode. It allows current to move in only one direction, as shown in Figure 10.3.1.

activity 10.1

praCtiCal eXperienCeS



194

The conventional current easily flows from p- to the n-type material when the positive terminal of a power supply is connected to the p-type end of the device. Reversing the connections of the power supply (positive connected to n-type and negative to p-type) results in a very tiny reverse current, the current is essentially stopped. The symbol for the diode is an arrow that indicates the direction of the conventional current when the positive and negative terminals of the power supply are connected to the p- and n-type materials respectively (see Figure 10.3.1b).

A key to understanding the physics of a p–n junction is to first examine the concept of diffusion. Free particles are in continuous random motion and move from areas of high to areas of low concentration. For example, a drop of ink placed in a beaker of water results in the ink particles slowly spreading outwards so that the water becomes uniform in colour. This occurs because the point of origin of the free particles is very small (drop of ink) compared with the rest of the medium (beaker of water). So it is more likely that the particles will move to the rest of the medium. This is diffusion and it is a property of all randomly moving particles. The energy for the movement of these particles (kinetic energy) comes from the ambient thermal energy. At a temperature of absolute zero there will not be any diffusion.

Similarly, the holes in the p-type material and the electrons in the n-type material are in constant random motion due to their thermal energies, resulting in a uniform density of the charge carriers across their respective materials. When the two crystals are brought into contact, electrons from the n-type crystal will naturally want to diffuse into the p-type crystal, because the electron density of its conduction band is lower. Conversely, the holes in the p-type region want to diffuse into the n-type crystal, as shown in Figure 10.3.2. This results in an electrical current, which is quickly stopped because an electric field builds up from the n to the p region, due to this charge separation, and stops the further flow of charge. A large part of this diffusion region results in the combination of electrons with holes, leaving it depleted of charge. This is known as the depletion region. However, there continues to be an excess of electrons and holes on the p and n sides respectively.

A depletion region is like two parallel plates with equal and opposite charges on them, resulting in an electric field in the space between them, which contains no charge.

The energy that created the electric field essentially came from the thermal energy of the charge carriers. At a temperature of absolute zero (0 K), there will not be enough energy for diffusion to occur and therefore no electric field and no depletion region will be created.

Electrons that diffused across and combined with holes cannot easily drift back since they lost energy as they fell into the holes. Similarly, holes that diffused combine with electrons from the n-type material. The trapping of the diffused holes and electrons has resulted in the conversion of thermal energy into the electrostatic energy stored in the electric field region.

As shown in Figure 10.3.3a, connecting the positive and negative terminals of a battery to the p and n sides of the junction respectively creates an electric field opposing that in the depletion region. This lowers the junction electric field strength and therefore allows further charge to diffuse across the junction. A thin p–n junction will enable the excess diffused charge to reach the

Figure 10.3.2 The electric field in the depletion region of the p–n region stops further diffusion of charge.

depletionlayer

freeholes

freeelectrons

fixednegative

ions

fixedpositive

ions

p-type n-type

EE

electricfield

Figure 10.3.1 (a) A forward-biased diode and (b) its symbol. (c) The graph shows the current versus voltage characteristics of a diode.

V

V

p n

p n

A

+

I

I

forwardbias

forwardbias

reverse bias

O

a

b

c

195


metal terminals on either side of the junction, thus resulting in the flow of current in the outside circuit. This type of connection to the p–n junction is known as forward bias. This also has the effect of reducing the width of the depletion region.

Conversely, connecting the negative and positive terminals of a battery to the p and n sides respectively, as shown in Figure 10.3.3b, increases the electric field strength in the junction region. This will cause some of the diffused electrons to travel back to the n region, setting up a small reverse current. This type of connection is known as reverse bias. It also has the effect of increasing the width of the depletion region.

RectifiersThe p–n junction or diode is used to convert AC to DC electricity in chargers or power supplies for electronic devices such as mobile phones and computers. A simple circuit, but not one that is used often, is shown in Figure 10.3.4. A sinusoidal alternating voltage, say from the AC mains supply, is applied to the p end of the diode. The output from the n end is the positive part of this signal. The negative part of the AC voltage is blocked. A component consisting of parallel plates (called a capacitor) at the output stores the positive charge and helps smooth the output signal so that a DC voltage will result.

Figure 10.3.3 (a) Forward-biased and (b) reverse biased p–n junctions

narrowdepletion layer

widedepletion layer

p-type n-type

p-type n-type

holeflow

temporaryhole flow

electronflow

temporaryelectron flow

electron flowelectron flow + –

+–

a

b

E

E

conductionband

valenceband

light emission

E = hƒ

E = hƒ

p-type n-type

hole flow electron flow

electronflow

electronflow

+ –

hole andelectron

recombine

Figure 10.3.4 A rectifier circuit can be used as a power supply.

Figure 10.3.5 A forward-biased light-emitting diode (LED)

C

V

tinput output

voltage smoothed

0

activity 10.2



Light-emitting diodes (LED) and laser diodesA current that passes through a forward-biased diode will lead to some recombination of electron and hole pairs. This recombination occurs when a conduction electron drops from the conduction to the valence band to occupy a vacancy (the hole), as shown in Figure 10.3.5. This change in energy can appear as heat given to the solid, or as the emission of light. The energy of the photon hf is equal to the energy lost by the electron, which is the energy gap Eg such that:

E hfhc

g = =λ

where Planck’s constant h = 6.63 × 10–34 J s, f is the frequency, c is the speed of light and λ is the wavelength. Manufacturers of LEDs can control the wavelength of the emitted light by controlling the size of the energy gap. Most remote-control units around the home have an LED on the end that you point towards a device. The LED looks like a very tiny plastic light blub, but works very differently. The wavelength of these LEDs is usually in the infra-red range of the spectrum and so is not visible to the eye. However, it is visible to most types of digital cameras such as those in mobile phones and video cameras.


196

phYSiCS FEATUREphOTOVOlTAiC CEllS—SOlAR CEllS

Figure 10.3.6 A solar cell is a p–n junction that absorb photons to create conduction electrons.

conductionband

valenceband

vlight absorption

absorbedphoton

n-type

hole flow electron flow

electronflow

electronflow

new hole andelectron created

loadresistor

p-type

Solar cells use the p–n junction to convert light directly into electricity. They have the more formal

name of photovoltaic cells (or PVs). Very little greenhouse gas emissions are associated with PVs, which makes them a strong candidate to replace coal-fired power stations for future power generation. Currently they have many applications that include the supply of electricity to remote locations or homes, and satellites.

In a PV, a valence band electron absorbs a photon and is excited up to the conduction band, thus contributing to an electrical current. In practice, PVs are simply p–n junctions. The photons absorbed in the p region produce conduction electrons near the depletion region. These electrons are swept to the n side of the junction by the electric field (see Figure 10.3.6). If the depletion layer is thin enough, then most of these electrons can reach the n region without recombining with a hole. The electrons in the n region have a much higher lifetime before recombining with a hole. Electrodes placed on the ends of the semiconductor collect this newly created current and deliver it to an external circuit. Electrons will travel around a circuit and back to the p region, where they will recombine with holes.

The actual structure of a solar cell is shown in Figure 10.3.7. The n-type

region is made thin enough so that most of the incoming light can be transmitted to the p-type region, where photons are absorbed to create the conduction electrons. These are swept up into the n region by the electric field at the p–n junction and are collected by the front contact metal grid, which can be connected to an external circuit. The size of the voltage created by solar cells depends on the potential difference across the depletion region.

The engineering of solar cells has advanced greatly in recent years so that greater efficiency can be obtained at lower cost. For example, an antireflective coating is placed on top so that very few photons are lost due to simple reflection. The size of the energy gap determines the wavelength range that will be absorbed to produce electrons—some parts of the sunlight spectrum deliver more energy than others. The maximum efficiency of 23% has been achieved in recent years but not on a commercial scale.

energyfrom Sun

solararrays

cover glass

n-type layer(semiconductor)

p-type layer(semiconductor)

junction

freed electrons holes filled by freed electrons

electronflow(current)

electricaltransmissionsystem

substrate

antireflectioncoating

transparentadhesive

frontcontact

current

backcontact

Figure 10.3.7 How a solar cell operates

activity 10.3



197


ChECkpOinT 10.31 Describe what is used to make a diode.2 Outline how a p–n junction is connected to a power supply in order for electricity to flow.3 Describe what happens to the distribution of electrons and holes when p- and n-type semiconductors are

brought into contact.4 Outline how a rectifier works.5 Explain why light is emitted in a LED.6 Describe the different parts of a solar cell.7 Explain why the n-type layer in a solar cell needs to be thin.

Figure 10.4.1 Thermionic devices (valves) were replaced by the much smaller transistors in modern-day electronics.

10.4 The control of electrical currentWe have seen that the diode (or rectifier) enables current to be conducted in one direction, for example, to convert AC to DC. Diodes are one of the most basic electronic components, but they have limited control over current. Remarkably, the first type of radio receiver used only a diode as a means of detecting a radio signal and obtaining the audible sound without the need for batteries or any other kind of power supply. This was called a crystal radio set and is still made today by enthusiasts to detect AM radio stations.

A great deal more control over current is required for the wide variety of electronic devices we use every day. For example, mobile phones have to transmit an electromagnetic wave at a precise frequency by making electrical current oscillate through an antenna. The phone also needs to detect a very weak electromagnetic wave, then amplify it (make it larger) and decipher the information contained within it such as voice, SMS, etc.

trY tHiS!Seeing infra-redGrab a remote control that you use for the TV set or DVD player. Look at the LED at the end of the remote control while pressing any button (for example, the volume or channel button). You should not see anything happening. Now point the remote control at your mobile phone camera or any other digital camera. While looking at the screen of the camera, press any button on the remote control. You should now see the LED flashing, because the camera is sensitive to the infra-red light from the LED but your eye is not.

Figure 10.3.8 The infra-red light from the LED of a remote control has been made visible by the digital camera that took this photograph.


198

A number of discoveries and inventions led to components that were used to accurately control the direction and magnitude of electrical current. The first of these were vacuum tubes known as thermionic devices or valves, which were later replaced by semiconductor technology such as diodes, transistors and integrated circuits.

Thermionic devicesDuring his research on incandescent lamps, Thomas Edison (1847–1931) discovered that a current can be made to flow between a heated filament and a metal electrode in a vacuum by applying a potential difference between them. The filament and electrode were the cathode and anode respectively. Heating the filament gives some electrons enough energy to escape the surface. These electrons are then accelerated to the anode.

John Ambrose Fleming (1849–1945) applied this effect to the construction of the first thermionic rectifier (diode) or valve, which allowed current to flow in only one direction. The name valve comes from an analogous mechanical device that allows fluids or gases to flow in one direction, such as the valve (or nozzle) on a bicycle or car tyre that enables air to be pumped in but does not let it out.

A thermionic diode consists of three components: the filament (or heater), the cathode and the anode (see Figure 10.4.2). The heater is used to heat the cathode, which releases electrons that are attracted to the anode (the plate). Lee De Forest (1873–1961) inserted a metal grid between the cathode and anode of a thermionic diode. The grid allows electrons to pass through (see Figure 10.4.3). This device is known as the triode. De Forest found that the electron current could be stopped by placing a negative voltage on the grid, or it could be allowed through by using a positive voltage. A voltage between these two extremes could be used to control the amount of current flowing through.

One application of the triode is the amplifier (see Figure 10.4.4). If the voltage on the grid is negative to the point that it only allows a very tiny current through, then a small AC signal applied to the grid will cause current flowing to the anode to vary in the same way. However, because of the large potential difference between anode and cathode, this current is larger than that in the original signal; that is, the triode has acted as a current amplifier. Moreover, this current could be passed through a resistor, resulting in a varying voltage across it. In this way, small voltage signals could be amplified. The invention of the triode was in part responsible for the revolution in electronics, such as in radio and television, in the early 20th century.

The transistorTransistors are semiconductor devices that control and amplify electrical current; they have largely replaced thermionic devices. Just as the p–n junction has replaced the thermionic diode, the transistor has replaced the thermionic triode. The transistor was invented in 1947 by John Bardeen and Walter Brattain, who with their boss William Shockley received the Nobel Prize for it in 1956.

In the early days of the transistor, germanium was used as the semiconductor because the methods existed for growing germanium crystals of high purity. Although silicon is more abundant and retains its semiconducting properties at higher temperatures, it needed a crystal of higher purity than could be achieved at the time. Germanium diodes and transistors were replaced with those made from silicon as soon as techniques for growing crystals improved and resulted in higher purity silicon.

Figure 10.4.2 A thermionic diode enables current to flow in one direction.

Figure 10.4.4 A small AC signal is amplified by a triode circuit.

Figure 10.4.3 A triode allows the electron current to be controlled electrically.

A

+

anode

heater

cathode

electrons

electrons

electrons

+150 V

0 V

output

anode

anode

cathode

cathode

electrons

grid

heater

heater (filament)

grid

199


The most common transistor is the bipolar (or junction) transistor. A schematic diagram of a bipolar transistor is shown in Figure 10.4.5. This transistor works in a similar way to the original transistor of Bardeen and Brattain, but its structure is very different.

The bipolar transistor (Figure 10.4.5) consists of a p-type semiconductor sandwiched between two n-type semiconductors. A lead (or wire) is connected to each layer of the material. The two outer layers are called the emitter and the collector; the central layer is called the base. This is often called an n–p–n transistor. The width of the base has been exaggerated for clarity, but is usually very thin.

An electron current will only flow from the emitter to the collector if there is a potential on the base with respect to the emitter (see Figure 10.4.6). Recall that in a p–n junction there is an electric field that points from n to p. Applying a positive voltage on the base of an n–p–n transistor with respect to the emitter reduces the size of this field to almost zero. This then allows electrons from the emitter to move into the base region. These electrons are then accelerated into the collector region by the electric field at the junction between base and collector. The electrons that moved into the collector can then flow into an external circuit.

Note that no current will flow without applying the potential to the base. The current to the base is much smaller than the current that it allows to flow from emitter to collector. Thus the transistor, like the triode, is a current amplifier. The circuit given in Figure 10.4.7 has a small AC voltage applied to the base of an n–p–n transistor, which causes a much larger current to flow into an external resistor between the emitter and the collector. The voltage across the resistor is larger than the input signal but varies in the same way. This circuit is a basic amplifier. Bipolar transistors also can be made in the p–n–p configuration.

Identify that the use of germanium in early transistors is related to lack of ability to produce other materials of suitable purity.

Figure 10.4.5 A schematic diagram of a bipolar transistor

collectorbase

emitter

collector

base

emitter

pn n

electrons

p

n

n

+ +

electrons

largecurrent

smallcurrent

electrons

Vbase Vcollector

emitter

base

collector

Figure 10.4.6 An electron current flows through an n–p–n transistor.

Figure 10.4.7 This small signal amplifier uses an n–p–n transistor.

loadresistorp

n

n

+ +

Vbase Vcollector

amplifiedoutputinput

A comparison of solid state and thermionic devicesSemiconductor devices such as diodes and transistors do the same job as their equivalent thermionic devices (the valve and the triode), but they have many advantages. The semiconductor devices are much smaller than their thermionic counterparts and use much less power. They are mechanically robust, whereas thermionic devices are made of glass and so are fragile, and semiconductors have longer operating lifetimes due to their lower temperature of operation.

Describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices.


200

phYSiCS FEATUREinTEgRATEd CiRCUiTS

Another type of transistor, called the field-effect transistors (FET), shown in Figure 10.4.8, consists

of two n regions embedded in a larger and lightly doped p substrate. The two n regions are called the source and the drain. An insulating layer made from an oxide of the semiconductor (such as silicon oxide) covers these, and a metal electrode (called the gate) is placed on top of this layer. Because the substrate is lightly doped, it is not very conducting, so that no current will flow naturally between the source and the drain. When a positive potential is applied to the gate, it causes electrons in the substrate to be attracted to the region between the two n regions and forms a conducting channel. In this way, current can be controlled by the voltage on the gate.

This metal–oxide–semiconductor field-effect transistor (MOSFET) is not as common as the bipolar transistor, but its construction makes it easy to place many MOSFETs on a single wafer of silicon. Combinations of MOSFETs, resistors and capacitors form very complicated circuits known as integrated circuits (IC). Integrated circuits containing hundreds of thousands of transistors may be as small as a few millimetres square. The IC is commonly known as the silicon chip, and is used in all modern-day electronics such as computers, televisions, mobile phones, etc.

There are many processes that go into producing an integrated circuit and they vary between different manufacturing plants and depend on the specifications of the required IC. The following description simply gives an indication of the many-step processes that go into making an IC.

An n-type layer is sandwiched between a lightly doped p-type silicon layer (the substrate) and a layer of silicon dioxide (SiO2) deposited on it. Photographic, chemical and ion-beam bombardment techniques are used to dissolve (etch) unwanted areas and deposit or dope other areas for the required patterns of transistors, resistors and capacitors.

Figure 10.4.8 A schematic diagram of a metal–oxide–semiconductor field-effect transistor (MOSFET)

few electronscan pass

substrate (p)

source (–) drain (+)

gate (small voltage)

nn

insulatinglayer

Figure 10.4.9 A wafer of p-type silicon with n-type and insulating layers is the starting point in the manufacture of an integrated circuit.

n-type

p

p

silicondioxidefilm

p-typesubstrate

ChECkpOinT 10.41 Describe the purposes of a diode and a triode.2 Referring to Figure 10.4.2, explain the need for the heater in a thermionic diode.3 Explain how a triode works.4 Outline how an amplifier works.5 Recall the reasons why germanium was originally used in semiconductor devices.6 Identify problems associated with the use of germanium in these devices.


201


CHapter 10This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTiViTY 10.1: ElECTROnS And hOlESUsing the computer simulation, investigate the behaviour of p-type and n-type semiconductors, and a p–n junction.

Discussion questions1 Describe the changes in the energy diagram when the semiconductors are

doped with p-type or n-type dopants.2 Use your observations of the simulator diode to explain what happens

inside a diode.

ACTiViTY 10.2: hiSTORY OF COMMUniCATiOnGather information about the history of the transistor and the development of telecommunications, and consider how the materials and technology available at the time influenced what could be developed.

Discussion questions1 Explain how the use of semiconductors changed telecommunications.2 Identify areas in which the introduction of the transistor has influenced

social behaviour.

ACTiViTY 10.3: SOlAR CEllSSolar cells are devices that can convert sunlight directly to electrical energy. Carry out an experiment to determine the effect of sunlight on a solar cell.Equipment: solar cell, fan or small motor, digital multimeter.

Discussion questions1 Determine the angle (with respect to the direction of the Sun) of the solar

cell that delivers the greatest amount of power to the motor, or the greatest voltage.

2 Determine the effect on power output of different wavelengths of light by placing, for example, coloured cellophane in front of the solar cell.

Perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor.

Gather, process and present secondary information to discuss how shortcomings in available communication technology lead to an increased knowledge of the properties of materials with particular reference to the invention of the transistor.

Identify data sources, gather, process, analyse information and use available evidence to assess the impact of the invention of transistors on society with particular reference to their use in microchips and microprocessors.

Identify data sources, gather, process and present information to summarise the effect of light on semiconductors in solar cells.

202

10 Semiconductors and the electronic revolution Chapter summary

review questions

• Theconductionofelectronscanbedescribedwithenergy band diagrams.

• Electronsinthevalencebandarenotfreetomoveanddo not contribute to conduction.

• Electronsintheconductionbandarefreetomoveandenable conduction to take place.

• Insulatorshavealargeenergygapbetweentheconduction and valence bands and not many valence electrons have enough energy to jump up to the conduction band.

• Metalshavenobandgapand,therefore,therearealwayselectrons in the conduction band.

• Theenergy-banddiagramofsemiconductorsissimilarto that of insulators but with a much smaller band gap.

• Semiconductorscanbemademoreconductingbyexciting electrons from the valence band to the conduction band by heat energy or a photon.

• Theconductivityofsemiconductorscanbeimproved by introducing impurity atoms into the crystal lattice.

• Asemiconductorwithimpuritiesthatintroduceextraelectrons into the lattice is known as an n-type semiconductor.

• Asemiconductorwithimpuritiesthatintroduceavacancyin the bonding is known as a p-type semiconductor.

• Theintroductionofimpuritiesintothesemiconductorcrystal lattice is known as doping.

• Theimpuritiesinsemiconductorsintroduceextraenergy levels in the energy gap region and so improve the conductivity of semiconductors.

• Semiconductordevicescanbemadebyusingtheproperties of the junction between p- and n-type semiconductors, and include the diode, the solar cell and the transistor.

• Transistorsreplacedthermionicvalvetriodesusedinthecontrol of electrical current.

• Transistorshavetheadvantageofbeingfaster,smallerand more energy efficient than thermionic valve triodes, and thousands can be produced on a small silicon wafer to make complicated integrated circuits.

phYSiCAllY SpEAkingCreate a visual summary of the concepts in this chapter by constructing a mind map incorporating the terms and equation in the following table.

p-type n-type Semiconductor

Energy bands Depletion zone Conductor

Valence band Forbidden gap Insulator

Thermionic devices Solid state Triode

Diode Transistor Solar cell

Holes Electrons Doping

REViEwing 1 Outline the difference between energy levels and

energy bands.

2 Describe the difference in terms of energy gaps between insulators, conductors and semiconductors.

3 Explain why it is wrong to say n-type semiconductors are negatively charged or that p-type semiconductors are positively charge.

4 Explain why the depletion zone occurs when p- and n-type semiconductors are placed in contact with each other.

5 Explain why the depletion zone will not form at a temperature of absolute zero.

6 Describe in terms of electric field strength why forward and reverse bias occurs.

7 Describe the path of an electron through a solar cell.

8 Give reasons why thermionic devices were not energy efficient.

9 Explain in terms of the electric field at a p–n junction how current can flow in only one direction.

203


10 Explain how a triode can be used to control the amount of current flowing in a circuit.

11 Compare and contrast a triode with a transistor.

12 Explain why silicon was not originally used in making semiconducting devices.

13 Explain the significance of integrated circuits on society.

SOlVing pROblEMS 14 A 1 milliwatt (10–3 W) red laser pointer outputs

a wavelength of 650 nm (nm = 1 × 10–9 m).a Calculate the energy of a photon emitted from

the laser.b Calculate the number of photons being emitted

every second. (Hint: Planck’s constant h = 6.63 × 10–34 J s.)

15 The electrical conductivity of undoped silicon can be increased by irradiating it with photons. This has the effect of exciting valence electrons into the conduction band. Given that the energy band gap of silicon is 1.14 eV, calculate the longest wavelength of a photon that can excite a valence electron to the conduction band. (Hint: Planck’s constant h = 6.63 × 10–34 J s.)

16 A photon of wavelength 3.35 μm (1 μm = 1 × 10–6 m) has just enough energy to raise an electron from the valence band to the conduction band in a lead sulfide crystal. Calculate the energy gap between these bands in lead sulfide. (Hint: Planck’s constant h = 6.63 × 10–34 J s.)

17 Figure 10.5.1 shows a sinusoidal input voltage Vi as a function of time t for the given circuit. Draw a similar graph of the output voltage Vo as a function of time.

Figure 10.5.1

Vi

Vi VoR

+v

0t

Revie

w Questions

11

204

Superconductivity

crystal, constructive interference, destructive interference, path length,

diffraction grating, Bragg law, phonons, critical temperature,

type-I superconductors, type-II superconductors,

critical field strength, vortices, flux pinning, BCS theory, Cooper pair,

coherence length, energy gap, spin

Surprising discoveryJust as an improved understanding of the conducting properties of semiconductors led to the wide variety of electronic devices, research into the conductivity of metals produced quite a surprising discovery called superconductivity. This is the total disappearance of electrical resistance below a certain temperature, which has great potential applications ranging from energy transmission and storage to public transport. An understanding of this phenomenon required a detailed understanding of the crystal structure of conductors and the motion of electrons through them.

11.1 The crystal structure of matterA crystal is a three-dimensional regular arrangement of atoms. Figure 11.1.1 shows a sodium chloride crystal (ordinary salt also called rock salt when it comes as a large crystal). The crystal is made from simple cubes repeated many times, with sodium and chlorine atoms at the corners of the cubes. Crystals of other materials may have different regular arrangements of their atoms. There are 14 types of crystal arrangements that solids can have.

The regular arrangement of atoms in crystals was a hypothesis before Max Von Laue and his colleagues confirmed it by X-ray diffraction experiments. William and Lawrence Bragg took this method one step further by measuring the spacing between the atoms in the crystal. Let us first look at the phenomenon

Figure 11.1.1 Crystal structure of sodium chloride. The red spheres represent positive sodium ions, and the green spheres represent negative chlorine ions.

205


of interference of electromagnetic radiation, and examine how this was applied to crystals using X-rays. Then we will see how the BCS theory of superconductivity made use of the crystal structure of matter. try thiS!

Crystals in the kitChenLook at salt grains through a magnifying lens. Each grain is a single crystal that is made from the basic arrangement of sodium and chlorine atoms shown in Figure 11.1.1. Although the grains mostly look irregular due to breaking and chipping during the manufacturing process, occasionally you will see an untouched cubic or rectangular prism that reflects the underlying crystal lattice structure.

CheCkpoInT 11.1Explain what is meant by the crystal structure of matter.

11.2 Wave interferenceThe wave nature of light can be used to measure the size of very small spaces. Recall that two identical waves combine to produce a wave of greater amplitude when their crests overlap, as shown in Figure 11.2.1a (see in2 Physics @ Preliminary sections 6.4 and 7.4). The overlapping waves will cancel to produce a resulting wave of zero amplitude when the crest of one wave coincides with the trough of the other (Figure 11.2.1b). This addition and subtraction is called constructive and destructive interference respectively and is a property of all wave phenomena.

As an example, two identical circular water waves in a ripple tank overlap (see Figure 11.2.2). The regions of constructive and destructive interference radiate outwards along the lines as shown. Increasing the spacing between the sources causes the radiating lines to come closer together (Figure 11.2.2b).

Figure 11.2.1 Two identical waves (red, green) travelling in opposite directions can add (blue) (a) constructively or (b) destructively.

Figure 11.2.2 Interference of water waves for two sources that are (a) close together and (b) further apart

t = 0 s

t = 1 s

t = 3 s

t = 4 s

t = 5 s

t = 6 s

t = 7 s

t = 0 s

t = 1 s

t = 3 s

t = 4 s

t = 5 s

t = 6 s

t = 7 s

a

b

lines of constructiveinterference

lines of destructiveinterference

ba

The interference of identical waves from two sources can also be represented by outwardly radiating transverse waves (see Figure 11.2.3). The distance that a wave travels is known as its path length. Constructive interference occurs when the difference in the path length of the two waves is equal to 0, λ, 2λ, 3λ, 4λ or any other integer multiple of the wavelength λ. Destructive interference occurs when the two waves are half a wavelength out of step. This corresponds to a path length difference of λ/2, 3λ/2, 5λ/2 etc.




wavesin phase

Figure 11.2.3 Constructive and destructive interference between identical transverse waves from two sources

Superconductivity11

206




wavesin phase

wavesin phase

wavesin phase

Figure 11.2.4 An interference pattern is formed by light passing through two narrow slits.

screen

doubleslit

Figure 11.2.5 The path difference between waves that produce constructive interference

Figure 11.2.6 Interference from multiple slits in a diffraction grating

try thiS!DiffraCtion grating in my stereoA commonly available diffraction grating is the humble CD or DVD. It has thousands of tiny and closely spaced pits that cause ordinary white light to break up into the colours of the rainbow because there is an angle for constructive interference for each wavelength contained in white light. A more clearly defined diffraction pattern can be made by shining a low power (less than 1 mW) laser pointer on the CD or DVD so that the beam is reflected to a nearby wall. You should notice that there are fainter dots near the reflected dot. These fainter dots arise from constructive interference of the laser light.

Light also has wave properties and produces interference effects when it is passed through two narrow and closely spaced slits (see Figure 11.2.4). Each slit acts as a source of light waves that are in step with each other. The resulting constructive and destructive interference pattern appears as bright and dark bands on a screen.

The position of the bright bands can be determined by applying the condition that the path length difference between the two waves must be an integer multiple of wavelengths (see Figure 11.2.5). This can be expressed by the following equation:

mλ = d sin θ

where m = 1, 2, 3 …, d is the spacing between the slits and θ is the angular position of the bright bands. The central bright band corresponds to m = 0. The first band on either side of it corresponds to m = 1, and so on.

Interference also occurs when many slits are used (see Figure 11.2.6). The bright bands become narrower and sharper on the edges, with increasing number of slits. In practice, the slits are parallel straight lines scratched, etched or moulded onto a glass plate. This arrangement of multiple slits is known as a diffraction grating.

A very good grating for visible light can have 2400 lines per millimetre. Such a grating provides very sharply defined bands whose

S1

slit 1

d

y

L

S2

screen

slit 2


θ

θ

θ

d sin θ

207


Figure 11.3.1 X-ray picture of a hand taken by Röntgen

CheCkpoInT 11.21 Recall the conditions for constructive interference.2 Describe how these differ from those for destructive interference.3 Compare the interference pattern produced by light and that produced by two interfering water waves.

11.3 X-ray diffractionX-rays are high-frequency electromagnetic waves first discovered by Wilhelm Conrad Röntgen (1845–1923). He was experimenting with cathode ray tubes and noticed that a fluorescent screen, some distance away from the tube, began to glow each time the tube was operated. Röntgen had discovered a new type of radiation, which he called X-rays to indicate that their nature was unknown. Röntgen also found that X-rays had great penetrating power and could be used for medical applications. Figure 11.3.1 shows one of the first X-ray pictures of a hand taken by Röntgen.

Several years of experimentation by scientists after Röntgen’s discovery showed that X-rays were electromagnetic waves like light but with an extremely short wavelength. A schematic diagram of an X-ray tube is shown in Figure 11.3.2. A beam of electrons strikes a metal target. The rapid deceleration of electrons on striking the target causes X-rays to be emitted. The wavelength of the most intense X-rays depends on the material from which the target is made.

A single narrow slit can also be used to obtain a blurry diffraction pattern. Using this crude method, rough estimates of 10–10 m were obtained for the wavelength of X-rays. It wasn’t physically possible for a proper diffraction grating with a well-controlled slit spacing of 10–10 m to be made, so accurate estimates of X-ray wavelength were not possible. Max von Laue (1879–1960) learned of the hypothesis that crystals were regular arrangements of atoms with a spacing between atoms of about 10–10 m. He realised that crystals were the gratings he needed for X-ray diffraction.

In 1912 W. Friedrich and P. Knipping followed von Laue’s suggestions and passed a narrow beam of X-rays through a zinc sulfide crystal. A photographic plate placed behind the crystal produced a regular pattern of spots (Figure 11.3.3), which was the interference pattern of the crystal which acted as a diffraction grating.

position can be accurately determined. The angular positions of the bright bands are determined by the formula given above for the double slit. Knowledge of the wavelength can be used to determine the slit separation and vice versa.

To obtain an interference pattern, the distance between slits must be close to the wavelength of the light falling on them. The interference pattern from a diffraction grating is also called a diffraction pattern.

cathode (electronsource)

electronbeam

vacuum

heavymetaltarget

metal rod(removes heat andelectrons)

cooling fins

X-rays

anode

Figure 11.3.2 An X-ray tube

Superconductivity11

208

CheCkpoInT 11.31 Recall Röntgen’s observations that led to the discovery of X-rays.2 Identify why X-rays are emitted when electrons strike a metal

surface.3 Explain the significance of crystals in determining the

wavelength of X-rays.

Figure 11.4.1 The different planes of a cubic crystal Figure 11.4.2 Crystal lattices consist of many parallel planes in many directions that can reflect X-rays.

Figure 11.4.3 The parallel planes reflect X-rays to produce constructive interference according to the Bragg law.

incidentangle

reflectionangle

dsinθ

θθ

θ

λ

θ

d

11.4 Crystal structureThe Laue diffraction patterns qualitatively showed that a crystal indeed consisted of a regular array of atoms. The idea of using the diffraction pattern to measure the atomic spacing came from the Australian-born physicist Sir William Lawrence Bragg (1890–1971), who, working with his father William Henry Bragg (1862–1942), realised that crystals can be considered as consisting of many planes oriented along different directions (see Figure 11.4.1).

Bragg realised that X-rays could penetrate the crystal structure and be reflected from a set of parallel planes (see Figure 11.4.2). Constructive interference is produced for certain reflection angles. In 1912, Bragg showed that the wavelength λ of X-rays was related to the spacing d between planes by modifying the double slit constructive interference equation as follows:

nλ = 2d sin θwhere n is an integer and θ is the angle of incidence measured between the X-ray beam and the crystal plane (see Figure 11.4.3). This is the Bragg law that governs all modern X-ray diffraction.

This pioneering work has now become a standard method of determining the crystal structure of materials.

Outline the methods used by the Braggs to determine crystal structure.

X-rays

leadcollimator

crystal

photographicfilm

a

Figure 11.3.3 Laue’s suggestion for X-ray diffraction

209


Which came first: λ or d ?From measurements of the diffraction angle θ, the Braggs were able to determine the wavelength λ of X-rays. This needed an accurate knowledge of the spacing d between planes in the crystal. However, this spacing was not known accurately, since there was no method available to measure it. How can you measure λ without knowing d and vice versa?

The Braggs solved this dilemma by using the diffraction pattern to determine the arrangement of the atoms, without the need to know the space between them. For example, it is possible to know that a crystal has a cubic arrangement from a Laue diffraction pattern, without the need to know the spacing of atomic planes or the wavelength of X-rays. Knowledge of the lattice geometry, the density of the crystal (mass/volume) and the mass (in grams) of the individual atoms enables the spacing between atoms to be calculated accurately. This calculated spacing d can now be used with the X-ray diffraction angle θ, to determine the wavelength λ from the Bragg diffraction equation. Today, we simply use the database of known crystal lattice spacings to determine X-ray wavelengths.

11.5 electrical conductivity and the crystal structure of metals

X-ray diffraction has shown that the atoms of most metals exist in one of three types of crystal lattices (see Figure 11.5.1). The crystal structure of metals can be viewed as a lattice of positive ions surrounded by a ‘sea’ of nearly free electrons, which makes metals such good electrical conductors. The binding mechanism in metals is the attractive force between the positive ions and the electron gas.

Remarkably, quantum physics predicts that there should be little or no resistance to the motion of electrons in a perfect crystal lattice since electrons behave like waves propagating through it. That is, the perfect regularity of the crystal enables the electrons to travel unimpeded, as a wave through the crystal.

Identify that metals possess a crystal lattice structure.

Describe conduction in metals as a free movement of electrons unimpeded by the lattice.

Identify that resistance in metals is increased by the presence of impurities and scattering of electrons by lattice vibrations.

CheCkpoInT 11.41 Outline the Braggs’ contribution to the understanding of X-rays.2 State the Bragg law.3 Outline how the Braggs determined the spacing of the atoms on crystals.

Figure 11.5.1 The structure of most metallic crystals can be (a) body-centred cubic, (b) face-centred cubic, or (c) hexagonal close-packed.

a b c

Superconductivity11

210

Figure 11.5.2 Electrical resistance is caused by electron collisions due to crystal lattice (a) impurities, (b) defects and (c) vibrations.

a b c

Anything that disturbs the regularity of the lattice results in electron collisions and contributes to the electrical resistance of the crystal.

In practice, metals do show electrical resistance, as evidenced by the increase in temperature of the metal wire in an electric heater or the heating element on a stove. Resistance in metals originates from collisions of electrons with irregularities in the crystal lattice. These can be caused by lattice vibrations, or impurities (a foreign atom substituted for one in the crystal lattice) and defects of the lattice (such as a missing atom) (see Figure 11.5.2). The crystal lattice of all metals above a temperature of 0 K consists of waves of lattice vibrations known as phonons. A phonon colliding with an electron causes it to lose energy and thus contributes to electrical resistance.

Real metal wires consist of many small crystals joined together and separated by irregular boundaries. The boundaries also serve as places where electron collisions take place and thus contribute to electrical resistance.

the Kelvin temperature Scale

The kelvin temperature scale differs from the Celsius scale by the following expression:

kelvin = Celsius + 273.15

The lowest possible temperature is 0 kelvin (or simply 0 K), which is –273.15°C. The kelvin is named after the British physicist William Thomson (1824–1907), who later was given the title Lord Kelvin. In 1900 he is reported to have said, ‘there is nothing new to be discovered in physics now. All that remains is more and more precise measurement.’ This was before the discovery of superconductivity, relativity, quantum physics, and all of the modern physics that has led to a radical transformation of our society through technology and our understanding of nature and the universe. Even great scientists can be short sighted. Figure 11.5.3 William Thomson

(Lord Kelvin)

CheCkpoInT 11.51 Outline the significance of X-ray diffraction to the structure of metals.2 Explain the effect on electrical resistance of irregularities that are introduced into a crystal.3 Give reasons why resistance in metals does not match the near-zero resistance predicted for crystal structures.4 Outline the role of a phonon in electrical resistance.

211


11.6 The discovery of superconductors

The phenomenon of superconductivity, in which the electrical resistance of certain materials completely vanishes at low temperatures, is one of the most interesting and sophisticated in condensed matter physics. It was first discovered by the Dutch physicist Heike Kamerlingh Onnes (1853–1926), who was the first to liquefy helium (which boils at 4.2 K at standard pressure). In 1911 Kamerlingh Onnes discovered the phenomenon of superconductivity while studying the resistance of metals at low temperatures. He studied mercury because very pure samples could easily be prepared by distillation.

The historic measurement of superconductivity in mercury is shown in Figure 11.6.1. The electrical resistance of mercury decreased steadily when it was cooled, but dropped suddenly to zero at 4.2 K. Soon after this discovery, many other elemental metals were found to have zero resistance when their temperatures were lowered below a certain temperature that is characteristic of the material. This is called the critical temperature Tc, some of which are given in Table 11.6.1.

Table 11.6.1 Some superconductors, their critical temperatures and critical magnetic fields

Class material CritiCal temperature (kelvin)

CritiCal magnetiC fielD strength (tesla)

Type I

(elements)

Tungsten 0.02 0.0001

Titanium 0.4 0.0056

Aluminium 1.18 0.0105

Tin 3.72 0.0305

Mercury (α) 4.15 0.0411

Lead 7.19 0.0803

Type II

(compounds and alloys)

Nb–Ti alloy 10.2 12

Nb–Zr alloy 10.8 11

PbMo6S8 14.0 45

V3Ga 16.5 22

Nb2Sn 18.3 22

Nb3Al 18.9 32

Nb3Ge 23.0 30

Type II

(high-temperature ceramic compounds)

YBa2Cu3O7 92Too high to measure, typically ~200 (estimated)

Bi2Sr2Ca2Cu3O10 110

Tl2Ba2Ca2Cu3O10 125

HgBa2Ca2Cu3O8 135

Superconductivity was an unexpected phenomenon. As shown in Figure 11.6.2, it was expected that by cooling a conductor the lattice vibrations (phonons) will be gradually reduced in amplitude. The reduction in lattice vibrations also reduces the number of collisions of electron with the crystal lattice and therefore reduces the electrical resistance. One would expect the resistance to gradually decrease to very low values at a temperature of 0 K. This is why it was surprising to find that the resistance dropped to zero at a relatively high temperature.

Process information to identify some of the metals, metal alloys and compounds that have been identified as exhibiting the property of superconductivity and their critical temperatures.

Figure 11.6.1 The resistance of mercury as measured by Kamerlingh Onnes

Temperature (K)

Res

ista

nce

(Ω)

0.15

0.10

0.05

0.004.1 4.2 4.3 4.4

Hg

Figure 11.6.2 The resistance of a normal conductor and a superconductor

superconductor

normalmetal

Temperature (K)0 TC

Ele

ctri

cal r

esis

tanc

e

Superconductivity11

212

11.7 The Meissner effectIn 1933, Walter Meissner and Robert Oschenfeld discovered that superconductors expel magnetic fields from their interiors in a way that is different from the behaviour expected of hypothetical perfect conductors. Figure 11.7.1 illustrates a thought experiment that highlights this difference. Imagine that both the ideal conductor and superconductor are above their critical temperature Tc; that is, they both are in a normal conducting state and have electrical resistance. A magnetic field Ba, is then applied, which penetrates both materials. Each sample is then cooled to below its critical temperature, so that they both have zero resistance. It is found that the superconductor expels the magnetic field from inside it, while the ideal conductor maintains its interior field. Note that the energy needed by the superconductor to expel the magnetic field comes from the superconducting transition, which is exothermic. Switching off the magnetic field induces currents in the ideal conductor that prevent changes in the magnetic field inside it, as stated by Lenz’s law (Module 2 ‘Motors and generators’). However, the superconductor returns to its initial state; that is, it has no magnetic field inside or outside it.

Figure 11.7.1 A thought experiment that illustrates superconductors are not the same as perfect conductors

T

Tc

Ba

0

Idealconductor

Super-conductor

CheCkpoInT 11.61 Recall how superconductivity was discovered.2 Define critical temperature.3 Compare the expected measurements of resistance as temperature is reduced with the experimental results.

CheCkpoInT 11.7Explain what happens to a magnetic field passing through an ideal conductor and a superconductor when the conductors are cooled to below their critical temperatures.

11.8 Type-I and type-II superconductors High magnetic fields destroy superconductivity and restore the normal

conducting state. Depending on the character of this transition, we may distinguish between type-I and type-II superconductors. The graph in Figure 11.8.1 illustrates changes in the internal magnetic field strength Bi (the field inside the superconductor) with increasing applied magnetic field. It is found that the internal field is zero (as expected from the Meissner effect) until a critical magnetic field Bc is reached, at which a sudden transition to the normal state occurs. This results in the penetration of the applied field into the interior.

Superconductors that undergo this abrupt transition to the normal state

213


above a critical magnetic field strength are known as type-I superconductors. Most of the pure elements listed in Table 11.6.1 tend to be type-I superconductors.

Type-II superconductors, on the other hand, respond differently to an applied magnetic field (see Figure 11.8.2). These superconductors have two critical field strengths, Bc1 and Bc2. As field strength is increased from zero, there is no change in the internal magnetic field of the superconductor until Bc1 is reached. At this field strength, the applied field begins to partially penetrate the interior of the superconductor. However, the superconductivity is maintained at this point. Superconductivity vanishes above the second, much higher, critical field Bc2. For applied fields between Bc1 and Bc2, the applied field is able to partially penetrate the superconductor, so the Meissner effect is incomplete and the superconductor is able to tolerate very high magnetic fields.

Type-II superconductors are the most technologically useful because the second critical field can be quite high, enabling high field strength electromagnets to be made out of superconducting wire. Most compounds listed in Table 11.6.1 are type-II superconductors. Wires made from, for example, niobium–tin (Nb3Sn) have a Bc2 as high as 24.5 tesla, though in practice it is lower.

There is a misconception among some non-specialists that the term type II refers to the copper oxide based high-temperature superconductors discovered in the late 1980s. Although these are type-II superconductors, so are many superconductors discovered before that time.

Figure 11.8.1 Type-I superconductors abruptly become normal conductors at field strengths above a critical magnetic field.

Inte

rnal

fie

ld B

i

External field Ba

Bc

Figure 11.8.2 Type-II superconductors have a partial penetration of the magnetic field between two critical fields.

Figure 11.9.1 A permanent magnet levitates above a superconductor due to the Meissner effect.

Inte

rnal

fie

ld B

i

External field Ba

Bc1 Bc2

CheCkpoInT 11.81 Describe the significance of internal magnetic fields and critical

magnetic fields.2 Distinguish between type-I and type-II superconductors.

11.9 Why is a levitated magnet stable?Figure 11.9.1 shows a spectacular demonstration of the Meissner effect in which a small permanent magnet floats on top of a high critical temperature superconductor (YBa2Cu3O7) cooled with liquid nitrogen (at 77 K). It demonstrates the repulsion of the magnetic field by the superconductor and thus the levitation of the magnet.

Eddy currents are created on the surface of the superconductor, and, consistent with Lenz’s law, these essentially produce a magnetic field that mirrors the field of the magnet, resulting in the repulsion and subsequent levitation of the magnet (see Figure 11.9.2). In reality, when the magnet is first placed over a small piece of superconductor it is unstable and falls off to the side. This is because the magnet will float over its ‘mirror image’ provided that image can keep moving with it. The superconductor is small and cannot produce a satisfactory magnetic field image near its edge, which results in ineffective repulsion.

So why does a levitating permanent magnet remain stable on top of a small superconductor? Even a little nudge causes the magnet to spring back to its original position as if somehow tied by invisible springs to that point. To explain

Superconductivity11

214

this, we need to expose a little secret used when demonstrating this levitation experiment. If the magnet is lightly placed over a newly cooled high-temperature superconductor, you should find that the magnet does not stay levitated for long and falls off very quickly. To get the magnet to stay, hold the magnet over the superconductor and, rather than letting it go, thrust it slightly towards the superconductor. This is a subtle movement and usually goes unnoticed by the audience. Now release the magnet and it will remain there stably. Incredibly, if the magnet is then removed and dropped back over the superconductor, it levitates stably without the need to thrust the magnet towards the superconductors. It is as if the superconductor has ‘remembered’ that the magnet was there. Moving the magnet back and forth parallel to the surface of the superconductor or allowing the superconductor to warm up above Tc and then cooling it down again will make the levitation of the magnet unstable once more. The magnet must again be thrust towards the superconductor to achieve stability. This is explained in the following section.

Vortex states and flux pinningStable levitation of a permanent magnet above a small flat superconductor only occurs with type-II superconductors. Certainly levitation occurs when using type-I superconductors but with a type-II superconductor the levitation is particularly stable and robust. The answer lies in the properties of type-II superconductors for an applied magnetic field between the two critical fields Bc1 and Bc2. Recall that for type-II superconductors, there is partial penetration of the magnetic field at field strengths between Bc1 and Bc2. This partial penetration is in the form of a regular array of normal conducting regions, as illustrated in Figure 11.9.3a. Techniques have been developed to photograph these regions, which are shown as a regular array of dark areas in Figure 11.9.3b.

These normal regions allow the penetration of the magnetic field in the form of thin filaments, usually called vortices. The vortices are aptly named because each is a ‘vortex’ or swirl of electrical current that is associated with this state (see Figure 11.9.3a). While in the vortex state, the material surrounding these normal regions can have zero resistance and partial flux penetration. Vortex regions are essentially filaments of normal conductor (non-superconducting) that run through the sample when an external applied magnetic field exceeds the lower critical field Bc1. As the strength of the external field increases, the number of filaments increases until the field reaches the upper critical value Bc2; the filaments then crowd together and join up so the entire sample becomes a normal conductor.

One can view a vortex as a cylindrical swirl of current surrounding a cylindrical normal-conducting core that allows some flux to penetrate the interior of type-II superconductors. Thrusting a permanent magnet towards a type-II superconductor will cause the applied magnetic field at the superconductor to be within the region of the two critical fields, which creates the vortex state. In principle, the motion of a levitating permanent magnet will cause these vortices to move. In practice, real materials (such as high critical temperature superconductors) have defects (missing or misplaced atoms, impurity atoms) in their crystal lattices. They are also composed of many crystals, all bound together, resulting in many crystal boundaries. The crystal defects and boundaries stop the motion of the vortices, which is known as flux pinning. This provides the stability of a levitating magnet. Pinning the motion of its magnetic field lines also means stopping the motion of the magnet. Flux pinning can only occur in type-II superconductors.

Figure 11.9.2 Eddy currents on the surface of the superconductor essentially create a mirror image of the magnet resulting in repulsion and levitation.

Figure 11.9.3 (a) Illustration and (b) photograph of a regular array of normal conducting regions (dark areas) in a type-II superconductor where the field penetrates the material.

permanentmagnet

image ofpermanentmagnet

high-temperaturesuperconductor

inducedshieldingcurrent

Ba

600 Å

a

b

215


CheCkpoInT 11.91 Explain what effect is being demonstrated by Figure 11.9.1.2 Explain the contribution of eddy currents to the levitation of a magnet over a superconductor.3 What are vortices?4 Explain how thrusting the magnet towards the superconductor increases the stability of the levitation.

11.10 BCS theory and Cooper pairsAccording to classical physics, part of the resistance of a metal is due to collisions between free electrons and the crystal lattice’s vibrations, known as phonons. In addition, part of the resistance is due to scattering of electrons by impurities or defects in the conductor. As a result, the question arose as to why this doesn’t happen in superconductors.

A microscopic theory of superconductivity was developed in 1957 by John Bardeen, Leon Cooper and J. Robert Schrieffer, and is known as the BCS theory (after their initials). The central feature of this theory is that two electrons in the superconductor are able to form a bound pair called a Cooper pair if they somehow experience an attractive interaction between them. At first this notion seems counterintuitive since electrons normally repel one another because of their like charges.

An explanation of the formation of Cooper pairs relies heavily on quantum physics; but here we present a classical picture of their formation (shown in Figure 11.10.1) and an explanation. An electron passes through the lattice and at some point the positive ions are attracted to it, causing a distortion in their nominal positions. The second electron (the Cooper pair partner) is attracted by the positive charge of the displaced ions. This second electron can only be attracted to the lattice distortion if it comes close enough before the ions have had a chance to return to their equilibrium positions. The net effect is a weak delayed attractive force between the two electrons.

This short-lived distortion of the lattice is sometimes called a virtual phonon because its lifetime is too short to propagate through the lattice like a wave, as a normal phonon would.

From the BCS theory, the total linear momentum of a Cooper pair must be zero. This means that the electrons travel in opposite directions, as shown in Figure 11.10.1. In addition, the nominal separation between the Cooper pair (called the coherence length) ranges from hundreds to thousands of ions! If electrons in a Cooper pair were too close, such as only a couple of atomic spacings apart, then the electrostatic (coulomb) repulsion would be much larger than the attraction from the lattice deformation and they would repel each other, and there would be no superconductivity. A current flowing in a superconductor just shifts the total moment slightly from zero so that, on average, one electron in a Cooper pair has a slightly larger momentum magnitude than its partner. They do, however, still travel in opposite directions.

The interaction between electrons in a Cooper pair is transient. Each electron in the pair goes on to form a Cooper pair with another electron, and this process continues with the newly formed Cooper pairs so that each electron

Describe the occurrence in superconductors below their critical temperature of a population of electron pairs unaffected by electrical resistance.

Discuss the BCS theory.

Figure 11.10.1 Classical description of the coupling of a Cooper pair. (a) The first electron approaches a section of the lattice and (b) deforms part of the lattice electrostatically. (c) A second electron is attracted to the net positive charge of this deformation.

e–

e–

e–

++

++

++

++

a

b

c

Superconductivity11

216

phYSICS FeATURehIgh-TeMpeRATURe SUpeRCondUCToRS: The eXCepTIonS To The RUle

In 1986 a class of materials was discovered by Bednorz and Müller

that led to the superconductors we use today on a bench top with liquid nitrogen to cool them. Bednorz and Müller received the Nobel Prize in 1987 for this work (the fastest ever recognition by the Nobel committee). The material we use mostly in school science labs is the yttrium–barium–copper oxide compound YBa2Cu3O7, otherwise known as the 1-2-3 superconductor. It is classified as a high-temperature (Tc) superconductor.

The critical temperatures of some high-temperature superconductors are given in Table 11.6.1. Critical temperatures as high as 135 K have been achieved. This has made experiments on superconductivity more accessible, since these need only to be cooled by liquid nitrogen (with a boiling point of 77 K), which is cheap and readily available. This is in contrast to the expensive and bulky equipment that uses liquid helium for cooling the traditional types of superconductors.

The crystal lattice structure of YBa2Cu3O7 is shown in Figure 11.10.3. Unlike traditional superconductors, conduction mostly occurs in the planes containing the copper oxide. It has been found that the critical temperature is very sensitive to the average number of oxygen atoms present, which can vary. For this reason the formula for 1-2-3 superconductor is sometimes given as YBa2Cu3O7 – δ where δ is a number between 0 and 1.

The nominal distance between Cooper pair electrons (coherence length) in these superconductors can be as short as one or two atomic spacings. As a result, the electrostatic repulsion force will generally dominate at these distances, causing electrons to be repelled rather than coupled.

For this reason, in these materials it is widely accepted that Cooper pairs are not caused by a lattice deformation, but may be associated with the type of magnetism present (known as antiferromagnetism) in the copper oxide layers. This means that high-temperature superconductors cannot be explained by the BCS theory, since that mainly deals with lattice deformations mediating the coupling of electron pairs. The research continues into the actual mechanism responsible for superconductivity in these materials.

Figure 11.10.2 Bednorz and Müller discovered high-temperature superconductors.

YBaCuO

Cu-O chains

Cu(2)

Cu(1)

CuO2layer

Figure 11.10.3 The crystal structure of YBa2Cu3O7, a high-temperature superconductor

217


11.11 Applications of superconductors The main advantage of superconductors is that there is no heat lost when

passing current through them, which means that currents can be made to persist indefinitely. For example, a superconducting electromagnet is made such that external power is applied for only a very short time. The electromagnet is then formed into a closed loop that enables the current (and field) to persist as long as the superconductor stays below its critical temperature; that is, the external power supply can be switched off!

The main disadvantage of superconductors is that they must be maintained at very low temperatures. This requires specialised vessels known as cryostats, which contain the cooling fluids such as liquid nitrogen and liquid helium. The brittle nature of high-temperature superconductors has limited their applicability. As a result, most superconducting applications today still use the more traditional superconductors such as niobium-tin (Nb3Sn), which can be made into flexible wires. The disadvantage is that they require cooling with liquid helium, which is much more expensive than liquid nitrogen and must be contained in more sophisticated cryostats.

Medical applicationsThe first large-scale commercial application of superconductivity was in magnetic resonance imaging (MRI). This is a non-intrusive medical imaging technique that creates a two-dimensional picture of, for example, tumours and other abnormalities within the body or brain. This requires a person to be placed inside a large and uniform electromagnet with a high magnetic field. Although normal electromagnets can be used for this purpose, their resistance would

Discuss the advantages of using superconductors and identify limitations to their use.

CheCkpoInT 11.101 Describe how classical physics explains resistance in metals.2 Outline how Cooper pair electrons form.3 Describe what is meant by ‘the total linear momentum of a Cooper pair must be zero’.

goes on to form Cooper pairs with other electrons. The end result is that each electron in the solid is attracted to every other electron, forming a large network of interactions. Causing just one of these electrons to collide and scatter from atoms in the lattice means the whole network of electrons must be made to collide into the lattice, which is energetically too costly. The collective behaviour of all the electrons in the solid prevents any further collisions with the lattice. Nature prefers situations that spend a minimum of energy. In this case, the minimum energy situation is to have no collisions with the lattice. A small amount of energy is needed to destroy the superconducting state and make it normal. This energy is called the energy gap.

In addition to having a linear momentum, each electron behaves as if it is spinning. This property, not surprisingly, is called spin. (The electron is not actually spinning, but behaves as though it does.) The BCS theory requires that the spins of Cooper pair electrons be in opposite directions.

Figure 11.11.1 Patients are located inside the bore of the superconducting magnet of an MRI machine.

Superconductivity11

218

dissipate a great deal of heat and have large power requirements. Superconducting magnets, on the other hand, have almost no power requirements apart from that required for cooling. Once electrical current flows in the superconducting wire, the power supply can be switched off because the wires can be formed into a loop and the current will persist indefinitely, as long as the temperature is kept below the transition temperature of the superconductor.

Superconductors can also be used to make a device known as a superconducting quantum interference device (SQUID). This device is extremely sensitive to small magnetic fields and can detect magnetic fields from the heart (10–10 tesla) and even the brain (10–13 tesla). For comparison, the Earth’s magnetic field is about 10–4 tesla. As a result, SQUIDs are used in non-intrusive medical diagnostics of the brain.

Scientific researchThe traditional use of superconductors has been in scientific research requiring high magnetic field electromagnets. One application of powerful superconducting electromagnets is in high-energy particle accelerators, such as the Large Hadron Collider at CERN (see section 15.4), in which beams of protons and other particles are accelerated to almost the speed of light and made to collide with each other to create more elementary particles. It is expected that this research will answer fundamental questions such as those about the origin of the mass of matter that makes up the universe.

A future use of superconducting electromagnets is in nuclear fusion energy generation using plasmas. A plasma is a fully ionised gas that is obtained by heating it to millions of degrees and trapping it inside a toroidal structure known as a tokamak by large electromagnets. The nuclei of the ions fuse together, producing energy. The operating gases of such reactors are deuterium and tritium, the isotopes of hydrogen. Deuterium is abundant in water, but tritium will be made inside the tokamak as a by-product of fusion reactions. There is no long-term radioactive waste with this process, which is why it is known as clean nuclear energy. Currently an international research plasma reactor—the International Thermonuclear Experimental Reactor (ITER)—is being built. The aim of the project is to demonstrate that energy production is possible with this method. A diagram of the projected reactor is shown in Figure 11.11.2. The magnetic field coils on the ITER will be made from superconductors and will need to only be powered once; the current through them will be sustained indefinitely, as long as the coils are kept cool.

Levitating trainsMagnetic levitation (maglev) trains have been built that use powerful electro-magnets made from superconductors. The superconducting electromagnets are mounted on the train and kept cool with liquid helium. As shown in Figure 11.11.3, normal electromagnets on a guideway beneath the train repel (or attract) the superconducting electromagnets to levitate the train while pulling it forwards. The superconducting electromagnets rely on the conventional ‘like-pole’ repulsion—not the Meissner effect—to achieve levitation.

Figure 11.11.2 ITER, a proposed test reactor for future clean energy production using nuclear fusion

Process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors and transmission of electricity through power grids.

219


vehicle

glidingskid

guidanceandbraking

levitationandpropulsionmagnet

guideway

armaturewindings(iron core)

guideway

superconductinglevitation magnet

superconductingpropulsion

magnet

propulsionmagnet

Although such a superconducting maglev train has been built and has demonstrated a top speed of 581 km h–1, there are several issues that limit its widespread commercial use. A strong magnetic field inside the train will exclude passengers with pacemakers or devices that have magnetic data storage including computers and credit cards. The powered conventional electromagnets on the guideway that levitate and propel the train are expensive to run over long distances, so alternative propulsion schemes may have to be used.

Power generation, transmission and storage Superconductors have the potential to make electricity generation and

transmission more efficient, which will reduce energy costs and the emission of greenhouse gases. Electricity generators used in coal-fired, nuclear and hydroelectric power plants use electromagnets, which heat up due to the resistance of their wires. Replacing these with superconducting wires will at least halve the amount of power lost in the electromagnets, even when the energy cost of making the liquid helium or nitrogen to keep the superconducting electromagnets in the superconducting state is taken into account.

The transmission of power to homes and businesses is carried out by high-voltage transmission lines. A high voltage enables a small current to be passed through the transmission lines, to minimise the amount of resistive heating in the wires. Nevertheless, there is some heating of the transmission lines and a substantial energy loss associated with it. Superconducting transmission lines will not suffer from such heating losses. Moreover, problems associated with high-voltage leakage of power by ionisation of the air can be overcome by reducing the voltage on the transmission lines and increasing the current through them.

Energy production at, say, coal-fired power plants varies, depending on the anticipated demand for electricity. At the moment, any overproduction of energy is stored by pumping water to higher levels in a large reservoir. Releasing this water into hydroelectric generators enables this stored energy to be retrieved as electricity. The problem with this method is that it is highly inefficient. A third of the excess energy is needed to operate the pumps to the reservoir.

Superconductors offer the possibility of storing an electrical current indefinitely in superconducting rings. This current could be retrieved at any time, provided the rings remain in the superconducting state. This technology may also solve problems such as the variability of supply by solar energy generation. Night-time power could be obtained from daytime storage of excess solar power.

CheCkpoInT 11.111 List the advantages and disadvantages of superconductors.2 State an application of superconductors and explain why it is an improvement on existing technology.

activity 11.1

practical eXperienceS


Figure 11.11.3 Levitation of a train using onboard superconducting magnets on a guideway that propels it with conventional electromagnets

practical eXperienceS

220

11 Superconductivity

chapter 11This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTIvITY 11.1: ApplICATIonS oF SUpeRCondUCTIvITYMagnetic levitation is the ability to control magnetic repelling forces in order to balance objects above each other. You will demonstrate that stable levitation of a permanent magnet can be achieved.Equipment: 2 circular magnets with holes in the centre, superconductor kit (optional), liquid nitrogen.

Discussion questions1 Describe the orientation of the poles of the magnets in order for levitation

to occur.2 Describe your attempts at achieving levitating magnets without the

supporting rod through the centre of the magnets. Discuss how this might differ for a levitating magnet over a superconductor.

Perform an investigation to demonstrate magnetic levitation.

Analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting.

Gather and process information to describe how superconductors and the effects of magnetic fields have been applied to develop a maglev train.

Process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors and transmission of electricity through power grids.

Figure 11.12.1 Set-up to have magnets levitate

221

chapter summary from ideaS toimplementation

• AnX-raydiffractionpatternresultsfromtheinterference of X-rays to form a pattern of dots on photographic film that can then be used to deduce the crystal structure of a solid.

• TheBragglawrelatestheX-raywavelengthλ to the spacing d between crystal planes and the diffraction angle θ by the equation:

nλ = 2d sin θ where n is an integer.• Resistanceinmetalsoriginatesfromcollisionsof

electrons with irregularities in the crystal lattice.• Crystallatticevibrationsareknownasphonons.• Superconductivityisthetotaldisappearanceofelectrical

resistance below a critical temperature Tc that depends on the material of the superconductor.

• TheMeissnereffectistheexpulsionofthemagneticfield from the interior of a superconductor.

• Magneticfieldscandestroysuperconductivityandrestore the normal conducting state.

• Type-Isuperconductorsundergoanabrupttransitiontothe normal state above a single critical magnetic field.

• Type-IIsuperconductorshavetwocriticalmagneticfields. The transition to the normal state occurs above the higher critical field.

• Type-IIsuperconductorsformregionsofnormalconductivity around which there are circulating electrical currents, known as vortices.

• Magneticfieldlinespenetratetheinteriorofatype-IIsuperconductor through the normal regions of the vortices.

• TheBSCtheoryexplainssuperconductivitybythecoupling of electron pairs, known as Cooper pairs, through an interaction with a lattice phonon.

• Theenergygapistheminimumenergyrequiredtodestroy the superconducting state.

• Applicationsofsuperconductorsincludeelectromagnetsfor MRI machines, and superconducting quantum interference devices (SQUIDs) for the detection of very small magnetic fields.

• Superconductors have potential applications in fusion energy research, levitating trains and power generation and transmission.

review questionsphYSICAllY SpeAkIngThe items in the columns are not in their correct order. Match each of the key concepts with its closest definition.

ConCept DefinitionSuperconductivity The point below which there is zero

resistance

Crystal lattice The array of dots on a photographic film created by X-ray diffraction

Laue pattern A burst of lattice vibrations

Phonon The state of matter in which electrical resistance is zero

Bragg law Allows a magnetic field to penetrate while maintaining the superconducting state

Tc The mathematical relationship between X-ray wavelength and crystal lattice spacing

Meissner effect The temporary attraction between electrons mediated by a lattice deformation

Type-II superconductor Expels magnetic fields from the interior

Cooper pairs The regular arrangement of atoms in a solid

RevIeWIng 1 Outline the work carried out by the Braggs in

understanding the wavelength of X-rays.

2 Describe the property of crystals that make them useful in understanding X-rays.

3 Explain the difference between constructive and destructive interference.

4 Outline how the approximate wavelength of X-rays was found. State the value achieved.

5 State the prediction by quantum physics of resistance in crystal structures and explain why these are not always seen.

6 Describe the role of phonons in superconductivity.

7 Outline the contributions of Kamerlingh Onnes to the understanding of superconductivity.

8 Explain how the drop in temperature allows for zero resistance.

9 Describe the Meissner effect.

222

11 Superconductivity

10 State the difference between a conductor and a superconductor when a magnetic field is applied and the temperature is reduced beyond the critical temperature.

11 Explain why the superconducting transition is exothermic.

12 Describe why a magnet levitates above a superconductor that is below its critical temperature.

13 Explain why there is the formation of eddy currents on the surface of superconductors.

14 Explain what is meant by a Cooper pair and describe how they are formed.

15 Outline an energy argument as to why electrons travel through a superconductor unimpeded when it is below its critical temperature.

16 Outline the BCS theory.

17 Create a table to compare the advantages and disadvantages of superconductors.

18 List applications of superconductors in everyday life.

SolvIng pRoBleMS 19 Calculate the separation of diffraction grating slits

when a helium–neon laser (λ = 633 nm) is shone on it. The angular position of the second-order bright band is 30°.

20 a Identify the critical temperature of each metal in Figure 11.12.2.

b Identify which of the two plots (diamonds or squares) shows a material with easily obtainable superconducting properties. State your reasons.

1000800600400200

00 50 100 150 200 250

Temperature (K)

Res

ista

nce

(Ω)

Figure 11.12.2

Revie

w Questions

223


phYSICS FoCUSSeMICondUCToRS To SUpeRCondUCToRS

Since the introduction of the commercial transistor in the late 1940s, the size and processing ability of computer chips has changed dramatically. According to Moore’s law, the number of transistors on a chip doubles every 2 years. But there is a limit—the physical size of atoms.

In the distant future, a new generation of computers that make use of quantum physics may overcome the size limitations of current (classical) computer technology. Quantum computers allow for processes to happen simultaneously rather than sequentially as in classic computers. For example, to factorise a 400 digit number would take a quantum computer a few minutes, but it would take current computers billions of years to do the same calculation.

Another distant technology that may speed up our current limitations on computation speed are the superconducting switching devices known as Josephson junctions. Although not the only technology being researched for future computers, this is one to watch.

1 Define what a semiconductor is. 2 Explain the significance of doping semiconductors. 3 State the significance of semiconductors in the

computing industry. 4 Outline the advantages of silicon over the

previously used germanium. 5 Explain why the p–n junction is important to

modern electronics. 6 Discuss the impacts the p–n junction has had

on society. 7 Analyse the implication of a limit to the size of

silicon chips. 8 Discuss the need to involve quantum mechanics

in the next generation computers. 9 Describe why quantum computers are so much

faster than standard computers.

ReSeARCh10 Research and outline how the Josephson junction

works.11 Research and describe the possible technologies

that can take over from semiconductors.12 Outline the role of nano-carbon tubes in speeding

up processing ability.

Figure 11.12.3 Moore’s law states that the number of transistors on a chip doubles every 2 years.

1950s

Silicon transistor

1960s

TTL Quad gate

1 transistor 16 transistors

1970s

8-bit Microprocessor

4500 transistors

1980s


275 000 transistors

1990s


3 100 000 transistors

2000s


592 000 000 transistors



224

3 The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 25 marks. It should take you approximately 45 minutes to complete this review.

multiple choice(1 mark each) 1 Predict the direction of the electron in Figure 11.13.1

as it enters the magnetic field.A Straight upB LeftC RightD Down

2 The diagrams in Figure 11.13.2 represent semiconductors, conductors and insulators. The diagrams show the conduction and valence bands, and the energy gaps. Which answer correctly labels each of the diagrams?

i ii iiiA Conductor Insulator Semiconductor

B Insulator Conductor Semiconductor

C Insulator Semiconductor Conductor

D Semiconductor Conductor Insulator

3 The graph in Figure 11.13.3 shows how the resistance of a material varies with temperature. Identify each of the parts labelled on the graph.

i ii iiiA Critical

temperatureSuperconductor material

Normal material

B Superconductor material


Normal material

C Critical temperature

Normal material Superconductor material

D Normal material Superconductor material


Figure 11.13.1 An electron in a magnetic field

Figure 11.13.2 Energy bands

Figure 11.13.3 Resistance varies with temperature

–

I II III

Temperature (K)

Res

ista

nce

(Ω)

I

IIIII

225


4 Experimental data from black body radiation during Planck’s time showed that predicted radiation levels were not achieved in reality. Planck best described this anomaly by saying that:A classical physics was wrong.B radiation that is emitted and absorbed is

quantised.C he had no explanation for it.D quantum mechanics needed to be developed.

5 Figure 11.13.4 shows a cathode ray tube that has been evacuated. Which answer correctly names each of the labelled features?

i ii iiiA Striations Cathode Anode

B Faraday’s dark space

Striations Cathode

C Crooke’s dark space

Anode Faraday’s dark space

D Cathode Faraday’s dark space

Striations

extended response 6 Explain, with reference to atomic models, why

cathode rays can travel through metals. (2 marks)

7 Outline how the cathode ray tube in a TV works in order to produce the viewing picture. (2 marks)

8 Give reasons why CRT TVs use magnetic coils and CROs use electric plates in order to deflect the beams, given that both methods work. (2 marks).

9 In your studies you were required to gather information to describe how the photoelectric effect is used in photocells.a Explain how you determined which material was

relevant and reliable.b Outline how the photoelectric effect is used in

photocells. (3 marks)

10 Justify the introduction of semiconductors to replace thermionic devices. (4 marks)

11 Magnetic levitation trains are used in Germany and Japan. The trains in Germany use conventional electromagnets, whereas the one in Japan uses superconductors. Compare and contrast the two systems. (3 marks)

12 a Determine the frequency of red light, which has a wavelength λ = 660 nm. (Speed of light c = 3.00 × 108 m s–1)b Calculate the energy of a photon that is emitted

with this wavelength. (Planck’s constant h = 6.63 × 10–34 J s) (4 marks)

Figure 11.13.4 An evacuated cathode ray tube

IIIII I

4 Quanta toQuarks

226

Figure 12.0.1 Bubble chamber tracks formed by the passage of ionising particles through a liquid that is kept at a temperature above its boiling point. The curved trajectories are the result of charged particles interacting with a magnetic field, and allow information concerning the charge and mass of the particles to be determined.

We all talk about ‘quantum leaps’, but did you know that Max Planck made the first quantum leap in 1901 when he introduced the idea of the quantisation of energy? At the beginning of the 20th century, scientists were confronted by an accumulation of experimental observations and explanations that lacked unification. Black body radiation, the photoelectric effect, radioactivity and the emission of sharp spectral lines by atoms in a gas discharge tube could not be adequately explained within the framework of Newtonian classical physics.

A new physics, quantum physics, was born. The story has some inspiring characters and storylines: how Niels Bohr synthesised the works of Planck, Einstein and Rutherford and proposed the now commonly recognised Rutherford–Bohr atomic model, and so provided an explanation for spectral lines; and how Louis de Broglie in 1924 took Planck’s idea, reversed it and proposed a ‘totally crazy idea’ that all matter has wave properties, which, in turn, gave birth to quantum mechanics.

The investigations into radioactivity led to the artificial manufacture of elements and the dawn of the atomic age. The development of particle accelerators often referred to as ‘atom smashers’ in the media, led to the discovery of a ‘particle zoo’, as a plethora of new subatomic particles were identified. Today, the quest to understand the building blocks and forces of nature continues, with the building of the Large Hadron Collider (referred to as the LHC), which is designed to produce conditions that mimic the environment present just after the birth of the universe.

Context

227

InquIry aCtIvIty

a ChaIn reaCtIon

Obtain some scrap paper! Gather together a crowd of people—your class will do— the more the better! And make sure you have safety glasses for all! Screw up the paper into hundreds of balls, each about the size of a ping-pong ball. Make sure everyone has at least six balls of paper and then gather everyone in close together.

These are the rules of the chain reaction game. 1 Safety glasses on at all times! 2 If you are hit by a paper ball you throw two balls of paper high up into the air.

Your teacher can lob in the first paper ball. It makes a great video clip! You can also vary the rules. For example, try throwing one paper ball when you are hit rather than two. Have fun!

Figure 12.0.2 The blue glow in the core of a water-cooler nuclear results from the radiation emitted when energetic charged particles travel faster than light through water.

12

228

From rutherford to Bohr

rutherford, Bohr, spectra, orbit, spectrum, quantum number, Planck’s constant, quanta, photon, absorption

spectra, emission spectra, Balmer series, rydberg’s constant, transition,

stationary state, Zeeman effect

Pieces of a jigsaw!By the beginning of the 20th century a large number of experimental facts had accumulated that could not be explained by existing theories: • thediscoveryoforderedseriesinatomicspectra• thephotoelectriceffect• radioactivity• evidencethattheatomhadinternalstructure.

12.1 atomic timelineIn the early 20th century many prominent scientists doubted the existence of atoms. Today atomic theory is widely accepted and taught throughout science curricula, forming the foundation upon which scientists, technologists and engineers understand the properties of matter. The ideas underpinning atomic models have changed over time, driven on by the interplay between available technology, theoreticians and experimentalists. The history of the ‘atom’ originates in Greece more 2000 years ago and the quest to reveal its inner structure continues today.

The atomic age was born, ushered in by the development and construction of atomic weapons, nuclear reactors and particle accelerators.

ErnEst ruthErFord‘Once, a distinguished

stranger, amazed by his unscholarly accent and appearance, mistook him for an Australian farmer.’ At the University of Manchester, Rutherford would proclaim to his recruits that: ‘all science is either physics or stamp collecting.’

Figure 12.1.1 Ernest Rutherford is most well known for his alpha-particle scattering experiments. He was awarded the Nobel Prize for Chemistry in 1908 for his work in investigations into the disintegration of the elements, and the chemistry of radioactive substances.

229

Quanta toQuarks

Table 12.1.1 Atomic timeline

Democritus (c 460 bce–c.370 bce)

He was a Greek philosopher who proposed that there was a limit to how small one could divide matter—the smallest indivisible particle was called an atom (atomos, Greek meaning ‘without slices’ or ‘indivisible’).

Aristotle (384 bce–322 bce)

He criticised Democritus, and proposed a model based upon four elements—earth, air, fire and water. His view held for some 2000 years.

John Dalton (1766–1844)

A Scottish teacher, Dalton in 1801 proposed his atomic model, based upon his studies in chemistry that:• Matteriscomposedofsmallindivisibleatoms.• Elementscontainonlyonetypeofatom.• Differentelementscontaindifferentatoms.• Compoundscontainmorethanonetypeofatom.

Henri Becquerel (1852–1908)

In 1896 he discovered that certain elements emitted radiation and decayed, suggesting that the atom was divisible.

JJ Thomson (1856–1940)

In 1904 he proposed the ‘plum pudding’ model of the atom in which electrons were embedded in a positive sphere like ‘plums’ in a pudding. This model was based upon Thomson’s experimental work and his discovery of electrons in 1897.

Ernest Rutherford (1871–1937)

In 1911 he proposed the Rutherford planetary model of the atom, based upon the results of Geiger and Marsden’s scattering experiments attheCavendishLaboratories.

Niels Bohr (1885–1962)

In 1913 Bohr proposed the Rutherford–Bohr model also commonly called Bohr’s model. Bohr provided a set of three postulates to address the issues raised by Rutherford’s earlier model, and this led to the development of a mathematical model to account for the spectra of the hydrogen atom.

Louis de Broglie (1892–1987)

In 1924 de Broglie introduced the concept of matter waves. This concept provided a mechanism for electrons to inhabit a stable orbit by having an integral number of wavelengths fitting around the circumference of the orbit, forming a standing wave.

James chadwick (1891–1974)

In 1932 he reported the discovery of the neutron. Rutherford some 12 years earlier had proposed its existence, and this discovery completed the constituents of the basic atomic model with which most people are familiar today.

ExpErimEntal surprisE

Rutherford’s radium produced alpha particles, and these

massive particles travelled at approximately 1.6 × 10+7m s–1. Rutherford in a later lecture described the extraordinary backscattering of alpha particles as ‘… the most incredible event that has ever happened to me in my life. It was almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you’.

CheCkPoInt 12.1Outline the main atomic models proposed between ancient times and 1913.

12.2 rutherford’s model of the atomIn 1907, New Zealander Ernest Rutherford (1871–1937) moved to the University of Manchester in England where, with Johannes Geiger (1882–1945) and Ernest Marsden (1889–1970), he continued his earlier work on firing alpha particles at metal foils.

They were shocked to find that approximately one alpha particle in every 8000 was deflected by the platinum and gold foils through angles greater than 90º. The Thomson (plum pudding) model of the atom (Figure12.2.1) predicted only small scattering because the atom had no large concentrations of charge or mass to deflect the massive and fast-moving alpha particles. A new atomic model was needed.

Figure 12.2.1 Thomson’s plum pudding atomic model

––

––

–

–

~10–10 m

positivelychargedmaterial

electron

From rutherford to Bohr12

230

+

+

––

–––+

++

+α

ααα

α

nucleus

a

b

StructureThe alpha particle experiments posed many questions. Rutherford hypothesised that for alpha particles to be deflected as observed, it would require a massive, but tiny, positively charged ‘charge centre’ (nucleus), approximately 10–15 m in diameter with a set of orbiting electrons (like planets orbiting the Sun). From Einstein’s analysis of Brownian motion, the radius of an atom was approximately 10–10 m, meaning that the tiny nucleus contains 99.9% of the mass and the atom is mostly empty space.

By 1910 Rutherford had formalised his atomic model with mathematical equations and directed Geiger and Marsden to thoroughly test his model. The series of experiments between 1908 and 1911 provided clear evidence that Thomson’s model was not correct. Interestingly, the scientific community was not persuaded or even interested. The two people who brought Rutherford’s model into mainstream science were Niels Bohr (1885–1962) and a young member of Rutherford’s team Henry Moseley (1887–1915).

LimitationsDespite the success of Rutherford’s planetary atomic model in explaining the scattering of alpha particles, the model failed to explain other important questions:• Whatisthenucleusmadeof?• Howaretheorbitsoftheelectronsarrangedaroundthenucleus?• Whatkeepsthenegativelychargedelectronsfromlosingenergyandspiralling

intothepositivenucleus?

Rutherford proposed that electrical attraction provided the centripetal force to keep electrons in orbit. However, orbiting electrons are accelerating, and Maxwell’s classical electromagnetic theory predicted that accelerating electrons would radiate away their energy in a fraction of a second and spiral into the nucleus, therefore Rutherford’s atomic model was not stable. Additionally, Rutherford’s planetary model could not explain the observed line spectra of excited gases.

Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits.

Figure 12.2.2 Scattering deflections predicted by (a) Thomson’s and (b) Rutherford’s atomic models

Figure 12.2.3 The components used by Geiger and Marsden to measure the deflection of alpha particles fired at thin metal-foil targets

++nucleus

a

b

source containing radon

α–particles

α–particlemetalfoil

viewing screen

231

Quanta toQuarks

CheCkPoInt 12.21 Explain why Thomson’s ‘plum pudding’ model only predicted small deflections in alpha particles passing through

a thin metal foil.2 Outline the significance of Marsden and Geiger’s scattering experiments to the development of the Rutherford

atomic model.

12.3 Planck’s quantised energy Max Planck (1858–1947) in 1901 proposed a theory to model the

spectrum of a black body (see section 9.2). This theory dictated that the energy of oscillations of atoms or molecules cannot have just any value; they can only possess a discrete amount of energy that is a multiple of the minimum value related to the frequency of oscillation by the equation:

E = hf

Planck’s assumption suggests that the energy of any vibration could only be a whole number multiple of hf:

E = nhf

Discuss Planck’s contribution to the concept of quantised energy.

ramp

stairs

Figure 12.3.1 This simple analogy shows the difference between continuous and quantised energy states. On the ramp the box can have any amount of potential energy, but the box on a staircase can only have discrete amounts of potential energy.

where n = 1, 2, 3…, E is energy (in joules), n is called a quantum number, h is Planck’s constant (with a value of 6.63 × 10–34 J s) and f is the frequency in hertz (s–1).

This would mean that energy was not a continuous quantity as had been believed, but rather that it was quantised into discrete packets. Interestingly, at the time of this work Planck considered this idea more of a mathematical device to obtain the ‘right answer’ rather than an actual physical reality of nature.

In 1905, Einstein extended the concept of quantisation and proposed a new theory of light. Einstein took Planck’s suggestion that the vibrational energy of atoms or molecules in a radiating object was quantised with energy. He argued that the vibrational energy of the atoms or molecules could only change by a multiple of hf and therefore proposed that light would be emitted in discrete packets (quanta) also obeying the equation E = hf.

Gilbert Lewis (1875–1946) in 1926 named these discrete packets of light photons (from photos, Greek meaning ‘light’).

Worked examplequestIonA photon has an energy of 2.8 eV.

a Calculate its energy in joules. b Calculate the frequency of the photon.

solutIona You will recall from section 10.2 that

1 eV = 1.602 × 10–19 joules

So: 2.8 × 1.602 × 10–19 = 4.5 × 10–19 joules

b Now rearranging E = hf and substituting in the energy and Planck’s constant we obtain:

fEh

= = ××

= ×−

−4 5 10

6 63 106 8 10

19

3414.

.. Hz


232

CheCkPoInt 12.31 Outline the idea Planck proposed to explain black body radiation.2 Explain the difference between the use of E = hf by Planck and Einstein.

QuantisEd EnErgy?

Why don’t we notice that energy comes in discrete packets or quanta in our everyday life? The value of Planck’s constant is very small;

therefore, for large everyday objects such as a cricket ball or a car, the energy appears to be a ‘continuous’ quantity. At the molecular and atomic level, however, the quantisation of energy becomes very apparent when energy is absorbed or emitted.

12.4 spectral analysisWhenlightispassedthroughaprismordiffractiongrating,theconstituentcolourspresent in the light are revealed in the spectrum produced. The prism separates the light using the property of refraction, whereas the diffraction grating uses the property of interference. The traditional device used by scientists to examine spectra is called a spectroscope. Modern instruments are called spectrometers.

There are two types of spectra: absorption spectra and emission spectra.• Absorptionspectra can be produced by passing white light (a continuous

spectrum) through a cool gas. The atoms or molecules in the gas will absorb certain specific wavelengths (colours) of light. The atom that absorbed the light is now in an excited state and will spontaneously emit a photon of light, usually in a different direction. Therefore the original beam of light will now have certain wavelengths depleted, and these will appear as a series of dark lines when observed through a spectroscope.

• Emissionspectra can be produced when a gas is excited. This can be achieved by heating the gas or by passing an electrical current through a low pressure gas. The light produced when viewed with a spectroscope will often be made up of a series of bright coloured lines.

spEctroscopEs

In a spectroscope, light from a source passes through a slit

and enters the collimator tube. The lens at the end of the collimator makes the light parallel before it illuminates the grating. The diffraction grating is viewed through a small telescope mounted on a rotatable platform so that the angle of the observed colour or spectral line can be measured.

Figure 12.4.1 A spectroscope showing the major components

source

slit lens

collimatorgrating

telescope

eye

θ

H alpha line656 nmTransition n = 3 to n = 2

400 nm 700 nm

Hydrogen absorption spectrum

Hydrogen emission spectrum

Figure 12.4.2 Absorption and emission spectra of hydrogen

233

Quanta toQuarks

Ancient peoples have observed the emission colours of a range of materials. A sprinkling of common salt (sodium chloride) into a flame produces an intense golden flame, and during the smelting of copper ore the flames in the furnace are often coloured an intense vivid green.

Spectra are a ‘window’ into the hidden atomic structure. Each element has its own unique spectral ‘fingerprint’. Chemists and students of chemistry still use the flame test to identify the constituent elements present in chemical samples.

A continuous emission spectrum can be produced by a hot object, forming a rainbow or continuum. For example, if you look at the light emitted from a white-hot piece of metal through a prism or diffraction grating you will see a full rainbow of colours.

Anders ångström (1814–1874) was the first to make detailed measurements of the visible spectrum of hydrogen, and in 1885 Johann Balmer (1825–1898) commenced a detailed study of the visible emission spectrum for hydrogen, which is now referred to as the Balmerseries.

Johannes Rydberg (1854–1919) generalised Balmer’s equation:

1 1 12 2λ

= −⎛

⎝⎜

⎞

⎠⎟R

n nf i

where λ is the wavelength of the spectral line, R is Rydberg’s constant (1.097 × 107 m–1), nf is the final state of the electron and ni is the initial state of the electron.

The emission spectrum in Figure 12.4.2 is the Balmer series emission spectrum for hydrogen corresponding to the values of nf = 2, and ni = 3, 4, 5, 6, 7 in the generalised Balmer equation.

The improvements in experimental techniques and technology for detecting infra-red and ultraviolet radiation led to the discovery of additional spectral lines and other spectral series. For example, Theodore Lyman (1874–1954) discovered the first spectral line in the Lyman series in 1906 and took another 8 years to detect the remaining lines. The generalised Balmer equation could be used to calculate accurately the wavelength of the spectral lines for all the newly discovered series.

Worked examplequestIonUsing the generalised form of Balmer’s equation, calculate the wavelength of the emitted photon for a transition between n = 5 and n = 3.

solutIonUsing the generalised form of Balmer’s equation to calculate 1

λ:

1 1 11 097 10

1

3

1

52 27

2 2λ= −

= × −.

= 7.80 × 105 m–1Rn nf i

Now take the inverse to determine the wavelength of the emitted photon:

λ = 1.28 × 10–6 m

Solve problems and analyse information using: 1 1 1

2 2λ= −

R

n nf i

Table 12.4.1 The Balmer series for hydrogen

SpEcTRAL LiNE BALmER’S gENERALiSED EquATioN

1 1 12 2λ

= −

R

n nf iNAmE coLouR λ (nm)

nf ni

Hα Red 656.3 2 3

Hβ Blue-green 486.1 2 4

Hγ Blue-violet 434.0 2 5

Hδ Violet 410.2 2 6

thE tEst oF timE

Rydberg’s constant R was initially determined empirically

from spectroscopy, and then theoretically by Niels Bohr, who used a mixture of classical and quantumideas.Lateritsvaluewasagain calculated using more fundamental constants in terms of quantum mechanics:

Rm e

h c∞ = e

4

o2 38ε


234

Table 12.4.2 Spectral series associated with the hydrogen atom

SERiES SpEcTRAL REgioN nf ni DiScovEREDLyman Ultraviolet 1 2, 3, 4, 5 … 1906–1914

Balmer Visible and UV 2 3, 4, 5, 6 … 1885

Paschen Infra-red 3 4, 5, 6, 7 … 1908

Brackett Infra-red 4 5, 6, 7, 8 … 1922

Pfund Infra-red 5 6, 7, 8, 9 … 1924

n = 2

n = 1

n = 3

n = 4n = 5

excitedstates

Paschenseries

Balmerseries

Lymanseries

ground state−13.6

−3.4

−1.5−0.85

−15

−10

−5

E = 0

ionised atom(continuous energy levels)

Ene

rgy

(eV)

Figure 12.4.3 Energy level diagram for the hydrogen atom with the Lyman, Balmer and Paschen series emission transitions and associated spectra for the Balmer series

Worked examplequestIon

a Predict the values for nf and ni for the transition of the second longest wavelength emission in the Lyman series.

b Calculate the wavelength for the second-longest wavelength in the Lyman series.

solutIona The second-longest wavelength in the Lyman series will correspond to the second-

lowest energy transition. In the Lyman series we know that nf is always 1. Thus ni = 2 will correspond to the lowest energy transition and ni = 3 will correspond to the second-lowest energy transition. Therefore nf =1 and ni = 3.

b Using the generalised form of Balmer’s equation to calculate 1λ

:

1 1 11 097 10

1

1

1

32 27

2 2λ= −

= × −.

Rn nf i

= 9.75 × 106 m–1

Now take the inverse to determine the wavelength of the emitted photon for the transition between ni = 3 and nf = 1.

λ = 1.03 × 10–7 m

235

Quanta toQuarks

Worked examplequestIonUsing the information in the table, construct an energy level diagram for n = 1 to n = 5 for the hydrogen atom.

a On your diagram, draw and label the Lyman, Balmer and Paschen series.

b Identify the values of ni and nf for a transition of an electron between energy levels that would absorb the highest frequency photon for the Paschen series, based on the energy values provided.

c Identify the values of ni and nf for the transition of an electron between energy levels that would emit the highest energy photon for the Balmer series.

solutIona Refer to Figure 12.4.3 and construct a similar diagram for the given values n = 1

to n = 5.

b From your diagram for part a, we see that for the Paschen series ni = 3 to nf = 5 is the greatest possible jump in energy, therefore this transition would absorb the highest energy photon.

c From your diagram for part a, we see that for the Balmer series ni = 5 to nf = 2 is the greatest possible jump in energy, therefore this transition would emit the highest energy photon.

CheCkPoInt 12.41 Recall the two types of spectra.2 List the names of five spectral series.3 Recall the basic process that produces the spectral series. 4 Explain the relationship between the energy of an emitted photon and quantum number n.5 Calculate the wavelength of a photon emitted by an electron jumping from n = 3 to n = 2.6 Outline the spectral series that corresponds to a transition from n = 3 to n = 2.

12.5 Bohr’s model of the atomNiels Bohr (1885–1962), a Danish physicist, spent a short time at the Cavendish Laboratory, Cambridge, working with JJ Thomson before joining Rutherford’s team at the University of Manchester in 1912. Bohr focused his attention on the planetary model and was equally impressed by both its successes and its obvious limitations. Bohr set to work and built upon Rutherford’s model by synthesising a mixture of Planck’s quantum theory and classical physics.

Bohr’s postulatesIn 1913 Bohr announced the revised planetary model of a hydrogen atom based upon the quantisation of energy and angular momentum of the electron. His new Rutherford–Bohr model included a set of three postulates to address the identified limitations of Rutherford’s model.

Figure 12.5.1 Niels Bohr

pRiNcipAL quANTum NumBER (n ) ENERgy (ev)1 –13.6

2 –3.4

3 –1.51

4 –0.85

5 –0.54


236

a sQuarE pEg in a round holE?

Despite Bohr showing all the signs of becoming a theorist

in an experimentalist laboratory, Rutherford responded to comments by saying, ‘Bohr’s different. He’s a football player!’

Bohr’s postulates for his atomic model1 Electrons exist in stable orbits. An electron can exist in any of several special

circular orbits with no emission of radiation. These orbits are called stationarystates.

2 Electrons absorb or emit specific quanta of energy when they transition between stationary states (orbits). In contradiction to classical electromagnetic theory, a sudden transition of an electron between two stationary states will produce an emission or absorption of quantised radiation (a photon), described by the Planck–Einstein relation.

3 Angular momentum of electrons is quantised. An electron in a stationary state (orbit) has a quantised angular momentum that can take only values of nh2π

where n is the principal quantum number.

Structure of the Rutherford–Bohr modelThe Rutherford–Bohr model has a small, positively charged nucleus that contains most of the atom’s mass. The electrons orbit the nucleus in classical circular paths. The electrons do not radiate energy continuously as predicted by Maxwell’s classical electromagnetic theory (i.e. accelerating charges will radiate electromagnetic waves, which would result in the electron spiralling into the nucleus), due to the quantisation conditions of energy associated with each electronorbit.Whenanelectronjumpstoahigherorlowerorbititwillabsorbor emit a quantum of energy in the form of a photon.

Despite being a hybrid theory that spanned both classical and the quantum physics, the Rutherford–Bohr model proved to be extremely successful in explaining many experimental observations.

Analyse the significance of the hydrogen spectrum in the development of Bohr’s model of the atom.

Define Bohr’s postulates.

12.6 Bohr’s explanation of the Balmer series

Diagrammatic illustration Let’sconsidertheBalmerseriesemissionspectraforahydrogenatom.When an excited electron in a stationary state (orbit) of ni > 3 jumps down to the stationary state (orbit) nf = 2, a photon is emitted. The energy of this photon will be equal to the energy difference ∆E between these two stationary states:

Energy of emitted photon = ∆E = Ei –Ef

This photon will have a characteristic frequency and wavelength, which can be calculated using the relationships E = hf and v = f λ (where v is c the speed of light) rearranged in the forms:

fEh

cf

= =and λ

Process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series.

CheCkPoInt 12.51 Explain the importance of stationary states to Bohr’s model.2 Summarise the role of Planck and Einstein’s concept of quantisation to the development of Bohr’s atomic model.

237

Quanta toQuarks

Bohr’s mathematical model Bohr’s model was not simply a diagrammatic representation of

an atom. He backed up his model with a mathematical framework that was based upon both classical and quantum ideas. Most importantly, the model was supported by experimental observations.

Quantised energy of the stationary states of the Bohr atom The expressions for the quantisation of the angular momentum and radii can be used to derive a quantised expression for energy:

En

En =12 1

Theoretically determining spectral linesNow recall Bohr’s second postulate concerning the emission or absorption of a specific quantum of energy when an electron transitions between stationary states (orbits). This obeys the Planck–Einstein relation hf = Ei – Ef.

If we now substitute in En

En =12 1 we obtain:

∆E Enbetween stationary states

i

= hf = × −12 1 E

n f

×12 1

If we take out the common factor of –E1 and make f the subject, we obtain the expression:

fE

h n n=−

−⎛

⎝⎜

⎞

⎠⎟1

2 2

1 1

f i

Using the relationship v = f λ where v is c the speed of light, we have fc

=λ

. Substituting this we obtain the expression:

c E

h n nλ=−

−⎛

⎝⎜

⎞

⎠⎟1

2 2

1 1

f i Dividing through by c we obtain a theoretically derived expression in a form

that is equivalent to the generalised Balmer equation:

1 1 112 2λ

=−

−⎛

⎝⎜

⎞

⎠⎟

E

hc n nf i

The expression –E1| hc corresponds to Rydberg’s constant and, when evaluated, agrees with the experimentally derived value. Bohr’s theory had successfully provided an explanation for some spectral phenomena and permitted the calculation of Rydberg’s constant. The theoretical derivation of Balmer’s equation was a major accomplishment and provided strong support for the Rutherford–Bohr atomic model of the hydrogen atom.

Describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum:

1 1 1

2 2λ= −

R

n nf i

Figure 12.6.1 illustrates the Rutherford–Bohr atomic model electron transitions for the Balmer spectral series.

activity 12.1

practical ExpEriEncEs


n = 6

n = 1

n = 2

n = 3

n = 4

n = 5

nucleus

Balmer series

Figure 12.6.1 Diagrammatic representation of Bohr’s explanation of the Balmer line series for the hydrogen spectrum


238

Worked examplequestIonAn electron makes a transition between the energy levels –0.85 eV and –3.40 eV in a hydrogen atom.

a Determine to which spectral series this photon would belong.

b Calculate the wavelength of the emitted photon without using Balmer’s equation.

c Predict if the spectral line associated with this transition is visible to the human eye.

d Evaluate the principal quantum numbers (n) to which each energy level corresponds.

e Substitute your values of ni and nf into the generalised Balmer equation and calculate the wavelength of the emitted photon. (This should be the same value as in part b.)

solutIona From Figure 12.4.3, the energy level –0.85 corresponds to n = 4 and energy level

and –3.40 eV corresponds to n = 2. The n = 2 tell us that this transition belongs to the Balmer series.

b The difference between these two energies corresponds to the energy of the emitted photon, E = 2.55 eV. Convert this to the SI unit joules (remember 1 eV = 1.602 × 10–19 J).

E = 2.55 × 1.602 × 10–19 = 4.09 × 10–19 J

Now, using E = hf, we can calculate the frequency of the emitted photon with this energy:

fEh

= = ××

= ×−

−4 09 10

6 626 106 17 10

19

3414.

.. Hz

Using the relationship v = f λ where v is c the speed of light, we have λ = c

f.

λ = = ××

= × −cf

3 0 10

6 17 104 86 10

8

147.

.. m

c Light of this wavelength is in the visible range.

d From Figure 12.4.3, the energy level –0.85 corresponds to n = 4 and the energy level –3.40 eV corresponds to n = 2.

e Using the generalised form of Balmer’s equation to calculate 1λ

:

1 1 11 097 10

1

2

1

42 05

2 27

2 2λ= −

= × −

=Rn nf i

. ××106 m–1

Now take the inverse to determine the wavelength of the emitted photon for the transition between ni = 4 and nf = 2.

λ = 4.86 × 10–7 m

CheCkPoInt 12.6Explain the relationships between Bohr’s atomic model and observed line spectra.

239

Quanta toQuarks

Figure 12.7.1 Emission spectra for (a) stronium, (b) barium and (c) calcium

12.7 limitations of the rutherford–Bohr model

The Rutherford–Bohr model provided a simple visual model and accurately predicted line spectra for the hydrogen atom. It synthesised classical mechanics with the concept of quantisation, and addressed a number of experimental observations. Despite its successes, it was nevertheless a hybrid model and many experimental observations remained unresolved.

The spectra of larger atoms The observed spectral series of larger atoms appeared to present patterns that

could be explained in a similar manner to the hydrogen atom. Despite the efforts of Bohr and his colleagues, they were unable to develop an arrangement of stationary states (orbits) that matched the experimental observations. Atoms larger than hydrogen have more than one electron, and these electrons were obviously interacting with each other in a complex manner. Bohr’s simple quantised planetary model explained only the hydrogen atom, the helium ion (He+) and the lithium ion (Li2+), which all have single electrons orbiting the nucleus.

Discuss the limitations of the Bohr model of the hydrogen atom.

Figure 12.7.2 Emission spectrum for hydrogen showing the differences in relative widths

The relative intensity of spectral linesThe spectra of the hydrogen atom and larger atoms all displayed three identifiable types of spectral lines based upon their width. These were categorised as sharp (s) lines, primary (p) lines and diffuse (d) lines. Also within these categories the intensity of the individual lines varied. Bohr’s postulates provided no explanation for these observations.

mEasuring thE magnEtic FiEld oF our sun

Astronomers use a spectroscope and the Zeeman effect to

measure the strength of the magnetic field associated with the surface structures of the Sun. The splitting of the lines in the hydrogen spectrum is proportional to the magnetic field strength.

a

b

c

Sr

Ba

Ca

400 450 500 550 600 650 700 750Wavelength (nm)

activity 12.2




240

univErsal units For lEngth and timE

The plaque on the Pioneer space probe used the hyperfine

transition of hydrogen, which is the most abundant element in the universe, to define a base unit for length and time.

CheCkPoInt 12.7Compare the origins and observed separations of the splitting in the ‘normal’ Zeeman effect and the hyperfine spectral lines.

The Zeeman effectIn 1862, Michael Faraday placed a sodium flame between the poles of a magnet to see if the bright sharp spectral lines were influenced by the magnetic field, but he observed no change. Some 30 years later (in the 1890s) Pieter Zeeman (1865–1943) repeated the experiment using both a more advanced spectroscope and a stronger magnetic field. Zeeman found that the spectral lines were indeed influenced and he observed that the previously single lines had each split into three. Classical physics and Bohr’s model could provide an explanation for these triplets. This was called the Zeeman effect. In 1897 Zeeman used an even stronger magnetic field and found that the triplet lines were also split. The additional splitting was also associated with the yet to be discovered spin property of electrons.

Bohr’s atomic model provided some explanation for the ‘normal’ Zeeman effect, but none for the ‘anomalous’ Zeeman effect.

Figure 12.7.4 A single spectral line splits into more lines when the source is subjected to a strong magnetic field.

Normal spectral line:With no magnetic field applied—single spectral line observed.

‘Normal’ Zeeman effect:Magnetic field applied—triple spectral line observed. Central line polarised parallel to applied magnetic field.Other two lines polarised perpendicular to applied field.

‘Anomalous’ Zeeman effect:When magnetic field applied—more spectral lines observed.

The existence of fine and hyperfine spectral linesThe fine structure of closely spaced spectral lines that are 0.1–0.5 nm apart is a result of an additional property of the electron called ‘spin’, which was proposed in 1925 by Ralph Kronig (1904–1995). In 1881, using interferometry, Albert Michelson (1852–1931) observed that some even finer spectral lines (called hyperfine) existed; these were about 0.001 nm apart.Theobservationwasnotaddresseduntil1924,whenWolfgangPauli(1900–1958) proposed the existence of a small nuclear magnetic moment caused by a non-spherical atomic nucleus. Bohr’s model did not provide any explanation for these observed phenomena.

Figure 12.7.3 A comparison of (a) fine structure and (b) hyperfine structure

0.1 nm

0.001 nm

0.1 nm

0.001 nm

a

b


241

Quanta toQuarks

chaptEr 12This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

aCtIvIty 12.1: hydrogen and the atomConnect a high-voltage source to a low-pressure hydrogen discharge tube, and observe the emitted light through a spectroscope. Identify the visible part of the hydrogen spectrum.Equipment:low-pressurehydrogen gasdischargetube,gasdischargetubehigh-voltagepowersupply,spectroscope.

Discussionquestions1 List the wavelengths of the

emission lines in the visible part of the hydrogen spectrum.

2 Identify the energy levels of the electron transition that creates these wavelengths.

3 Outline Bohr’s explanation of the Balmer series of emission lines.

Perform a first-hand investigation to observe the visible components of the hydrogen spectrum.

Process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series.


2 2λ= −

R

n nf i

Analyse secondary information to identify the difficulties with the Rutherford–Bohr model, including its inability to completely explain:•thespectraoflargeratoms•therelativeintensityof

spectral lines•theexistenceofhyperfine

spectral lines•theZeemaneffect.

Figure 12.8.1 Experiment set-up and the visible emission spectral lines of hydrogen

aCtIvIty 12.2: ProBlems wIth the rutherford–Bohr modelIn this activity you will look at the difficulties with the Rutherford–Bohr model in explaining certain observations.

Discussionquestions1 Identify the ways in which this model is an improvement over the

previously accepted model of the atom.2 Explain the limitations of this model.

242

12 From rutherford to Bohr chapter summary

• AtomictheoryhaditsoriginsinGreecemorethan 2000 years ago with ideas put forward by Democritus and Aristotle.

• JJThomsonin1904proposedthe‘plumpudding’model comprising a positive sphere in which the electrons were distributed like ‘plums in a pudding’.

• ErnestRutherfordin1911usedtheresultsofGeigerand Marsden’s scattering experiments to propose a ‘Rutherford’s planetary model’. Rutherford’s model comprised a massive, positively charged nucleus with a set of orbiting electrons (like planets orbiting the Sun), inferring that the atom is mostly empty space.

• DespitethesuccessofRutherford’splanetaryatomicmodel in explaining the scattering of alpha particles, the model failed to explain: — what the nucleus is made of — how the orbits of the electrons are arranged around

the nucleus— what keeps the negatively charged electrons from

losing energy and spiralling into the positive nucleus.

• Planckin1901proposedatheorytomodelthespectrum of a black body and introduced the concept of quantisation E = nhf where n = 1, 2, 3, 4 …

• Aspectroscopeusesaprismordiffractiongratingtoreveal the constituent colours present in the light.

• Therearetwotypesofspectra:— emission spectra— absorption spectra

• Spectraarea‘window’intothehiddenatomicstructure.Each element has its own unique spectral ‘fingerprint’.

• Rydberg’sgeneralisedBalmerequationallowsyoutocalculate the wavelength of the photon emitted or absorbed for any transition of an electron between stationary states:

1 1 12 2λ

= −

Rn nf i

where λ is the wavelength of the spectral line R is Rydberg’s constant 1.097 × 107 m–1

nf is the final state ni is the initial state.

• HydrogenhasseveralspectrallineseriesincludingtheLyman, Balmer, Paschen, Brackett and Pfund series.

• Bohrin1913describedtherevisedplanetarymodelof a hydrogen atom based upon the quantisation of energy and angular momentum of the electron.

• Bohr’sthreepostulatesforhisatomicmodel:1 Electrons exist in stable orbits called stationary

states. 2 Electrons absorb or emit specific quanta of energy

when they transition between stationary states (orbits), described by the Planck–Einstein relation hf = Ei – Ef.

3 Angular momentum of electrons is quantised.• TheRutherford–Bohrmodel,despiteitssuccesses,was

nevertheless a hybrid model and many experimental observations still remained unresolved including:— the spectra of larger atoms— the relative intensity of spectral lines— the existence of hyperfine spectral lines— aspects of the Zeeman effect.

243

Quanta toQuarksreview questions

PhysICally sPeakIngThefollowingtableisalljumbledup.Copythetableintoyourworkbookandcorrectlyarrangetheinformation.Youmayalsoadd additional features associated with each atomic model. For each atomic model, draw and label a pictorial representation.

SciENTiST AND DATE FEATuRES oF THE ATomic moDEL DRAw A picToRiAL REpRESENTATioN oF THE moDELDemocritus

c 400 bce

• Planetaryorbitsoftheelectrons

• Threepostulates

Aristotle

c 300 bce

A positive sphere with electrons embedded in it

Dalton

1801

• Planetaryorbitsoftheelectrons

• Smallpositivelychargednucleus

Thomson

1904

Limittohowsmallyoucoulddivide matter; the smallest is called an atom

Rutherford

1911

Electronsactlikewaves

Bohr

1913

There are four elements: earth, air, fire and water

de Broglie

1924

• Elementsonlycontainonetype of atom

• Differentelementscontain different atoms

• Compoundscontainmorethan one type of atom

244

12 From rutherford to Bohr

revIewIng 1 Explain why flame tests of unknown samples are so

useful to scientists.

2 CompareMaxPlanck’sandAlbertEinstein’sideasrelating to ‘quanta’.

3 Outline Rutherford’s model of the atom.

4 Explain how emission line spectra are formed.

5 Identify the major events that led Niels Bohr to propose his model of the atom.

6 Define Bohr’s postulates associated with his 1913 model of the atom. Describe how these postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum.

Describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum: 1 1 1

2 2λ= −

R

n nf i

7 Recall the relationship between energy levels in the Bohr model of an atom and the observed Balmer series spectra.

8 Discuss the limitations of the Bohr model of the hydrogen atom.

9 When excited atoms were placed in a strong magnetic field some previously observed single spectral lines split. Explain why the Rutherford–Bohr model could not completely explain the observed phenomena.

10 Marsden and Geiger in 1909 fired alpha particles at thin platinum and gold foils.a Recall what the then current model of the atom

proposed by JJ Thomson theoretically predicted.b Recall the name of the scientist they were working

with during these experiments.c Outline their experimental findings. d Discuss the implications these experiments had

for the development of a new model of the atom.

Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits.

11 Recall why the Rutherford–Bohr model could not explain spectra of atoms larger than hydrogen.

12 Compare the Zeeman effect and hyperfine spectral lines.

solvIng ProBlems 13 A photon has an energy of 6 eV. Calculate:

a its energy in joulesb the frequency of the photonc the wavelength of the photon.

14 Using the formula En

En = ×12 1 and given that the

energy of the first stationary state (orbit) in the hydrogen atom is –13.6 eV, calculate:a the energy of the fourth stationary state in the

hydrogen atomb the energy difference between the fourth and first

stationary statesc the frequency of a photon emitted by an electron

transitioning from stationary state n = 4 to n = 1d the wavelength of a photon for this transitione the spectral series to which this photon belongs.

15 An electron makes a transition between the –3.40 eV and –13.6 eV energy levels of a hydrogen atom.a Calculate the wavelength of the emitted photon

without using Balmer’s equation.b Predict if the spectral line associated with this

transition is visible to the human eye.c Evaluate the principal quantum numbers (n) of

each energy level.d Substitute your values of ni and nf into the

generalised Balmer equation and calculate the wavelength of the emitted photon. (This should be the same value as part a.)

e Identify the spectral series to which this photon would belong.

16 Using the Bohr model, construct diagrams of the following electron transitions and describe the role of the photon in these transitions:a an electron being excited from the ground state

(n = 1) to the fourth excited state (n = 5)b an excited electron dropping from the n = 6

energy level to the n = 2 energy levelc an excited electron dropping from the n = 4 to

n = 3 and then back to the ground state (n = 1)

17 Using the information in the following table, construct an energy level diagram for n = 1 to n = 5 for the hydrogen atom.

pRiNcipAL quANTum NumBER (n) ENERgy (ev)1 –13.6

2 –3.4

3 –1.51

4 –0.85

5 –0.54

245

Quanta toQuarks

a Identify the electron transition between the energy levels in your diagram that would absorb the highest frequency photon.

b Identify which transition of an electron between energy levels would emit the highest energy photon.

c Demonstrate on your energy level diagram how the Balmer series of spectral emission lines is produced.

d Identify the transition between energy levels in the Balmer series that would produce the longest wavelength photon.


2 2λ= −

R

n nf i

18 Rydberg generalised Balmer’s equation for the hydrogen atom:

1 1 12 2λ

= +⎛

⎝⎜

⎞

⎠⎟R

n nf i

a Recall what each term in the equation represents.b Calculate the wavelength of the emitted photon for

a transition between n = 6 and n = 1.c Predict the values for nf and ni for the transition

of the second-longest wavelength in the Balmer series.

d Calculate the wavelength for the second-longest wavelength in the Balmer series.

19 A photon is emitted from a hydrogen atom with a wavelength of 410.12 nm. The electron associated with the transition that produced the photon is now in energy level nf = 2.a Using Balmer’s generalised equation, calculate

the value of ni.b Construct a diagram showing the transition and

identify all relevant features.c Identify the series associated with this transition.

20 TheLymanseriesforahydrogenatomhasthefollowing energy levels: E1 = –13.6 eV, E2 = –3.40 eV, E3 = –1.51 eV, E4 = –0.85 eV, E5 = –0.54eV, E6 = –0.38 eV.a Construct a diagram using the Bohr model to

represent the series.b ConstructanenergyleveldiagramfortheLyman

series.c Calculate the energy of the photon emitted in the

transition from n=5intheLymanseries.d Calculate the wavelength for this emission.–

Revie

w Questions

246

12 From rutherford to Bohr

PhysICs foCusquanta to quarks tImelIne

Physics is not just about learning facts. Physics is a human journey of exploration seeking to explain the universe in which we live. Understanding the history of events and people will allow you to ‘stand upon the shoulders of giants’ and continue this quest. Often the ‘stories’ told in modern textbooks skip over the years of frustration and confusion that scientists faced and present only the milestones and important discoveries.

When you study the history of scientific advancement, it can become quite confusing if you do not see the situation in the context of the period of history.

For example, many of the experiments carried out from the late 1800s to the modern particle accelerators relied on vacuum pump technology. How did this technology frustrate the pioneers who worked withlow-pressuregasexperiments?Considerhowtheability for the scientist to produce a strong magnetic field may have influenced the outcome of an experiment. For example in 1862, Michael Faraday placed a sodium flame between the poles of a magnet to see if the bright sharp spectral lines were influenced by the magnetic field; Faraday observed no change. Some 30 years later in 1896, Pieter Zeeman repeated the experiment using both a more advanced spectroscope and a stronger magnetic field, and found that the spectral lines were indeed influenced. Zeeman also observed that some previously single lines had each split into three.

Maybe you are now thinking that the interplay between technology, people, events and ideas might be quite interesting.

Construct a timeline that includes people, experiments, advances in technology, models and theories for the period from 1885 and Balmer’s observations of the hydrogen spectrum through to 1913 when Niels Bohr proposed his atomic model. You can include pictures of people, models and experiments. Investigate and explore the relationships between events. You might construct your own or, as a class, build a huge timeline and display it in your laboratory.

Some technologies you might research:• Advancesinvacuumpumptechnology• Advancesinspectroscopetechnology• Sourcesofelectricity

extensIonSome additional technologies you might research:• Methodsofdetectingthenon-visible

electromagnetic spectrum, including infra-red, ultraviolet light, X-rays and gamma rays

• Typesofmagnetsavailable• Electricandelectroniccircuitry

As you progress further through this module extend and include more details on your timeline.

H1. Evaluateshowmajoradvancesinscientific understanding and technology have changed the direction or nature of scientific thinking


Figure 12.8.2 JJ Thomson in his Cavendish laboratory

13

247

Birth of quantummechanicsA new paradigmNiels Bohr’s hybrid atomic model, which combined classical and quantum concepts, held centre stage from 1913 through to the early 1920s. From a historical perspective we must remember that World War I, 1914–1918, disrupted research laboratories throughout Europe. The 1920s saw the move away from classical ideas and culminated at the 1927 Solvay Conference in Copenhagen where modern day quantum mechanics was born. In the years leading up to this conference Louis de Broglie proposed that all matter has wave properties, Erwin Schrodinger developed a mathematical model for describing wave mechanics, Heisenberg revealed the statistical nature of quantum theory and formulated his most famous discovery—the Heisenberg uncertainty principle—and Pauli developed his exclusion principle.

13.1 The birthNewtonian classical physics was under attack throughout the 1920s and the outcome, which was based upon theoretical and experimental evidence, gave birth to quantum mechanics. The major debates came to a head at the 1927 Solvay Conference in Copenhagen where the old deterministic view of matter was cast aside in favour of a world that was ruled by probabilities. This outcome, however, was certainly not unanimously accepted and most notably Bohr and Einstein continued to debate the interpretations for more than 20 years.

A GreAt Loss

Harry (Henry) Moseley was one of the brilliant young scientists who

worked with Rutherford. In about 1910, using X-ray diffraction of crystals, he discovered a systematic relationship between wavelength of characteristic X-rays and atomic number, which is now known as Moseley’s law. When World War I broke out, he enlisted and joined the British Royal Engineers. He was killed in action by a sniper shot in 1915 at Gallipoli. The loss of Moseley and other scientists during this war prompted British and other world governments to implement policy to not allow scientists to enlist for combat postings.

CheCkpoinT 13.1Outline how quantum mechanics was started.

quantum mechanics, de Broglie, Schrodinger, wave mechanics, diffraction, quantum numbers,

wave function, heisenberg, uncertainty principle, pauli, exclusion principle

Birth of quantum mechanics13

248

13.2 Louis de Broglie’s proposalBy 1915, William Henry Bragg (1862–1942) had provided convincing evidence that X-rays have particle properties; his son William Lawrence Bragg (1890–1971), an Australian, had developed an equation based upon the wave nature of X-rays that enabled a detailed analysis of X-ray diffraction patterns.

In the early 1920s, Prince Louis de Broglie (pronounced ‘de broy’), with the knowledge that X-rays and light possessed both wave and particle properties, began to consider the wave–particle duality as a natural symmetry.

Matter wavesIn 1923, Louis de Broglie began with a supposition that was based upon the Planck–Einstein equation linking energy quanta to frequency. He equated Einstein’s special relativity energy–mass relationship E = mc2 with the Planck–Einstein equation E = hf:

mc2 = hf

He rearranged it to derive an expression for the momentum of a photon:

mchfc

=

Now mc is mass times velocity, hence the photon’s momentum is:

phfc

=

Using the relationship λ =cf

in the form fc

=λ

and substituting, he obtained:

ph

=λ

Figure 13.1.1 Participants at the 1927 Solvay conference Back row: A Piccard, E Henriot, P Ehrenfest, E Herzen, T De Donder, E Schrödinger, JE Verschaffelt, W Pauli, W Heisenberg, RH Fowler, L Brillouin Middle row: P Debye, M Knudsen, WL Bragg, HA Kramers, P Dirac, A Compton, L de Broglie, M Born, N Bohr Front row: I Langmuir, M Planck, M Curie, HA Lorentz, A Einstein, P Langevin, CE Guye, CTR Wilson, OW Richardson

249

quAntA toquArks

He had derived an expression for the momentum of a photon in terms of its wavelength.

In 1924, de Broglie proposed the concept of matter waves by applying his belief in symmetry to propose that all ‘particles’ of energy should also possess an associated wavelength. At this time, particles of matter such as electrons, atoms and alpha particles were known to possess the properties of mass and charge, but there was no experimental evidence to indicate that these particles exhibited wave phenomena.

De Broglie’s proposed matter waves could be calculated mathematically by using what is now known as the de Broglie equation:

λ λ= =hp

hmv

or

where λ is the wavelength, h is Planck’s constant, m is the mass of the particle, v is the velocity of the particle and p is the momentum of the particle. (Remember p = mv.)

Louis de Broglie presented his hypothesis in his doctorial dissertation. Unfortunately the combination of the involvement of de Broglie and other French scientists in unfriendly debates with Niels Bohr and his supporters and the experimentally unsupported proposition that particles such as electrons possessed wave properties meant that the majority of scientific community did not take de Broglie’s ideas seriously.

Worked exampleQueSTion

a Calculate the wavelength of an electron moving with a velocity of 6.0 × 105 m s–1.

b Compare this calculated wavelength with the wavelength of visible light (400–650 nm).

SoLuTiona Using de Broglie’s equation λ = h

mv and substituting in the following values:

h = 6.63 × 10–34 J s

me = 9.11 × 10–31 kg

v = 6.0 × 105 m s–1

we obtain

λ = ×× × ×

=−

− −6 63 10

9 11 10 6 0 101 2

34

31 5 1

.

. ..

J s

kg m s11 10 9× − m

Therefore the wavelength of the de Broglie matter wave is 1.21 × 10–9 m.

b The visible spectrum ranges from violet light, with the shortest wavelength (~400 nm), to red with a wavelength of 650 nm. For the purpose of this comparison we will use the average wavelength of these extremes 525 nm, which corresponds to a yellow/green colour.

First we need to convert 525 nm into metres (5.25 × 10–7 m). Now, comparing the ratio of the wavelength of the electron (1.21 × 10–9 m) with the average wavelength of visible light (5.25 × 10–7 m), we discover that the wavelength for an electron travelling at 6.0 × 105 m s–1 is approximately 430 times smaller than the wavelength of visible light.

Figure 13.2.1 Louis de Broglie

Solve problems and analyse information using: λ = h

mv


250

ALWAys keep your eye on the physics

An important point to consider when you are calculating the de Broglie wavelength for macroscopic objects such as cars and balls is that these are

not discrete fundamental particles. Rather they are a ‘bulk’ collection of such. Therefore, if you calculate the de Broglie wavelength for an object at near rest (very close to zero velocity) you would obtain a large wavelength, which of course is not what we observe. This is because you have not considered that the atoms and molecules making up the object are actually vibrating at high velocities due to their thermal energy. Also it is important to remember that, even at absolute zero (–273°C), all atomic particles including atoms and electrons will still have some motion and therefore cannot have zero velocity.

Initially Einstein was the only high-profile physicist to support de Broglie’s hypothesis. Erwin Schrodinger, a loner, who was then based at the University of Zurich, was intrigued and took up the idea proposed by de Broglie. He became the architect of wave mechanics, a fundamental component of quantum mechanics.

CheCkpoinT 13.21 Outline the significance of the work of the Braggs in Louis de Broglie developing his theorical research.2 Determine the wavelength of an electron moving at 3.0 × 105 m s–1.3 Calculate the momentum of a proton that has a wavelength of 1.01 × 10–9 m.

13.3 Diffraction A defining property of waves is their ability to ‘flare’ or seemingly bend

around corners. Waves in water, sound waves and light can all exhibit this property called diffraction. Let’s first look at water waves. If you generate a set of plane waves in a shallow trough of water using a ruler, you will see the waves propagate down the length of the trough. Now, if you place an object such as an edge, a block, a gap or a narrow opening in the path of the waves, you will observe a strange behaviour. Instead of ‘casting a shadow’, each wavefront bends around the object and enters the shadow region.

Let’s now observe the behaviour of light as it passes an edge, through a narrow slit and a small aperture (Figure 13.3.1). Again you can see that the light does not cast a sharp shadow, instead it forms a series of bright and dark lines. These patterns are known as diffraction patterns.

Define diffraction and identify that interference occurs between waves that have been diffracted.

Describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties.

251

quAntA toquArks

a b c

Figure 13.3.1 Light passing (a) around a razor blade, (b) through a slit and (c) through a small aperture

try this!Observing a diffractiOn patternSimply take two pens or pencils (you can even use your fingers). Place them very close together, almost touching, to form a narrow slit. Now hold them up in front of one eye (about 3–4 cm in front works well) now close the other eye. Look through the slit at a bright light source such a fluorescent light. Vary the width of the slit and observe carefully! What do you see?

you cAn’t escApe them

Sometimes when you are sitting back relaxing or looking upwards into a clear blue sky, you

might occasionally see tiny fuzzy, out-of-focus ‘hairs’ or ‘blobs’. These are called floaters and they are, literally, that. They are little bits of jelly broken off the vitreous humour (the clear gel that fills the space between the lens and the retina of the eye) floating just in front of your retina. The fuzziness is caused by the diffraction of light around the edges of these floaters.

CheCkpoinT 13.31 Define diffraction.2 Sketch a diffraction pattern that would result as light moves around a hair.

13.4 Confirming de Broglie’s hypothesis

Clinton Davisson (1881–1958) and Lester Germer (1896–1971), using electron scattering, reported the experimental discovery of electron waves in 1927. Davisson and Germer’s discovery was preceded by an accident in the laboratory in which an explosion shattered the evacuated glass vessel holding the nickel target. The introduction of air oxidised the surface of the nickel target, which then had to be degassed by heating it to a high temperature. This process allowed a number of large nickel crystals to form. Experiments with this target produced a new set of data in which sharply defined currents of electrons were present. Davisson and Germer quickly realised that these new results were a result of the recrystallisation, and they constructed new nickel targets of single crystals.

Describe the confirmation of de Broglie’s proposal by Davisson and Germer.


252

Figure 13.4.2 Davisson and Germer’s apparatus and results. With the single crystal nickel target, Davisson and Germer detected several peaks in scattered electrons.

λ

incident wavesin phase

scattered wavesin phase

d

electrongun

power supply

a b

c

I

50°

50°

V = 54 V

0 15° 30° 45° 60° 75° 90° θ

crystalsurface

electron beam (in vacuum)

electrondetector

nickel crystal

Up until this time Davisson and Germer had been unaware of de Broglie’s hypothesis, but during a meeting at Oxford in 1926 Davisson heard of de Broglie’s proposed matter waves. He immediately realised that the new results he and Germer had observed were similar to those of X-ray diffraction patterns. With this insight, Davisson and Germer quickly identified that the scattered electrons were indeed producing diffraction patterns.

At the same time, George Thomson (1892–1975) conducted a different experiment in which he fired a beam of electrons through a thin gold foil and observed diffraction rings. Therefore, in 1927, two different research teams using different experimental techniques had verified Louis de Broglie’s hypothesis concerning the wave nature of electrons.

in the fAmiLy

George P Thomson, the co-discoverer of the wave

properties of electrons, was the son of JJ Thomson the discoverer of the electron.

Figure 13.4.1 GP Thomson

CheCkpoinT 13.41 Outline how matter waves were first observed.2 Discuss what other evidence has been observed to support Louis de Broglie’s hypothesis.

13.5 electron orbits revisitedBohr’s first postulate stated that an electron can exist in any of several circular orbits with no emission of radiation. However, he provided no supporting reasoning to explain why the orbiting electron would not radiate away its energy, as predicted by Maxwell’s classical electromagnetic theory, and simply spiral into the nucleus.

When de Broglie formulated his concept of matter waves, he envisioned that electron orbits were standing waves. Each orbit (stationary state) was occupied by an integral number of wavelengths that fitted around the circumference of the orbit.

The circumference of a circular orbit is 2πR and hence an integral number of wavelengths:

nλ = 2πR

Activity 13.1

prActicAL eXperiences


253

quAntA toquArks

Figure 13.5.1 Comparison of standing waves wrapped in a circle and on a rope. For a standing wave to be produced in a circle, a whole number of matter waves must fit into the circumference.

λ

λ

λ

λ

n = 1

λ

n = 2

λ

n = 3

λn = 4

try this!standing wavesUsing a slinky spring or a rope, generate standing waves by having one person hold one end firmly while the other person shakes their end. Once you get a stationary wave you will notice that if you stop shaking, the standing wave is quite stable and continues to vibrate. On a rope or slinky you can generate standing waves other than those shown in Figure 13.5.1. Why are these other standing wave patterns not suitable for forming circular standing waves? (Hint: Consider interference.)

CheCkpoinT 13.51 Outline how de Broglie’s matter waves help to explain the stable orbits of electrons.2 Calculate the radius of the smallest orbit of an electron with wavelength 1.21 × 10–9 m.

13.6 Further developments of atomic theory 1924–1930

Quantum mechanics, the modern version of quantum theory, was developed between 1925 and 1930. During this period, physicists moved from the hybrid mixture of classical and quantum concepts to a fully mathematical model, to describe the behaviour of matter. The strangeness of wave–particle duality and the many associated paradoxes were sidelined as a new physics was born.

Erwin Schrodinger (1887–1961)Schrodinger’s interests were wide and spanned most of modern physics including statistical mechanics, X-ray diffraction, relativity and field theory. After World War I, Schrodinger moved between several academic posts and in 1921 he settled for 6 years at the University of Zurich. Here Schrodinger, encouraged by Einstein,

where n is 1, 2, 3, 4, 5 …, λ is the wavelength and R is the radius of the orbit. To visualise these standing waves we first imagine a standing wave on a

slinky spring or rope, obeying the condition nλ where n can have the values n = 1, 2, 3 etc. When you shake a rope or a slinky spring you can produce standing wave patterns that contain whole or integral numbers of wavelengths.

Now we can take these standing waves and wrap them end-to-end to form a circular standing wave that corresponds to the circular Bohr orbit. These values of n are called the principal quantum numbers and they tell you the number of wavelengths fitting into the Bohr orbit.

Remember that the standing waves in Figure 13.5.1 represent the matter wave nature of the electron. Often students imagine the electrons travelling in wavy orbital paths around the nucleus—this is not the correct interpretation.

Explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis.


254

Figure 13.6.1 Erwin Schrodinger

A fAmous cAt And inspirinG A neW cAreer pAth to dnA

Everyone has heard of Schrodinger’s cat—the one that

is both alive and dead—well that is until you open the box and see. But a little less known fact is that in 1944 Schrodinger published a little book entitled What is Life?, which looked at the new field of molecular biology. Francis Crick the co-discoverer of the DNA double helix said that this book influenced him to change his career from physics to molecular biology.

developed wave mechanics by synthesising the works of de Broglie, Planck, Einstein, ideas from Hamilton and the fundamental differential equations from optics. In 1926 Schrodinger published the equation known by all physicists as ‘Schrodinger’s equation’:

− + =hm x

V x E2 2

22d

d

ψψ ψ( )

When solved, the equation provides a wave function. The wave function contains all the measurable information about a particle. If you square this function you obtain a probability density, which allows you to predict the likelihood of finding a particle. This very quickly successfully addressed a large variety of atomic and molecular problems. The price paid for this new way of looking at matter was that the classical and deterministic view of the atomic world was totally replaced by a system in which you could only calculate the probability of finding a particle in a certain place and time.

Werner Heisenberg (1901–1976)Heisenberg met Bohr in 1922, he was just 20 years old and working toward his PhD. At the end of a lecture Heisenberg raised an objection to what Bohr had been discussing. Bohr was so impressed that he invited Heisenberg on a walk and they discussed the difficulties of quantum theory for more than 3 hours. Bohr liked physical models of atoms, but Heisenberg thought it was nonsense to talk about electron orbits when it was obvious that you could not observe the electrons or their orbital paths.

Heisenberg went on to discover the first complete version of quantum mechanics. The new theory that was developed in parallel with Schrodinger had taken a completely different path. Heisenberg’s theory had abandoned any attempt to create a visualisation of the atom, ignored the ideas of waves and particles, and in itself was purely mathematical.

Heisenberg took de Broglie’s proposal and Schrodinger’s wave equation and was able to ascertain that there was a limit to how precisely you would measure pairs of quantities. The pair he identified was that of position and momentum. In 1927 Heisenberg announced his uncertainty principle, which is expressed mathematically as:

Δ Δ Δ Δx ph

E th

≥ ≥2 2π π

and

where ∆x is the uncertainty in the position of a particle, ∆p is the uncertainty in the momentum of a particle and h is Planck’s constant.

You can gain some insight into understanding the uncertainty principle by considering the following example. If you want to know the position of a particle, you will need to use a high-energy photon that has a small wavelength as a high-resolution probe to locate your particle. Now consider that the photon ‘bounces’ nicely off the particle and you detect the photon and calculate the position of your particle. You have located the particle very precisely, but the particle is now recoiling and you will have very little information about its momentum.

An even more bizarre implication of Heisenberg’s uncertainty principle (pointed out by Einstein) is that particles (i.e. energy) can appear out of the nothingness of space for a very brief period of time, and then disappear.

255

quAntA toquArks

These particles are called virtual particles and we will discuss some aspects of them in Chapter 15 ‘The particle zoo’.

In summary, Heisenberg’s contributions to atomic theory include:1 the development of his matrix mechanics, which, like Schrodinger’s wave

equation, allowed the nature of the atom and especially spectra to be explained2 the uncertainty principle, which set limits on the precision of measurements.

Wolfgang Pauli (1900–1958)Pauli and Heisenberg were great friends and studied as research students under Arnold Sommerfield at the University of Munich. Pauli’s introduction to quantum theory was as a student listening to Sommerfield’s lectures. Sommerfield had extended the Rutherford–Bohr model to include elliptical orbitals, and the Bohr–Sommerfield theory attempted to describe the hydrogen molecule.

During the decade after 1913, physicists focused on the role of quantum numbers. A major outstanding question was ‘How many quantum numbers were required to account for the observed chemical and physical behaviour of atoms?’

Pauli responded by presenting an ad hoc solution to address this puzzle. He found that by assigning four quantum numbers to each electron in an atom, combined with a set of rules, he was able to produce a system that explained a number of outstanding issues including the structure of the periodic table.

Several prominent scientists including Heisenberg, Bohr, Compton and Pauli had considered that a fourth quantum number related to electron spin might be required to describe the behaviour of electrons in an atom. The more experienced physicists hesitated, but two Dutch graduate students George Uhlenbeck and Samuel Gouldsmit published and described the fourth quantum number as spin on the basis of the ideas presented in a paper Pauli had published on his exclusion principle.

In summary, Pauli made several important contributions to atomic theory:1 He proposed that each electron in an atom would be described by four

quantum numbers.2 He proposed, through his exclusion principle, that no two electrons in an

atom could have a set of four identical quantum numbers.3 His exclusion principle and rules provided a system to explain the

arrangement and number of electrons in each atomic orbital, thus providing an explanation for the structure of the period table.

4 He proposed that a neutrino was also emitted in beta decay, which provided an explanation for the spectrum of beta-particle energies observed.

Figure 13.6.2 Werner Heisenberg

Figure 13.6.3 Wolfgang Pauli

CheCkpoinT 13.61 State what Schrodinger’s equation allows you to find.2 Outline Heisenberg’s contributions to quantum mechanics.3 Recall how Pauli’s work explained outstanding problems in the theories.

Gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory.

prActicAL eXperiences13 Birth of quantum mechanics

256

chApter 13This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTiviTy 13.1: ATomiC TheoryResearch and present a 5 minute speech assessing the contributions of Heisenberg and Pauli to the development of atomic theory.

Discussion questions1 Outline the contributions of Heisenberg and Pauli to atomic theory.2 Assess how these contributions have affected our understanding of

the atom.

Gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory.

chapter summary• LouisdeBroglieproposedthatallmatterhaswave

properties. These are called matter waves.• ThewavelengthofadeBrogliematterwavecanbe

determined by using de Broglie’s equation:

λ λ= =hp

hmv

or

where λ is the wavelength h is Planck’s constant m is the mass of the particle v is the velocity of the particle p is the momentum of the particle.

• Diffractionisapropertyofwavesanditischaracterisedby the waves bending around corners or spreading out after passing through a narrow slit.

• Diffractionpatternscanonlybeproducedbywaves.• In1927twodifferentresearchteamsusingdifferent

experimental techniques had verified Louis de Broglie’s hypothesis concerning the wave nature of electrons:– Clinton Davisson and Lester Germer, using electron

scattering– George Thomson firing a beam of electrons through

a thin gold foil and observing diffraction rings.

• LouisdeBroglieexplainedthestabilityoftheelectronorbits in the Bohr atom by stating that each orbit (stationary state) was occupied by an integral number of wavelengths that fitted around the circumference of the orbit.

• Heisenberg’scontributionstoatomictheoryinclude:1 the development of his matrix mechanics2 the uncertainty principle, which set limits on the

precision of measurements.• Pauli’scontributionstoatomictheoryinclude:

1 proposing that each electron in an atom would be described by four quantum numbers

2 proposing through his exclusion principle that no two electrons in an atom could have a set of four identical quantum numbers

3 his exclusion principle and rules provided an explanation for the structure of the periodic table

4 proposing that a neutrino was also emitted in beta decay, which provided an explanation for the spectrum of beta-particle energies observed.

257

quAntA toquArksreview questions

phySiCALLy SpeAkingCopy and complete the following text.

______________ atomic model published in ______________ was a mix of ______________ and ______________ concepts. In ______________ Louis de Broglie proposed that particles had both ______________ and ______________ properties. ______________ developed the idea of ______________ waves, which, when applied to the situation of an electron ______________ in Bohr’s atomic model, provided a mechanism to explain the stability of ______________. The concept of a ______________ wave in which the ______________ of the orbit was equal to the de Broglie ______________, calculated using λ =

hmv

.

For the hydrogen atom you can use the equation En

En =12 1 where E1 = –13.6 eV

to calculate the ______________ of the electron in a given orbit (n). You can then rearrange the relation E = hf to calculated the frequency. Substituting the frequency into λ = c/f you can calculate the ______________ associated with the electron for the orbit.

You calculate the wavelength for n = 1 and the first orbit (n = 1) has a circumference equal to one ______________. You recalculate the wavelength for n = 2 and the second orbit (n = 2) has a ______________ equal to ______________ wavelengths.

In 1927 ______________ and ______________ fired electrons at a ______________ target and observed ______________ phenomena that verified the ______________–______________ duality of electrons.

reviewing 1 Define diffraction.

2 Describe how interference can be achieved.

3 Recall instances in which diffraction occurs.

4 Outline an experiment you could conduct to observe interference.

5 In terms of the Bohr atomic model, explain how matter waves were used to account for the stability of electron stationary states.

6 Describe the experiment performed by Davisson and Germer in 1927 and outline its significance.

7 Louis de Broglie proposed that any kind of particle has both wave and particle properties. Recall the response given by the scientific community.

8 Recall and list the contributions made by Heisenberg to the development of atomic theory.

9 Recall and list the contributions made by Pauli to the development of atomic theory.

10 If the speed of an electron is increased, predict what would happen to its de Broglie wavelength.

11 In terms of de Broglie’s hypothesis, discuss the implications for locating an electron in the ground state.

Define diffraction and identify that interference occurs between waves that have been diffracted.

Explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis.

Describe the confirmation of de Broglie’s proposal by Davisson and Germer.

258

13 Birth of quantum mechanics

Revie

w Questions

SoLving proBLemS 12 Louis de Broglie equation is λ = h

mv.

a Recall what each term represents.b Calculate the wavelength of an electron in a television tube with

a velocity of 3.0 × 107 m s–1.

13 Consider an electron of wavelength of 1.0 m.a Calculate its velocity. b Is this wavelength possible?

14 Using de Broglie’s equation, calculate the wavelength of:a an electron travelling at 6.0 × 105 m s–1

b a proton travelling at 6.0 × 105 m s–1.

15 Calculate the velocity of a proton with a wavelength of 1.38 × 10–12 m.

16 a Calculate the wavelength of: i a person (80 kg) walking at 2 m s–1

ii a soccer ball (0.43 kg) kicked at a speed of 5 m s–1

iii a car (1000 kg) travelling at 80 km h–1.b Assess the validity of your calculations against real-world observation.

17 Consider an electron at rest.a Calculate the wavelength of the electron.b Is this situation possible?

18 A car of mass 850 kg is travelling very slowly at 7.0 × 10–37 m s–1.a Calculate the wavelength associated with the car.b Will you see the wave properties of the car?

Solve problems and analyse information using: λ = h

mv

259

quAntA toquArks

phySiCS FoCuSThe miCroSCope: how SmALL CAn we See?

The maximum magnification of microscopes in most schools is usually about 400–600 times. More expensive research microscopes can achieve magnifications of up 2000 times. You would think that if you made better lenses and optics you could increase your magnification to whatever you wanted. Yet this is not the case and the reason for this limit is related to the wavelength of the light that is illuminating the object you are examining.

You can think of the wavelength of the light as your probe, and any details smaller than your probe would be difficult to see (or resolve). A useful analogy is to consider that you are blindfolded and asked to examine an unknown object using only the palms of your hands instead of your fingers. Your fingers are smaller and therefore able to detect finer details. The same is true for the wavelength you use to look at an object through a microscope.

Now consider using something that has a shorter wavelength than light to illuminate the object. You might consider using X-rays or gamma rays, which certainly have shorter wavelengths. Unfortunately you cannot easily build an optical system that focuses these forms of radiation, and if you could the radiation tends to have such high energies that it would just pass straight through the object. Of course various types of X-ray machines use these attributes to image bones and airport luggage.

Now consider using a beam of electrons. They have an associated de Broglie wavelength that can be many times smaller than the wavelength of visible light.

You can easily focus and manipulate the beam using electrostatic and electromagnetic lenses. You can also easily ‘see’ the electrons using detectors ranging from a simple fluorescent zinc sulfide screen to specialised complex semiconductor arrays.

1 Discuss how the wavelength of the radiation that illuminates the object affects the detail you can see with a microscope.

2 Determine the size of the smallest feature you could visualise using a light microscope.

3 Research to find the maximum magnification possible using an electron microscope.

4 Research and investigate the different types of electron microscopes, including:a transmission electron microscopeb scanning electron microscopec scanning transmission electron microscoped reflection electron microscope.

exTenSion5 Outline the different ways in which specimens are

prepared if they are to be examined with an electron microscope.

6 Discuss why the way in which most specimens need to be prepared differs from they way they are prepared when we use an optical light microscope.

7 Investigate the advantages and disadvantages of using an electron microscope.


Describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties.

a b

Figure 13.7.1 (a) Light microscope, (b) electron microscope

14

260

20th century alchemists

neutron, beta decay, atomic mass, nucleons, strong nuclear force,

atomic number, nuclides, transmutation, alpha decay, decay series, atomic mass unit, fission,

nuclear reactor, controlled nuclear reaction, atomic pile, critical mass,

uncontrolled nuclear chain reaction, neutron scattering

Base metals to goldThe idea of changing base metals into gold has been an age-old quest of alchemists. Their attempts were not successful because all the types of reactions the alchemists performed were merely chemical reactions. To achieve their dream, they needed to change the identity of the atom by changing the number of protons in the nucleus. In this chapter we will trace the history of nuclear physics, the ‘modern alchemy’, from the early 1930s through to the construction of the first man-made nuclear reactor.

14.1 Discovery of the neutronRutherford proposed the idea of the neutron in 1920. At the time he considered the neutron to be a system comprising a proton and an electron tightly bound together. This system provided an explanation for the ejection of electrons from the nucleus during beta decay. James Chadwick (1891–1974) set to work to find the neutron.

Chadwick set up two main experiments in which he fired alpha particles at a beryllium target. He allowed the unknown radiation generated in the first experiment to pass through paraffin blocks and in the second experiment to pass through nitrogen gas.

He then applied the conservation laws of momentum and energy, by setting up simultaneous equations that involved the known masses and velocity measurements of the alpha particle and ejected nuclei from both experiments.

Chadwick solved these equations in order to determine the likely mass of the ‘unknown radiation’. A month after commencing these experiments, Chadwick submitted for publication a paper entitled ‘Possible existence of a neutron’ in which he reported the neutron to be just slightly more massive than a proton.

Figure 14.1.1 James Chadwick

261

quanta toquarks

Many late nights

CP Snow, a research student at the Cavendish Laboratory, recalled that Chadwick worked night and day for 3 weeks

in his quest to identify the neutron.

particles

Befoil paraffin

protons

(hydrogennuclei)

?α

Befoil

particles

N2 gas

nitrogennuclei

?α

Figure 14.1.2 Chadwick’s two main experiments

With the discovery of the neutron, the nucleus of the atom was considered to consist of protons and neutrons; the number of protons equalling the number of electrons in the neutral atom. The number of neutrons could vary to account for the observed differences in the atomic mass of the atom. The proton and the neutron are referred to as nucleons, when they are in a nucleus.

14.2 The need for the strong forceThe new model of the nucleus posed a new problem. What forces held the protons and neutrons together to form a stable nucleus?

The new model of the nucleus posed a new problem. What forces hold the protons and neutrons together to form a stable nucleus? Is it gravity? Assuming the size of the nucleus is roughly ~1 × 10–15 m, the gravitational force acting on a proton in the nucleus of a helium-4 atom due to the other proton and 2 neutrons is approximately 5.6 × 10–34 N (can you calculate it?)

The force of Coulomb (electrostatic) repulsion between the two protons in the nucleus is approximately 200 N, which is ~1035 times larger! (See Physics Phile ‘Repulsive protons’.)

Evaluate the relative contributions of electrostatic and gravitational forces between nucleons.

CheCkpoinT 14.11 Describe Rutherford’s vision of the neutron.2 Outline how Chadwick determined the mass of the unknown radiation from his experiment.3 Describe the structure of the atom prior to and after the discovery of the neutron.

Table 14.1.1 The properties of nucleons

ProPerty Proton neutronCharge +1.6 × 10–19 C Neutral

Mass 1.673 × 10–27 kg 1.675 × 10–27 kg

Discuss the importance of conservation laws to Chadwick’s discovery of the neutron.

Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties.

repulsive protons

The electrostatic force between charged particles follows Coulomb’s law, an almost identical law to Newton’s gravity, except that it’s

much stronger and it can be repulsive. Let’s calculate the repulsive force between two protons in a helium nucleus:

F kq q

delectrostatic

C

=

≈

= ×× ×−

1 22

919

9 0 101 6 10 1 6

.( . )( . 110

1 0 10230

19

15 2

−

−×

C

N

)

( . )

where k is the electrostatic constant, q1 and q2 are the charges of the protons and d is the distance between them.

20th century alchemists14

262

14.3 Atoms and isotopesThe number of protons in a nucleus is referred to as the atomic number (Z). The total number of protons and neutrons is called the atomic mass number (A).

TransmutationArtificial transmutation

The process of changing one element into another is called transmutation. Transmutation occurs naturally in stars through the process of nuclear fusion and here on Earth via natural radioactivity by which certain elements decay spontaneously.

In 1919, Rutherford fired alpha particles into nitrogen gas and detected a highly energetic particle that he identified to be a proton. This was actually the first artificially induced transmutation—the alpha particle had collided with the nitrogen nucleus to produce an oxygen nucleus and a highly energetic proton.

Rutherford initially detected the production of very high energy protons and, after further experiments in which he used a cloud chamber, he identified that this was not just a simple collision. Rather his calculations indicated that the nitrogen nucleus had absorbed the alpha particle and then ejected a high-energy proton. Rutherford then showed that the new nucleus was oxygen. He had changed nitrogen-14 into oxygen-17, as prescribed by this equation:

Natural transmutationToday we know that nuclei with more than 83 protons (Z > 83) or atomic mass numbers greater than 209 (A > 209) are unstable and decay. In these atoms, the repulsive electrical forces between the protons overcome the strong nuclear force. We also know that instability due to odd number pairing of protons and neutrons can occur in smaller nuclei. The two most common natural decay processes are the emission of alpha and beta particles.

Account for the need for the strong nuclear force and describe its properties.

CheCkpoinT 14.21 What is the difference in the magnitude of the electrostatic and gravitational forces between protons in the nucleus?2 List the properties of the strong nuclear force.

Figure 14.2.2 Force acting on a positive charge as it approaches the nucleus

Figure 14.2.1 Nuclear force versus separation between nucleons

attr

acti

onre

puls

ion

Forc

e

1 22 3 4 Separation of nucleons× 10–15 m

attr

acti

onre

puls

ion

Forc

e

Distancefrom nucleusrange of the

nuclear force

nuclear attraction

Coulomb repulsion

+

Clearly, the force of gravity was too weak to overcome the huge electrostatic repulsive force produced by the interaction between protons—a third fundamental force of nature was required to explain the stability of the nucleus. Experiments over the period from 1930 to 1950 showed that this new strong nuclear force possessed the following properties.1 The strong nuclear force does not depend on the charge, therefore all

nucleons (protons and neutrons) bind together with the same force. This is supported by evidence that protons and neutrons are equally likely to be ejected from a nucleus in a collision.

2 The strong nuclear force acts over short distances of about 1 × 10–15 m, the diameter of a nucleus, and within this range the force is much stronger than the electrostatic forces. The evidence supporting this is that otherwise the nucleus would have a tendency to attract more protons and neutrons.

3 The strong nuclear force between the nucleons acts only between adjacent nucleons. The evidence supporting this is based upon the observed stability of the nucleus.

Figure 14.2.1 shows the forces between two nucleons, and from the graph you can see that the strong nuclear force is at a maximum at a distance of approximately 1.3 × 10–15 m. If the nucleons are less than 0.5 × 10–15 m apart, a repulsive force is present.

If we consider the resultant force acting on a proton or an alpha particle approaching a nucleus, we see that the Coulomb repulsion increases and then sharply decreases once the particle is within the range of the strong nuclear force (see Figure 14.2.2).

Interactive

Module

263

quanta toquarks

14.4 TransmutationArtificial transmutation

The process of changing one element into another is called transmutation. Transmutation occurs naturally in stars through the process of nuclear fusion and here on Earth via natural radioactivity by which certain elements decay spontaneously.

In 1919, Rutherford fired alpha particles into nitrogen gas and detected a highly energetic particle that he identified to be a proton. This was actually the first artificially induced transmutation—the alpha particle had collided with the nitrogen nucleus to produce an oxygen nucleus and a highly energetic proton.

Rutherford initially detected the production of very high energy protons and, after further experiments in which he used a cloud chamber, he identified that this was not just a simple collision. Rather his calculations indicated that the nitrogen nucleus had absorbed the alpha particle and then ejected a high-energy proton. Rutherford then showed that the new nucleus was oxygen. He had changed nitrogen-14 into oxygen-17, as prescribed by this equation:

24

714

817

11He N O H+ → +

Natural transmutationToday we know that nuclei with more than 83 protons (Z > 83) or atomic mass numbers greater than 209 (A > 209) are unstable and decay. In these atoms, the repulsive electrical forces between the protons overcome the strong nuclear force. We also know that instability due to odd number pairing of protons and neutrons can occur in smaller nuclei. The two most common natural decay processes are the emission of alpha and beta particles.

Define the term ‘transmutation’.

CheCkpoinT 14.3Copy and complete the following table to identify the number nucleons and electrons in elements.

element mass number atomic number Protons neutrons electronsC

He

F

Xe

The number of neutrons (N) can be calculated by subtracting the atomic number (Z) from the mass number (A).

N = A – ZThe nuclear structure of an element is represented by Z

A X where X is the element symbol.

Atoms of each element have the same number of protons in their nuclei, but the number of neutrons can vary, and these atoms are called isotopes. There are 91 naturally occurring elements and we have identified more than 2500 different nuclei (or nuclides) associated with these. Of these isotopes, 90% are unstable and often quickly decay into other nuclei.

Describe nuclear transmutations due to natural radioactivity.


264

Alpha decayAlpha-particle emission usually occurs with large unstable nuclei. An alpha particle is identical to a helium nucleus 2

4He . After the nucleus has emitted the alpha particle, the atomic mass A decreases by four, corresponding to the loss of two protons and two neutrons, the atomic number Z decreases by two, corresponding to the loss of two protons, and the number of neutrons N decreases by two, corresponding to the loss of two neutrons.

The general equation for this process is:

ZA

ZAX Y→ +−

−24 α or more formally Z

AZAX Y He→ +−

−24

24

X represents the nuclei of the parent element and Y represents those of the daughter element.

A common naturally occurring alpha decay is that of uranium-238 decaying to thorium-234:

92238

90234U Th→ + α or more formally 92

23890

23424U Th He→ +

Beta decayThere are three different types of beta-decay processes: beta-minus (β–), beta-plus (β+) and electron capture.

Beta (minus) decay

The beta-minus particle is identical to an electron. Beta-minus decay occurs when the ratio of neutrons to protons is too high, and involves the transformation of a neutron into a proton, an electron and an antineutrino νe .

The transformation process is represented as:

n p e→ + − +β ν

The generic equation for this process is:

ZA

ZAX Y e→ + ++

−1 β ν or more formally Z

AZ

AX Y e e→ + ++ −1 10

00ν

The atomic mass Z remains the same, as the total number of nucleons is unchanged. The atomic number A increases by one and the number of neutrons N decreases by one, corresponding to the transformation of a neutron into a proton.

A commonly occurring beta-minus decay is thorium-234 to protactinium-234:

90234

91234

eTh Pa→ + +−β ν or again formally 90234

91234

10

00

eTh Pa e→ + +− ν

Beta (plus) decay

The beta-plus particle is identical to an anti-electron commonly called a positron. The process of beta-plus emission involves the transformation of a proton into a neutron, positron and a neutrino. The transformation process is represented as p n e→ + ++β ν .

The general equation for this process is:

ZA

ZAX Y e→ + +−

+1 β ν or more formally Z

AZ 1

A10

eX Y e→ + +− + ν

A commonly occurring beta-plus decay is the transformation of neon-19 to flourine-19:

1019

919

eNe F→ + ++β ν or again formally 1019

919

10

00

eNe F e→ + ++ ν

Figure 14.4.1 This Segrè chart is a plot of number of neutrons (N ) against the atomic number (Z ) for the uranium-238 decay series.

Z

N

146

145

144

143

142

141

140

139

138

137

136

135

134

133

132

131

130

129

128

127

126

125

12480 81 82 83 84 85 86 87 88 89 90 91 92 93Hg Tl Pb Bi Po At Rn Fr Ra Ac Th Pa U Np

238U

234Th

230Th

234Pa234U

226Ra

222Rn

218Po

214Pb

210Pb

210Tl

214Bi

210Bi

210Po206Pb

214Po

α decay

β decay

4.47 × 109 y

2.46 × 105 y

7.54 × 104 y

1.637 × 10–4 s

1600 y

3.82 d

3.10 min

27 min

19.9 min

1.30 min

22.3 y

5.01 d

19.9 min

138.38 d

24.10 d

70 s

the unseen

In many parts of the world the bricks, cement and soil contain significant

quantities of isotopes from the uranium-238 series. Radium-226 decays to radon-222—a colourless, odourless, inert radioactive gas that is heavier than air and often settles in poorly ventilated cellars and basements. Now consider breathing in a radioactive gas! It decays by ejecting a high-energy alpha particle that can easily damage lung cells. But that is not the worst of it! A radioactive nucleus of polonium-218 remains; it is not a gas and is very likely to lodge itself in your lung tissue and continue to decay down the 238U series. So avoid hiding in the cellar or basement!

265

quanta toquarks

CheCkpoinT 14.41 Define a transmutation and outline how Rutherford made this happen.2 Identify how Rutherford showed that the high-energy protons were not the result of a collision.3 List the properties of an atom that will naturally transmute.4 List the types of decay that can occur to help stabilise a nucleus and the properties each possess.

Figure 14.4.2 Plot of atomic mass (A ) against the atomic number (Z ) for the uranium-238 decay series

Z

A238

234

230

226

222

218

214

210

20681 82 83 84 85 86 87 88 89 90 91 92Tl Pb Bi Po At Rn Fr Ra Ac Th Pa U

238U

234Th

230Th

234Pa234U

226Ra

222Rn

218Po 218At 218Rn

214Pb

210Tl 210Bi 210Po

206Tl206Tl 206Pb

α decay

β decay

214Bi 214Po

24 d 6.7 h

4.5 x 109 y

2.5 x 105 y

7.5 x 104 y

1600 y

3.8 d

3.1 min0.04 s1.6 s

3.1 min 1.6 s

27 min

1.3min

4.2min

22 y 5 d

20 min20 min 1.6 x 10–4 s

5 d 138 d

210Pb

Electron capture

This is the process in which an electron from an inner shell is captured by the nucleus, resulting in the conversion of a neutron to a proton, for example:

417

37

eBe e Li+ → +− νThe three types of beta decay are mediated by the fourth force of

nature called the weak nuclear force, which was proposed by Enrico Fermi in 1934 to explain beta decay.

As Rutherford and Soddy’s 1903 paper stated, often a series of decays occurs. For example, in nature the decay series for urainium-238 is common. Figure 14.4.1, a Segrè chart, and Figure 14.4.2, a plot of atomic mass against atomic number, both show the decay series for uranium-238 ( 92

238U), which undergoes eight alpha decays and six beta decays, terminating at the stable isotope lead-206 ( 82

206Pb). The original nucleus for each transition is called the parent and the resultant nucleus is called the daughter.

Gamma radiationAfter the emission of an alpha or beta particle, the daughter nucleus is sometimes left in an excited state. The change in energy from this excited energy state of the nucleus to the ground state of the nucleus will result in the emission of a very high energy photon called a gamma ray and is represented by the Greek letter γ. Gamma radiation does not change the atomic number (Z) or the atomic mass (A) and therefore it is not an example of transmutation.

A general equation for gamma emission can be written as:

ZA

ZAX X* → + γ

The * denotes the nucleus is in an excited state.

Interactive

Module

activity 14.1



14.5 The neutrinoIn alpha-particle decay, the energies of the ejected alpha particles have well-defined values of kinetic energy, whereas in beta-particle decay a broad spectrum of kinetic energies for the beta particles is observed. Many scientists including Otto Hahn (1879–1968), Lise Meitner (1878–1968) and Chadwick had all carefully studied beta decay. By 1930 experiments clearly showed that the spectra of the kinetic energies of ejected beta particles when graphed was a smooth and continuous curve ranging from just above zero to a maximum

Figure 14.5.1 Energy distribution for beta particles

Kinetic energy of the β-particles (MeV)

Rel

ativ

e nu

mbe

r of

β p

arti

cles 9

876543210

1 2 3 4 5 6 7 8 9 10 11 12


266

value that was dictated by the parent nuclei. Refer to Figure 14.5.1 for a typical energy distribution for beta particles.

In 1930, Wolfgang Pauli (1900–1958), in a letter addressed to Lise Meitner and Hans Geiger, proposed that a new particle was also emitted during the beta-decay process. Pauli proposed that this new particle had no charge and would only very weakly interact with matter. This newly proposed particle would allow the distribution of energies to be explained. The energy released during a decay could be shared between the beta particle and the neutrino in any ratio; thus, the spectrum of energies observed could be explained.

Enrico Fermi formally proposed a theory for beta decay whereby a neutron in the nucleus of an atom was transformed into a proton, an electron and an antineutrino. His theory also proposed a fourth force of nature, the weak nuclear force (commonly referred to as the ‘weak force’). The scientific community accepted the theory, even though Reines and Cowan did not experimentally detect the neutrino until 1953.

Discuss Pauli’s suggestion of the existence of the neutrino and relate it to the need to account for the energy distribution of electrons emitted in β-decay.

the elusive neutrino

The thermonuclear reactions in the core of the Sun create an environment that produces streams of

neutrinos. As you stand at midday with the Sun overhead, approximately 1013 neutrinos per second pass through your body. At midnight when the Sun is on the other side of the Earth, amazingly still about 1013 neutrinos per second pass through your body. They have passed straight through the Earth! From this you can see how weakly neutrinos interact with matter, and shows the extent of the task physicists had to surmount to detect these elusive particles.

are you nuClear Free?

Did you know that one very widespread application of nuclear physics is the smoke detector? The

detector commonly used in houses has a small amount of americium-241, which emits alpha particles, which ionise the air between two charged electrode plates. A small current normally flows between the plates; but if smoke enters the detector, it disrupts the normal current and the alarm sounds.

CheCkpoinT 14.5Explain what led to the need for the neutrino.

14.6 Was einstein right?Using a mass spectrometer the masses of electrons, protons and nuclei (as ions) can be determined by examining their radius of curvature as they move through a magnetic field. The common unit used to measure these extremely small masses on an atomic scale is the atomic mass unit (amu), which is used rather than the kilogram (kg), the SI unit for mass. Using this unit the neutral carbon-12 atom has a mass of exactly 12 amu. It is easy to convert between atomic mass units (amu) and kilograms (kg).

1 amu = 1.6605 × 10–27 kg

Worked exampleQuesTion

a Convert 4.05 amu to kilograms.

b Convert 5.023 × 10–27 kg to amu.

soluTiona 4.05 × 1.6605 × 10–27 = 6.73 × 10–27 kg

b 5 023 10

1 6605 103 025

27

27

.

..

××

=−

− amu

Einstein was right! Using his energy–mass equivalence E = mc2 allows us to also convert an energy value into an equivalent mass or, vica versa, a mass into an energy. The common unit of energy in atomic physics is the MeV (mega-electron volts 1 × 106 eV). Converting between amu and MeV is also easy: 1 amu = 931.5 MeV.

267

quanta toquarks

Worked exampleQuesTionCalculate the equivalent energy contained in 0.0015 amu.

soluTion0.0015 × 931.5 = 1.4 MeV

Mass defect The total mass of a stable nucleus is less than the total sum of the masses

of the protons and neutrons it contains. How can this be? Einstein linked mass to energy in his equation E = mc 2. In our case, when a nucleus forms, some energy is radiated away. This loss of energy reduces the mass of the nucleus. Conversely, when the unstable nucleus of a large atom splits into two or more fragments, the combined mass of the daughter nuclei (and possible neutrons) is less than the mass of the parent nucleus. This loss of mass is caused by some of the original parent mass being transformed into the kinetic energy of the fragments and also possibly gamma radiation.

The difference in mass between the total original mass and the total final mass is called the mass defect.

Worked exampleQuesTionCalculate the mass defect for a helium-4 atom ( 2

4He), given the rest mass of a helium atom is 4.002602 amu.

soluTionA helium-4 atom consists of a nucleus (2 protons and 2 neutrons) and 2 electrons.

Calculate the total mass of these constituent particles:

• Massofprotons=2× 1.007276 amu• Massofneutrons=2× 1.008665 amu• Massofelectrons=2× 0.000549 amu• Thetotalmassoftheconstituentparticlesis4.032980amu.

Calculate the difference in masses by subtracting the total mass of the constituent particles from the actual mass of the atom:

4.032980 amu – 4.002602 amu = 0.030378 amu

Therefore the mass defect for the helium-4 atom is 0.030378 amu.

Explain the concept of a mass defect using Einstein’s equivalence between mass and energy.

Solve problems and analyse information to calculate the mass defect and energy released in natural transmutation and fission reactions.

Figure 14.6.1 The mass of constituent parts is greater than the mass of the whole nucleus.

CheCkpoinT 14.61 State what amu stands for and give reasons for its use.2 Convert 3.1 amu to kilograms.3 Calculate the number of MeV from 0.3 amu.4 Define mass defect.


268

14.7 Binding energyAnother useful way of interpreting this mass defect in atoms and nuclei is to look at it in terms of energy. We can use Einstein’s equivalence of mass and energy to convert this mass defect into what is referred to as the binding energy of the nucleus. The binding energy tells you how much energy you would need to separate the nucleus of the atom back into separate protons and neutrons.

Worked exampleQuesTionCalculate the binding energy for the helium-4 ( 2

4He) nucleus.

soluTionIn our previous calculation we determined the mass defect for the helium-4 atom ( 2

4He) to be 0.030378 amu.

Using Einstein’s equivalence between mass and energy we simply convert the mass defect into the units of energy by using the relationship 1 amu = 931.5 MeV.

The binding energy for helium-4 nucleus ( 24He) = 0.030378 × 931.5 = 28.3 MeV.

To investigate the stability of various nuclei, the binding energy (Eb) per nucleon can be calculated by dividing the binding energy by the total number of nucleons (protons and neutrons) in the nucleus.

= b=E

ABinding energy per nucleon

binding energy

atomic mass number

Worked exampleQuesTionCalculate the binding energy per nucleon for helium-4 nuclei ( 2

4He).

soluTionIn our previous calculation we determined the binding energy for the helium-4 atom ( 2

4He) to be 28.3 MeV. We know that the atomic mass number for helium-4 is 4.

Therefore: Binding energy per nucleonbinding energy

atomic mass number= b=

EA

28.3 MeV7.075 MeV

4= =

We can see from Figure 14.7.1 that elements with atomic mass numbers between 40 and 80 have nuclei that are tightly bound. The binding energy of elements with atomic mass number greater than 80 is slightly less, and the binding energy of elements with atomic mass number less than 40 decreases sharply. You will notice that 2

4He has quite a high binding energy per nucleon, which makes it very stable and explains why alpha particles 2

4He rather than single protons are ejected from nuclei in alpha decay.

The graph also shows that if you fuse light nuclei together you would increase the binding energy per nucleon, thus energy would be released. This process, called fusion, is the nuclear reaction that provides the energy source for stars. You can see that if the nucleus of a heavy atoms splits, the binding energy per nucleon will increase. This process is called fission, and is the energy source used in nuclear reactors.

269

quanta toquarks

14.8 nuclear fission Between 1934 and 1938, Enrico Fermi and his fellow researchers fired

neutrons at a variety of target elements and produced many new unstable radioactive nuclei. In the majority of cases some nuclei in the target would absorb a neutron and then emit a beta particle. This beta-decay process transforms the neutron into a proton, an electron and an antineutrino, thus increasing the atomic number by one. Fermi predicted that he should be able to produce transuranic elements (elements beyond uranium) with atomic numbers greater than 92 (Z > 92). Fermi bombarded uranium with neutrons and tested the properties of the new radioactive elements. He was sure he had produced element number 93.

Otto Hahn (1879–1968), Lise Meitner (1878–1968), Otto Frisch (1904–1979), Fritz Strassman (1902–1980), Irene Joliot-Curie and Pavle Savitch (1909–1994) all repeated Fermi’s experiments in order to identify the new isotopes. With Hitler’s invasion of Austria in 1938, Lise Meitner (an Austrian Jew) emigrated to the safety of Sweden. Hahn and Strassman continued the work and, a few months later, Hahn wrote to Meitner outlining the final results of their research, which indicated that smaller nuclei were present. Further research supported these results, confirming that the decay products of the bombardment of uranium were not transuranic elements but rather unstable isotopes that included barium (Z = 56), radium (Z = 88) and lanthanum (Z = 57), all of which have nuclei with atomic numbers much less than that of uranium (Z = 92).

Describe Fermi’s initial experimental observation of nuclear fission.

CheCkpoinT 14.71 Define binding energy.2 Referring to the graph (Figure 14.7.1), explain how binding energy can help you understand when energy is

released or absorbed.

UHg

fission

KrFe

greateststability

Ca20Ne16O

12C4He

fusion

0 50 100 150 200Mass number (A)

Bin

ding

ene

rgy

per

nucl

eon

Eb/

A (

MeV

/nuc

leon

) 9

8

7

6

5

4

3

2

1

0

Figure 14.7.1 Binding energy is greatest for elements with atomic mass numbers between 40 and 80.


270

14.9 Chain reactionsLeo Szilard (1898–1964) realised that if the fission fragments included two or more neutrons, these neutrons could cause other uranium nuclei to split, resulting in a self-sustaining chain reaction. Further experiments by American and French scientists showed that indeed the fission of a uranium nucleus releases between two and four energetic neutrons. There are many dozens of possible decay pathways, for example:

92238

92239

3690

56146 3U n U Kr Ba n+ → → + +

The scientists working on this research realised that this chain reaction if controlled could be used as a useful power source, but if it was allowed to proceed uncontrolled, it could produce a huge explosion. In August 1939, war with Germany seemed imminent and the American-based scientists felt that America should investigate the possibility of building a bomb. Leo Szilard, Eugene Wigner and Edward Teller drafted a letter that was signed by Einstein and sent to the US President Franklin D Roosevelt.

It was known that 92238U would not support a chain reaction because the

neutrons released during the fission process did not have the right energy to cause other 92

238U nuclei to split. However, the rarer naturally occurring isotope

92235U acted differently and was known to fission when struck by a slow neutron.

In 1935 Enrico Fermi and his team set about designing and carrying out an experiment to see if a controlled nuclear reaction was possible. In 1942 Fermi’s team designed and commenced building a fission reactor in a squash court at Chicago University. Some 40 000 graphite bricks, weighing 350 tonnes, were used as a moderator to slow down the high-energy neutrons. They also provided the reactor with a structural component that gave the reactor its name, ‘the pile’. Natural uranium in the form of uranium oxide, which contains about 0.7%

92235U , provided the fuel for this reactor, and a set of control rods made from sheets of cadmium nailed to sticks of timber acted as ‘neutron sponges’. These could be withdrawn or inserted into the ‘pile’ to control the number of neutrons and thus the rate of the fission reactions.

The atomic pile took 6 weeks to construct, and in early December 1942 the control rods were withdrawn 15 cm at a time. With each withdrawal the neutron counters would climb and level off. At 2:00 pm, Fermi announced that, based upon his calculations, the next withdrawal would result in a self-sustaining reaction. As predicted, the neutron intensity increased more and more rapidly. After 4

12 minutes Fermi raised his hand and announced: ‘The pile has gone

critical’. The control rods where pushed back in and the reactor shut down.

Describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942.

earth’s oWn Fission nuClear reaCtor

Long ago, actually about 1.7 billion years ago in what is now

Gabon, Africa, there was a large natural uranium deposit. At that time in geological history there was more urainium-235 present in ore bodies; approximately 3% compared to today’s average 0.7%. The scene was set and nature provided the final ingredient: a flowing aquifer provided a water source that could act as a natural moderator. In the 1970s this ore body was mined and the ratio of 235U to 238U was found to be significantly lower than normal, revealing the telltale fingerprint of this ancient natural nuclear reactor.

CheCkpoinT 14.81 Identify what a transuranic particle is.2 Outline the experiment performed by Fermi which resulted in nuclear fission and not the production of

transnuranic elements.

271

quanta toquarks

Controlled and uncontrolled nuclear chain reactions A chain reaction requires:

1 fuel that is capable of fission; referred to as fissile material2 an amount of fuel that produces sufficient neutrons to cause new fission;

this is referred to as the critical mass3 neutrons with an energy that allows them to be absorbed by other fuel

nuclei. Slow (or ‘thermal’) neutrons are often best, as the de Broglie wavelength is larger at low speeds. This larger wavelength increases the likelihood of a neutron interacting with nearby nuclei and hence the likelihood of it being absorbed.

The conditions for a controlled nuclear chain reaction are such that the available neutrons, which cause the fission, are regulated. The control mechanisms in a nuclear reactor include the use of neutron-absorbing materials within the physical structure and the manipulation of control rods.

In an uncontrolled nuclear chain reaction, such as in a nuclear reactor meltdown or in a fission atomic bomb, the production of neutrons goes unchecked and the fission reactions increase at an accelerating rate. This process releases an enormous amount of energy in a very short period of time and results in a significant explosion.

Compare requirements for controlled and uncontrolled nuclear chain reactions.

Figure 14.9.1 Fermi’s atomic pile

CheCkpoinT 14.91 Explain how a chain reaction can be sustained in nuclear fission.2 Outline Fermi’s experiment to show nuclear fission.3 Give reasons why artificial nuclear fission is a desired process.


272

14.10 neutron scatteringThe discovery of the neutron and the further development of nuclear reactors provided scientists and industry with a powerful analysis tool. Chadwick’s work had revealed that neutrons were subatomic particles with a mass approximately the same as that of the proton, and no net electric charge. From de Broglie’s work we also know that neutrons will exhibit wave properties and their de Broglie wavelength will vary, depending on their velocity. Interestingly, although neutrons have no net charge, it was discovered they do have the property of spin; they therefore possess a magnetic property and act like little subatomic magnets. These properties make the neutron an excellent tool with which to probe aspects of atomic structure.

There are four main applications of neutron scattering:• Spacing: The de Broglie wavelength associated with slow (thermal) neutrons

allows investigations to determine the structure and spacing between atoms in solids and liquids.

• Motion: The energy of slow (thermal) neutrons is similar to the energies associated with the movements of atoms in solids and liquids, allowing researchers to investigate atomic vibrations and the forces between atoms.

• Magnetic structure: The neutron acts like a subatomic magnet, which allows researchers to examine at the atomic level the structures of semiconductors and magnetic materials.

• Inner structure: X-rays and electrons are scattered by atomic electrons, but neutrons are scattered by atomic nuclei and, unlike alpha particles, are not influenced by electrostatic forces. This allows neutrons to penetrate dense materials to depths of several centimetres, therefore making it possible for researchers to study material deep inside large pieces of equipment (such as aircraft engines), and inside containment vessels that have varying conditions of pressure, temperature and environment.

The first neutron-scattering experiments were carried out by Ernest Wollan and Clifford Shull in 1945 using another graphite pile reactor built at Oak Ridge. Today neutron diffraction is used in combination with X-rays to investigate the structures and properties of a wide range of materials.

The Bragg Institute at the Australian Nuclear Science and Technology Organisation (ANSTO) OPAL reactor facility, located just south of Sydney, currently houses eight neutron-beam instruments. The neutron-scattering and X-ray techniques are used to solve complex research and industrial problems across a wide range of fields including engineering, pharmaceuticals, mining, plastics and biology.

Describe how neutron scattering is used as a probe by referring to the properties of neutrons.

CheCkpoinT 14.101 Outline how neutrons can be used to probe atomic structure.

2 List applications of neutron scattering.


273

quanta toquarks

Chapter 14This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTiviTy 14.1: DeTeCTing rADiATionUsing different pieces of detection apparatus, determine the type of radiation and the penetration ability of the radiation being emitted. Equipment list: alpha, beta and gamma radiation sources, aluminium foil of varying thicknesses, paper, 2 mm thick lead sheets, Geiger–Müller tube, spark counter, cloud chamber, dry ice, methylated spirits, ruler.

Discussion questions1 List the apparatus best suited to detect each form of radiation. 2 In order of penetrating ability, list each radiation. 3 Identify the properties that make each radiation identifiable.

Perform a first-hand investigation or gather secondary information to observe radiation emitted from a nucleus using the Wilson Cloud Chamber or similar detection device.

274

14 20th century alchemists Chapter summary

• Rutherfordproposedtheideaofthe‘neutron’in1920.• In1934Chadwickfiredalphaparticlesataberyllium

target and allowed the unknown radiation generated to pass through paraffin blocks and nitrogen gas. Chadwick then applied the conservation laws of momentum and energy to determine the mass of the neutron.

• Theprotonandtheneutronarereferredtoasnucleonswhen they are in a nucleus.

• Athirdfundamentalforcecalledthestrongforcewasrequired to explain the stability of the nucleus, as the force of gravity was too weak to overcome the huge electrostatic repulsive force produced by the interaction between protons.

• Thestrongnuclearforcehasthefollowingproperties:– It does not depend on the charge, therefore all

nucleons (protons and neutrons) bind together with the same force.

– It acts over a short distance, in the order of the size of a nucleus, and the force within this range is much stronger than the electrostatic forces.

– It acts only between adjacent nucleons. • Anelement’snuclearstructureisrepresentedbyZ

A X where X is the element symbol.– The number of protons in a nucleus is referred to as

the atomic number (Z). – The total number of protons and neutrons is called

the atomic mass number (A).– The number of neutrons (N) can be calculated by

subtracting the atomic number (Z) from the mass number (A).

• Isotopesareatomsofanelementthathavethesamenumber of protons but a different number of neutrons in their nuclei.

• Theprocessofchangingoneelementintoanotheriscalled transmutation.

• Thefirstverifiedartificiallyinducedtransmutationwasreported in 1919 by Rutherford when he fired alpha particles into nitrogen gas.

• Thetwomostcommonnaturaldecaytransmutationprocesses are the emission of alpha and beta particles.

• Alphadecayusuallyoccurswithlargeunstablenucleiand an alpha particle is identical to a helium nucleus

24He .

• Therearethreedifferenttypesofbeta-decayprocesses:beta-minus (β–), beta-plus (β+) and electron capture.

• In1930Pauliproposedtheexistenceoftheneutrinotoaccount for the energy distribution of electrons emitted in beta decay.

• Thetotalmassofastablenucleusislessthanthetotalsum of the masses of the protons and neutrons it contains.

• Thedifferenceinmassbetweenthetotaloriginalmassand the total final mass is called the mass defect.

• Thebindingenergytellsyouhowmuchenergyyouwould need to separate the nucleus of the atom back into separate protons and neutrons.

• Fermi’sinitialnuclearfissionexperimentsinvolvedfiringneutrons at a variety of target elements and producing many new unstable radioactive nuclei.

• In1942Fermi’steamdesignedandbuiltafissionreactor (an atomic pile) in a squash court at Chicago University.

• Achainreactionrequires:1 fuel that is capable of fission 2 a ‘critical mass’ of fuel, which produces sufficient

neutrons to cause new fission3 neutrons with an energy that allows them to be

absorbed by other fuel nuclei. • Acontrollednuclearchainreactionrequiresthenumber

of available neutrons that cause the fission to be regulated.

• Anuncontrollednuclearchainreactionoccurswhentheproduction of neutrons goes unchecked and the fission reactions increase at an accelerating rate.

• Neutron-scatteringtechniquesareusedtosolvecomplexresearch and industrial problems across a wide range of fields including engineering, pharmaceuticals, mining, plastics and biology.

275

quanta toquarksreview questions

physiCAlly speAkingPretend you are Charles Wilson, the inventor of the cloud chamber, shown in Figure 14.11.1. Name the components of your new invention. Prepare a short scientific report on your new invention for presentation to your class.

a Discuss the features of the apparatus; you could also include a description of what happens when you place a magnet on top of the apparatus.

b Complete your report, outlining the importance of it to reveal the nature of matter.

revieWing 1 Name the scientist who predicted the existence of

the neutron.

2 Copy and complete the following table for the properties of the nucleons of an atom.

ProPerties neutron ProtonSymbol

Charge

Mass (kg)

3 Define the term transmutation.

Discuss the importance of conservation laws to Chadwick’s discovery of the neutron.

4 Recall how the conservation laws used by Chadwick allowed him to identify the neutron.

5 Assess why the invention and development of the Wilson cloud chamber was important in identifying transmutation.

6 Describe the properties of the strong force.

7 Copy and complete the following table.

Process a brief descriPtion of each ProcessFission

Fusion

Alpha emission

Beta emission

Describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942.

Compare requirements for controlled and uncontrolled nuclear chain reactions.

8 Outline Fermi’s 1942 demonstration of a controlled nuclear reaction.

9 Describe the requirements for:a a controlled chain reactionb an uncontrolled nuclear reaction.

Figure 14.11.1 Wilson cloud chamber

B

E

C

A

DF

276

14 20th century alchemists

10 Copy and complete the following table which outlines the events associated with the discovery and identification of the neutron.

scientist(s) exPeriment contribution1920 Firing alpha particles into nitrogen gas Observed transmutation and also predicted the

existence of the neutron

1930 Bothe and Becker Identified the production of a weak but penetrating ‘radiation’

1932 Joliot Curie and ______________

Fired alpha particles into a beryllium target and allowed the ‘radiation’ to pass through a paraffin block

Published a paper describing …

1934 ______________ Fired ______________ into a beryllium target and allowed the radiation to pass through a ______________ and ______________

Used the conservation laws on his measurement observations and identified that the ‘radiation’ was most probably composed of ______________

11 Referring to Figure 14.11.2, describe in words the force acting on a proton or an alpha particle that is fired directly toward the nucleus.

Discuss Pauli’s suggestion of the existence of the neutrino and relate it to the need to account for the energy distribution of electrons emitted in β-decay.

12 This question relates to the graph in Figure 14.11.3.a Explain why the distribution of kinetic energies

for the beta particles was so baffling for the scientists.

b Name and recall the properties of the new particle suggested by Pauli.

c Account for the energy distribution of the electrons emitted in beta decay.

13 Describe the following processes associated with nuclear transmutations due to natural radioactivity. Include any other particles that may be present in the process and write word equations for each process.a Alpha emission (α-decay)b Beta emission (β-decay)c Emission of gamma radiation (γ)d Another naturally occurring transmutation

solving proBlems 14 Consider a helium nucleus where the distance

between the protons is 1 × 10–15 m and given that

F kq q

dF G

m m

delectrostatic gravitationaland= =1 2

21 2

2

a Calculate the electrostatic force between the two protons.

b Calculate the gravitational force between the two protons.

c Account for the need of the strong force.

Figure 14.11.2

distancefromnucleus

Coulomb repulsion

nuclear attraction

repulsion in the nucleus

range of thenuclear force

attr

acti

onre

puls

ion

Forc

e

Figure 14.11.3

KE of beta particles

N

277

quanta toquarks

Solve problems and analyse information to calculate the mass defect and energy released in natural transmutation and fission reactions.

15 Calculate the mass defect for a lithium-6 atom (36Li),

given the rest mass of a lithium-6 atom is 6.015122 amu and the following rest masses:

proton = 1.007276 amu neutron = 1.008665 amu electron = 0.000549 amu

16 Helium ( 24He) has a binding energy of 7.1 MeV

per nucleon. Given that the mass of an electron is 0.000549 amu, a proton is 1.0073 amu and a neutron is 1.0087 amu, calculate:a the total binding energy in the 2

4He nucleus in MeV

b the mass defect of the 24He nucleus in amu

c the mass of a 24He atom in amu.

17 Figure 14.11.4 shows a graph of average binding energy versus mass number.a Define the term binding energy.b Construct an argument based on the information

in Figure 14.11.4 why alpha particles rather than just protons are ejected from some radioactive nuclei.

c Estimate the binding energy of iron.

18 Consider this fission reaction of the uranium-235 nucleus:

92235

01

53131

01U n I ? 3 n+ → + +

a Identify the missing product in the above equation.b From the information in Figure 14.11.4, estimate

the binding energies of the original uranium-235 and the two daughter nuclei.

19 Identify the missing components in the following reactions:

a 01

2965

2966n Cu ?+ →

b 1122

10?

10Na ? e→ + ++ ν

20 Tritium and deuterium are isotopes of hydrogen and the symbols T and D are often used when writing nuclear reactions.

Consider the following fusion reaction:

13

12

24

01T D He n+ → + + γ

Given that the mass of deuterium is 2.014 amu, tritium is 3.016 amu and helium is 4.003 amu, calculate the energy of the γ-ray.

Figure 14.11.4

0 50 100 150 200 250

Atomic mass (A)

Bin

ding

ene

rgy

per

nucl

eon

(MeV

)

9

8

7

6

5

4

3

2

1

0

235U

208Pb56Fe16O

12C

9Be6Li

3H3He2H1H

4He

2.9 MeV/nucleon

1.15 MeV/nucleon

Revie

w Questions

278

14 20th century alchemists

physiCs FoCusproBing ATomiC sTruCTureIn 1934 Ernest Rutherford’s prediction of neutrons was proved by his associate James Chadwick, work that gained Chadwick the Nobel Prize in Physics in 1935. More recently, Bertram Brockhouse and Clifford Shull shared the Nobel Prize for Physics in 1994 for pioneering contributions to the development of neutron-scattering techniques for studies of condensed matter. This is the modern application of neutron scattering—determining the arrangement and motion (structure and dynamics) of atoms and molecules, both of which can determine the physical properties of the material they form.

Today, these methods are practised at neutron-scattering centres such as the Bragg Institute at the Australian Nuclear Science and Technology Organisation (ANSTO) located at Lucas Heights, south of Sydney. Neutrons are used to study atomic positions, motions and magnetic properties of materials. Neutron scattering provides unique information about a material that cannot be obtained by other methods.

1 Discuss why neutrons are so useful in probing the structure of materials.

2 Discuss the role of neutrons in fission reactors and outline how they could be used as a neutron source.

3 You are a physicist at ANSTO, describe how you would construct apparatus to provide your researchers with a beam of neutrons.

exTension4 Investigate and report on the ways in which you

could detect scattered neutrons.5 Neutron scattering is very sensitive to different

isotopes of an element (atoms that have nuclei with the same number of protons but different numbers of neutrons). Outline some reasons why this would be the case.


Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties.

Figure 14.11.5 Dr Vanessa Peterson is changing a sample in a high-temperature furnace near Echidna, the high-resolution neutron powder diffractometer at ANSTO. In this experiment, the structure of the material was determined, as it changed with temperature.

279

15 The particle zoo The atomic ageThe unleashing of the power stored in the nucleus of an atom dramatically shaped the world we live in today. The detonation of two atomic bombs dropped on Japan in 1945, and the post-World War II threat of atomic weapon proliferation, leading to the Cold War between the Soviet Union and the West, changed the lives of millions of people. On the scientific front, the development of particle accelerators allowed the three constituents of matter known in the 1930s to grow to several hundred by the mid-1970s. The secrets of the atom were being uncovered and the actual fabric of our universe was being unveiled. The giant accelerators were recreating the processes that occurred during the first few fractions of a second during the birth of time and space—the Big Bang.

15.1 The Manhattan Project The United States of America’s top-secret nuclear bomb research project,

code named the Manhattan Project, ushered in the atomic age. The project comprised many sub-projects including Fermi’s ‘atomic pile’ in Chicago. Another reactor was built near Richland in Washington to produce plutonium, and a gaseous diffusion plant in Oak Ridge Tennessee was constructed to extract the fissile isotope 92

235U from uranium ore. At the famous Los Alamos facility in New Mexico, the theoretical and experimental work for the design, development and testing of the atomic bomb was carried out.

At Los Alamos, research into the construction of two types of atomic bombs was carried out. A uranium-235 bomb named ‘Little Boy’ was never tested prior to being dropped on Hiroshima on 6 August 1945. A more complex plutonium-239 bomb, which was tested at the Trinity site in the New Mexico desert, resulted in the construction of the bomb named ‘Fat Man’, which was dropped on Nagasaki on 10 August 1945.

The driving fear that fuelled the Manhattan Project centred on the possibility that Germany was building its own bomb. After World War II had ended, it was found that Hitler’s military had invested the majority of its scientific research and development into conventional weapons and rocketry. They had access to a heavy water plant in Norway, but very little progress had been made toward the construction of an atomic bomb.

Manhattan Project, isotope, Los Alamos, fusion, fission, fuel rods, core, moderator, control rods, coolant,

radiation shielding, radioisotopes, half-life, particle accelerator, cyclotron,

positron, radiopharmaceuticals, radiotherapy, irradiated, cosmic rays,

muons, synchrotron, linear accelerators, quarks, hadrons, bosons,

pions, Standard Model, mesons, baryons, leptons

Activity 15.1



The particle zoo 15

280

The Manhattan Project: its legacy The Manhattan Project left a legacy that shaped the 20th century in ways

that few could have imagined. The project led to the post-World War II nuclear arms race between the Soviet Union and the West. After the test detonation of the Soviet Union’s first bomb in 1949, relations between the Soviet Union and the West deteriorated. This tension became known as the Cold War and continued until the early 1990s; it ended with the dissolution of the Soviet Union. The United States of America, Britain and the Soviet Union developed fusion bomb technology, known as the thermonuclear bomb, the H-bomb or the hydrogen bomb. By the mid-1960s France and China had tested atomic weapons, and today Israel, India, Pakistan and North Korea also possess atomic weapons. Despite the Nuclear Non-proliferation Treaty, which pursued a reduction in weapon stockpiles, today there is a worldwide arsenal of approximately 20 000 nuclear weapons.

In tandem with the military developments, civil projects using nuclear energy have also proliferated. Today there are 440 nuclear power plants worldwide, which supply 15% of the world’s electricity demand, and there are more than 200 research reactors. The use of reactors for nuclear medicine, neutron scattering and other industry-based applications highlights the peaceful endeavours pursued by scientists and engineers today.

Gather, process and analyse information to assess the significance of the Manhattan Project to society.

CheCkPoinT 15.11 Outline the parts and purposes of the Manhattan project.2 Justify the project being ‘fuelled’ by the race to beat Germany to building the bomb.3 What was the Cold War?4 List developments that have come about due to the knowledge gathered from this project (other than weapons).

15.2 nuclear fission reactorsNuclear reactors have a variety of applications in modern society. They are used as an energy source to generate electricity for power grids and to power naval vessels. Research laboratories and industry use reactors as a neutron source to probe the structure and properties of materials, and irradiate materials to make a range of industrial, agricultural and pharmaceutical isotopes.

Components of a typical nuclear fission reactor Fission reactors are carefully designed to control the rate of fission

(splitting) of large nuclei to ensure their safe operation. The fission process releases energy, which manifests itself as heat and high-energy gamma radiation. For the purpose of this course we will limit our discussion to uranium fuel nuclear fission reactors, which comprise six key components: fuel rods, core, moderator, control rods, coolant and radiation shielding.• Fuel rods: These are tubes filled with pellets of uranium oxide. The fuel rods

are located in the centre of the reactor known as the core.• Core: The core of the reactor houses the fuel rods, control rods, a coolant

system and a moderator material (which sometimes also acts as a coolant).

Explain the basic principles of a fission reactor.

281

quANTA ToquARkS

• Moderators:These materials slow down neutrons, improving the chance of the neutron being captured by a nucleus. Commonly used moderators include ordinary water, heavy water and graphite. The neutrons are slowed by colliding inelastically with the moderator nuclei. For water, the neutron has about 20 collisions, which reduce its kinetic energy from about 2 MeV to 1 eV.

• Control rods: These are usually made from boron or cadmium and are located between the fuel rods. The control rods absorb neutrons and are adjusted so that the chain reaction proceeds at a constant rate. A constant rate is achieved when, for the average fission, each ejected neutron initiates one new fission. If the reactor needs to be shut down for service or in an emergency, the control rods are inserted fully into the core. These capture the neutrons and the chain reaction ceases.

• Coolant: The reactor generates heat and the coolant transfers this heat away from the core. In some reactors normal or heavy water is used as a coolant and also serves as a moderator. In nuclear electrical power stations, the heat is used to make steam, which powers a turbine and generator.

• Radiation shielding:Reactor cores emit large quantities of gamma radiation and neutrons. Lead and graphite are used to absorb and reflect radiation, which protects the containment walls of the reactor vessel. The walls themselves are made from high density concrete. These radiation shields are designed to protect people and the environment and prolong the working life of the reactor facility.

The nuclear reaction resulting from the absorption of a slow (or thermal) neutron by uranium-235 has many possible pathways for the decay process to proceed.

Following are three examples:

92235

92236

55140

3793 3U n U Cs Rb n+ → → + +

92235

92236

57147

3587U n U La Br 2n+ → → + +

92235

92236

3690

56143U n U Kr Ba 3n+ → → + +

containment building

controlrod

pump

fuelrod

pressuriser (prevents water boiling)

boiler

water acts as coolant and moderator in reactor core

primary coolantcircuit (high pressure water)

sea water to condense steam

water

electricityto gridcondenser

turbines and generator

Figure 15.2.1 A schematic diagram of a pressurised water reactor. A typical 1000 MW power plant consumes about 6 000 000 tonnes of black coal each year, or about 25 tonnes of enriched uranium that has been obtained from around 75 000 tonnes of ore.

Figure 15.2.2 The energy stored in the nuclear fuel undergoes a number of transformations.

Nuclear energyof fuel

Heat energy of coolant

Kinetic energy of steam and turbine

Electric energy

Light, heat,sound, etc.

The particle zoo 15

282

CheCkPoinT 15.21 Define fission.2 Construct a table that lists the six key elements in a nuclear reactor and outline their purpose.

15.3 RadioisotopesRadioisotopes are increasingly being used in medicine, scientific research and industry. They provide a unique tool that is often cheaper and more effective than alternative techniques and processes. Radioisotopes are atoms that have an unstable ratio of protons to neutrons and will decay via alpha or beta decay to attain a more stable configuration. Some may also emit gamma radiation. Each radioisotope has a specific half-life—the time it takes half of the radioisotope sample to decay. For example, if a sample contains 12 grams of a radioisotope with a half-life of 5 hours, in 5 hours’ time half of the radioisotope will have decayed and 6 grams will remain. Five hours later there will be 3 grams of the original radioisotope present in the sample, and so on.

Radioisotopes occur naturally or can be produced by changing the ratio of nucleons in the nuclei of a target material in a nuclear reactor or a particle accelerator. A nuclear fission reactor provides a source of neutrons that can be used to irradiate a target material. A particle accelerator such as a cyclotron can accelerate protons or ions and fire them into a target material. Both processes are required to produce the range of radioisotopes used in medical and industry-based applications.

In Australia, the Australian Nuclear Science and Technology Organisation (ANSTO) facilities in Sydney, produces a range of radioisotopes. ANSTO operates Australia’s only nuclear research reactor, OPAL, at Lucas Heights and the National Medical Cyclotron at The Royal Prince Alfred Hospital. It is necessary to have both the research reactor and the cyclotron, as each produces different types of radioisotopes. Pharmaceutical companies, research centres and hospitals operate other medical and industrial cyclotrons.

Radioactive tracersOften radioisotopes are used in medical and industrial applications to track the movement, flow or absorption of materials. The radioisotope can be incorporated into a molecule or introduced into the system. These isotopes then act as tracers and allow the position, flow or absorption to be mapped. In complex biological systems it is possible to map the pathway of specific chemicals being transported and used by cell tissue. For example, glucose is metabolised by brain tissue and so the radioisotope carbon-11 can be placed into the glucose molecule to act as a tracer in medical imaging. The carbon-11 via beta-decay is a positron emitter, and the radioactive labelling of the glucose molecules can be tracked and imaged using positron emission tomography (PET). An industrial application is the use of a radioactive tracer in mapping the dispersion of a sewerage ocean outfall.

283

quANTA ToquARkS

Medical radioisotopesNuclear medicine has been used routinely since the 1970s. Radioisotopes that decay quickly are said to have short half-lives, and are often used in diagnostic procedures. Radioisotopes with longer half-lives are often used therapeutically to target diseased organs and tumours.

Radiopharmaceuticals Radiopharmaceuticals can be classified as diagnostic or therapeutic. The

overarching process on which this pharmacology is based is the specific uptake and absorption of chemicals by organs and specific body tissues. This provides an ideal mechanism that enables radioisotopes to be incorporated into or attached to molecules, which are then tracked or used to target specific tissues.

Diagnostic radiopharmaceuticals are used to assess the functioning of organs including the lungs, heart, liver and brain; identify bone fractures not visible in X-rays; and assess the flow of fluids such as blood. The decay of the diagnostic pharmaceuticals can be monitored by detectors that may comprise a simple device such as a Geiger counter or a complex array of detectors that convert the information into an image. Diagnostic radiopharmaceuticals subject the body to a very low radiation dose, usually comparable to a routine diagnostic X-ray.

Therapeutic radioisotopes generally contain radioisotopes with longer half-lives and once absorbed by the specific tissue or organ will deliver a target dose of radiation.

Radiotherapy In radiotherapy the radiation emitted from a radioactive source is directed

at an area of diseased tissue. This procedure is referred to as teletherapy when the source is located outside the body. In brachytherapy a radioactive implant is used to target the specific tissue.

Positron emission tomography (PET)This imaging technique also relies on the principle that specific molecules are absorbed by specific organs or tissues. In this imaging procedure the radioisotope used will decay and produce a positron via beta decay. The positron (an antimatter electron) will collide with an electron and the two will annihilate, producing two gamma rays. These are detected and triangulated with other events by a computer, and the location and activity of the targeted organ or tissue is imaged. (See Chapter 19 ‘Imaging with gamma rays’ for more details.)

Industrial radioisotopesRadioisotopes are used widely in industry across applications as diverse as checking the structural integrity of bridges, determining wear in engine components, examining welds in gas lines, imaging internal structures in jet aircraft engines, and thickness control. Radioisotopes commonly used in measuring the thickness of materials include iridium-190 and cobalt-60. The most common (often unrecognised) application of a radioactive isotopes is smoke detection, and americium-241 is the radioisotope most commonly used in home-based detector systems.

Describe some medical and industrial applications of radioisotopes.

Figure 15.3.1 Brachytherapy for prostate cancer can be administered using ‘seeds’, small radioactive rods implanted directly into the tumour.

The particle zoo 15

284

Agricultural radioisotopes Radioactive isotopes are used in the agricultural industry as tracers in

plants to explore chemical and biological processes. As in medical applications, specific chemicals are taken up and used by specific parts of plants. Radioisotope tracers are attached or incorporated into molecules, which are then taken up by plants. For example, the radioisotopes phosphorous-32 or nitrogen-15 can be added to fertilisers and the uptake of the fertiliser measured by a sensitive Geiger counter. Similarly, toxic heavy metal compounds such as those of mercury or cadmium can be introduced into a test soil. The uptake of these tracer radioisotopes can be used to investigate potential problems associated with livestock feed or food for human consumption when planted in areas containing these toxins.

Food irradiation In many countries certain foods are irradiated in order to increase the

shelf life and make some foods safer to eat. The exposure of the food to gamma radiation targets disease-causing bacteria or those that cause spoilage. A common, widely used radioisotope is cobalt-60. The gamma rays passing through the food possess enough energy to destroy many bacteria without changing the texture or flavour of the food. The food never comes into contact with the radioactive source or other forms of radiation and therefore is not at risk of itself becoming radioactive. Examples of irradiated foods found worldwide include potatoes and onions in which sprouting is reduced, grains, meats including poultry and some fish, many spices and dried herbs, and fresh fruits.

Identify data sources, and gather, process, and analyse information to describe the use of:• anamedisotopeinmedicine• anamedisotopeinagriculture• anamedisotopeinengineering.

Figure 15.3.2 Commonly used radioisotopes and the organs in which they act

Pancreasselenium-75Pancreasselenium-75

Placentaiodine-123iodine-131carbon-11

Lymphgold-198Lymphgold-198

Bladdergold-198

Prostategold-198indium-111

Kidneytechnetium-99mmercury-197iodine-131

Kidneytechnetium-99mmercury-197iodine-131

Eyesphosphorus-32

Thyroidtechnetium-99miodine 123iodine-131iodine-132selenium-75

Lungsxenon-127xenon-133nitrogen-13oxygen-15carbon-11indium-113mtechnetium-99m

Brainiodine-131mercury-197technetium-99moxygen-15carbon-11

phosphorus-32arsenic-74indium-133miron-18

Heartrubidium-81mthallium-201cesium-137

Spleenchromium-51rubidium-81technetium-99m

Bloodiron-59Bloodiron-59

Livertechnetium-99miodine-131gold-198

Bonestrontium-85strontium-87fluorine-18iron-52phosphorus-32technetium-99m

Kneeyttrium-90rhenium-186

Kneeyttrium-90rhenium-186

Activity 15.2



285

quANTA ToquARkS

PhYSiCS FeATURehYdRogen AS An eneRgY CARRieR

Recent scientific interest in hydrogen as an alternative energy carrier to carbon-based fossil

fuels such as petrol has generated questions that can be answered using neutrons generated by a fission reactor, and neutron-scattering techniques. One of the problems with using hydrogen as an energy carrier is the difficulty in storing such a small molecule. Before we can study new hydrogen-storage materials, we need to know where the hydrogen is in the material and how it is interacting with the storage material.

At Australia’s OPAL research reactor, neutron diffractometers such as ANSTO’s high-resolution neutron powder diffractometer (Echidna) can be used to ‘see’ hydrogen. The protium (1H) in the material is replaced with deuterium (2H), and the neutron scattering by 2H that is obtained in a neutron-diffraction experiment allows the position of the deuterium in the material to be established.

For example, this technique was used to determine nine positions on the inner surface (referred to as adsorption sites for molecular deuterium) within the porous material Cu3(1,3,5-benzenetricarboxylate)2, which adsorbs hydrogen. The neutron scattering arising from the deuterium molecules in this material is shown in yellow in Figure 15.3.3.

Neutron spectrometers give information about the dynamics and motions of atoms and molecules. Neutron spectroscopy takes advantage of the exceptionally strong scattering from hydrogen 1H, which dominates the spectrum obtained. The scattering from 1H gives information about the dynamics (motions) of the hydrogen and can be used to understand the types of interactions that the hydrogen has with the host material. In the case of Cu3(1,3,5-benzenetricarboxylate)2, some of the nine sites for molecular hydrogen have a stronger adsorption enthalpy with the material than others.

Figure 15.3.3 Neutron scattering was used to identify the hydrogen sites in the porous material Cu3(1,3,5-benzenetricarboxylate)2.

CheCkPoinT 15.31 Define the term radioisotope.2 Explain what half-life is.3 Outline how radioisotopes can be made.4 Recall what is a radioactive tracer and what it is used for.5 Explain the significance of using different half-lives in medical radioisotopes.6 Outline what radiopharmaceuticals are used for.7 Compare and contrast radiotherapy and brachytherapy.8 Recall one use of radioisotopes in industry and one in agriculture.9 Why can radiation be used to sterilise food safely?

The particle zoo 15

286

15.4 Particle accelerators Prior to the development of particle accelerators, scientists had observed

showers of particles caused by cosmic rays colliding with the Earth’s atmosphere. By the late 1940s, Carl Anderson (1905–1991) had identified positrons and muons created by cosmic rays. Cosmic rays bombard the Earth continuously, and they provided many challenges to scientists, who had to use balloons to lift equipment high into the atmosphere to carry out many experiments. This was not very convenient and what was required was the development of a more controllable ground-based laboratory. The subsequent development of a variety of particle accelerators has provided scientists with the tools to probe and investigate the structure of matter.

Basic accelerator design and principles All particle accelerators have three basic components:

• asourceofchargedparticles(ionsorelementaryparticles)• atubeorchamberthatishighlyevacuatedsothattheparticlescantravel

without colliding with air molecules• amechanismtoaccelerateandcontrolthetrajectoryoftheparticles.

There are a number of common designs for particle accelerators including electrostatic, linear, cyclotron, betatron, synchrotron and storage-ring colliders for particle accelerators. Modern accelerator facilities often use a combination of these; for example, a Van de Graaff generator is often used to initially accelerate and inject a particle beam into a larger linear or synchrotron accelerator.

Electrostatic acceleratorsThere are two common electrostatic accelerators, which were developed in the 1930s: the Cockroft–Watson accelerator and the Van de Graaff accelerator. You maybe familiar with the school version Van de Graaff generator (200 kV). The early accelerators were scaled-up versions capable of accelerating particles across potentials of 1.5 MV. The Science Museum in Boston, Massachusetts, now houses the (physically) largest air-insulated Van de Graaff ever built. At the top it has two joined domes, one housing the top of the belt mechanism and the other housing the top of the original accelerator tube. The museum now uses the generator to produce bolts of artificial lightning. Modern Van de Graaff accelerators are smaller and use a tandem system that can accelerate ions across a potential of 20 MV.

Linear acceleratorsLinear accelerators comprise a very long straight evacuated tube that contains a set of hollow metal drift cylinders (Figure 15.4.1). Each alternate cylinder is electrically connected to an alternating power supply, which provides an alternating potential difference. An ion is injected into the tube and is initially attracted toward the first cylinder. Once the ion is in the cylinder, the electric field drops to zero and the potential polarity is reversed, so that when the ion emerges from the cylinder it is repelled from the first cylinder and attracted to the second. When the ion enters the second cylinder, it enters a region of zero electric field and drifts at a constant velocity through the cylinder. The polarity again reverses and when the ion emerges it is repelled by the second cylinder and

Identify ways by which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter.

287

quANTA ToquARkS

CyclotronIn 1932 Ernest Lawrence (1901–1958) and Stanley Livingston (1905–1986) developed a small, compact accelerator that electrostatically accelerated ions or elementary particles across a small gap between two semicircular D-shaped hollow metal cavities called ‘dees’ (see Figure 15.4.3). Once accelerated across this gap, the particle is turned through 180º in a near semicircular path by a constant uniform strong magnetic field. The particle is again accelerated across the gap and again turned through 180º. The two dees are connected to a constant high-frequency alternating power source, which provides the electric field that accelerates the particle across the gap. The polarity across the gap is reversed as the particles are being turned. The electric field inside the semicircular hollow metal dee is zero, and therefore does not affect the path of the particle. As the velocity of the particle increases, the radius of its orbit also increases, resulting in an outward spiral trajectory. One of the dees has an exit point and a beam of particles is produced. A design limit is imposed upon the cyclotron because, if you attempt to accelerate particles beyond energies of 20 MeV, the increase in relativistic mass causes the particles to become out of phase with the constant high-frequency source. At the ANSTO-operated National Medical Cyclotron at The Royal Prince Alfred Hospital, Sydney, the particle beam is used to produce radioisotopes.

attracted to the third. This process is repeated for the entire journey of the ion down the tube. The alternating potential remains at a constant frequency and, therefore, as the ions increase in velocity, the lengths of the hollow metal drift cylinders are increased. The Stanford Linear Accelerator Center (SLAC) houses the most well-known and longest linear accelerator (Figure 15.4.2). It is 3.2 km long.

Figure 15.4.1 Structure of linear accelerator components

Figure 15.4.3 (a) Schematic diagram of a cyclotron. (b) Particles are accelerated across the gap by an electric field, so that each semicircular spiral has a larger radius. (c) By the time the particle reaches the gap again, the voltage has reversed and the particle is accelerated. These diagrams show the acceleration of positive ions.

Figure 15.4.2 The Stanford linear accelerator accelerates alternate bunches of positrons (e+) and electrons (e–) to 50 GeV. Magnets separate the particles and bend them in arcs to produce head-on e+– e– collisions. The damping rings and focusing magnets help bunch and focus the beam.

source+ – + +–

D1

D2

D1

D2

+

–+

–

B

high frequencyalternatingvoltage

ion or electronsource

E E

a b c

acceleratedparticle beam

d

positronelectron

focusingmagnets

collisiondetector

transverse magnetic field

bending magnets

3 km

positron source

damping rings

electron injector

The particle zoo 15

288

Figure 15.4.5 The Fermilab accelerator accelerates protons in one ring and antiprotons in the other. The beams are made to collide with a total energy of ~2000 GeV. Superconducting magnets force the particles to stay in the rings.

+ proton source main ring

collisiondetector

Tevatron ring

magnets arecoaxial withrings

– antiprotonsource

Figure 15.4.4 A Van de Graaff accelerator injects particles into the synchrotron at high speeds. The magnets force the particles into a curved path. Each time the particle passes the high-frequency alternating voltage, it is accelerated.

target

deflectingmagnets

deflectingmagnets

high frequencyalternating

voltage

injectionpoint

Van de Graaffaccelerator Synchrotron

The modern accelerators that strive to produce the highest energy collisions are based upon the synchrotron design (see Figure 15.4.4). Charged particles are injected into a circular evacuated tube that is shaped into a ring. The ring has a series of magnets whose fields can be adjusted to maintain a circular trajectory for the accelerated particles. The ring also has one or more locations where an adjustable high-frequency (radio frequency) alternating power source provides a region in which the particle is accelerated. Both the alternating power source and the magnets are adjusted to increase the speed of the particles and maintain a circular orbit inside the evacuated tube. Modern synchrotrons use superconducting magnets to maintain the trajectories of the accelerated charged particles. The synchrotron at the Fermilab Tevatron has a circumference of 6.28 km (see Figure 15.4.5) and the newly commissioned Large Hadron Collider (LHC) at CERN has a circumference of 27 km (see Figure 15.4.6). The LHC has 1232 superconducting magnets and will accelerate protons up to 10 TeV and collide them head-on, travelling in opposite directions.

Using the accelerated particle beamsThe high-energy particles can either be fired at a stationery target or collided with other high-energy particles travelling in the opposite direction. Both collisions are governed by the laws of conservation of energy and momentum. Fixed target collisions can involve a wide range of target materials, and they produce secondary beams of particles that can be used in a variety of experiments to probe the nature of matter. In the late 1970s scientists at CERN developed a way to collide protons and antiprotons—a collision that has nearly zero overall momentum. This allowed the total energy of the collision to be available to produce the new varieties of particles that these types of experiments were uncovering.

CERN

ALICE

Switzerland Geneve France

ALICE

CMS

CMS

LHC–B

LHC-B

ATLAS

ATLAS

Point 1 Point 2

Point 5

Point 8

SPS

LEP/LHC

TI 8 TI 2

Figure 15.4.6 Overall view of the LHC experiments Figure 15.4.7 An engineer checks one of the magnets in the tunnel of the LHC.

289

quANTA ToquARkS

muon chamberhadron calorimeter

E-M calorimetertracking

e–

e+

Figure 15.4.8 The diagram shows typical interactions with sub-detectors for some common particles.

trackingdevice

electromagneticcalorimeter

hadroniccalorimeter

particleidentification

detector

photons

electrons orpositrons

muons

pions orprotons

neutrons

Particle detectors Particle detectors are designed to record and provide a way of ‘seeing’ the

explosion of particles that can be generated as a result of the collision. Modern particle detectors are made up of a series of layers of sub-detectors, each specialising in identifying particular types of particles or specific properties (see Figure 15.4.8). These sub-detectors can be classified into three main types:• Trackingdevices detect and reveal the trajectory of a particle. Often a

magnetic field is used to bend the trajectory of charged particles into a curve. The curvature provides physicists with information about the momentum of the particle. Early detectors such as the cloud chamber and the bubble chamber provided a direct visual record of the paths of particles. Modern tracking devices produce electrical signals that are processed by computers to reconstruct the path of particles. The outer layer of many detectors contain a muon chamber that detect these weakly interacting particles, which can travel through many metres of dense material.

• Calorimeters used in detectors are of two types: the electromagnetic calorimeter, which is positioned close to the collision, and the hadronic calorimeter. These sub-detectors are designed to stop, absorb and measure the energy of a particle. The electromagnetic calorimeter measures the energy of lighter particles such as electrons and photons. Hadronic calorimeters absorb the energy of heavier particles, which contain quarks (these types of particles are called hadrons, and include pions, protons and neutrons). The interactions inside both types of calorimeters often cause successive cascades or ‘showers’ of new particles and with each interaction or collision the momentum and energy is shared between more particles, until eventually all are slowed and stopped. To accurately measure the total energy requires the detector to stop nearly all particles, and thus explains the huge overall size of the detectors used. Calorimeters can stop most known particles except muons and neutrinos.

• Particleidentificationdetectors identify the type of particle by various techniques. Two commonly used methods to detect the nature of charged particles are Cherenkov radiation, produced when charged particles travel faster than the normal speed of light of that medium, and transition radiation, produced when a charged particle crosses the boundary between certain materials.

Figure 15.4.9 Schematic design for a typical particle detector

The particle zoo 15

290

PhYSiCS FeATUReThe woRLd’S LARgeST MiCRoSCoPe

How do we study the fundamental building blocks of matter, and the forces that

determine how they interact with each other, at the very tiniest of distances? Ironically, we do it by building the world’s largest ever particle accelerator, which acts a bit like a very sophisticated microscope. A 100 years or so ago, Ernest Rutherford showed us that we could learn about what was inside an atom by bombarding it with an energetic ‘probe’, in this case an alpha particle, and observing what happens to it. This principle has served us well, and by applying it physicists have burrowed down into the nuclei of atoms to find protons and neutrons, and into the protons and neutrons to find quarks and gluons. Many other particles, some of which can only be made in accelerators, have turned up along the way. The ones which we view, at least today, as truly fundamental have been identified, and two of the forces of nature, the electromagnetic and the weak, have been unified.

On the outskirts of Geneva, in a 27 km circumference tunnel 50–100 m underground, the Large Hadron Collider (LHC) has the task of colliding two beams of protons at very high energies, in order to allow us to take the next step along our road to understanding. Using Einstein’s famous principle relating energy and mass, the two colliding protons are destroyed and some of the energy they carry is transformed into the mass of new particles. We can understand why the protons need to have large energy and momentum by recalling de Broglie’s idea that particles can be assigned a wavelength that is inversely proportional to their momentum. The higher the momentum and the shorter the wavelength, the shorter the distances we can study. Many of the particles produced in the collisions live for only the tiniest fraction of a second, and their existence must be inferred from the products of the collision (i.e. their decay). To observe all of the products of the collisions in detail, very complex detectors the size of several-storey buildings have been constructed.

These detectors ‘track’ the emerging charged particles, bending them in magnetic fields to work out their charge and momentum from their trajectories. Further out from the collision point, calorimetry is employed to absorb particles such as photons and electrons and measure their energy.

Mounting scientific experiments on this scale takes the combined efforts of thousands of physicists from all over the world, including Australia. In Figure 15.5.10, postgraduate student Jason Lee from the University of Sydney can be seen standing in front of the giant ATLAS detector in its final stages of construction. ATLAS will scour the products of the proton collisions at its centre to search for things such as the ‘Higgs field’, which is believed to be involved in giving mass to the other fundamental particles; yet-to-be discovered ‘supersymmetric’ particles that are predicted by some new theories; and possibly even hints of extra dimensions. Jason completed high school in Sydney and an undergraduate degree at Oxford University before returning to Sydney to do his postgraduate studies. Like other Australian students in experimental high-energy physics, he spends some of his time at CERN, where he can interact with physicists from all over the world and gain experience of the LHC project first hand. For his postgraduate studies he is investigating how well ATLAS can identify electrons and their antimatter partners,

Figure 15.4.10 Jason Lee at the site of the giant ATLAS detector, which is in its final stages of construction and testing

291

quANTA ToquARkS

the positrons, and trying to improve the ability of the detector to pick them up.

Electrons and positrons turn up in the expected decay products of many of the new particles being hunted for. An example of what a Higgs particle decay might look like in the ATLAS detector can be found in Figure 15.4.11. In this example, the products of the decaying Higgs particle travel upwards in the picture and a spray of other particles, known as a ‘jet’, travels downwards (the yellow cone of particles that is absorbed

in the calorimeters, ending in blue and orange ‘blobs’). The Higgs particle decays to two other short-lived particles known as Z bosons, with one Z boson decaying to an electron and a positron (shown in blue travelling upwards) and the other to a muon and antimuon (shown in red). The other particles in the picture (in yellow) are the debris from the two protons that collided. The human shown in the figure gives an indication of scale. She would not be standing there if the LHC was operating!

Figure 15.4.11 ATLAS detector schematic with a simulated detection of a Higgs boson

The particle zoo 15

292

15.5 The Standard ModelIn the early 1930s the list of identified sub-nuclear particles comprised the electron, the proton and the neutron; the positron and the neutrino had been proposed by Pauli. It was not long before the muon and the positron were identified in cloud chambers while examining cosmic showers, and by the 1960s a hundred other particles had been identified. The physicists were confronted with this ever-increasing ‘zoo’ of particles, as the energy of the collisions increased. They had no overall unifying theory to explain the behaviour or adequately classify these particles. The quest to uncover which of these particles were truly fundamental became a primary area of research.

A few of the major discoveries that led to the development of the Standard Model of matter are discussed here. In 1964 Murray Gell-Mann (1929–) and George Zweig (1937–) proposed the existence of quarks to explain the properties of a family of particles called hadrons. The hadrons are further subdivided into two types: mesons, those containing two quarks (comprising a quark and antiquark pair), and baryons those containing three quarks. In 1967 the electro–weak theory, unifying the electromagnetic and weak nuclear force, was proposed by the collaboration of Steven Weinberg (1933–) and Sheldon Glashow (1932–), and independently by Abdus Salam (1926–1996). In 1969 the first evidence of quarks was reported by Jerome Friedman (1930–), Richard Taylor (1929–) and Henry Kendall (1926–1999). Then, during the period of 1970–1973, the Standard Model was formulated. All the components of the model except the Higgs boson were verified over the next 25 year period.

Today the Standard Model of Matter provides a mechanism to classify and explain the nature of matter. It includes three of the four fundamental forces but excludes gravity (Figure 15.5.1). The model has three main components: quarks, leptons and force carriers. The experiments at Fermilab and the new CERN LHC to search for the Higgs boson will either confirm the Standard Model’s validity or send the experimentalists and theorists back to formulate a new or revised model. Either way, the major challenge facing particle physicists is the need to develop a grand unified theory that will unite all four of the fundamental forces.

Discuss the key features and components of the Standard Model of matter, including quarks and leptons.

CheCkPoinT 15.41 List the three basic parts of an accelerator.2 Identify each of the types of accelerators and their uses.3 List the types of collisions that can be used to identify information about particles.4 Create a table that identifies the types of detectors that can be used and specify what they will show.

293

quANTA ToquARkS

Quarks There are six varieties (commonly called flavours) of quarks and in all there

are 12 types of quarks. The first generation of quarks are named up and down, the second generation are charm and strange, and the third are top and bottom. Each quark is considered to be a fundamental point-like particle that carries the properties of mass, charge and colour. The electric charge of a quark is fractional and is either –1/3e or +2/3e, and they possess a ‘colour charge’ that can be either red, blue or green. Quarks also have antimatter counterparts that possess the same magnitude of electric charge with the opposite sign, and colours that are antired, antiblue and antigreen. Quarks with different colour charge attract each other, as do quark–antiquark pairs of the same colour–anticolour (e.g. red and antired). The colour charge is associated with the strong force and is constantly changing as they interact with gluons.

Quarks do not exist as bare entities, and only form composite particles called hadrons. Within this classification, quarks can either pair with an antiquark to form a meson, or three quarks can combine to form a baryon. All hadrons have a total electric charge of 1e, and the total colour charge will sum to white (achieved by the quarks in mesons having a colour and an anticolour, or in baryons by the combination of all the quarks having different colours).

W–W+

Z0

matter

atom electron

proton

neutron

nucleus

Matterparticles

Forceparticles

All ordinaryparticles

belong tothis group

Theseparticles

existed justafter the

Big Bang.Now they are

found onlyin cosmicrays and

accelerators

Theseparticles

transmit thefour

fundamentalforces

of naturealthoughgravitons

have so farnot been

discovered

LEPTONS QUARKS

quarks

1st

FA

MIL

Y2

nd FA

MIL

Y3

rd FA

MIL

Y

ElectronResponsible for electricityand chemical reactions:it has a charge of –1

Muon neutrinoCreated along with muonswhen some particles decay

Tau neutrinoRecently discovered

Electron neutrinoParticle with no electric charge, and possibly no mass; billions fly through your body every second

UpHas an electric charge ofplus two-thirds; protons contain two,neutrons contain one

MuonA heavier relative of theelectron; it lives for two-millionths of a second

TauHeavier still; it is extremely unstable. It was discoveredin 1975

GluonsCarriers of thestrong forcebetween quarks

Felt by:quarks

Felt by:quarks and charged leptons

Felt by:quarks and leptons

Felt by:all particles with mass

The explosive release of nuclear energyis the result of the strong force

PhotonsParticles thatmake up light;they carry theelectromagneticforce

Electricity, magnetism and chemistry areall the results of electromagnetic force

Intermediatevector bosonsCarriers of theweak force

Some forms of radioactivity are theresult of the weak force

GravitonsCarriers ofgravity

All the weight we experience is theresult of gravitational force

CharmA heavier relative of the up;found in 1974

TopHeavier still

DownHas an electric charge of minusone-third; protons contain one,neutrons contain two

StrangeA heavier relative of the down;found in 1964

BottomHeavier still; measuringbottom quarks is an importanttest of electroweak theory

Figure 15.5.1 A tabular presentation of the three main components of the Standard Model and the three generations of particles classified as quarks, as well as leptons and a description of the gauge bosons that mediate the four forces of nature

The particle zoo 15

294

quARk RoAd RuLES

When remembering what quarks are in protons and neutrons, think of overtaking lane signs. The passing lane is a proton—two up quarks

and one down quark. No overtaking is a neutron—one up quark and two down quarks.

Table 15.5.1 Properties of quarks

Symbol Charge maSS Colour ChargeUp u +2/3e Red, blue or green

Down d –1/3e Red, blue or green

Charm c +2/3e Red, blue or green

Strange s –1/3e Red, blue or green

Top t +2/3e Red, blue or green

Bottom b –1/3e Red, blue or green

Table 15.5.2 Properties of hadrons

hadron example Symbol ConStituent quarkS

Meson (2 quarks)

Positive pion π+ u d

Neutral kaon K0 ds

Baryon (3 quarks)

Proton p uud

Neutron n udd

Sigma plus ∑+ uus

Note: u d is an up–antidown pair.

LeptonsThere are six varieties and 12 types of leptons. The most commonly known is the electron. All leptons appear to be fundamental point-like particles. The electron, the heavier muon and the massive tau all have an electric charge of –1e and their antimatter counterparts have a charge of +1e. Each of these leptons has an associated neutrino that is named accordingly; for example, the electron (e) has an electron neutrino (νe ).

CheCkPoinT 15.51 Draw a chart that shows the classes and subclasses of particles, starting with hadrons.2 List the properties of quarks, hadrons and leptons.

proton uud

Passing

neutron udd

No passing


295

quANTA ToquARkS

CHAPTER 15This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTiviTY 15.1: The MAnhATTAn PRojeCTResearch the Manhattan Project and the events that surrounded it. Assess the impact it has had on society then and now.

Discussion questions1 Outline the discoveries that were made during this project.2 Describe the effect of the outcome of the project on the war.3 Assess the significance to society today.

ACTiviTY 15.2: iSoToPeS in SoCieTYResearch a series of isotopes that have been developed for medicine, agriculture and engineering purposes. List their advantages over prior methods.

Discussion questions1 Name an isotope that is used in: a medicine b agriculture c engineering.2 Describe the purpose of one of these isotopes and how it is used.3 Explain how the use of that isotope has replaced previous technology.

Gather, process and analyse information to assess the significance of the Manhattan Project to society.

Identify data sources, and gather, process, and analyse information to describe the use of:• anamedisotopeinmedicine• anamedisotopein

agriculture• anamedisotopein

engineering.

296

15 The particle zoo Chapter summary

• TheManhattanProjectwasthecodenamefortheUSA’s top-secret nuclear bomb research project.

• Nuclearfissionreleasesenergythatmanifestsitselfasheat and high-energy gamma radiation.

• Fissionreactorsarecarefullydesignedtocontroltherateof fission (splitting) of large nuclei to ensure their safe operation.

• Uranium-fuellednuclearfissionreactors,comprisesixkey components: – Fuel rods are tubes filled with pellets of uranium

oxide and are located in the centre of the reactor known as the core.

– The core of the reactor houses the fuel rods, the control rods, a coolant system and a moderator material (which sometimes also acts as a coolant).

– Moderators slow down neutrons and so improve the chance of the neutrons being captured by a nucleus. Commonly used moderators include ordinary water, heavy water and graphite.

– Control rods absorb neutrons and are adjusted so that the chain reaction proceeds at a constant rate. The rods are usually made from boron or cadmium and are located between the fuel rods. If the reactor needs to be shut down for service or in an emergency, the control rods are fully inserted into the core.

– The coolant transfers the heat generated by the reactor away from the core. In nuclear power stations the heat is used to make steam, which powers a turbine and generator.

– Radiation shielding protects people and the environment and prolongs the working life of the reactor facility. Lead and graphite are used to absorb and reflect the large quantities of gamma radiation and neutrons emitted by the core.

• Radioisotopesareatomsthathaveanunstableratioofprotons to neutrons. They will decay via alpha or beta decay to attain a more stable configuration; some may also emit gamma radiation.

• Radioisotopesoccurnaturally,ortheycanbeproducedby changing the ratio of nucleons in the nuclei of a target material in a nuclear reactor or a particle accelerator.

• Radioisotopeshaveaspecifichalf-life,whichdescribesthe time it takes half of the radioisotope sample to decay.

• Radiopharmaceuticalscanbeclassifiedasdiagnosticortherapeutic. – Diagnostic radiopharmaceuticals are used to assess

the functioning of organs including the lungs, heart, liver and brain; identify bone fractures not visible in X-rays; and assess the flow of fluids such as blood.

– Therapeutic radioisotopes generally contain radioisotopes with longer half-lives, and once absorbed by the specific tissue or organ will deliver a target dose of radiation.

• Radioisotopeshaveapplicationsasdiverseascheckingthe structural integrity of bridges, determining wear in engine components, examining welds, imaging internal structures in jet aircraft engines, and thickness control.

• Particleacceleratorsprovidescientistswiththetoolstoprobe and investigate the structure of matter.

• Particleacceleratorshavethreebasiccomponents:– a source of charged particles (ions or elementary

particles)– a tube or chamber that is highly evacuated so that

the particles can travel without colliding with air molecules

– a mechanism to accelerate and control the trajectory of the particles.

• Commonparticleacceleratordesignsincludeelectrostatic, linear, cyclotron, betatron, synchrotron and storage ring colliders.

• Particledetectorsaredesignedtorecordandprovide a visualisation of the ‘explosion’ of particles that can be generated as a result of the collision.

• Thefourfundamentalforcesinnaturearegravity,theelectromagnetic force, the strong nuclear force and the weak nuclear force.

• TheStandardModelincludesthreeofthefourfundamental forces; it excludes gravity. The model has three main components: quarks, leptons and bosons (the force carriers).

297

quANTA ToquARkSReview questions

PhYSiCALLY SPeAkingVery quarky

Across

1 Discovered the neutron

6 Slows things down

9 Made up from any three quarks

10 Atom ‘smasher’

12 Planck has one

15 Same number of protons

17 A type of scattering

18 Keeps things cool

19 Type of reactor

Down

2 Type of accelerator

3 Family name for the carrier of all the forces

4 The anti one often accompanies the beta particle

5 A project

7 You will find them in a nucleus

8 An electron is one

11 Lived 1871–1937

12 The mass you require to start a chain reaction

13 A constant R

14 They are the reason for the strong force

16 Fermi’s first name

1 2

3 4 5

6

7

8

9 10 11

12

13

14 15 16

17

18

19

298

15 The particle zoo

Reviewing 1 Recall three of Enrico Fermi’s contributions to physics.

2 Define the term critical mass.

Explain the basic principles of a fission reactor.

3 Describe the roles of the following components in a fission nuclear reactor.a fuelb control rodsc moderatord coolante radiation shields

4 Explain using a schematic sketch the basic processes involved in a nuclear-powered electricity station.

5 Name two materials that could be used in a fission reactor as: a a moderator b control rodsc shielding.

6 Define the term radioisotope.

Describe some medical and industrial applications of radioisotopes.

7 Name and describe the use of a radioisotope used in:a medicineb agriculturec engineering.

Describe how neutron scattering is used as a probe by referring to the properties of neutrons.

8 Neutron scattering is used to probe structures of many materials. Explain how each of the following properties is useful.a the wave nature of the neutronb the magnetic moment of the neutronc strong interaction of the neutron with nucleid the range of energies the neutron can possess

9 Draw, label and describe the main features of the following particle accelerators.a Van de Graaffb linear acceleratorc cyclotrond synchrotron

10 Discuss the purpose of accelerating particles and colliding them with a target (or their antiparticle) in terms of:a the de Broglie wavelengthb Einstein’s equation E = mc2

c probing matterd producing new types of particles.

11 Identify the ways by which physicists continue to develop their understanding of matter using accelerators.

Discuss the key features and components of the Standard Model of matter, including quarks and leptons.

12 Name the six quarks.

13 Name three leptons.

14 Compare fermions and bosons.

15 Recall the constituents of these particles.a mesonb baryonc protond neutrone pion

16 Complete the following table.

typeS of boSon role in Standard modelPhoton

Strong force

Graviton

17 Describe the purpose of the Manhattan Project.

18 Describe the long-term ramifications of the Manhattan Project on society.

19 The Standard Model classifies the constituents of matter into three families.a Name the three families.b Describe the properties of each family.

20 Explain why the top quark was the last of the six quarks to be found.

21 Recall why quarks are never found as individual free particles.

Revie

w Questions

299

quANTA ToquARkS

PhYSiCS FoCUSLinking The veRY big wiTh The veRY SMALLThe very big and the very small are inextricably linked as cosmology and particle physics enter a new era of research. Many of the recent developments in cosmology have depended on advances in high-energy physics.1 Explain why it is necessary to accelerate particles

to extremely high velocities.2 Describe how high-energy accelerators such as the

Large Hadron Collider can test aspects of theories such as the Big Bang.

exTenSion3 Research how high-energy physics was and is still

carried out using cosmic rays.

H1. Evaluates how major advances in scientific understanding and technology have changed the direction or nature of scientific thinking

Identify ways by which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter

4

300

The review contains questions that address the key concepts developed in this module and will assist you to prepare for the HSC Physics examination. Please note that the questions on the HSC examination that address the option modules are different in structure and format from those for the core modules. Past exam papers can be found on the Board of Studies NSW website.

Multiple choice(1 mark each) 1 Which of the following lists only contains people who

provided science with atomic models in the period up until 1913?A Bohr, Democritus, Heisenberg, SchrodingerB Aristotle, Bohr, Pauli, RutherfordC Bohr, Dalton, Rutherford, ThomsonD Bohr, Einstein, Rutherford, Schrodinger

2 Which of the following scientists provided an explanation for the spectrum of beta-particle energies observed, and proposed that each electron in an atom could be described by four quantum numbers and a system that provided an explanation for the structure of the period table?A Erwin SchrodingerB Werner HeisenbergC Wolfgang PauliD Louis de Broglie

3 In 1924, de Broglie proposed the concept of matter waves. Which of the following statements is false in relation to de Broglie’s ideas at this time?A All particles of matter would have an associated

wavelength.B An electron in orbit around the nucleus would

follow a ‘wavey’ path.C The majority of the scientific community did not

take de Broglie’s ideas seriously.D There was no experimental evidence to support

de Broglie’s ideas.

4 A nuclear chain reaction requires certain conditions. Which of the following statements provides the best description?A Fuel that is capable of fissionB An amount of fuel that produces sufficient

neutrons to cause new fissionC Neutrons with an energy that allows them to be

absorbed by other fuel nuclei.D All of the above

5 Today the Standard Model of Matter provides a mechanism to classify and explain the nature of matter. Which of the following statements is false in relation to the model?A Quarks are the subatomic particles that make up

all matter.B The model includes three of the four fundamental

forces but excludes gravity.C The model has three main components: quarks,

leptons and force carriers.D Hadrons are subdivided into two types: mesons

and baryons.

301

quANTA ToquARkS

Short response 1 In 1913 Niels Bohr published a paper ‘On the

Constitution of Atoms and Molecules’, which described the planetary model of a hydrogen atom.a In terms of the Rutherford–Bohr model, present

a diagrammatic representation of how ONE of the Balmer series spectral lines is produced. (2 marks)

b Calculate the wavelength of the electromagnetic radiation emitted from a hydrogen atom when an electron drops from the third shell to the ground state. (3 marks)

c Outline the ONE spectral observation that the Rutherford–Bohr model could not completely explain. (1 mark)

2 State the two laws of physics used by James Chadwick when he discovered the neutron. (1 mark)

3 The Standard Model has the components shown in the table.

a Identify ONE way in which physicists have developed their understanding of matter using accelerators. (2 marks)

b Compare neutrons and protons in terms of their constituents. (2 marks)

4 The diagram below shows the kinetic energy distribution of electrons for the beta decay of bismuth-214.

Account for the energy distribution of the electrons emitted in beta decay. (3 marks)

5 Naturally occurring polonium-218 emits an alpha particle to produce a new element X. This new element emits a beta particle to become bismuth-214.

Po X→ + +α γ

a Identify the new element X. (1 mark)b Describe, using an equation, the decay of

element X to produce bismuth-214. (2 marks)c Calculate the number of protons and neutrons in

element X. (1 mark)

quark up down Strange Charm bottom topCharge

+23

e −13

e −13

e +23

e −13

e +23

e

Mass GeV/c2 0.005 0.01 0.2 1.5 4.7 180

KE

N

Extended response 6 You have gathered, processed and analysed

information to assess the significance of the Manhattan Project.a Describe how you ensured that the information

you gathered was reliable. (2 marks)b Assess the impact the work carried out at

Los Alamos has had on society and the environment. (5 marks)

5 medical physics

The human body is a complex machine and many things can go wrong with it. However, unlike a machine, we cannot do a total shutdown to locate and correct a problem. Most of the functions of our body need to continue working when we are looking for a problem and when we are trying to fix it. This is where physics comes to the aid of modern medicine by providing the tools to investigate and, in some cases, repair problems in the living body.

We ‘see’ using light, and optical-fibre technology offers a minimally invasive way of looking inside the body to actually see what may have gone wrong. Listening is also an important way for doctors to find evidence of problems but, remarkably, we can now use sound to look deep within the body without ever breaching the surface.

Although they present some potential damage to human tissues, X-rays can also give us vital information of ‘structural’ problems below the surface. For more than a century, X-rays offered the only ‘non-invasive’ way to see inside the body. Now, when combined with enhanced computer technologies, X-rays provide powerful tools for creating images of the structures within the body.

Modern physics also offers magnetic resonance imaging (MRI) as a powerful method to investigate both the form and function of individual tissues. Carefully chosen

radioisotopes offer similar possibilities, because they can be used to identify where atoms are accumulating within our

bodies. Other radioisotopes provide a therapeutic tool to help combat certain problems.

None of these techniques will reveal everything, but physics has provided

doctors with a range of diagnostic and therapeutic tools that earlier generations could never have imagined.

Context

Figure 16.0.1 This image shows the pattern of brain tissue loss in methamphetamine users, relative to healthy adults, mapped using MRI scans.

Areas of greatest lossemotion, reward (limbic system)memory (hippocampus)

0% Loss 3% Loss 5% Loss

302

303

InquIry ACtIvIty

WhAt mAkes you tICk?

The ancient Egyptians knew that the heart was centre of blood supply in the human body. It’s pretty obvious when you look inside the body. However, just how it worked remained a mystery, because looking inside dead bodies was not permitted in many societies. Great advances were made by Andreas Versalius (1514–1564) in the 16th century precisely because dissection was permitted in Italy in some circumstances. In 1543 he published De Humani Corporis Fabrica (On the Fabric of the Human Body), which marks the foundation of the scientific study of human anatomy.

The function of many of the organs of the human body has been poorly understood until recently—none more so than the brain. However, modern imaging techniques such as PET and MRI are slowly revealing the operation of the brain by uncovering what parts are involved in different actions or affected by different diseases.

In this exercise you should seek out reliable sources of information on the web to investigate the difference between modern structural and functional imaging of the brain.1 Define what is meant by the structural and functional imaging.2 How would you classify the images in Figures 16.0.1 and 16.0.2—are they

structural or functional?3 What imaging techniques allow structural differences to be seen in the brain?4 What diseases might be revealed by structural images of the brain?5 What imaging techniques allow functional differences to be seen in the brain?6 Outline one study of the brain that might be conducted using functional images.7 How reliable are your sources? Why do you expect them to be reliable?

Figure 16.0.2 A drawing of the human brain by Vesalius

16

304

Imaging with ultrasound

transducer, sonogram, Doppler ultrasound, piezoelectric transducer,

piezoelectric effect, convex array transducer, sector scan, linear array

transducers, resolution, acoustic impedance, impedance matching,

acoustic coupling, A-mode scan, B-mode scan, M-mode scan, 2D real-time

scan, 3D ultrasound, 4D ultrasound, bone density, dual X-ray absorptiometry,

Doppler ultrasound, colour Doppler imaging, Doppler effect, echocardiography

See like a batBats, dolphins and some birds have evolved the ability to produce and sense sound waves that are reflected from objects in their surroundings, enabling them to hunt prey and to navigate in the dark. The principle used by these animals has been applied to radar (using radio waves) and sonar (using sound waves).

Since the 1960s, the principles of sonar have also been applied for medical purposes. Ultrasound imaging is a medical procedure used to produce pictures of the inside of the body by using sound waves with a frequency much higher than that audible to humans.

16.1 What is ultrasound? Sound is a vibration in an elastic, mechanical medium that can be solid, liquid or gas. A sound wave is made up of periodic pressure variations in the medium through which it travels. It propagates as a longitudinal or pressure wave of alternating compressions and rarefactions (see Figure 16.1.1 and in2 Physics @ Preliminary section 5.5).

The speed of sound is different in different media, as shown in Table 16.1.1.

As well as speed, you should recall three other important properties of sound: frequency (the number of vibrations per second), wavelength (the distance between adjacent compressions) and amplitude (the maximum displacement of the particles in the medium from their equilibrium position). The greater the amplitude of a sound wave, the more energy the wave carries.

Humans can hear sounds with frequencies between about 20 Hz and 20 kHz. Ultrasound is any sound that has a frequency greater than this upper limit. Other mammals such as bats and dolphins use

Figure 16.1.1 A longitudinal wave with compression and rarefaction pressure variations graphed versus distance

high

Distance

Air

pre

ssur

e

low atmosphericpressure

rarefaction

compression

lowpressure

highpressure

lowpressure

highpressure

305

medicalphysics

ultrasound waves with frequencies as high as 125 kHz for navigating and sonar visualisation.

Ultrasound used for medical purposes has frequencies that range from 500 kHz to 30 MHz, but for most imaging applications, the ultrasound used is in the range from 3.5 MHz to about 10 MHz. The exception is ultrasound used for imaging blood vessels in the body. Tiny probes on the end of a catheter (tube) are inserted into the blood vessels and operate at frequencies up to 30 MHz.

The choice of ultrasound frequency that is most suitable for producing a medical image is a compromise between conflicting criteria. Higher frequency ultrasound produces images with a better resolution of detail, but they have poorer penetration through tissues. Imaging depth into tissue is limited by attenuation (absorption) of the ultrasound waves, and this increases as the frequency is increased.

Table 16.1.1 Velocity of sound in air and ultrasound in different human tissues

Material Velocity of sound (m s–1)Air 330

Fat 1450

Water 1480

Brain 1540

Liver 1550

Kidney 1560

Blood 1570

Muscle 1580

Lens of eye 1620

Skull bone 4080

Average soft tissue 1540

Identify the differences between ultrasound and sound in normal hearing range. CheCkpoInt 16.1

1 Explain how sound is made by moving air particles.2 Outline the difference between sound and ultrasound.3 List the advantages of low frequency and high frequency ultrasound.

16.2 principles of ultrasound imaging

In ultrasound imaging, ultrasound waves are emitted and detected by a device called a transducer. A transducer is any device that converts energy from one form to another. For example, a loudspeaker is a transducer that converts electrical energy to sound energy; a microphone does the reverse. The ultrasound transducer combines the functions of a speaker and a microphone. It can transmit and detect ultrasound.

The ultrasound transducer emits pulses of ultrasound for only ~1% of the time and detects for ~99% of the time. The transducer detects ultrasound waves that are reflected from boundaries between different tissues in the body. The amount of reflection depends on differences in the properties of the two tissues at the boundary.

The intensity of the reflected wave returned to the transducer is determined by:1 the difference in the acoustic properties of the tissues at the boundary—

the greater the difference, the more energy is reflected2 the characteristics of the intervening tissue—some tissues absorb more energy

than others3 the angle at which the ultrasound is reflected from the boundary between

two tissues—the production of an image is easiest when the reflection occurs perpendicular to the tissue boundary.

A strong reflection from a tissue boundary in the body produces a sharply defined image. A poor image will be produced if the reflection is weak or if tissues in the body absorb the energy before it returns to the transducer.

activity 16.1

pRacTical eXpeRieNces


seeing with ultrasound16

306

Ultrasound imaging is one of the safest and least expensive medical imaging technologies. It is useful for producing images of soft tissues and organs. Two major advantages of ultrasound imaging are that it is non-invasive and it does not involve use of ionising radiation (as in X-rays) and it is therefore very safe to patients. More invasive procedures carry the risk of infection, and ionising radiation damages DNA.

Ultrasound imaging is commonly used to produce images of a developing foetus during pregnancy, where its use can reveal foetal development and movement. Ultrasound imaging is also used to produce images of the kidneys, liver, pancreas and heart. Doppler ultrasound imaging is a special type of imaging used to measure blood flow rate to diagnose heart disease and blocked blood vessels.Figure 16.2.1 Principles of using ultrasound as a diagnostic tool

diskstorage

keyboard/cursor

printer

transducer

display

PZT crystals

sound backing

CPU

Transducerpulse controls • frequency • duration • scan mode

Figure 16.2.3 Colour ultrasound of a kidneyFigure 16.2.2 An ultrasound examination of a pregnant woman. Note the transducer in the technician’s hand and in contact with the patient’s abdomen.

To produce an ultrasound image or sonogram, a sonographer applies a gel to the area of the patient’s skin overlying the region to be imaged and the transducer is moved across the skin. For example, to obtain images of the heart, the transducer is placed in contact with the chest.

A computer processes information from the reflected waves received by the ultrasound transducer to produce real-time images, which are displayed on a screen (Figures 16.2.1 and 16.2.2).

The patient does not feel any discomfort. Because the images are produced while the procedure is being done, the sonographer can manipulate the transducer to obtain the best images for the doctor to examine. A variety of different types of images can be produced, depending on the way the data were collected and how they were processed (Figures 16.2.3 and 16.2.4).

307

medicalphysics

physICs FeAtureoBstetrICs

Probably the most well-known use of medical ultrasound imaging is in obstetrics. It began in

1966, but now most women have ultrasound imaging at some time during their pregnancy. Among the things that can be determined using ultrasound are the number, position and size of babies, if a baby’s internal organs are growing normally and how much fluid is around the baby. Measuring the head size is used to help determine the age of the foetus. Figure 16.2.4 illustrates ultrasound imaging of a foetus.

Some doctors have expressed possible concerns about the safety of using ultrasound in making three-dimensional images (movies) of foetuses simply so parents can have images of the baby before it is born. Although the risk is low, they argue that if the procedure is not necessary for a medical diagnosis, it should be avoided.

4. Implications of physics for society and the environment

Figure 16.2.4 Ultrasound of a foetus

Ultrasound can be classified as low or high intensity. For sound, intensity corresponds to the loudness of a sound, and it is a measure of the amplitude and the energy of a wave. Low-intensity ultrasound has a minimal effect on tissues as it passes through them. High-intensity ultrasound heats the tissues it passes through.

The intensity of ultrasound used to produce images is kept as low as possible to minimise the risk to the patient. Low-intensity ultrasound is considered safe enough to produce images during pregnancy and to produce images of vital organs such as the heart. The use of ultrasound to produce images is referred to as a diagnostic application.

High-intensity ‘therapeutic ultrasound’, on the other hand, affects the medium and can be used to heat injured muscles or to disintegrate kidney stones.

Limitations of ultrasound imagingHigh-frequency sound waves cannot effectively penetrate bone or air, so ultrasound cannot be used to produce images of parts of the body if there is bone or air between the ultrasound transducer and the body part of interest. As a result, ultrasound cannot image the adult brain and parts of the body obscured by gas in the intestines, and produces poor quality images of the lungs. Ultrasound can pass through parts of the skeleton of a foetus before the bone calcifies and hardens.



308

Table 16.2.1 Advantages and disadvantages of ultrasound

adVantages disadVantages It is non-invasive and does not require surgical procedures. Many reflections occur within the body and therefore good imaging

is operator dependent.

Patients can be examined without sedation, and relatively quickly and conveniently.

An ultrasound image is harder to interpret than an X-ray image.

Since sound is low energy and doesn’t ionise the tissue, it does not damage DNA, cells and tissues.

It is difficult to produce clear images in obese patients because of attenuation and reflection from fat.

It is relatively cheap compared with other scanning technologies. The presence of air and bone pose problems because their acoustic impedances are so different from those of soft tissue.

CheCkpoInt 16.21 Explain the purpose of the transducer in ultrasound imaging.2 Define what is meant by ‘intensity’ of ultrasound imaging.3 Explain how high frequency sound waves can make images of internal organs.4 State the difference in the effects of low- and high-intensity ultrasound.

16.3 piezoelectric transducers Ultrasound scanners use a piezoelelectric transducer made from a crystal

that responds to an oscillating voltage placed across it by oscillating slightly in size. The oscillating crystal produces a sound wave. It also works in reverse, converting the sound energy into electrical energy. The process is called the piezoelectric effect, and was discovered in 1880 by Pierre Curie (1859–1906) and his brother Jacques (1856–1941).

A ceramic called lead–zirconate–titanate (PZT) is commonly used to make piezoelectric transducers. This material responds to relatively small voltages or forces; that is, it has high electromechanical efficiency. The properties of PZT can be altered by modifying the ratio of zirconium to titanium, or by adding small amounts of other materials such as lanthanum. Figure 16.3.2 shows how PZT rods are embedded in plastic, which is then used to construct different types of transducers for ultrasound machines.

In ultrasound imaging, a convex array transducer is used to produce a sector scan. The scanning surface of these transducers is curved outwards

Describe the piezoelectric effect and the effect of using an alternating potential difference with a piezoelectric crystal.

Figure 16.3.1 Apply force to a piezoelectric crystal and it will produce a voltage; apply a voltage and the crystal will expand (or contract) slightly.

Figure 16.3.2 (a) PZT rods are embedded into plastic to construct transducers of various shapes, such as (b) a convex array transducer.

force

force

– voltage

+ voltage

PZT rods

epoxy resinmatrix

a b

309

medicalphysics

Figure 16.3.4 The head of a foetus in the womb. Note the small acoustic window— the area through which the ultrasound enters the body at the top of the image, which is the abdomen of the mother.

Figure 16.3.3 A convex array transducer

sector scan

convexarray

data processingcomputer

monitor

organs

Table 16.3.1 Frequency and resolution for linear array transducers with parallel beams

MHz axial resolution

lateral resolution

3.0 1.1 mm 2.8 mm

4.0 0.8 mm 1.5 mm

5.0 0.6 mm 1.2 mm

7.5 0.4 mm 1.0 mm

10.0 0.3 mm 1.0 mm

CheCkpoInt 16.31 Define the term transducer.2 Outline the energy transformations that are happening in an ultrasound transducer.3 Define the piezoelectric effect and describe how it is useful in ultrasound imaging.4 Recall the advantages of convex sector systems.

(convex), resulting in a divergent beam (Figure 16.3.3). This is different from earlier linear array transducers, which had parallel beams.

Convex array transducers use as many as 512 piezoelectric elements to display a wedge or pie-shaped image. The size of the area of skin through which the ultrasound enters the body (the acoustic window) depends on the curvature of the transducer surface. There is generally a narrow window at the skin surface, while still providing a wide angle deeper beneath the skin (Figure 16.3.4). The narrow field of view close to the transducer makes structures near the transducer difficult to view.

The smallest size of objects that can be imaged (resolution) using ultrasound depends on the wavelength (or frequency). All waves have the property of being ‘diffracted’ by objects; that is, their paths are bent around the object. After passing around the object, the paths of the waves recombine and undergo interference (see in2 Physics @ Preliminary section 7.4), which tends to ‘smudge’ the reflection of the object.

If the size of the object is approximately one wavelength or less, the smudging becomes so significant that the individual object is not visible, limiting the ability to resolve fine detail. This ‘diffraction limit’ applies to all waves. In ultrasound, this limits resolution both in the direction along the beam (axial) and across the beam (lateral) and is summarised in the Table 16.3.1.

Low frequency waves are more effective in penetrating human tissue. This results in a dilemma for ultrasound imaging. In order to produce a clear image of an object deep in the body, low frequency waves should be chosen over high frequency waves. However, to obtain a clearer image of smaller structures, a high frequency needs to be used. Clearly the final choice is a compromise.


310

16.4 Acoustic impedance The ability of a medium to transmit sound varies between materials. Even for a single material, it also varies with the frequency of the sound. The acoustic impedance describes how readily a specific sound frequency will pass through a material. It is defined as the product of the density ρ (in units of kg m–3) of the material and the acoustic velocity u (in m s–1) in the material:

Z = ρυZ therefore has units of kg m–2 s–1, which is called a rayl.

Worked example questIonBone has a density of 2 × 103 kg m–3. The speed of sound in bone is 4080 m s–1. Calculate the acoustic impedance of bone.

ρ = 2 × 103 kg m–3

u = 4080 m s–1

solutIonZ = ρu = 2 × 103 kg m–3 × 4080 m s–1

Z = 8.16 × 106 rayl

Acoustic impedance differs for different tissues in the body because of differences in their densities and the speed of ultrasound in the tissue. Acoustic impedance can be used to predict the proportion of the energy of an ultrasound wave that is reflected from a boundary between two tissues.

Reflection of ultrasound at tissue boundaries When ultrasound waves meet a boundary between tissues with different

acoustic impedances, some of the ultrasound energy (intensity) will be reflected and some will be transmitted. Conservation of energy requires that the sum of the reflected and transmitted energies must be equal to the original energy incident on the boundary. Note that for any reflection to occur, the acoustic impedances of the two media Z1 and Z2 must be different. A large difference in impedance results in a large reflection.

The incident ultrasound intensity can be represented by Io, with the reflected intensity Ir and the transmitted intensity It, as illustrated in Figure 16.4.1. At a tissue boundary, the ratio of the reflected intensity of ultrasound to the original intensity of the ultrasound is equal to the ratio of the square of the difference of the acoustic impedances to the square of the sum of the acoustic impedances, as represented by the equation:

I

I

Z Z

Z Z

r

o=

−

+

2 12

2 12

Define acoustic impedance: Z = ρu

and identify that different materials have different acoustic impedances.

Solve problems and analyse information using: Z = ρu

Define the ratio of reflected to initial intensity as: I

I

Z Z

Z Z

r

o=

− +

2 12

2 12

Identify that the greater the difference in acoustic impedance between two materials the greater the reflected proportion of the incident pulse.

Figure 16.4.1 Reflection occurs at a boundary between two materials with different acoustic impedance.

IO

Ir

Z1ρ1υ1= Z2

ρ2υ2=

It = IO – Ir

311

medicalphysics

Table 16.4.1 Acoustic impedance of ultrasound in human tissue

substancedensity(kg m–3)

ultrasound Velocity (m s–1)

iMpedance(rayl)

Water 1000 1540 1.54 × 106

Fat 920 1450 1.33 × 106

Lens of eye 1100 1620 1.78 × 106

White matter (brain) 1040 1541 1.60 × 106

Grey matter (brain) 1040 1541 1.60 × 106

Muscle 1040 1585 1.65 × 106

Kidney 1040 1561 1.62 × 106

Blood 1060 1570 1.66 × 106

Liver 1065 1549 1.65 × 106

Bone (other than skull) 1810 4080 7.38 × 106

Skull bone 1910 4080 7.79 × 106

Identify that different materials have different acoustic impedances.

Worked examplesquestIonCalculate the percentage of ultrasound intensity reflected at a boundary between fat and muscle.

Fat: Z1 = 1.334 × 106 rayl

Muscle: Z2 = 1.648 × 106 rayl

solutIon

I

I

Z Z

Z Z

r

o=

− +

= × − ×2 12

2 12

61 648 10 1 334 1[( . ) ( . 00

1 648 10 1 334 100 011

6 2

6 6 2

)]

[( . ) ( . )].

× + ×=

Hence only 0.011 of the incident intensity is reflected. This is 1.1% of the incoming intensity.

questIonCalculate the percentage of ultrasound intensity reflected at a boundary between air and muscle.

Air: Z1 = 0.0004 × 106 rayl

Muscle: Z2 = 1.648 × 106 rayl

solutIon

II

Z Z

Z Zr

o=

− +

=×( )− ×

2 12

2 12

6 6

6 6

1 648 10 0 0004 10 2

21 648( ). .10 0 0004 100 999

( )

× + ×( ) =

. ..

Hence 99.9% of the incoming intensity is reflected. Since intensity is proportional to energy, we can say ‘99.9% of the energy’ is reflected.

Bone has a much higher acoustic impedance than other tissues, so most of the ultrasound energy that encounters bone is reflected. For this reason, ultrasound cannot image the internal structure of bone or a brain inside an adult skull. Some ultrasound penetrates the skull of a foetus, because the bone has not yet calcified— it is cartilage, similar to the tissue that gives ears their semi-rigid nature.

If ultrasound travelling from air or other gas strikes a soft-tissue boundary of any type, then most of the ultrasound energy is reflected. For this reason, when an ultrasound is being carried out, a gel is placed on the skin and the ultrasound probe is moved over this gel. The most important function of the gel is to exclude any air from the region between the transducer and the skin. The gel has an acoustic

Solve problems and analyse information using: I

I

Z Z

Z Z

r

o=

− +

2 12

2 12

Describe how the principles of acoustic impedance and reflection and refraction are applied to ultrasound.


312

16.5 types of scansTo perform an ultrasound scan, a transducer is placed against the patient’s skin, directly over the region to be imaged. The transducer sends a very brief pulse of ultrasound into the tissue. The pulse travels into the body as a beam, similar to a searchlight. Interfaces along the way reflect some of the ultrasound energy back to the transducer. The transducer converts the energy of the reflected echo into electrical signals, which are sent into amplifiers and signal-processing components of the ultrasound machine. The exact delay between when the transducer first emits the ultrasound pulse and when it picks up an echo allows the machine to calculate how far the reflecting interface is from the transducer, given the known speed of ultrasound in the tissue.

Ultrasound can be used in different ways, called modes, which produce different data and images.

A-mode scans When a single-element transducer is used and the transducer is

stationary, only a single dimension can be recorded. That dimension is the distance of the echoing interface from the transducer. A typical A-mode scan (or A scan) is shown in Figure 16.5.1. Returning echoes produce a vertical displacement of the signal display. The amount of displacement is proportional to the amplitude. This was the first use of ultrasound as it is relatively simple and requires little signal processing to display the data. New ultrasound devices use computers to analyse the various time values.

Typically the A scan is used to examine midline structures in the brain, solid or cystic structures, foreign bodies in the eye, abnormal fluid accumulation around the heart or lungs, or to guide biopsy and amniocentesis needles.

Describe the situations in which A scans, B scans and phase and sector scans would be used and the reasons for the use of each.

impedance similar to that of human tissue. The term used to describe this process is impedance matching or acoustic coupling. Impedance matching between two media results in most of the energy being transmitted through the interface, with almost no reflection.

A further complication arises at the boundary between two different tissues, because ultrasound is also refracted at a boundary (see section 18.1 and in2 Physics @ Preliminary section 8.3). This bending of the sound path can complicate the analysis of the data.

Solve problems and analyse information to calculate the acoustic impedance of a range of materials, including bone, muscle, soft tissue, fat, blood and air and explain the types of tissues that ultrasound can be used to examine.

CheCkpoInt 16.41 Describe the conditions necessary for reflection of ultrasound waves to occur at a boundary.2 Calculate the percentage of ultrasound energy that reflects from a muscle and bone boundary. (Use values in

Table 16.4.1 to help.)3 Explain why it is necessary to use gel on the transducer in order to get clear images.

Figure 16.5.1 An A-mode scan signal

high-frequency signalabsolute value of signalHibert envelope of signal

Am

plit

ude

Time of flight (µs)

1200

800

400

–400

–800

–1200

0

2 3 4 5 6 7

313

medicalphysics

Figure 16.5.2 The timing of a return echo indicates depth in the body. The intensity of reflection may be indicated as a signal versus time in A-mode scans or as a bright spot in B-mode scans.

pulse

echo

to scandisplay

transducer

A scandisplay

B scandisplay

Ech

o st

reng

th

Time

organ vertebraB-mode scansThe most common type of medical ultrasound is called a B-mode scan (or brightness-mode scan). In a B-mode scan the returning echoes are displayed on the screen as dots, rather than as a signal plotted versus time, as in A-mode scans. The intensity or brightness of the dot is proportional to the amplitude of the reflected pulse (Figure 16.5.2).

In the simplest scanning mode, a single-element transducer can be moved around to provide different lines probing through the body. Like the beam of a searchlight, it picks out structures along the line of the beam. An image can be built up from these scans (Figure 16.5.3).

M-mode scansA single-ultrasound beam can be used to produce an M-mode image in which movement of an individual structure such as a heart valve can be displayed versus time. An M-mode scan uses a high-sampling frequency (up to 1000 pulses per second) to capture the rapid motion. This display is essentially a series of B-mode scans lined up side-by-side in time (Figure 16.5.4).

2D real-time scansUsing a convex-array transducer with many piezoelectric elements, beams of ultrasound can be launched in different directions into the body. Each beam produces a B-mode scan. Once all the echoes from the first beam have been picked up, the transducer sends a second pulse along a slightly different beam direction into the tissue. Echoes are again picked up and sent for processing to produce an image, and another pulse is launched in another direction, and so on. Like the beam of a searchlight swept across the night sky, the pulsed beam from an ultrasound transducer is swept through the body, mapping out reflecting surfaces and forming a 2D real-time scan. The echo’s image is two-dimensional and shows detail along

Figure 16.5.3 B-mode scan with a moving transducer can be used to reconstruct a two-dimensional ultrasound image of a foetus.

probe placentalegs

spinefoetalskull probe

probe

ultrasound imagesynthesisedfrom scans

Figure 16.5.4 An M-mode scan through the left ventricle of the heart allows measurement of the ventricle size (green markers) as the heart beats.


314

one slice through the body, originating at the transducer. Beams are swept and ultrasound images are formed very rapidly in ‘real-time’ images. As the operator holds the transducer in contact with the skin, the image appears live on a video monitor. Real-time images are usually made with a sector scanner, so the images will be pie shaped.

The B-mode scan is continuously updated and displayed in real time, with the image changing as the tissues move in the field of the ultrasound beam. The movement of the object being scanned can be monitored on the image.

Interpreting ultrasound images requires considerable skill and knowledge, as the images in Figures 16.5.5 and 16.5.6 demonstrate. An untrained observer would have difficulty making sense of the image without explanation.

Where to from here?Complex ultrasound transducers are used for producing real-time 3D ultrasound images. These simultaneously produce sector scans in many planes (Figure 16.5.7). Combined with tilting or moving the transducer, this yields ultrasound data of body structures from many angles. High-speed digital signal processing permits the large amount of gathered data to be assembled into a 3D image (Figure 16.5.8).

Figure 16.5.6 This image of a kidney reveals a large, dark cyst that shows no internal echoes because of its homogeneous structure.

Figure 16.5.5 This sector scan image of a gall bladder reveals a strong (bright) echo from an abnormally thick wall.

Figure 16.5.8 Example of a 3D ultrasound image of a foetus

CheCkpoInt 16.51 Summarise the process of (a) A-mode scanning, (b) B-mode

scanning and (c) M-mode scanning.2 Outline examples of when each mode of scanning would be used.

Figure 16.5.7 Examples of scanning patterns produced by modern transducers. Each small blue triangle indicates a sector scan (a) moved in an angle or (b) rotated relative to the transducer.

a b

Ultrasound using this 3D imaging technique is sometimes used to produce a movie. This is usually called 4D ultrasound. There is a section of the ultrasound industry that promotes 4D ultrasound for non-medical reasons—trying to persuade parents that it is fun to have such a record of the child before it is born.

315

medicalphysics

16.6 ultrasound at work Bone-density measurementsOsteoporosis literally means ‘porous bones’. Bones that once were strong typically become fragile as a person gets older, due to loss of calcium. The problem is common in postmenopausal women, but a significant number of men also suffer from a loss of bone density. A person may not realise they have osteoporosis until they suffer a vertebral fracture when doing ordinary activities such as lifting a bag of groceries, or break a hip in a fall.

To determine if a person has osteoporosis, it is necessary to measure bone density (or bone mineral density). Bone-density testing can measure the amount of bone in different parts of the skeleton, and can predict the risk of future fractures and monitor changes in bone-mineral density due to medical conditions or therapy.

There are several ways to measure bone density, most of which involve the use of X-ray radiation. X-ray based methods usually take a bone-density measurement of the hip, spine, forearm or heel by passing two X-ray beams of different energy through the bone. The technique is called dual X-ray absorptiometry (DXA or DEXA).

Alternatively, ultrasound can be used to estimate the bone density of the heel, as the heel bone contains a high percentage of the kind of bone most affected by osteoporosis. During an ultrasound examination, two soft rubber pads are placed in contact with either side of the heel. Transducers within these pads send and receive ultrasound waves through the heel bone. The speed and absorption of ultrasound vary with bone density. The test takes about a minute and is performed in a seated position (Figure 16.6.1). No injections or invasive procedures are necessary and the test results are processed immediately.

What are the relative advantages of ultrasound imaging and DXA? • UltrasoundsystemsaresmallerandlessexpensivethanDXAsystems.• Ultrasoundrequiresnoexposureofthebodytoionisingradiation(although

X-ray levels are very low with DXA—lower than in most other X-ray procedures).

• StudiesshowthatultrasoundmainlymeasuresthebonemasswhileDXAisbetter able to predict bone strength.

• ReproducibilityofresultsusingultrasoundisnotquiteasgoodaswithDXA.• Relativeriskforhipfractureispredictedaswellbyultrasoundoftheheelas

by DXA of the hip.

Blood flow and the Doppler effect Doppler ultrasound scanning is used to image any structure or tissue that

pulsates or moves. It is based on the principle that whenever the reflector surface moves with respect to the transducer, there is a shift in the frequency of the ultrasound received by the transducer compared with the frequency it emitted (the Doppler effect; see in2Physics @ Preliminary section 13.8). This concept is illustrated for blood flow in Figure 16.6.2, while Figure 16.6.3 illustrates a case in which Doppler imaging reveals abnormal blood flow in a heart. Colour Doppler imaging is colour-coded, using redder colours to show flow or movement towards the transducer and bluer colours to show flow away from it. (Interestingly, this is the opposite of the choice an astronomer would make in presenting Doppler images.)

Identify data sources, gather, process and analyse information to describe how ultrasound is used to measure bone density.

Describe the Doppler effect with respect to sound and how it is used in ultrasonics to obtain flow characteristics of blood flow through the heart.

activity 16.2



Figure 16.6.1 A machine used to measure the bone density of your heel


316

The Doppler shift effect has been used for a long time in foetal heart rate detectors. Further developments in Doppler ultrasound technology in recent years have resulted in its increased use in obstetrics to assess and monitor the well-being of a foetus.

Outline some cardiac problems that can be detected through the use of the Doppler effect.

Figure 16.6.2 Different transducers (A–D) in this system ‘see’ blood flowing in different directions—with a component towards or away from the transducer. This produces a Doppler effect that can be used to calculate a velocity at that point.

A

vesselflow

A B C D

B C D

beam direction

skin

Sonogram

EchocardiographyEchocardiography is the name given to ultrasound imaging of the heart. It is used to investigate heart-valve function, ventricular function, congenital heart diseases (including holes in the septum separating the left and right halves of the heart), cardiac tumours and obstructions in cardiac blood vessels.

Pulsed ultrasound from a transducer is used and a continuous recording is made of the echoes received from the various parts of the heart, showing the motion of the various parts. A standard transthoracic echocardiogram can be conducted through the thorax (chest wall). Higher resolution imaging can be achieved by a trans-oesophageal echocardiogram in which a transducer on a specialised probe is passed down the patient’s oesophagus, thereby getting much closer to the heart.

Blood-flow characteristics in the foetal blood vessels can also be examined using Doppler ultrasound. Diminished flow, particularly in the diastolic phase of a pulse cycle (as the heart muscle relaxes) is associated with problems with the circulation of blood in the foetus. Colour Doppler can also be used to assess the success of coronary bypass surgery, and is particularly useful in the diagnosis and assessment of congenital heart abnormalities such as atrial or ventricular septal defects (a hole between the atria or the ventricles; see Figure 16.6.3).

doppleR agaiN aNd agaiN

The Doppler effect (see in2 Physics @ Preliminary

section 13.8) is one of the most versatile techniques for measuring speed. As well as its use in Doppler ultrasound, it is used by astronomers to measure the motion of stars and by police radars to detect speeding drivers.

Figure 16.6.3 Colour Doppler ultrasound of a newborn’s heart, revealing blood flow (green) through a hole in the septum (grey) between the ventricles

CheCkpoInt 16.61 Discuss why the heel is an ideal place to use ultrasound to determine bone density.2 Explain how the Doppler effect can help diagnose the well-being of a foetal heart.3 Outline how echocardiography works.

activity 16.3




317

medicalphysics


ACtIvIty 16.1: ultrAsounD ImAgesCollect some images made using ultrasound and interpret information from the image.

Discussion questions1 Interpret the grey scale that accompanies most ultrasound images.2 Determine what parts of the body are best imaged with ultrasound.3 List things that can be determined from ultrasound images.

ACtIvIty 16.2: Doppler ultrAsounDUse several data sources to find and observe a Doppler ultrasound image that shows the flow of blood through the heart.

Discussion questions1 Define the Doppler effect.2 Explain how different directions in the video image are shown.3 Outline what can be determined from these images.

ACtIvIty 16.3: Bone DensItyFind out how bone density measurements are made using ultrasound and present the information in a 5 minute talk to the class.

Discussion questions1 Define bone density measurements and discuss why they are needed.2 Outline how the measurements are taken.3 Explain why ultrasound is a good method for obtaining these

measurements.

Gather secondary information to observe at least two ultrasound images of body organs.

Identify data sources and gather information to observe the flow of blood through the heart from a Doppler ultrasound video image.

Identify data sources, gather, process and analyse information to describe how ultrasound is used to measure bone density.

318

16 seeing with ultrasound chapter summary

Review questions

• Ultrasoundcantravelthroughthebodybutisreflectedfrom interfaces between different tissues.

• Formostmedicalimagingapplications,ultrasoundinthe range from 3.5 MHz to about 10 MHz is used.

• Thereflectedsoundcanbedetectedattheskinsurface,allowing imaging of structures deep within our bodies in a non-invasive way with little stress to the patient.

• Higherfrequencyultrasoundproducesimageswith a better resolution, but has poorer penetration through tissues.

• Lowerfrequencyultrasoundhasbetterpenetrationthrough tissues, but produces images with a lower resolution.

• Theintensityofthereflectedwaveisdeterminedbythedifference in the acoustic properties of the tissues and the absorption and angle of the intervening tissue.

• Ultrasoundwavesareemittedanddetectedby a transducer.

• Apiezoelectric transducer uses a crystal that responds to an oscillating voltage by oscillating slightly in size and responds to the pressure of a sound wave by producing an oscillating voltage.

• Aconvex-arraytransducerusesadivergingbeamtoproduce a sector scan.

• Acousticimpedancedescribeshowreadilyaspecificsound frequency will pass through a material. Acoustic impedance: Z = ρυ

• Whenultrasoundwavesmeetaboundarybetweentissues with different acoustic impedances, some of the ultrasound energy will be reflected and some will be transmitted. The greater the difference in acoustic impedance, the greater the intensity of the reflected signal.

I

I

Z Z

Z Z

r

o=

−

+

2 12

2 12

• Impedancematching(acousticcoupling)betweentwomedia results in most of the energy being transmitted through the interface, with almost no reflection.

• AnA-modescanshowsreturningechoesasaverticaldisplacement, proportional to the amplitude of the reflected signal.

• AB-modescanshowsthereturningechoesasdotswithintensity proportional to the returning echo amplitude. These dots are used to construct a 2D real-time scan.

• Amoviemadeusing3Dultrasoundimagesiscalled 4D ultrasound.

• Bonedensityscansusethefactthatthespeedandabsorption of ultrasound vary with bone density.

• DopplerultrasoundscanningusestheDopplereffecttoimage any structure or tissue that moves.

• ColourDopplerimagingiscolour-codedtoshowflowor movement towards or away from the transducer.

• Echocardiographyisultrasoundimagingoftheheart.

physICAlly speAkIngBelow is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

Ultrasound Transducer Doppler effect

Frequency Piezoelectric Acoustic impedance

Reflection Linear array Impedance matching

Bone-density measurement Echocardiography A-mode scan

B-mode scan Sector scan Blood flow

319

medicalphysics

revIeWIng 1 Compare quantitatively the frequencies of medical

ultrasound with sound in the normal hearing range of humans.

2 Contrast medical ultrasound and the sound that humans hear in terms of the media in which they mainly occur.

3 Figure 16.7.1 is a colour ultrasound image. Identify the organ that is imaged here and outline the main benefit of this type of image.

4 Outline the piezoelectric effect and its place in ultrasound imaging.

5 Assess the advantages of ultrasonic examination of a pregnant woman against the dangers associated with an ultrasound examination.

6 Define acoustic impedance in words (without using equations).

7 Identify the units for the quantities represented in the equation Z = ρu that defines acoustic impedance.

8 Account for the fact that different materials have different acoustic impedances.

9 The acoustic impedance of two tissues in the body is the same, and yet one has a greater density than the other. Explain how this is possible.

10 Identify the quantities represented by each symbol in the equation

II

Z Z

Z Zr

o

=− +

2 12

2 12

State the units for each quantity.

11 Describe the effect of increasing the difference in acoustic impedance at a tissue boundary on the proportion of the incident ultrasound energy that is reflected at the boundary.

12 Describe the conditions that must be satisfied for reflection to take place at a tissue boundary in the body.

13 Explain why ultrasound imaging is not used for investigating the function and structure of the brain.

14 Outline how refraction complicates the process of obtaining clear images of organs in the ultrasound process.

15 Discuss current issues associated with the use of obstetric ultrasound imaging.

16 Describe the Doppler effect and outline how this effect may be observed using sound waves.

17 Explain how ultrasound can be used to obtain information about the flow of blood through the heart.

18 Describe how ultrasound is used to measure bone density.

19 Compare the nature and use of A-mode and B-mode ultrasound scans.

20 Explain why ultrasound is not used to scan the lungs.

21 Assess the statement ‘Tissues having the same density have the same acoustic impedance’.

solvIng proBlems 22 Bone has a density of 2 × 103 kg m–3. The speed of

sound in bone is 4080 m s–1. Calculate the acoustic impedance of bone.

23 Calculate the density of kidney tissue, given that the acoustic impedance is 1.623 × 106 rayl and that the ultrasound velocity in kidney tissue is 1561 m s–1.

24 Calculate the percentage of the ultrasound energy that would be reflected from an interface between blood and brain tissue.

Figure 16.7.1 Example of an ultrasound image

Revie

w Questions

17

320

Imaging with X-rays

X-rays, cathode, radiographs, tomography, computed axial

tomography (CAT), characteristic X-rays, Bremsstrahlung, anode, radiographer,

filter, collimator, grid, fluorescent, contrast agents

The invisible reveals the hidden Wilhelm Röntgen (1845–1923) was studying cathode rays in a Crookes tube wrapped in cardboard, when he noticed a faint green glow from the fluorescent screen painted with barium platinocyanide—the rays emitted from the tube were passing through the cardboard. He found the rays could also pass through other papers and books. This phenomenon intrigued Röntgen and he dedicated the next few months to systematically investigating the properties of these rays. He named this new type of radiation X-rays. On 22 December 1895, Röntgen took the first X-ray, an image of the hand of his wife Anna. After this, many scientists and companies began investigating X-rays. Thomas Edison began a rigorous investigation into fluorescent materials. He developed the fluoroscope using calcium tungstate, which became the standard for medical X-ray examinations. Today, with the development of semiconductor detectors and image-processing software, X-rays are used in computed axial tomography (CAT scans) to provide medical doctors with high-resolution 3D images of skeletal, vascular and soft tissue.

17.1 Overview and history: types of X-ray images

X-rays are a form of electromagnetic radiation that lies between ultraviolet and gamma radiation on the electromagnetic spectrum. X-rays are able to pass through our body. In medicine, they are used for diagnosis of medical problems and injuries, and for therapeutic purposes such as treatment of certain tumours.

From 1896, many applications of X-rays were immediately recognised and generated a wide range of interest. Companies began researching and developing a range of specialised versions of Crookes tubes to produce X-rays. However, Crookes tubes are classified as cold cathode tubes and were often unreliable, as they needed to contain a small quantity of gas to operate. In 1913 William D Coolidge (1873–1975) invented the Coolidge X-ray tube, which permitted a stable and controllable continuous production of X-rays. Coolidge drew upon the thermionic diode (vacuum tube), which used a hot cathode (a filament) that permitted current to flow in a high vacuum. Most modern X-ray tubes are variations on Coolidge’s basic design.

321

medIcalphysIcs

A range of film-based and ‘fluoroscope’ devices were developed and refined, providing the familiar two-dimensional X-ray radiographs. In 1930, Alessandro Vallebona (1899–1987) proposed the idea of tomography in which a single slice of the body could be recorded on radiographic film. In the 1970s the development of affordable computers capable of processing image data, and the parallel development of semiconductors and image intensifiers allowed a new generation of X-ray devices to be developed. The technology was now available to re-engineer Vallebona’s tomography. Allan McLeod Cormack (1924–1998) developed the theory and Godfrey Hounsfield (1919–2004) independently built a prototype of the new generation of X-ray tomography scanner. This became known as computed tomography (CT) or computed axial tomography (also called CAT).

Today, with the development of more sensitive detectors and advances in image processing software, CAT scans can provide two- and three-dimensional images with resolutions of 1 cm. The combination of traditional radiographs and CAT scans provides the medical profession with important therapeutic and imaging diagnostic tools.

Figure 17.1.1 Wilhelm Röntgen’s first X-ray taken on 22 December 1895. It is the hand of his wife Anna Berthe Röntgen and you can see her wedding ring.

Figure 17.2.1 Schematic representation of a rotating-anode diagnostic X-ray tube

rotor

rotor/anodesupport

bearings

stator of induction motorrotating anodeglass envelope

cathodeblock

focusingcup electrons exit window

targetfilament

CheCkpOinT 17.11 Describe the properties of X-rays.2 Construct a timeline that shows the development of technology in

X-ray devices.

17.2 The X-ray tube X-rays are emitted when high-energy electrons strike a target. There are

two processes by which X-rays can be produced: characteristic X-rays and Bremsstrahlung. The efficiency of both processes in converting the kinetic energy of the electrons into X-rays is often less than 2%; the remainder of the energy is converted into heat.

The Coolidge X-ray tube developed in 1913 has remained the basis for the majority of today’s diagnostic devices. Several improvements have been made to increase efficiency, control and cooling of the anode target. Figure 17.2.1 shows a schematic representation of a modern rotating-anode diagnostic X-ray tube.

Describe how X-rays are currently produced.

Imaging with X-rays17

322

The tube is highly evacuated and generally constructed from glass. It contains an anode and a cathode. A potential difference of between 20 000 and 300 000 volts is applied between the cathode and the anode. The cathode has a small filament (similar to an incandescent lamp filament), which is heated to a dull red glow by a small current. The hot filament readily releases electrons and the high potential difference then accelerates these electrons from the cathode towards the target anode. The cup shape of the cathode behind the filament focuses the electron beam towards a point on the target anode. When the electrons hit the angled tungsten anode, X-rays are emitted through the exit window at the side of the X-ray tube. Because of the inefficiency of the process, a huge amount of thermal energy is produced. To ensure the tungsten anode does not melt, it is often rotated to more evenly distribute the energy. The anode is often mounted on a copper heat-sink that may also have an additional cooling system attached. Shielding, to reduce the emission of X-rays in unwanted directions, surrounds the tube itself.

CheCkpOinT 17.21 Name the two types of X-rays.2 Discuss the energy conversion of electrons as they strike a surface to produce X-rays.3 Describe the structure of a Coolidge tube.

17.3 Types of X-rays X-rays are often categorised as ‘hard’ and ‘soft’ X-rays, descriptive terms

that indicate the relative penetrating ability of the X-ray beam. ‘Soft’ X-rays do not penetrate through body tissue and are absorbed easily. These absorbed X-rays are not useful for producing radiograph images and they pose an increased risk to patients, of genetic mutations and cancers. X-ray machines have aluminium filters that attenuate these lower energy ‘soft’ X-rays from the X-ray beam. ‘Hard’ X-rays have a higher energy and greater penetration into and through tissue, thus producing a sharper image.

Changing the potential difference between the cathode and anode can vary the distribution of ‘hard’ and ‘soft’ X-rays produced by the X-ray tube. Increasing the voltage to the tube will produce more ‘hard’ X-rays. The radiographer can adjust the voltage, to obtain the best quality image of the particular organ, tissue, vessel or bone under examination.

Table 17.3.1 Properties of X-rays

‘Hard’ X-rays • Higher potentials are applied to the X-ray tube• Produced by the impact of high-energy electrons onto the anode• Shorter wavelength ~0.01 nm• Higher penetration (the X-ray photons have more energy)• Produce higher resolution images

‘soft’ X-rays • Lower potentials are applied to the X-ray tube• Produced by the impact of lower energy electrons onto the anode• Longer wavelength ~1 nm• Lower penetration (the X-ray photons have less energy)• Produce lower quality images

Compare the differences between ‘soft’ and ‘hard’ X-rays.

323

medIcalphysIcs

Characteristic X-rays The accelerated electrons can interact with and knock out an electron from

the inner electron shell of the anode. If an outer shell electron drops down and fills the vacant position, an X-ray photon of a specific energy and wavelength, peculiar to the elements of the target will be produced. In the total X-ray spectrum, the characteristic X-rays appear as sharp well-defined peaks.

Figure 17.3.1 shows the characteristic peaks for X-rays produced by a molybdenum target anode bombarded by electrons accelerated through a potential of 35 kV. The details of the process are shown in Figure 17.3.2.

Figure 17.3.1 Characteristic X-rays show up as distinct sharp peaks within the X-ray spectrum.

Figure 17.3.2 X-rays can be produced by firing high-speed electrons at a metal target. These electrons can eject electrons from the inner shells of the atoms of the target. Vacancies will be quickly filled by electrons dropping down from higher levels, emitting X-rays with specific defined wavelengths. Occasionally an emitted characteristic X-ray’s photon is re-absorbed by another electron, which is then ejected from the atom (an ‘Auger electron’).

3

2

1

Rel

ativ

e in

tens

ity

Brehmsstrahlungcontinuum

characteristicX-rays

X-rays from amolybdenumtarget at 35 kV

Kβ

Kα

0.02 0.04 0.06 0.08 0.10 0.12

Wavelength (nm)

X-ra

y in

tens

ity

X-ray wavelength

incidentelectron ejected

electrone–i

e–e–

e–e–e

e–

e– e–

2p

2p

2p

2s

2s

2s

1s

1s

1s

a gap remains

2 posible pathsto fill the gap

characte

ristic

X-ray

characteristic X-ray

Augerelectron

Figure 17.3.3 Bremsstrahlung (braking radiation) is characterised by a continuous distribution of wavelengths. The curves in this graph are based on the 1918 paper of Clayton Ulrey (1884–1963).

10

8

6

4

2

Rel

ativ

e in

tens

ity

0.02 0.04 0.06 0.08 0.10

X-ray continuum radiation(Bremsstrahlung)

Wavelength (nm)

50 kV

40 kV

30 kV

20 kV

Bremsstrahlung The word Bremsstrahlung means ‘braking radiation’ in German. It is used

to describe the radiation emitted when electrons are decelerated or ‘braked’, when fired into a metal target. When an electron rapidly changes its velocity, the lost kinetic energy is converted into photons of electromagnetic radiation. When the energy of the incident electrons is high enough, the radiation emitted will be X-rays. Decelerating electrons from the beam interact and produce a continuous distribution of radiation.

Figure 17.3.3 shows a set of curves for a tungsten target with electron beams of four different energies. The details of the Bremsstrahlung process are shown in Figure 17.3.4.

Figure 17.3.4 The processes associated with high-energy and low-energy Bremsstrahlung

hard X-ray

soft X-ray

e–i

e–i

e–i

e–i

X-ra

y in

tens

ity

Wavelength (nm)

incident electron incident electron


324

Figure 17.4.2 (a) An incomplete (greenstick) fracture of the radius and ulna of the lower arm and (b) a simple fracture of the ulna

17.4 production of X-ray images The key components of a medical X-ray machine are an X-ray tube, a filter, a collimator, a platform to situate the patient, a grid and a detector. Figure 17.4.1 shows a typical X-ray machine.

The filter is designed to remove ‘soft’ X-rays, which pose a risk to the patient and are not useful for imaging purposes. The collimator provides a mask to shape the output X-ray beam to minimise unwanted X-ray exposure to the patient. The grid situated just in front of the detector absorbs scattered secondary X-rays that would make the image fuzzy.

Conventional radiographsIn traditional radiographs, fluorescent screens absorb the energy in the X-ray beam that has penetrated the patient. This energy is converted into visible light that has same information as the original X-ray beam. This light is then used to expose the X-ray film. The more efficient this light conversion, the less the patient needs to be exposed to X-ray radiation. In most systems, the film is sandwiched between two fluorescent screens in a cassette so that the film emulsion is exposed from both sides. In modern X-ray machines, film is replaced by arrays of electronic detectors.

Conventional X-ray systems are used to image a variety of patient conditions including suspected skeletal bone fractures (Figure 17.4.2), cancer in the breast (Figure 17.4.3) and lung conditions.

Figure 17.4.1 A basic diagnostic X-ray machine

CheCkpOinT 17.31 Distinguish between hard and soft X-rays.2 Create a table to list the properties of hard and soft X-rays.3 Explain how characteristic X-rays are produced.4 Explain how Bremsstrahlung X-rays are produced.

a b

325

medIcalphysIcs

Imaging vascular and hollow structuresX-ray imaging does not show vascular (veins and arteries) or hollow soft tissue structures such as the digestive tract in great detail. To provide radiologists with a mechanism to reveal these structures, non-toxic contrast agents containing barium or iodine, which absorb X-rays, are used. These contrast compounds can be ingested, or injected into an artery or vein.

These procedures were first carried out between 1906 and 1912 and allowed blood vessels, the stomach, digestive tract, the gall bladder and bile ducts to be seen in situ for the first time. Today the process is very similar. An initial X-ray is taken, then contrast material containing iodine is injected into the bloodstream and a second X-ray taken. These two images are then digitally subtracted, leaving only the image of the blood vessels (Figure 17.4.4).

Another once common procedure is called a barium meal X-ray (Figure 17.4.5). This procedure involves the patient drinking a suspension of barium sulfate and then a series of radiographs are taken of the oesophagus, stomach and duodenum. Barium meal tests are declining with the increasing use of endoscopy (section 18.2), which allows the doctor to directly visually inspect the oesophagus, stomach and duodenum.

Figure 17.4.3 This breast X-ray, called a mammogram, shows (a) a healthy breast and (b) a breast with a tumour (arrowed).

Figure 17.4.4 Renal artery angiogram showing blood vessels, a kidney and the spine

Figure 17.4.5 Barium swallow X-ray of the large intestine of a patient with cancer of the sigmoid colon

CheCkpOinT 17.41 List the components of an X-ray machine and their functions.2 List the possible uses of conventional X-rays.3 Outline how it is possible to image hollow structures within the body.

a b


326

17.5 X-ray detector technologyIn the early 1900s a head X-ray would require the patient to be exposed to 10 minutes of radiation. Today a similar higher resolution image would expose the patient to one-fiftieth of the radiation used 100 years ago. Technological advances over the last 100 years have improved the quality of X-ray images and dramatically reduced the X-ray exposure to patients.

Originally radiographic images were produced on glass photographic plates. These were replaced by film cassettes with scintillating screens which intensified the images. In 1955, the X-ray image intensifier was developed and X-ray images were displayed on a television monitor. The image intensifier allowed doctors to image blood vessels and the heart in real time.

New digital technology and semiconductor detectors (silicon or germanium doped with lithium) began to be developed in the 1970s. Today, with the advent of large semiconductor array detectors, it has become possible to collect digital data that can be processed into high resolution images. These images can be enhanced and processed, transmitted for remote diagnosis and easily stored and retrieved. Over the next 15 years most conventional X-ray systems will be upgraded to all digital technology.

Figure 17.6.1 A 64-slice CAT scanner

CheCkpOinT 17.51 Discuss how image intensifiers have changed what doctors can do in diagnosis.2 List the benefits of digital technology.

17.6 production of CAT X-ray images A computed axial tomography (CAT) or computed tomography (CT)

scanner is a diagnostic device that uses an X-ray tube that is rotated around the patient. A detector collects the X-rays and digitises the information. A series of linear scans is processed by a computer and a series of two-dimensional slices or a three-dimensional rotatable image can be constructed. The development of CAT scan technology was enabled by the development of microcomputers and detector technology that could digitise data. In medical circles, the term CT is more commonly used than CAT, but they are interchangeable.

Basic designThere are several designs of CAT scanners in use. One commonly used scanner has the patient placed in a fixed position on a bed. The X-ray tube and detector array are attached to a movable C-shaped gantry that is rotated 180º around the patient and then progresses forward and records the next set of scans. Another more modern system has a moveable bed is that moved slowly through a fixed toroidal structure (a donut) that houses the X-ray tube and detector. As the bed progresses, the X-ray tube and detectors continually rotate around the toroidal structure, scanning the patient in a spiral motion.

Explain how a computed axial tomography (CAT) scan is produced.

327

medIcalphysIcs

a long WaIt

Godfrey Hounsfield’s original CT scan in 1972 took

several hours to acquire a single slice of image data and then more than a day to reconstruct the data into a single image. Today a CAT scan can construct an image in less than 1 second.

The scanning processIn older style devices (Figure 17.6.2a), a linear scan of data is taken at a large number of positions, producing the digitised data for that slice through the patient’s body. The source and detector are then rotated to produce a new scan.Modern devices speed up this process and direct a narrow fan-beam through the patient. The intensity of the emerging transmitted radiation is recorded by a line of adjacent detectors (Figure 17.6.2b and c). Fast CAT scans can now also be performed with a cone-shaped X-ray beam and an area-array detector panel.

ring ofstationarydetectors

a

b c

detector

detectorarray

collimatorX-ray source

X-raysource

X-raysource

X-raytube scan 90 scan 135sc

an 4

5

0° 180°

detectors

fanbeam

patient

direction ofX-ray tubemotion

Figure 17.6.2 The history of CAT: (a) A pencil-beam X-ray source and a detector scan linearly across the patient from a series of angles. (b) A fan-beam X-ray source and an arc of detectors rotate around the patient. (c) Only the fan-beam X-ray source rotates; the detector ring is stationary.

Constructing the two-dimensional slice imageTo form an image, we need to calculate the radiation that is absorbed at each small volume element (voxel) pixel within the scanned structure. These scans are three-dimensional versions of two-dimensional picture elements (pixels) that make up a picture from your digital camera. With this information we can construct a digital image. To assist in the explanation of the process used in CAT scans, we will use the simplified example shown in Figure 17.6.3.

In this example, two linear scans of a hollow cross have been made (Figure 17.6.3a). One linear scan has been made from the left and the other from the top. The intensities for each beam have been measured by the detectors opposite the X-ray sources. The computer then translates the information from the detectors into the total amount of the incident X-ray radiation that was absorbed as it passed through the structure. These values have been recorded down the right-hand side and along the bottom edge of the diagram (Figure 17.6.3a). This is the same information as obtained in a normal radiograph.

Now to obtain specific information about each voxel, we can sum the vertical and horizontal absorption values and record the value in each corresponding voxel (Figure 17.6.3b). A grey scale can then be overlaid, with the value ‘10’ being white, ‘0’ being black and the other values appropriate shades of grey. Figure 17.6.3c shows this grey-scale overlay.

Figure 17.6.3 Mechanism for constructing the 2D image slice

54245

54245

10 1099 98 86

9 98 86

7

7 76 64

9

10 109 7 9

X-rays

X-rays

a

b

c

d


328

The image generated shows some sort of cross structure but it is fuzzy. Now if we reassign the grey scale and assign light grey to all voxel values greater than 6 and a dark grey to all voxels with values of 6 or less, we obtain a clearer image of the original structure (Figure 17.6.3d).

This is the basic process of a CAT scan. For a real CAT scan in which there are complex sets of absorption calculations, a lot of data processing is required to recreate a high-resolution image from the data within a second. Figure 17.6.4 shows a single two-dimensional slice from a scan.

Because the data contains three-dimensional information, the imaging software can combine a series of slices to produce a three-dimensional image. Computer imaging software can also remove distracting tissues from the scan images and generate an image that shows the specific organ, tissue or structure under examination. Figure 17.6.5 shows a CAT angiogram of the blood vessels in the brain, with the surrounding tissue removed.

Common applications for CAT scans include:• identifyingtraumainjuriestothelungs,heart,spleen,kidneysandliver• planningforandassessingtheresultsofsurgery• planningradiationtreatmentsfortumours• detectingosteoporosisbymeasuringbonedensity.

Figure 17.6.4 A CAT scan of the lungs with the window level set to demonstrate the vessels and airways. This is used to look for diseases such as pneumonia or lung cancer.

Figure 17.6.5 An intracranial angiogram showing the blood vessels of the brain

CheCkpOinT 17.61 Outline how CAT images are produced.2 Describe how a two-dimensional scan is produced.3 Outline the benefits of a CAT scan.4 Outline the uses of a CAT scan.

329

medIcalphysIcs

17.7 Benefits of CAT scans over conventional radiographs and ultrasound

Conventional radiographs provide an image of all the structures present, superimposed on each other. For example a chest X-ray of the lungs will show all the ribs. The ribs therefore obscure details associated with the lungs. A CAT scan allows the ribs to be removed from the image and provides the doctor with a full image of the lungs. When the head is imaged, CAT scans provide a detailed image that is not as affected by the skull bone as a radiographic image. A CAT scan of the head, for example, can distinguish between the white matter, grey matter and spinal fluid. This enhanced discrimination allows the doctor to more easily identify vascular problems, tumours and subtle abnormalities. CAT scans contain the information to produce a three-dimensional image, allowing many angles of a particular area of interest to be viewed by the doctor. Software can be used to remove distracting tissues or structures.

Modern ultrasound equipment can also produce three-dimensional images (see section 16.5). Ultrasound can provide an initial diagnosis but cannot clearly identify the level of tissue damage or sites of internal bleeding. It is unable to penetrate bone or gas, and therefore tissues beyond these regions cannot be imaged. For this reason, ultrasound cannot be used to examine the brain, which is surrounded by the skull, or tissue masked by gas pockets in the abdominal region. A CAT scan can image all these regions, with resolution that is superior to that of ultrasound imaging.

Describe circumstances where a CAT scan would be a superior diagnostic tool compared to either X-rays or ultrasound.

CheCkpOinT 17.71 Identify the uses of CAT for which a radiograph can not be used.2 List problems that CAT scans can overcome that ultrasound imaging can not.

activity 17.1

pRactIcal eXpeRIences


pRactIcal eXpeRIences17 Imaging with X-rays

330

chapteR 17This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACTiviTy 17.1: CAT versus X-rAysFind a CAT scan and an X-ray radiograph of the same part of the body and compare the information that can be determined from each.

Discussion questions1 Outline characteristics that allow you to determine that an image is

a CAT scan.2 Outline characteristics that allow you to determine that an image is an

X-ray radiograph.3 Identify applications that would make a CAT scan more appropriate than

an X-ray radiograph.4 Identify applications that would make an X-ray radiograph more

appropriate than a CAT scan.

Gather information to observe at least one image of a fracture on an X-ray film and X-ray images of other body parts.

Gather secondary information to observe a CAT scan image and compare the information provided by CAT scans to that provided by an X-ray image for the same body part.

331

medIcalphysIcschapter summary

Review questions

• X-raysareaformofelectromagneticradiationthatliesbetween ultraviolet and gamma radiation on the electromagnetic spectrum.

• X-raysareusedforthediagnosisofmedicalproblemsand injuries, and for therapeutic purposes such as the treatment of certain tumours.

• ThemodernX-raytubeisbaseduponthe1913designby William Coolidge, which used a hot cathode and permitted a stable and controllable continuous production of X-rays.

• TheCATscanwasindependentlyconceivedby Allan McLeod Cormack and Godfrey Hounsfield.

• X-raysareemittedwhenhigh-energyelectronsstrike a target.

• TherearetwoprocessesbywhichX-rayscanbeproduced:characteristic X-rays and Bremsstrahlung.

• X-raysareoftencategorisedbytheterms‘hard’X-raysand ‘soft’ X-rays, which are descriptive terms indicating the relative penetrating ability of the X-ray beam.

• ThebasiccomponentsofanX-raymachineareanX-raytube, a filter, a collimator, a platform to situate the patient, a grid and a detector.

• Contrastcompoundscanbeingestedorinjectedtoassist X-ray imaging of vascular (veins and arteries) or hollow soft tissue structures.

• BenefitsofdigitaltechnologytoallX-raysystemsinclude:– lower X-ray doses to patients– images that can be enhanced and processed,

transmitted for remote diagnosis and easily stored and retrieved.

• Acomputedaxialtomography(CAT)orcomputedtomography (CT) scanner is a diagnostic device that uses an X-ray tube that is rotated around the patient. A detector collects the X-rays and digitises the information.

• ThedevelopmentofCATscantechnologywasenabledby the development of microcomputers and detector technology that could digitise data.

• CATscannersusepowerfulimage-processingsoftwareto recreate images from the scan data collected.

• CommonapplicationsforCATscansinclude:– identifying trauma injuries to the lungs, heart,

spleen, kidneys and liver– planning for and assessing the results of surgery– planning radiation treatments for tumours– detection of osteoporosis by measuring bone density.

• CATscansgenerallyprovidehigherresolutionimagesacross a greater range of body tissues than either radiographs or ultrasound.

physiCAlly speAkingThis is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

X-ray Hard Soft Characteristic

BremsstrahlungPenetration ability

Resolution Diagnostic

Benefits 2D scanImage intensifier

332

17 Imaging with X-rays

reviewing 1 Outline two ways in which X-rays are produced in an

X-ray tube.

2 Describe the differences between characteristic X-rays and Bremsstrahlung.

3 Using a chest X-ray as an example, recount how an X-ray image of the body is produced.

4 Compare the wavelengths of ‘hard’ and ‘soft’ X-rays.

5 A student wrote the statement: ‘Soft X-rays could also be called ultraviolet rays’. Assess this statement.

6 Assess the impact on society of doctors being able to examine images such as Figure 17.4.2, showing part of a patient’s arm.

7 For more than 100 years, X-ray images have been used to examine one type of tissue in the body more than any other. Identify this tissue and outline why X-rays are so successful in producing images of this type of tissue.

8 Figure 17.8.1 shows the chest area of two patients. Identify the types of X-ray images shown and compare the information provided by each image.

9 Explain why it is sometimes necessary for radiographers to take two radiographs of the chest area, one from the front and one from the side to provide the same information that can be seen in a single CAT scan of the chest.

10 Compare the effectiveness of CAT scans and radiographs to resolve soft tissue in the body of a patient.

11 Compare the advantages of the 2D and 3D images.

12 Compare the passage of the X-rays through a patient given a conventional chest X-ray (radiograph) and a CAT scan of the chest region.

13 Explain how a CAT scan is produced.

14 Outline the main technological development that took place in the 1970s that made CAT scans possible.

15 Assess the impact of particular advances in physics on the development of CAT scans.

Figure 17.8.1

Revie

w Questions

333

18Imaging with light

endoscope, endoscopy, total internal reflection, critical angle, optical fibres, core, cladding, coherent fibre bundle,

non-coherent fibre bundle, biopsy

Look and seeOptical fibres have had a great impact on society, although the fibres themselves are usually hidden from view. One of those impacts is in a doctor’s endoscope, a device to peer inside the human body without major surgery. Optical fibres allow light to be directed onto an area of interest and then lenses relay an image back to the doctor. Combined with tiny surgical instruments, these devices allow ‘minimally invasive’ surgery to reduce the trauma of surgery, having a radical effect on the treatment and recovery of patients with problems ranging from a torn knee cartilage to leaky heart valves.

18.1 Endoscopy An endoscope is a medical device used to shine visible light into a patient’s

body and relay an image out, to allow organs, tissues and cavities to be seen. Light is directed to the area of interest through a flexible bundle of optical fibres. Lenses are used to focus an image of the area of interest onto a separate bundle of optical fibres that transfer the image out of the body. The image is usually processed electronically to display it on a screen in real time or to record still or moving images for later use.

The process of using an endoscope to examine a patient is called endoscopy. The tube-like part of the endoscope, which enters the body, is typically less than a centimetre in diameter. Depending on the purpose of the endoscopy, the tube is inserted via a natural opening of the body, or through a small cut made in the skin. Miniature surgical instruments can be attached to the part that enters the body and these are controlled from outside the body by a doctor using control wires that pass through another narrow tube that is also a part of the endoscope. Tissue samples can be removed from the patient and surgical procedures can be carried out.

The main advantages of endoscopy are that it is minimally invasive and provides real-time images in true colour. The operator can manipulate the endoscope to obtain the best view of the area of interest inside the patient.

Figure 18.1.1 A gastric ulcer seen through an endoscope

Imaging with light18

334

PHYSICS FEATURE

Figure 18.1.2 Light moving from high to low refractive index materials may be totally internally reflected.


air

water

air

water

air

water

air

water

normal normal normal

90°

normal

nr

r

i ii c i c r i

r

ni

nr

ni

nr

ni

nr

ni

a b c d

core

jacket

cladding

cladding

core

Figure 18.1.3 A cross-section of an optical fibre

Figure 18.1.4 Light is transmitted along an optical fibre by total internal reflection.

OPTICAL FIbRES

You will recall learning about the properties and behaviour of light (see in2 Physics @ Preliminary

Chapter 8). The path of a light ray meeting a boundary between two transparent materials changes if there is a difference in the refractive indices of the two media. This process is called refraction.

Total internal reflection may occur at a boundary if the light is travelling from a medium with a higher refractive index ni to one having a lower refractive index nr and the angle of incidence exceeds the critical angle (θc) given by:

sin θc = nn

i

r

Total internal reflection (Figure 18.1.2) is the basic principle underlying the operation of relatively large diameter optical fibres.

Optical fibres are usually made of glass with two layers—the light-carrying, higher index inner core and the surrounding lower index cladding (Figure 18.1.3). Light entering the core at one end of the

optical fibre is transmitted along the optical fibre by total internal reflection, as illustrated in Figure 18.1.4.

Most optical fibres used for communications purposes transmit light in the near infra-red region of the electromagnetic spectrum at wavelengths of 800–1600 nm. The glass used in optical fibres has maximum transparency at these wavelengths. (Remember that the wavelength of visible light is between about 400 and 700 nm.)

Fibres used in medical endoscopy must be able to transmit light in the visible region of the spectrum. Because of the short distances involved (usually less than about 2 m), the fibres do not need to have to have the same transparency as fibres used for communication technology, nevertheless endoscopic fibres are made from high quality, homogeneous, high transparency glass. Homogeneity is an important property of the fibre because any irregularities would distort and degrade the image.

Explain how an endoscope works in relation to total internal reflection.

335

medIcalphysIcs

The main parts of an endoscope are shown in Figure 18.1.5. The endoscopic tube that is inserted into the patient typically contains the following parts:• abundleofopticalfibrestotransmitlighttothepoint

of observation• acoherentbundleofopticalfibrestocarrytheimageof

the tissue to the observer• asystemoflensestofocusanimageofthetissuesunder

examination onto the optical fibre bundle. At the viewing end of the optical fibre bundle, more lenses allow direct viewing of the image by eye, or connect to a camera that feeds a video screen or computer

• suctiontubetoremovebloodandotherloose,obscuringtissue material from the area under inspection

• aninletandanoutlettopermittheareaunderobservationto be flushed with clear saline (salt water) solution to increase visibility

• controllinestomanipulatethetubeinsidethepatient.These vary in complexity, depending on the task undertaken. A simple gastroscopy may not require complex controls. Endoscopes used for surgical procedures may require fine control over the position of the lens at the tissue end

• miniatureremote-controlledsurgicalinstrumentsmaybepresent, ranging from simple suction tools to more elaborate surgical tools used in operations.

Figure 18.1.5 Gastroscopy is the examination of the upper gut (oesophagus, stomach and duodenum) using a flexible fibre optic endoscope.

light tosee inside

Various devices can be passed down side channels. These can be manipulated by the doctor to takespecimens etc.

oesophagus

stomach

duodenum

doctor looksdown endoscpe

power sourcevideo out etc.The endoscope

is a thin,flexible fibre-optic ‘telescope’.

endoscope passed downoesophagus into stomach

Worked example QUESTIOn Calculate the critical angle for a glass optical fibre with a core refractive index of 1.48 and a cladding refractive index of 1.46.

SOLUTIOnThe critical angle will be a value of θi such that the angle of refraction is 90°.

ni = 1.48, nr = 1.46, θr = 90°

Snell’s law states:ni

sin θi = nr sin θr

Rearrange this to make θi the subject:

θθ

ir r

i=

−sinsin1 nn

Substitute values:

θi =×

= °−sin

. sin.

.1 1 46 901 48

80 5

Coherent and non-coherent bundles of fibres If all of the fibres in an optical fibre bundle are parallel along the full

length of the bundle, so that there is a uniform one-to-one correspondence between the positions of the fibres at one end of the bundle and the positions of the opposite ends of each fibre at the other end of the bundle, then the bundle

activity 18.1

pRacTIcal eXpeRIeNces


Imaging with light18

336

is said to be a coherent fibre bundle. The light from an object projected onto the ends of the fibre bundle by a simple lens will travel along the fibres and emerge as a corresponding image of the object at the other end of the bundle (Figure 18.1.6). This is the principle of the medical endoscope.

In a non-coherent fibre bundle, one or more of the fibres swap positions relative to each other at opposite ends of the fibre. Typically, the arrangement of fibres along the bundle is random, so that although the light travels along each fibre, no clear image is produced at the exit end (Figure 18.1.7). A non-coherent fibre bundle is adequate to simply transmit light to the point where observations are being made with the endoscope. White light is used so that the doctor can observe the tissues in true colour.

Discuss differences between the role of coherent and non-coherent bundles of fibres in an endoscope.

Figure 18.1.6 The arrangement of fibres in a coherent bundle allows an image to be transmitted by the bundle.

object

image

12345

12345

Figure 18.1.7 A non-coherent bundle does not transmit a sensible image.

object

no image

12345

24513

CHECkPOInT 18.11 Outline how an optical fibre allows you to see something inside the body.2 Define total internal refraction.3 Explain why infra-red light is used in communication applications but visible light is used in medical applications.4 Identify the difference between coherent and non-coherent bundles of fibres.

18.2 Medical uses of endoscopesEndoscopes are used to visually examine the inside of a patient’s body. Being able to see tissues allows doctors to diagnose diseases such as ulcers and tumours, and also to determine the nature and extent of injuries such as damaged cartilage and ligaments in joints.

Endoscopes used for specialised purposes have different names. An arthroscope is used to examine joints, a bronchoscope to view the lungs, a laparoscope to view female reproductive organs and an otoscope is used to view the ear.

A disadvantage of endoscopic examination is that it only allows the surface of tissues to be viewed, and so its use is limited to problems that cause visible changes to the surfaces of tissues. It can, however, be linked with ultrasound imaging to produce better results than ultrasound scans from outside the body.

337

medIcalphysIcs

developmeNTs IN eNdoscopy

An endoscopic capsule (Figure 18.2.3) is an endoscope with

no optical fibres! This small capsule can be swallowed by the patient and contains a wireless camera that can pass through the intestinal system and report via video link what is observed. This will improve endoscopic observation of the digestive tract.

A risk in using an endoscope is that the part of the endoscope inside the body can tear tissues while it is being moved about. Endoscopic examination of the bowel presents a particular risk, because the bowel contains bacteria, which, if they enter the bloodstream, can produce a fatal infection. However, the risk in using an endoscope is much less than the risk encountered if the abdomen had to be opened up in conventional surgery. The fact that a patient usually has to be sedated or anaesthetised presents another minor risk to the patient, although this type of risk is the lesser problem when weighed against the alternative of an undiagnosed or untreated problem.

Endoscopes that have been modified with surgical instruments can be used to remove tissue samples for testing. This process is called a biopsy, and this is one of the most common endoscopic procedures.

A common example of a biopsy is the removal of polyps or other growths from the intestine for further examination and testing. Endoscopy reduces this risk because the incisions and amount of cutting is minimised. Figure 18.2.1 shows a biopsy being performed within the abdomen. The sample of tissue cut off can then be placed or sucked into a tube attached to the endoscope and withdrawn from the body.

Minimally invasive surgery is conducted using optical fibre instruments that are often an integral part of the endoscope. Surgery that is commonly carried out with the aid of an endoscope includes removal of the gall bladder and the prostate, and repairs to damaged tissues in joints. A common joint operation is the repair of the anterior cruciate ligament in the knee (Figure 18.2.2). This part of the anatomy is frequently torn in sports such as netball and football, which involve vigorous twisting forces on the knees. People once condemned to months off the sporting field by knee injuries are now returning to their sport within weeks, because of endoscope-aided surgery.

Explain how an endoscope is used in: • observinginternalorgans• obtainingtissuesamplesof

internal organs for further testing.

Figure 18.2.2 External view of surgery to repair an anterior cruciate ligament

CHECkPOInT 18.21 List the advantages and disadvantages of endoscopy.2 Outline how a biopsy is done.

optical dome

lens holder

illuminating LEDs

lens

battery

antenna

Figure 18.2.3 An endoscopic capsule

Figure 18.2.1 A biopsy from within the abdomen

pRacTIcal eXpeRIeNces

338

18 Imaging with light


ACTIvITY 18.1: OPTICAL FIbRESMany light shops sell products known generally as ‘optical fibre lights’, which consists of numerous fibre optic tubes through which coloured light is passed via some colour-changing mechanism. Obtain one of these tubes and use a LED or laser light to shine light through the tube. Change the direction of the illumination by moving the end of the tube in different directions. Equipment: optical fibre (e.g. from an optical fibre lamp), light source (LED), power supply.

Discussion questions1 Identify what optical fibres are made of and explain how something

so brittle can be made so flexible.2 Explain how light is transferred down an optical fibre.3 Explain how optical fibres are used in an endoscope to transfer images

from inside the body.

Perform a first-hand investigation to demonstrate the transfer of light by optical fibres.

Gather secondary information to observe internal organs from images produced by an endoscope.

Figure 18.3.1 An optical fibre lamp

339

chapter summary medIcalphysIcs

Review questions

• Anendoscopeisanopticalinstrumentthatallowsreal-time observation of internal organs.

• Lightistransmittedthroughopticalfibresbytotalinternal reflection.

• Cheapernon-coherentbundlesoffibrescarrythelightinto the body.

• Moreexpensivecoherentbundlescarrytheimageout of the body.

• Advantagesofendoscopy– The tissues and organs are seen in real colour.– Imaging is in real time, enabling the doctor to

respond to what is seen.

– It allows minimally invasive tissue sampling and minor surgery that is safer and cheaper, with quicker recovery than open surgery.

– The process uses non-ionising radiation, namely light, an advantage over X-rays.

• Disadvantagesofendoscopy– It is more time consuming than ultrasound and

X-rays.– It presents minor risks to the patient, especially if

an anaesthetic is required. Operations on the bowel involve a risk of infection.

– Only the surface of tissues is visible.

PHYSICALLY SPEAkIngUse some of the chapter key words to complete the following paragraph.

The medical technique known as ________________ allows

minimally invasive procedures such as ________________

to be performed via a body opening or a small incision.

The ________________ allows a surgeon to view internal tissues

through the ________________ while it is illuminated via the

________________. An ________________ works because light is

confined to the ________________ enclosed by the

________________ of a fibre by ________________.

REvIEwIng 1 Explain the importance of total internal reflection to

the operation of an endoscope.

2 Compare the structure of coherent and non-coherent fibre bundles.

3 Compare the function of coherent and non-coherent fibre bundles in an endoscope.

4 Explain how an endoscope is used to observe internal organs.

5 Explain how an endoscope is used to obtain tissue samples from the stomach of a patient.

6 Recall an investigation that you carried out to demonstrate the transfer of light by optical fibres.

7 Assess the advances in medical techniques as the result of the use of endoscopes.

8 Explain why endoscopic surgery is often referred to as ‘keyhole surgery’.

SOLvIng PRObLEMS 9 Determine the angle of refraction of light that passes

from water (n = 1.33) to glass (n = 1.48) at an incident angle of 30º.

10 Determine the critical angle for a material with refractive index of 1.4 that is immersed in: a glass (n = 1.48)b water (n = 1.33)

11 A critical angle of 48.75° is measured at the boundary between air (n = 1) and another medium. Calculate the refractive index of the medium. Can you identify the probable medium?

Revie

w Questions

19

340

Imaging with gamma rays

radioactive decay, radiation, radioactive, radioisotopes, nucleons,

atomic number, mass number, isotopes, alpha decay, alpha particle, beta decay,

beta particle, positron decay, antiparticle, positron emission tomography (PET),

gamma decay, half-life, radiopharmaceuticals, nuclear reactor, cyclotron, gamma camera, bone scan,

collimator, scintillator, single-photon emission computed tomography (SPECT)

Radioactivity can be good!Images made using ultrasound, X-rays and visible light can show anatomical structures rather well. They all involve sending various forms of energy into the body. A rather different approach is to introduce radioactive elements into a person’s body and study the radiation that emerges. Images can be made of the bones as well as soft tissues including the brain, heart, liver and thyroid. Rather remarkably, this approach allows the production of images from outside the body that show how a person’s organs are functioning. Here we consider two of these minimally invasive but powerful diagnostic tools: bone scans using radioactive tracers and positron emission tomography.

19.1 Isotopes and radioactive decayFor the 30 or so lightest elements, the number of protons is roughly the same as the number of neutrons in the nucleus in most of their naturally occurring isotopes. These isotopes are stable. However, many elements have isotopes whose nuclei have too few or too many neutrons. These isotopes are unstable and undergo radioactive decay in which they change and emit radiation. The type of radiation that is emitted depends on the nature of the decay (see in2 Physics @ Preliminary section 15.5).

There are 82 elements that have at least one stable isotope. The stability depends on the ratio of protons to neutrons. As the atomic number increases, the ratio of neutrons to protons needed for stability also increases.

Many elements have naturally occurring unstable isotopes. These are called radioactive isotopes or radioisotopes. The nucleus of a radioisotope (the parent nucleus) usually transforms itself into another nucleus (the daughter nucleus) by emitting particles and energy. It will decay repeatedly until it forms a daughter nucleus that is stable.

Outline properties of radioactive isotopes and their half-lives that are used to obtain scans of organs.

341

medIcalphysIcs

Isotopes

An atomic nucleus consists of nucleons—

protons and neutrons. The number of

protons in the nucleus is called the atomic number, while the total number of nucleons is called the mass number. Atoms of the same element with different numbers of neutrons are called isotopes of that element. Many isotopes occur naturally, but some are made artificially. In section 15.4 of in2 Physics @ Preliminary we represented this information in a compact form. For example, an important isotope of fluorine is:

Mass numberAtomic number 9

18FIt is called fluorine-18, with 18 being the mass number. Other isotopes that are important in medicine include carbon-14, iodine-131, phosphorus-31 and technetium-99. Hydrogen is the only element that has special names for its three isotopes: hydrogen, deuterium and tritium.

Figure 19.1.1 Reducing the danger from radiation involves increasing distance, maximising shielding and reducing the time of exposure.

increasing distance

maximumshielding

reducing time of exposure

Figure 19.1.3 Uranium-238 undergoes alpha decay

parent nucleusU-238

alpha particle(helium nucleus)

daughter nucleusTh-234

4He2

Alpha decay Some unstable nuclei decay by emitting a particle that

contains two protons and two neutrons in a process known as alpha decay. The remaining nucleus has a mass number that is reduced by 4 and an atomic number that is reduced by 2. This particle emitted from the nucleus is called an alpha particle (α-particle). Alpha particles are helium nuclei ( 2

4He) and they rapidly become helium atoms, as they gain electrons from the surroundings. Such reactions are the source of most of the helium on Earth.

For example, radioactive uranium-238 undergoes alpha decay to produce thorium-234. The daughter nucleus has 2 protons less than the parent nucleus and so it is a different element. In a nuclear reaction, both mass number and charge are conserved, and the decay process can be described by an equation:

92238

90234

24U Th He→ +

or illustrated by a diagram (see Figure 19.1.3).

An alpha particle can only travel a few centimetres in air before it loses its kinetic energy and gains electrons to become a helium atom. In living tissue the range is about 50 µm, about half the width of a human hair. Due to their relatively large mass, alpha particles carry a lot of energy and have a high ability to ionise the surrounding medium, making them very dangerous to living cells. They are not used very much in medicine.

Beta decay When a radioactive nucleus undergoes beta decay, a

neutron changes into a proton, releasing a high-energy electron in the process. The electron is ejected from the nucleus with such a high velocity that it totally escapes the atom. An electron (represented as e– or −1

0ee) emitted from the nucleus in this way is called a beta particle (β-particle). An electron has only 1/1836

electron

electron

electronneutron

2 neutrons

proton

proton

proton

1H hydrogen1

3H tritium1

2H deuterium1

Figure 19.1.2 Isotopes of hydrogen—each with one proton and one accompanying electron

Imaging with gamma rays19

342

of the mass of the proton, and its mass number is 0. Its negative charge gives it an atomic number of –1.

Beta decay increases the atomic number of the nucleus by one but the mass number does not change. There is one less neutron but one more proton and hence the total number of nucleons does not change.

Carbon-14 decays by beta decay to nitrogen-14. The process can be described using an equation or a diagram (Figure 19.1.4).

614

714C N + e→ −1

0

The two products are a stable nitrogen nucleus and a beta particle (electron).

Beta particles have a range that depends on their energy. In air they travel several metres, but they travel only a few millimetres in human tissues.

Beta decay: positron productionSome nuclei of radioactive elements are unstable because they have too many protons in the nucleus, relative to the number of neutrons. These become more stable by a form of beta decay called positron decay. A positron (represented as e+ or β+ or sometimes +1

0e ) is the antiparticle of the electron. It has the same mass as an electron and the same magnitude charge as the electron, but it is positively charged. In positron decay, a proton in the nucleus decays to a neutron and a positron. The atomic number of the decaying nucleus decreases by one but the mass number remains the same. The positron is ejected with high kinetic energy from the nucleus.

The most important positron-emitting isotope is fluorine-18, an artificially produced isotope. Its decay equation is:

918

818

10F O e→ + +

Carbon-11 and oxygen-15 are also artificially produced positron emitters. When a positron and an electron collide, they mutually annihilate

(see Physics Phile ‘Evil twins’ p 73). Their total mass is converted into energy, producing two identical gamma (g) ray photons with a total energy consistent with Einstein’s famous equation E = mc2 where m is the total mass of the two particles (the mass of each electron is 9.1 × 10–31 kg) (see section 3.4). The energy of each gamma ray is

E = 9.1 × 10–31 × (3 × 108)2 = 8.2 × 10–14 J = 511 keV

The two gamma rays emerge in opposite directions to conserve momentum (Figure 19.1.5). In positron emission tomography (PET) (section 19.5) it is these gamma rays, not the positrons, that are detected and used to produce medical images.

Both kinds of beta decay are always accompanied by the creation of neutrinos, which are almost undetectable and not relevant to medical applications.

Gamma ray emittersThe most widely used medical isotope is technetium-99. It is a beta emitter that decays to ruthenium-99. The higher energy form of technetium-99, called metastable technetium-99 or technetium-99m, is also unstable and becomes more stable by emitting electromagnetic radiation. At the energy involved, this radiation is a gamma ray. The gamma ray carries away no charge or mass and so the nucleus remains technetium-99. It is the gamma decay of technetium-99m that is the medically important decay.

Identify that during decay of specific radioactive nuclei positrons are given off.

Discuss the interaction of electrons and positrons resulting in the production of gamma rays.

Figure 19.1.4 Carbon-14 decays to nitrogen-14 and an electron.

parentnucleus

betaparticle

daughternucleus

14N 7

14C 6

Figure 19.1.5 A positron and an electron annihilate to produce two gamma rays.

e+

e–e+

positronemitter

gamma rayphoton gamma ray

photon

343

medIcalphysIcs

Gamma decay also frequently accompanies alpha or beta decay. Cobalt-60 is a well-known beta and gamma emitter.

Gamma rays have no charge and so can pass easily through matter. To achieve the 95% absorption of the gamma energy from a cobalt-60 source, the rays must pass through 60 mm of lead, 100 mm of iron, or 330 mm of concrete.

atom-sIzed energy unIts

Particles or gamma rays produced in radioactive decay

carry away energy. The SI unit for energy is the joule, but that’s a lot of energy for a tiny particle or even a single gamma ray. A more ‘atom-sized’ unit of energy is the electron volt (eV)—the energy acquired by an electron in accelerating through a potential difference (voltage) of one volt. More generally, the energy acquired by a particle of charge q is given by:

E = qV So the energy acquired by the electron is: E = (1.6 × 10–19 C)(1 V) = 1.6 × 10–19 JSo 1.6 × 10–19 J = 1 eV As with other units, a large amount of energy can be expressed as keV (kilo-electron volts) and MeV (mega-electron volts) and so on.

Figure 19.1.6 The penetrating ability of different forms of radiation

human hand aluminium thinlead

thickerlead

concrete

alphabeta

X-ray (medical)gamma ray

neutrons

19.2 Half-life The time it takes for half the mass of the parent isotope to decay into

daughter nuclei is defined as the half-life of the isotope. After one half-life, only 50% of the original parent isotope remains; 50% of that remaining amount decays after another half-life, leaving just 25% of the original parent isotope and so on.

The half-life of a radioactive isotope can be deduced from a graph showing the mass of the remaining radioactive atoms of the element plotted against time. In Figure 19.2.1, the time taken for 2000 g of strontium-90 to be reduced to 1000 g is 28.1 years, the half-life of strontium-90. The daughter isotope is yttrium-90, which rapidly decays to zirconium-90, which is stable.

This mathematical model, called exponential decay, is applicable to all forms of radioactive decay. The rate of radioactive decay of an isotope is not affected by changes in physical conditions such as temperature or pressure. The decay rate is unchanged by any chemical reactions (or compounds in which the radioactive isotope may be incorporated). Every radioisotope has its own half-life.


CHECkPoInT 19.11 Describe what happens to an unstable nucleus.2 Identify two naturally occurring isotopes of the same element.3 Compare an electron with a positron.

Figure 19.2.1 The mass of strontium-90 remaining versus time, from an original sample of 2000 g

28.1 56.2 84.3 140.5 168.6 196.7 224.8 252.9 281.0 309.1

Time (years)112.4

2000

1800

1600

1400

1200

1000

800

600

400

200

0

Am

ount

(gr

ams)

0.0

Strontium-90 decay


344

CHECkPoInT 19.21 Define half-life.2 Outline how you would find the half-life of an isotope from a graph showing its mass in a sample versus time.

19.3 Radiopharmaceuticals: targeting tissues and organs

Radioisotopes are used to produce functional images of the body. They are used to examine blood flow to the brain, to assess functioning of the liver, lungs, heart or kidneys, to assess bone damage, and to confirm other diagnostic procedures. In contrast to the imaging techniques already discussed—ultrasound images, X-ray images and CAT scans—the use of radioisotopes can show how the body is functioning, rather than simply showing detailed images of tissues and organs in the body. Diseases such as cancer alter chemical processes in the body and images produced using radioisotopes can reveal these changes. Radioactive chemicals used in medicine are called radiopharmaceuticals.

Choose your element Different isotopes of the same element have identical chemical properties.

This is important in medical applications because radioactive isotopes can be substituted for non-radioactive atoms normally used by the body. Radiopharmacologists are able to attach various radioisotopes to biologically active substances and introduce them into the body. The radioisotope is then incorporated into the normal biological processes.

Radiopharmaceuticals are chosen so that when they enter the body they will circulate around the body and be absorbed by the organ of interest. The radiopharmaceutical may be a substance that is specifically used by a particular organ or it may be part of a molecule that is used by the organ to be imaged.

Radiopharmaceuticals are prepared by replacing one of the atoms in the molecules of that substance with a radioactive atom. This process is sometimes referred to as ‘tagging’ or ‘labelling’. When the radiopharmaceutical is placed in the body, it accumulates in the target organ and so the radiation that is emitted from that organ during the imaging process will be greater than the amount emitted from other organs and tissues.

The choice of a radioisotope used for medical imaging is based on the following criteria. It must:• producegammarays(directlyorindirectly)sinceonlygammaradiationis

likely to leave the body • haveahalf-lifethatislongenoughforthemoleculetoenterthemetabolic

processes, yet short enough to minimise the radiation dose to the patient• betakenuprapidlyintothedesiredtissue—thisisachievedbyincorporating

the radioisotope into an appropriate compound that is metabolised by the target tissue in the body

• berapidlyexcretedfromthebody.Appropriatelychosencompoundsarebroken down and excreted as part of the normal body chemistry.


Describe how radioactive isotopes may be metabolised by the body to bind or accumulate in the target organ.

345

medIcalphysIcs

Radiopharmaceuticals are used in very small quantities for diagnostic work. Just enough is administered to obtain the required information before the radiopharmaceutical decays, therefore minimising cell damage from the radiation. The radiation dose received is similar to that from diagnostic X-rays.

Where to get your radioisotopesRadioisotopes are produced in two main ways: in a nuclear reactor or in a cyclotron particle accelerator.

In a nuclear reactor, fission (splitting) of heavy nuclei produces large numbers of neutrons. The target element to be converted into a radioisotope is placed in the path of these neutrons and the nuclei absorb one or more neutrons, producing an unstable isotope with an excess of neutrons. These typically decay via beta decay.

Different isotopes can be produced in a cyclotron. In a cyclotron, protons are accelerated in a vacuum and fired into the nucleus of a target atom, to create isotopes that have an excess of protons. These isotopes typically decay via positron emission.

In Australia, most radioisotopes for medical purposes are prepared at the Australian Nuclear Science and Technology Organisation (ANSTO) OPAL nuclear reactor at Lucas Heights in Sydney and at the National Medical Cyclotron at the Royal Prince Alfred Hospital in Sydney (see Physics Focus p 353). The ANSTO reactor and the cyclotron make various radiopharmaceuticals, including some of those listed in Tables 19.3.1 and 19.3.2.

Nuclear reactors and cyclotrons are expensive to build and operate. The cost of these machines and the increasing demand for radioisotopes has added significantly to health costs in Australia and this has an impact on society through increased medical insurance costs. However, it is not practical to import many medical radioisotopes from overseas because of the short half-lives and the difficulty of transporting them safely.

Table 19.3.1 Examples of radioisotopes produced in a nuclear reactor

Isotope half-lIfe emIssIon UsesCobalt-60 5.3 years β, g External beam radiotherapy

Iodine-131 8 days β, g Cancer diagnosis and imaging of the thyroid gland

Phosphorus-32 14.3 days β Treatment of excess red blood cells

Molybdenum-99 2.75 days ‘Parent’ isotope used in a generator to produce technetium-99m, the most widely used isotope in nuclear medicine

Technetium-99m 6 hours g To image the skeleton and heart muscle in particular, but also for brain, thyroid, lungs, liver, spleen, kidney, gall bladder, bone marrow, salivary and lacrimal glands, heart blood pool, infection and numerous specialised medical studies

Table 19.3.2 Examples of radioisotopes produced in a cyclotron

Isotope half-lIfe emIssIon UsesCarbon-11 20.3 minutes β+ In PET for studying brain physiology and pathology, for investigating

epilepsy; have a useful role in cardiology

Nitrogen-13 10 minutes β+ In PET scans to tag ammonia and trace protein metabolism

Oxygen-15 2.03 minutes β+ In water for PET scan on blood flow in muscle tissue

Fluorine-18 109.8 minutes β+ For investigating tumours in the breast, prostate, liver and brain

Iodine-124 4.5 days β+ To investigate cancer, notably in the thyroid gland

Figure 19.3.1 The National Medical Cyclotron Facility at Royal Prince Alfred Hospital


346

CHECkPoInT 19.31 Explain the process of tagging.2 Discuss why only small amounts of radioisotopes are used in patients.3 Outline how radioisotopes are made.4 Explain why we do not just import all radioisotopes from overseas.

PHYSICS FEATURETECHnETIUm-99m

The radioisotope technetium-99m (Tc-99m) is used in more than half of all nuclear medicine

procedures. It is an isotope of the artificially produced element technetium and is almost ideal for use in nuclear medicine scanning.

Technetium-99m has a half-life of 6 hours and when it decays it emits a single gamma ray with energy of 140 keV. It does not emit any alpha or beta particles—an advantage since these cause biological damage. A further advantage of Tc-99m is that it has several electron valence states, so it has versatile chemical properties and can be used to produce a wide variety of pharmaceuticals.

A technetium generator, consisting of a lead vessel enclosing a glass tube containing the radioisotope molybdenum-99, is supplied to hospitals by ANSTO. Molybdenum-99 (Mo-99) has a half-life of 66 hours and it decays to Tc-99m by normal beta decay. The

Tc-99m is washed out of the lead or glass pot as required by passing a saline solution through the glass tube, a process called elution. After about 2 weeks the generator is returned for recharging.

The two nuclear reactions involved in Tc-99m production can be represented as:

4299

10Mo e→ + −43

99 Tcm half-life 2.75 days

4399

4399Tc Tcm → + γ half-life 6.01 hours

Technetium-99m has a half-life of 6 hours, which is long enough to examine metabolic processes, yet short enough to minimise the radiation dose to the patient. It is usually injected into the patient’s body and eliminated from the body via the kidneys. Technetium-99m allows the diagnosis of many conditions in the brain, lungs, heart, bones and other organs, avoiding the use of surgery.

Figure 19.4.1 Principle of a gamma camera

digital signal processing unit

photomultipliersNaI (Tl) detector

lead collimator

radioactive tracer in body

gamma photons

19.4 The gamma cameraA gamma camera (Figure 19.4.1) detects gamma rays emitted by a radiopharmaceutical in the patient’s body. The camera reveals the distribution of radioactive material in a patient and this distribution is determined by the uptake of the radiopharmaceutical, which is dependent on the functions—normal or otherwise—of the body.

In a bone scan, for example, technetium-99m methylene diphosphonate (Tc99m-MDP) is usually injected into the patient. The radioactive phosphate travels through the blood and is metabolised and accumulates in the bones.

The patient lies under a gamma ray camera. The gamma rays coming from the radioactive tracer in the body travel out of the body in all


347

medIcalphysIcs

directions. Unlike visible light, gamma rays cannot be focused to form an image. Instead a lead collimator with parallel holes passing from the bottom to the top is used to allow only those gamma rays directed through the holes to reach the detector itself.

The detector is a scintillator, made of a crystal of sodium iodide containing traces of thallium. The thallium impurity causes the crystal to scintillate (i.e. to emit a small flash of light) when a gamma ray enters the crystal. The flash is very faint, so sensitive devices called photomultipliers are used to detect the flash and amplify its effects, converting the result into an electrical signal. A computer processes the signals to produce an image that may be viewed directly on a screen, printed or stored electronically, since the information is all digitally encoded.

Regions of bone having a high metabolic activity appear darker on the bone scan because more of the radiopharmaceutical is absorbed (Figure 19.4.2). Conditions that may cause increased metabolic activity include cancer cells multiplying in the bone, healing of a fracture or damage from arthritis.

A gamma camera is also used in single-emission computed tomography (SPECT). The camera acquires two-dimensional images from multiple directions. These are then assembled into a three-dimensional image in a computer, in a similar way to other three-dimensional tomographic imaging techniques.

CHECkPoInT 19.41 Outline how a gamma camera creates an image of the body using

gamma radiation.2 Explain how the ‘grey scale’ in the bone scan (Figure 19.4.2) can

be interpreted.

Figure 19.4.2 This bone scan shows numerous hotspots along the spine, pelvis, shoulders and ribs that indicate that in this patient the breast cancer has spread to the bone.

activity 19.1

practIcal eXperIences


19.5 Positron emission tomographyPositron emission tomography (PET) is a ‘functional’ imaging technology that allows physicians to assess chemical changes related to how the body is functioning. In this way, PET is like a bone scan. PET imaging can analyse sugar metabolism, blood flow, oxygen use and a long list of other vital physiological processes. It is often used to investigate brain function and diseases affecting the brain, such as epilepsy, schizophrenia and Parkinson’s disease, as well as investigating which parts of the brain are active during specific activities.

PET imaging uses positron-emitting radiopharmaceuticals to obtain images. Positron-emitting isotopes exist for carbon, oxygen, nitrogen, fluorine and others, which allows these radioisotopes to be substituted into many naturally occurring substances used by the body (e.g. proteins, water and sugars—especially glucose). An increased level of metabolic activity in a part of the body associated with the chosen radiopharmaceutical causes a higher radiopharmaceutical concentration in that part and hence the gamma rays from those parts are more intense than those from parts in which there is less metabolic activity.


348

For example, if glucose labelled with fluorine-18, a positron emitter, is given to a patient who is then asked to perform a task, the regions in the brain in which the isotope concentrates most are the regions that are actively involved in that behaviour. Figure 19.5.1 shows PET images that reveal the different active parts of the brain of a person reading (seeing) written words or hearing the same words spoken by another person.

The radiopharmaceuticals are usually injected into a patient, although oxygen-15 can be inhaled and absorbed into the body through the lungs. The radioactive atoms decay, emitting positrons that travel about 2 mm before encountering an electron. Mutual annihilation occurs and two gamma rays are emitted in opposite directions. In a PET scanner these gamma rays trigger pairs of gamma-ray detectors that are arranged in a ring around the patient (Figure 19.5.2).

When two detectors directly opposite each other record photons within a few nanoseconds of each other (called a coincidence pair), the data are stored for processing to produce the image. Non-coincident hits on a single detector are ignored because they do not provide useful information about the location of the annihilation that produced them.

Signals reaching opposite detectors are analysed by a computer, which determines the position of the source of the annihilation event using the intersections of the gamma ray trajectories. Regions that emit the most gamma-ray photons produce a stronger signal than regions that emit fewer photons. A large number of measurements from the ring of detectors surrounding the patient is required to compute the distribution of the radioisotope in the body.

PET images don’t show anatomical structures well, so they are usually shown superimposed onto other medical images such as CAT scans, to clarify the location of the PET signal (Figure 19.5.3).

Figure 19.5.1 This PET image contrasts the visual and auditory stimulation of the brain.

seeing words

hearing words

Figure 19.5.2 (a) A PET scanner showing coincidence detection of gamma rays by opposite detectors and (b) an example of a PET scanner

detectors

coincidenceunit valid

event

a b

349

medIcalphysIcs

Figure 19.5.3 In this false colour PET scan (centre) two regions of high metabolic activity typical of cancerous tumours are indicated by the bright white hot-spots (in the spine and the prostate). This image is combined with a CAT image (left) to produce the fused image (right) that accurately indicates their location.

CHECkPoInT 19.51 Explain what features in the body PET scans can be used to identify that X-rays and ultrasound can not.2 Outline how the radioisotopes for a PET scan can be introduced into the body.3 Explain how coincidence pairs in a PET scanner allow an image to be constructed.

PET is a valuable diagnostic tool because the images that it produces show differences in chemical activity in different parts of the body. These differences may result from normal processes or they may be the result of diseases such as cancer or other functional abnormalities in the body. The great diagnostic benefit of PET results from the fact that it is possible to incorporate positron emitters into a wide variety of radiopharmaceuticals which can be produced to target specific organs or chemical processes in the body.

PET has provided new knowledge, in particular about the functions of the normal brain, as well as of diseases that affect the brain. This knowledge has also greatly modified brain surgery.

Describe how positron emission tomography (PET) technique is used for diagnosis.

activity 19.2




350


ACTIvITY 19.2: HEAlTHY oR dISEASEd?Typical images of healthy bone and cancerous bone are shown. The tumours show up as hot-spots. Use the template in the activity manual to research and compare images of healthy and diseased parts of the body.

Discussion questions1 Examine Figure 19.4.2. There is a hot-spot that is not cancerous near the

left elbow. Explain.2 In the normal scan (Figure 19.6.2a), the lower pelvis has a region of high

intensity. Why is this? (Hint: It may be soft tissue, not bone. Looking at Figure 19.6.2b might help you with this question.)

3 State the differences that can be observed by comparing an image of a healthy part of the body with that of a diseased part of the body.

Gather and process secondary information to compare a scanned image of at least one healthy body part or organ with a scanned image of its diseased counterpart.


ACTIvITY 19.1: BonE SCAnSA bone scan is performed to obtain a functional image of the bones and so can be used to detect abnormal metabolism in the bones, which may be an indication of cancer or other abnormality. Because cancer mostly involves a higher than normal

Perform an investigation to compare a bone scan with an X-ray image.

Figure 19.6.1 Comparison of an X-ray and bone scan of a hand

Figure 19.6.2 Bones scans of (a) a healthy person and (b) a patient with a tumour in the skeleton

rate of cell division (thus producing a tumour), chemicals involved in metabolic processes in bone tend to accumulate in higher concentrations in cancerous tissue. This produces areas of concentration of gamma emission, indicating a tumour.

Compare the data obtained from the image of a bone scan with that provided by an X-ray image.

Discussion questions1 Identify the best part of the body for each of these

diagnostic tools to image.2 Compare and contrast the two images in terms of

the information they provide.

a b

351

chapter summary medIcalphysIcs

review questions

• Thenumberofprotonsinanucleusisgivenbytheatomic number, while the total number of nucleons is given by the mass number.

• Atomsofthesameelementwithdifferentnumbersofneutrons are called isotopes of that element.

• Manyelementshavenaturallyoccurringunstableradioisotopes.

• Inalphadecayanunstablenucleusdecaysbyemittingan alpha particle (α-particle).

• Inbetadecay,aneutronchangesintoaprotonand a high-energy electron that is emitted as a beta particle (β-particle).

• Inpositrondecay,apositron—theantiparticleoftheelectron—is emitted.

• Whenapositronandanelectroncollide,theirtotalmass is converted into energy in the form of two gamma-ray photons.

• Ingammadecayagammaray (g) is emitted from a radioactive isotope.

• Thetimeittakesforhalfthemassofaradioactiveparent isotope to decay into its daughter nuclei is the half-life of the isotope.

• Artificialradioisotopesareproducedintwomainways:in a nuclear reactor or in a cyclotron.

• Agammacameradetectsgammaraysemittedby a radiopharmaceutical in the patient’s body.

• PETimagingusespositron-emittingradiopharmaceuticals to obtain images using gamma rays emitted from electron–positron annihilation.

PHYSICAllY SPEAkIngBelow is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

Radioactive decay

Radiation Radioisotope Nucleon

Neutron Proton Isotope Alpha decay

Beta decayGamma decay

Antimatter PET

Half-life Bone scanPositron decay

Scintillator

REvIEwIng 1 Recall how the bone scan produced by a radioisotope

compares with that from a conventional X-ray.

2 Analyse the relationship between the half-life of a radiopharmaceutical and its potential use in the human body.

3 Explain how it is possible to emit an electron from the nucleus when the electron is not a nucleon.

4 Assess the statement that ‘Positrons are radioactive particles produced when a proton decays’.

5 Discuss the impact that the production and use of radioisotopes has on society.

6 Describe how isotopes such as Tc-99m and F-18 can be used to target specific organs to be imaged.

7 Use the data in Table 19.6.1 to answer the questions:a Which radioactive isotope would most likely be

used in a bone scan? Justify your choice.b Propose two reasons why cesium-137 would not

be a suitable isotope to use in medical imaging.

Table 19.6.1 Properties of some radioisotopes

RadIoactIve soURce RadIatIon emItted half-lIfeC-11 β+, g 20.30 minutes

Tc-99m g 6.02 hours

TI-201 g 3.05 days

I-131 β, g 8.04 days

Cs-137 α 30.17 years

U-238 α 4.47 × 109 years

352


8 Figure 19.6.3 shows a bone scan and an X-ray image. Identify which is which and justify your choice.

9 Compare the processes used to produce a bone scan image and an X-ray image.

10 Compare the radiation used to produce a bone scan image and an X-ray image.

11 Compare the resolution of a bone scan with that of an X-ray.

12 Explain why the resolution achieved in a bone scan is dependent on the design of the lead collimator used in the gamma camera.

13 Explain why tumours are visible in the patient in Figure 19.6.4b.

Figure 19.6.6 Decay curve of a radioisotopeYears

50.0%

25.0%

12.5%

6.3%3.1%

0 2 4 6 8 10 12 14 16

14 Describe the interaction between electrons and positrons.

15 Explain how a PET scan is produced and compare the image with that obtained from a CAT scan.

16 Identify the key characteristic of a PET image that gives it a diagnostic advantage over a CAT image.

Solving problemS 17 Complete the decay series of U-238 shown in Figure

19.6.5 by including the decay product—an α, β or γ particle—in each step, as shown for the α-emission seen in the first step.

18 Figure 19.6.6 shows the percentage of a radioactive isotope that remains as a function of time. a Estimate the half-life of this isotope.b Would the isotope represented by this decay

curve be suitable for medical imaging? Justify your answer.

Revie

w Questions

Figure 19.6.3 Two different images showing the chest region of the human body

a b

Figure 19.6.4 A bone scan of (a) a healthy patient and (b) a patient with tumours

a b Figure 19.6.5 The decay series of U-238

Z

N

146

144

142

140

138

136

134

132

130

128

126

124

1228078 82 84 86 88 90 92 94

238U234Th

230Th

234Pa

234U

226Ra222Rn

218Po218At214Pb

210Pb

210TI

214Bi

210Bi

210Po206Pb

214Po

α

353

medIcalphysIcs

PHYSICS FoCUSAnSTo RAdIoPHARmACEUTICAl PRodUCTIonlUCAS HEIgHTS PRodUCTIon FACIlITY

The Lucas Heights Facility in Sydney, operated by ANSTO, manufactures radiopharmaceuticals such as molybdenum-99 generators for technetium-99m and iodine-131, together with other associated products including chromium-51, thallium and gallium. The facilities are ideal for simultaneous production of large quantities of different isotopes used for diagnosis and therapy. An average of 3500 medical isotopes and 2500 packages are dispatched per month. The packaging, transport and delivery of all products comply with strict national and international regulations.

molYBdEnUm PlAnT

ANSTO Health is in the process of commissioning a new plant that will be used to manufacture molybdenum-99. ANSTO is currently importing this product from overseas to meet domestic needs; however, this is costly (as a heavy airfreight product) and shipment is not always reliable, which emphasises the importance of local Australian production. Molybdenum-99 is used as a raw material for 80% of nuclear medicine procedures performed around the world.

Molybdenum-99 is formed by the fission of uranium-235 that is itself formed by irradiation of low enriched uranium (LEU) in ANSTO’s nuclear reactor. The molybdenum-99 is then used as a generator for technetium-99m, which can then be formulated into a plethora of radiopharaceuticals.

The new plant is located behind shielding because of radiation, and has been carefully designed to protect workers and the external environment during and after processing.

nATIonAl mEdICAl CYCloTRon FACIlITY (nmC)

The NMC was established in 1991 and produces specialised isotopes for SPECT (single photon emission computed tomography) imaging. The facility houses a 30 MeV cyclotron, the largest in Australia.

Located at Sydney’s Royal Prince Alfred Hospital, Camperdown, the cyclotron accelerates protons to the required energy and then fires them at different targets to generate the radioisotope required. The radioactive isotope is then recovered and purified for dispensing into the finished dosage at Lucas Heights. The products produced at the cyclotron facility are gallium-67, thallium-201, iodine-123 and iodine-123.

1 Discuss why the correct packaging and transport of radiopharmaceuticals is important.

2 Discuss why it is important to have facilities to produce radiopharmaceuticals in Australia.

3 Explain why molybdenum-99 is an especially important radioisotope.

ExTEnSIon4 The OPAL nuclear reactor opened in 2007,

replacing the earlier HIFAR reactor. Some critics of the nuclear reactor program claim a new reactor was unnecessary.

Investigate this issue and discuss your conclusions in terms of:a the need for radiopharmaceuticalsb the safety of a nuclear reactor.

4. Implications of physics for society and the environment


20

354

Imaging with radio waves

spin, magnetic moment, parallel, antiparallel, precession, Larmor frequency, radio frequency (RF),

magnetic resonance imaging (MRI), resonance, relaxation, longitudinal

relaxation time constant, transverse relaxation time constant,

RF transceiver coils, gradient coils, functional MRI,

Hydrogen callingMagnetic resonance imaging (MRI) uses radio waves emitted by hydrogen atoms to produce high-resolution images of the body. Magnetic resonance has been used as a tool for studying atomic structure since 1946, but it was not until 1973 that medical imaging was suggested, and another 15 years before useful images were obtained. Like CAT scans, MRI can produce tomographic images (slices) of the body, enabling detailed two- and three-dimensional images to be constructed from the data. Magnetic resonance imaging was initially used to produce images showing structural detail; however, recent developments have led to functional MRI, which has greatly advanced knowledge of how the brain and other organs work.

20.1 Spin and magnetism Particles such as protons, neutrons and electrons have a property called

spin, which can be visualised as a rotation (spinning) of the particles. However, spin is a quantum-mechanical property of the particles and the classical view of a spinning ball is a useful ‘model’, but it is not reality. Protons are not simply tiny spinning charged balls, but some of their properties are similar.

For protons, neutrons and electrons the spin property can only have one of two possible directions: up or down (sometimes called ‘spin up’ and ‘spin down’). If the total number of protons and neutrons in a nucleus is even, then their spins align in pairs in opposite directions so that the net spin of each pair and the whole nucleus is zero (Figure 20.1.1). If there is an odd number of nucleons, then the nucleus has a net spin, since there must be one unpaired nucleon. Nuclei having a net spin include hydrogen, phosphorus-31, fluorine-19, nitrogen-15 and carbon-13.

Identify that protons and neutrons in the nucleus have properties of spin and describe how net spin is obtained.

Figure 20.1.1 Using our model of spin, we can visualise two particles of opposite spin resulting in zero net spin.spin vector

protons

opposite spins resultingin zero net spin

spin vector

355

medIcalphysIcs

Spin produces a magnetic effect called a magnetic moment or magnetic dipole. You can think of this as being similar to the magnetic field produced by moving charge (a current) around the wire of a solenoid (see Figure 4.1.10). Therefore, charged particles such as electrons and protons behave like tiny magnets. Even uncharged neutrons have a magnetic moment, because they contain fundamental charged particles called quarks (see section 15.5). Nuclei with an odd number of nucleons will therefore have a net spin and always have an unpaired magnetic moment and so behave like tiny magnets. When placed in a strong magnetic field these nuclei behave a little like a compass needle in the Earth’s magnetic field. A compass needle always points north, but the nuclei magnetic moments may assume one of two possible alignments—they can align parallel or antiparallel to the magnetic field. External magnetic fields or the magnetic field of electromagnetic radiation may affect the nuclei because of this magnetic moment.

Hydrogen is the most important element in the process of magnetic resonance imaging because of the magnetic properties of the nucleus and the fact that it is present in water molecules, proteins, fats and carbohydrates—in fact, in most of the molecules in the body. Since it normally has just a single proton forming the nucleus, a hydrogen nucleus will clearly have a magnetic moment. The magnetic properties of hydrogen are not normally evident, because the magnetic moments of the nuclei are randomly aligned, resulting in a zero net magnetic effect (Figure 20.1.2).

Identify that the nuclei of certain atoms and molecules behave as small magnets.

Explain that the behaviour of nuclei with a net spin, particularly hydrogen, is related to the magnetic field they produce.

Figure 20.1.2 The net spin of hydrogen results in a magnetic moment that is often not apparent in large groups of hydrogen nuclei.

south

magneticmoment

north

CHeCkpoInt 20.11 Outline what is meant by the ‘spin’ of a particle.2 Describe the situations in which an atomic nucleus has zero net spin.3 Define the term magnetic moment.4 Describe the two possible alignments of the magnetic moment of a proton.

20.2 Hydrogen in a magnetic fieldWhen placed in a very strong magnetic field, the proton will tend to align with the applied field, because of the interaction between the external field and the magnetic properties of the proton. In a medical context, the strength of the magnetic field used is typically between 0.5 and 5 T, although fields used in research applications may be more than 10 T. The ‘parallel’ and ‘antiparallel’ alignments to the field are not strictly parallel to the field. They are actually angled as illustrated in Figure 20.2.2. The parallel configuration has a slightly lower energy than the antiparallel configuration; just how much lower depends on the strength of the magnetic field. In a 1 T field, the difference is just 0.18 µeV (0.18 × 10–6 eV or 2.8 × 10–26 J). As a lower energy state, it is

Describe the changes that occur in the orientation of the magnetic axis of nuclei before and after the application of a strong magnetic field.

Imaging with radio waves20

356

therefore slightly preferred—but only about 0.0003% more protons favour the parallel configuration. Injecting the energy difference that corresponds to a photon with a radio frequency of 42.6 MHz can cause the proton to flip from the parallel to the antiparallel configuration. This same energy will be released if it flips back.

When the applied external field is removed, the protons become randomly orientated again because of collisions occurring between the randomly moving atoms.

Precession When you have a magnetic moment angled to the direction of a magnetic

field, the field will exert a force and hence a twist (a torque) on the magnetic moment. This will cause the direction of the magnetic moment to rotate about the magnetic field direction. This is called precession. It is similar to the precession of a toy spinning top caused by the force of gravity. In our model of a spinning proton, the direction of the magnetic moment corresponds to the imagined spin axis of the proton, which is analogous to the spin axis of the top (Figure 20.2.2).

The frequency of precession of a magnetic moment such as a proton is called the Larmor frequency. It depends on the magnetic properties moment of the proton (µp) and is proportional to the magnitude of the external field (B). It is given by:

fB

hLarmorp

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

2μ

where h is Planck’s constant.For a proton in a 1 T magnetic field,

the Larmor frequency is 42.6 MHz, corresponding to radio frequency (RF) electromagnetic waves.

Define precessing and relate the frequency of the precessing, (i.e. Larmor frequency), to the composition of the nuclei and the strength of the applied external magnetic field.

Figure 20.2.1 Randomly orientated hydrogen nuclei align to an externally applied magnetic field.

parallel

antiparallel

no external field

external magnetic field

Figure 20.2.2 Precession of a hydrogen nuclei around the applied magnetic field

precession

external magnetic field

Try ThIs!A spinning topWhen you try to push over a child’s spinning top, it doesn’t fall over. Try it! Instead, the top’s axis of rotation itself starts to rotate around the vertical axis—it ‘precesses’ (at least until frictional forces cause it to tip too far and it hits the ground).

The force of gravity produces a twisting effect (a torque) on the top. If the top weren’t spinning, this would cause it to simply fall over. Precession is the motion that results when the top is spinning. The axis of rotation of the top sweeps out a conical motion.

Figure 20.2.3 Precession of a spinning top

force of gravity

precession

rotation

357

medIcalphysIcs

20.3 tuning in to hydrogenExcitation

In magnetic resonance imaging (MRI) a person is placed in a strong magnetic field and a pulse of RF electromagnetic radiation is sent into their body. The frequency is tuned to match the Larmor frequency of protons within the field.

The RF pulses have two effects on the protons in the person’s body. First, they cause a ‘spin flip’ to occur. A few of the protons aligned parallel to the external field absorb the RF energy because it matches the energy necessary to flip from the parallel to the slightly higher energy antiparallel configuration. Before the RF pulse there was a small imbalance in numbers, with slightly more protons in the parallel configuration. The net magnetic effect of the alignment of protons along the magnetic field direction (usually called the longitudinal magnetisation) was not zero. The RF pulse results in the number of parallel and antiparallel protons becoming more nearly equal, causing the net magnetic effect in this direction to become zero (Figure 20.3.1). This direction is along the head-to-toe axis of the person in the MRI machine.

The second effect of the RF pulses is to cause the protons to precess in step (‘in phase’) with each other. This can be compared to soldiers all marching in step. The protons precessing in phase with each other create a net magnetic effect, Bxy, in the plane perpendicular to the field (Figure 20.3.2). This net magnetic effect rotates at the Larmor frequency.

The RF wave is used to transmit energy into the tissues to be imaged. The energy of the RF electromagnetic waves is absorbed effectively by protons in hydrogen, because the frequency of the radiation and the precession frequency of the protons are the same. Said another way, the energy of the RF photons matches the energy difference between the parallel and antiparallel configurations. This energy exchange in which properties are matched is called resonance.

Discuss the effect of subjecting precessing nuclei to pulses of radio waves.

Figure 20.3.2 (a) The protons precess in phase in response to the RF pulse. (b) The net magnetic field is in the xy plane.

Bxy

N

S

N

S

Figure 20.3.1 (a) Parallel (blue) and antiparallel (red) protons are aligned within a strong magnetic field. (b) RF radiation causes some protons to jump to the antiparallel configuration.

Y

N

S

Z

X

N

S

CHeCkpoInt 20.21 Explain what happens to a proton in an external magnetic field.2 Using the photon energy given by E = hf, demonstrate that the frequency f required to cause a proton to flip its

alignment in a 1 T magnetic field is 42.6 MHz.3 Explain how the frequency required to flip a proton will change if the applied magnetic field is doubled.

a

a

b

b

activity 20.1

pracTIcal eXperIeNces



358

Relaxation After the RF pulse, the protons return to their original state

and, as they do so, they re-emit the energy they absorbed from the radio wave. They do this over a period varying from about 0.01 to 0.1 s. The emitted energy is again an RF wave that is detected using an antenna. The signals emitted by the protons are used to create the MR image. The return of the proton to the less excited state after absorbing the RF energy is called relaxation.

The emission of the energy from the hydrogen nuclei returning to the lower energy state is described by two time constants, T1 and T2, associated with two different electromagnetic processes (Table 20.3.1).

The longitudinal relaxation time constant T1 is a measure of the time taken for protons to return to their normal ratio of parallel and antiparallel configurations relative to the field. In this process, the longitudinal magnetisation returns to its non-zero strength. The signal that is detected as a result of this process increases in strength from zero as the number of protons aligned parallel to the external field approaches its equilibrium state (Figure 20.3.4).

The time constant T1 is defined as the time taken for the signal from the protons aligned in this state to reach a particular value (about 63% of the final equilibrium value).

T1 is dependent on the interaction between the precessing protons and the other atoms in the material. It is sometimes called the spin–lattice interaction, because the process was first observed in crystals, which have a lattice structure.

resoNaNce

Resonance occurs between two oscillating objects if they have the same natural frequency and if there is some

means by which energy can be transferred from one object to the other. The everyday experience of pushing a child on a swing demonstrates the principle. To make the child swing higher, you must push at the same frequency as the child is swinging—push every time they get to the top of the back swing. If you push at the wrong time, you will just interfere with the swing (and hurt yourself!). Resonance can also be observed in a guitar when pressing down on the fifth fret of the first string and plucking the second string. If the guitar is tuned correctly, the first string will also vibrate. This occurs because under these conditions the two strings have the same natural frequency and the energy can be transferred from one string to the other via the body of the guitar.

nut

1st and 2nd strings

1st fret

5th fret

Figure 20.3.3 Resonance can be observed in a guitar.

Figure 20.3.4 The relationship between the longitudinal relaxation time constant T1 and the increasing longitudinal magnetisation

T1 Time

100%

Long

itud

inal

mag

neti

sati

on

Table 20.3.1 Relaxation times for various tissues in a 1 T field

tissue t1 (ms) t2 (ms)Brain grey matter 520 100

Brain white matter 390 90

Cerebrospinal fluid 2000 300

Fat 180 90

Muscle 600 40

Blood 800 180

Water 2500 2500

359

medIcalphysIcs

The transverse relaxation time constant T2 is a measure of the time taken for the magnetic component perpendicular (transverse) to the external field to return to zero. This happens as the precessing protons become out of phase with each other (in the analogy, this is like the soldiers getting out of step). As a result of the loss of phase between the precessing protons, the transverse magnetisation, Bxy , decreases exponentially to zero (see Figure 20.3.5).

The time constant T2 is dependent on the interaction between the precessing protons and is sometimes referred to as the spin–spin relaxation time. Figure 20.3.5 The relationship between the transverse

relaxation time constant T2 and the decreasing transverse magnetisation. The arrows in the circles represent the components of the magnetic properties of the protons becoming progressively more random, causing the transverse magnetisation, and the RF signal associated with it, to decrease to zero.

T2 Time

100%

Tran

sver

se m

agne

tisa

tion

Bxy

BxyBxy

CHeCkpoInt 20.31 Outline the sequence of events during an MRI scan.2 Explain the two consequences of applying correctly tuned RF

pulses to the hydrogen in the body.3 Outline the general concept of resonance.4 Define the relaxation times T1 and T2.

20.4 It depends on how and where you look

Each type of tissue in the body contains different amounts water and therefore different amounts of hydrogen. Hydrogen is also present in carbohydrates, proteins and fats, and the amounts of these also vary in different tissues. The amplitude of the radio signal emitted as the relaxation processes occur increases as the number of protons in the tissue increases. An MRI can be created which shows these differences in proton density and thereby highlights the difference between different tissues.

Each tissue has unique values for T1, and T2. Magnetic resonance images can also be obtained to show specific tissues more clearly using the differences in the values of T1 and T2 that are characteristic of each tissue type. If the contrast in an image results mainly from differences in the T1 values, the images are called ‘T1 weighted’. Images dominated by T2 differences are called ‘T2 weighted’. Figure 20.4.1 illustrates the difference in contrast between T1 and T2 weighting.

Explain that the amplitude of the signal given out when precessing nuclei relax is related to the number of nuclei present.

Figure 20.4.1 Contrast differences in (a) T1 weighted and (b) T2 weighted images centred on the knee of a 9-year-old boy. The arrows indicate the lesion responsible for his knee pain.

a b


360

The difference in relaxation times T1 and T2 is significantly greater for tissues containing relatively large amounts of water. As a result, MRI is very sensitive to variations in the water content of tissues (see Table 20.4.1) and this is a significant factor in MRI’s ability to produce high-resolution, high-contrast images (see Figure 20.4.2).

Because tumours are characterised by rapid cell division and high growth rates, they typically have a higher percentage of water than similar non-cancerous tissue, and can thus be clearly imaged using MRI. Haemoglobin molecules in red blood cells also provide a clearly identifiable resonance signal and so MRI can be used to compare the blood content of different tissues. This is often greater in cancerous tissue, again because of the high growth rates.

MRI contrast agents are unique in radiology because it is not the chemical of the contrast agent that is detected, but rather the effect that the chemical has on the protons in nearby hydrogen atoms.

Gadolinium (bonded to a carrier molecule) is a contrast agent that, when injected into the bloodstream, shortens the relaxation time T1 of protons near it, making the tissues appear brighter in a T1-weighted image. Because gadolinium normally stays in blood vessels, it has the effect of making these blood vessels and and areas in the body where blood leakage is occurring appear brighter.

Another powerful tool used to enhance image contrast is to vary the time intervals between successive repetitions of a sequence of applied RF signals. This changes the weighting of the relaxation times and can improve the contrast in the image between different tissues.

Explain that large differences would occur in the relaxation time between tissue containing hydrogen-bound water molecules and tissues containing other molecules.

CHeCkpoInt 20.41 Account for the different MRI signal amplitudes observed from

different tissues in the body.2 Explain how different parts of the body can be highlighted in an

MRI image.3 Outline why tumours are so clear on scans.4 Explain why gadolinium is injected during an MRI scan.

Figure 20.4.2 MRI scans of the brain show more contrast and detail than conventional X-ray or CAT scans because MRI is more sensitive to the differences in water content of grey matter and white matter.

Table 20.4.1 Water content of various tissues

tissue WAter content (%)Brain grey matter 70.6

Brain white matter 84.3

Heart 80

Blood 93

Bone 12.2

20.5 the MRI scannerThere are approximately 200 MRI scanners in Australia. All major hospitals and many private imaging facilities in Australia have one, although they cost more than $2 million to purchase and about $1 million per year to operate and staff. Part of this cost is for liquid helium to cool the superconducting electromagnets. These produce the strong magnetic fields—up to 5 T—required for MRI. Compare this with the field of about 50 × 10–6 T at the Earth’s surface.

The patient is placed in the MRI scanner tunnel and the current in the superconducting coils is turned on to produce the strong magnetic field that will align the proton spins. These coils are shown in Figure 20.5.1, but not in Figure 20.5.2. When the patient lies in the MRI scanner, the strong magnetic field is

activity 20.2



361

medIcalphysIcs

Figure 20.5.1 Cutaway view of an MRI scanner

patienttable

patient

scanner

superconductingmagnet coils

gradientcoils

RFtransceiver

Z

Y

X

Figure 20.5.2 Gradient in an MRI scanner

Z coil

Y coil

X coil

patient

RF transceiver

Z

Y

X

CHeCkpoInt 20.5Explain why the cost of running MRI scanners is so high.

Z

Y

X

Figure 20.5.3 Image slices and voxels in an MRI scan

activity 20.3



parallel to a line running in the head-to-toe direction, which is normally referred to as the longitudinal axis or the z direction.

The pulse of radio waves from the RF transceiver coils is transmitted through the patient’s body. This flips the magnetic axes of some of the protons, resulting in a zero magnetic field from the protons in the z direction. It also causes the protons to precess in phase with each other. As some of the protons flip back and return to precession that is out of phase, the protons in each volume element (voxel) emit radio waves with an intensity proportional to the number of hydrogen atoms in the voxel.

Gradient coils produce a small systematic variation in the net magnetic field throughout the patient’s body. The Z coils (Figure 20.5.2) produce variation in the z direction, which defines a slice through the body, represented by the blue region in Figure 20.5.3. The X and Y coils change the net magnetic field in their respective directions so that its strength is different in each voxel. As a result of these variations in the imposed field, the protons precess at slightly different frequencies. These frequencies allow the positions of the protons emitting RF signals to be determined.

Hundreds of pulse cycles are used to obtain proton density data from each voxel in the plane of interest.

The RF signals emitted by the protons provide data that is processed by a computer to produce an image for each slice through the body, and hence to create a 3D image that can be manipulated and viewed from any angle.


362

20.6 Applications of MRI One of the first highly successful applications of MRI was its use in diagnosing multiple sclerosis, an incurable disease that affects the nervous system. MRI clearly shows the scar-like plaque resulting from this disease, marking the destruction of the insulating myelin sheaths around the axons of nerve cells (Figure 20.6.1), something other imaging technologies are unable to do. Other examples of MRI imaging are shown in Figures 20.6.2 and 20.6.3.

MRI has changed scientific thinking about the progress of diseases such as cancer and multiple sclerosis, because these diseases can now be detected earlier and different treatments can then be used. MRI also allows the progress of a disease and its treatment to be monitored more safely, increasing the chances of survival.

Advantages of MRI• MRIisnon-invasiveandhasnoknownsideeffects,althoughafewpatients

experience anxiety in the confined, noisy space of the MRI scanner.• MRIprovidesexcellentsoft-tissueimaging,providingbettercontrastthanCAT

or conventional X-rays and much better resolution than ultrasound.• ImagecontrastcanbealteredbychangingtheT1 and T2 weightings.• TherearefewcasesofadversereactiontoMRIcontrastagents.

Disadvantages of MRI• Patientswithheartpacemakersormetalpartsintheirbodiescannotbeimaged

because of the strong magnetic fields.• Scansmaytakeuptoanhourtocomplete.• MRIscannersareveryexpensiveandtherunningcostsarehigh.• MRIscannersareverysensitiveandmustbescreenedfromoutsideradioand

magnetic interference.• Thestrongmagneticfieldproducedcaninterferewithnearbyelectronic

equipment and is dangerous in the presence of iron tools or furniture.

Most current MRI images show structure within the body. An important area of technical development is the production of functional MRI images that reveal differences in chemical processes happening in the body. Other developments include the use of even stronger magnetic fields, allowing better resolution and contrast, and ultimately imaging using elements other than hydrogen.

The possibility of future ‘high’ temperature superconductors, requiring only liquid nitrogen cooling, could impact significantly on MRI technology by removing the need for expensive, non-renewable liquid helium. Research and development has already produced portable MRI scanners for specialised applications. These machines do not require the use of superconducting magnets, greatly reducing the size and cost of the machine.

Figure 20.6.1 False-colour MRI of the brain of a person with multiple sclerosis shows regions of myelin destruction (black–orange).

Figure 20.6.2 MRI shows blood flow in the arteries of the brain as white, relative to stationary brain tissue.

Figure 20.6.3 MRI shows cancer (colour coded orange) that has spread from the lungs to the brain.

CHeCkpoInt 20.61 Describe what is seen in an MRI scan in a patient that has

multiple sclerosis.2 Create a table to contrast the advantages and disadvantages of

MRI and CT scans.


363

medIcalphysIcs

chapTer 20This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACtIvIty 20.1: MAgnetIC ReSonAnCe IMAgeSLook at the MRI scans and list their characteristics. Use images of healthy and damaged tissue and identify what characteristics allow you to distinguish between the two.

Discussion questions 1 List characteristics of an MRI scan.2 Outline how damaged tissue can be identified on an MRI scan.3 Give reasons why MRIs are used to identify areas of high blood flow.4 Construct a table that lists the parts of an MRI, identifying their

functions.

ACtIvIty 20.2: A CoMpARISon oF SCAnnIng teCHnIqueSResearch each of the diagnostic methods and compare the advantages and disadvantages of each technique. Prepare a summary of your findings.

Discussion questions1 Identify the most appropriate application for each of the methods listed.2 List advantages for each of the diagnostic methods listed.3 Identify limiting factors for each method.

ACtIvIty 20.3: MedICIne And pHySICSResearch how medical applications of physics have affected society and write a report to assess these impacts.

Discussion questions1 Recall some medical applications of physics.2 Identify the areas of society these applications have affected.

Perform an investigation to observe images from magnetic resonance image (MRI) scans, including a comparison of healthy and damaged tissue.

Identify data sources, gather, process and present information using available evidence to explain why MRI scans can be used to:• detect cancerous tissues• identify areas of high blood

flow• distinguish between grey and

white matter in the brain. Gather and process secondary

information to identify the function of the electromagnet, radio frequency oscillator, radio receiver and computer in the MRI equipment.

Identify data sources, gather and process information to compare the advantages and disadvantages of X-rays, CAT scans, PET scans and MRI scans.

Gather, analyse information and use available evidence to assess the impact of medical applications of physics on society.

364

Imaging with radio waves20 chapter summary

review questions

• Particleshaveapropertycalledspin,whichcanbevisualised as a rotation (spinning) of the particles, although this is not reality.

• Protons,neutronsandelectronscanonlyhavetwospinstates: ‘spin up’ and ‘spin down’.

• Ifthereisanoddnumberofnucleonsinanucleus,thenthe nucleus has a net spin.

• Spinproducesamagneticeffectcalledamagnetic moment or magnetic dipole.

• Nuclearmagneticmomentsmayalignparallelorantiparallel to a magnetic field. The parallel configuration has a slightly lower energy and is therefore slightly preferred.

• Injectingenergyequaltotheenergydifferencebetweenthe spin states can cause the proton to flip from the parallel to the antiparallel configuration. This same energy will be released if the proton flips back.

• Amagneticmomentangledtoamagneticfielddirectionexperiences a force and hence a twist (a torque) that causes precession of the direction of the magnetic moment.

• Thefrequencyofprecessionofamagneticmomentiscalled the Larmor frequency given by:

fB

hLarmorp

=⎛

⎝⎜⎜

⎞

⎠⎟⎟

2μ

where B = magnetic field and h = Planck’s constant.

• Magneticresonanceimaging(MRI)involvessendingapulse of radio frequency (RF) radiation into a person’s body while located in a strong magnetic field. The frequency is tuned to match the Larmor frequency of protons within the field.

• Thereturnoftheprotontothelessexcitedstateafterabsorbing the RF energy is called relaxation.

• Thelongitudinalrelaxationtimeconstant (T1) is a measure of the time taken for protons to return to their normal ratio of parallel to antiparallel configurations relative to the field.

• Thetransverserelaxationtimeconstant(T2) is a measure of the time taken for the magnetic component perpendicular to the external field to return to zero.

• MRIisverysensitivetovariationsinthewatercontentof tissues.

• MRIcontrastagentsaffecttheprotonsinnearbyhydrogen atoms.

• Gradientcoilsproducesmallvariationsinthenetmagnetic field throughout the patient’s body.

• FunctionalMRIimagesrevealdifferencesinchemicalprocesses happening in the body.

pHySICALLy SpeAkIngComplete the passage below by filling in the missing words from the list provided.

antiparallel, field direction, energy, flip, higher energy,

intensity, relaxation time constants, magnetic resonance

imaging, quantum mechanical, parallel, protons, model,

radio, randomly, reality, rotation, magnetic field, spin

MRI stands for _____________. This diagnostic tool uses a

property of atoms called _____________ which can be

visualised as a _____________ of the particles. However, spin is

a _____________ property of the particles and the classical

view of a spinning ball is a useful _____________, but is not

_____________. Single _____________ (hydrogen nuclei)

generally all have their spins oriented _____________. When a

_____________ is applied, they all line up _____________ or

_____________ to the _____________. _____________ waves

transmitted through the patient’s body can cause protons to

momentarily _____________ to the _____________ antiparallel

state. As they return to their original orientation, they re-emit

the _____________. The _____________ of the emission tells us

how many protons there are in that area of the body. The

_____________ tell us about the environment of the protons in

that area of the body.

365

medIcalphysIcs

RevIewIng 1 Explain why fluorine-19 has a nuclear magnetic

moment, but nitrogen 14 does not.

2 Identify the property of tissues that permits images like those in Figure 20.7.1 to be produced using an MRI machine.

3 Describe the effect on the energy of protons in an MRI scanner that allows an image to be produced.

4 Outline why cancerous tissues usually have a greater amount of blood flowing to them than normal tissue of the same type.

5 Describe the behaviour of a proton in a strong magnetic field.

6 Describe the process of precession in a mechanical system such as a child’s toy top.

7 Describe the process of precession involving protons in a strong magnetic field.

8 Summarise the conditions required for radio waves to cause protons precessing in a strong magnetic field to undergo a ‘spin flip’.

9 Justify the need for superconducting electromagnets in an MRI scanner.

10 Describe the function of the radio frequency transceiver coils in an MRI scanner.

11 Explain the role of the computer in an MRI machine.

12 Describe the purpose of the gradient coils in an MRI machine.

13 Describe what is meant by the term relaxation time.

14 Compare the advantages and disadvantages of MRI and CAT scans.

15 Briefly assess the impact of MRI scanners on medicine.

16 Assess the impact of superconductivity on the development of MRI.

17 Identify possible future directions of physics research that would contribute to improvements in MRI.

SoLvIng pRobLeMS 18 When a 1 T field is applied to a portion of the body

containing 2 million protons, estimate how many extra protons favour the parallel configuration over the antiparallel configuration.

19 Calculate the magnetic moment of a proton, given that the Larmor frequency for a proton in a 1 T magnetic field is 42.6 MHz.

20 Calculate the Larmor frequency of an electron, given that the magnetic moment for an electron is –9284.764 × 10–27 J T–1.

Figure 20.7.1 MRI images show a normal brain (left) and the brain of a patient with a tumor.

Revie

w Questions

366

5multiple choice(1 mark each) 1 Two differences between the normal hearing range

of sound and ultrasound are:

normAl sound ultrAsoundA Not easily scattered by

human tissue, 20 Hz to 20 kHz

Easily scattered by human tissue, above 20 kHz

B Shorter wavelength, audible

Longer wavelength, inaudible

C 20 Hz to 20 kHz, longer wavelength

Above 20 kHz, shorter wavelength

D Above 20 kHz, audible 20 Hz to 20 kHz, inaudible

2 An advantage of a CAT scan over an X-ray as a diagnostic tool would be:A seeing broken ribs more clearly.B the images contain three-dimensional information

that allows for greater discrimination of structure.C the ability to carry out a biopsy.D lower cost.

3 Figure 20.8.1 shows two images of the same patient’s brain, using two different medical imaging techniques. The two techniques were:A Left: Ultrasound; Right: MRIB Left: Ultrasound; Right: CATC Left: CAT; Right: MRID Left: MRI; Right: CAT

4 Using the data in the table below, determine which answer is closest to the percentage of ultrasound intensity that is reflected from a boundary between kidney and fat.

substAnce density (kg m–3)

ultrAsound velocity (m s–1)

impedAnce (rayl)

Fat 920 1450 1.33 × 106

Kidney 1040 1561 1.62 × 106

A 0.29% C 12%B 1% D 18%

5 In medical imaging, the term relaxation time refers to:A how long it takes for an MRI machine to

recharge.B the time for proton spin to return to equilibrium.C the minimum time between scans.D the maximum time before a radioisotope decays

to untraceable amounts.

The review contains questions that address the key concepts developed in this module and will assist you to prepare for the HSC Physics examination. Please note that the questions on the HSC examination that address the option modules are different in structure and format from those for the core modules. Past exam papers can be found on the Board of Studies NSW website.

Figure 20.8.1

367

medIcalphysIcs

short response 6 The ultrasound image in Figure 20.8.2 shows an

organ in a child that is not usually imaged in adults. Identify the organ and describe the features of the image that make it recognisable as an ultrasound image. (4 marks)

7 Explain how an endoscope can be used to obtain tissue samples from the bladder of a patient. (2 marks)

8 For each of the radioactive isotopes in the table below explain why it would or would not be a good choice for use in a bone scan. (5 marks)

rAdioActive source

rAdiAtion emitted

HAlf-life

C-11 b+, g 20.30 minutes

Tc-99m g 6.02 hours

I-131 b, g 8.04 days

Cs-137 a 30.17 years

U-238 a 4.47 × 109 years

9 The nucleus of has magnetic properties that enable it to be used for MRI.a Explain why is not used. (2 marks)b Identify the primary factor that determines the

amplitude of the signal produced by nuclei after an RF pulse at the resonant frequency has been applied to a part of a person’s body. (2 marks )

extended response 10 Assess the impact of MRI on diagnosis of disease,

in comparison to other medical imaging techniques. (5 marks)

1H1

4He2

1H1

Figure 20.8.2

368

6 Astrophysics

Figure 21.0.1 The 2-degree field ‘top end’ of the Anglo-Australian telescope (AAT) uses optical fibres to direct light from up to 400 astronomical targets to spectrographs.

CONTEXT ‘I render infinite thanks to God, for being so kind as to make me alone the first observer of marvels kept hidden in obscurity for all previous centuries.’

Galileo Galilei in Siderius Nuncius (Sidereal Messenger)

The Italian scientist Galileo Galilei (1564–1642) was not quite right in his claim to be the first to see the universe revealed by the newly invented telescope. That accolade may belong to Englishman Thomas Harriot (1560–1621). However, it was certainly Galileo’s short record of his observations in Siderius Nuncius (Sidereal Messenger), published in March 1610, that announced a revolution in astronomy to the educated world in Europe. Observations with telescopes revealed previously unknown objects, unimagined by thousands of years of naked-eye observers.

Nonetheless, astronomy remained largely a ‘collector’s science’ for another two centuries. Observers used telescopes of increasing size to discover objects and note their positions and details, but knew little of their true nature. It was not until the early years of the 19th century that physicists discovered lines in the spectrum of sunlight, and later realised that they were the fingerprints of the elements making up the Sun and the stars. Although Newton’s law of universal gravitation explained motion in the sky, it was the understanding of spectral lines that really put physics into astronomy. Spectroscopy remains at the heart of modern astrophysics.

In this module we will explore some aspects of both gravity and spectroscopy in astrophysics, leading to our current understanding of the life history of stars.

368

INQUIRY ACTIVITY

LEARN AbOUT A sTAR

A lot of astronomical data gathered by astronomers is now available on-line in various catalogues. One important catalogue of stars was produced by the Hipparcos satellite in the early 1990s. It observed the positions and basic observable properties of more than 100 000 stars with high precision and more than 2.5 million stars to lower precision.

In this exercise, you will use the Hipparcos catalogue to find out about one of the bright stars of the Southern Cross (the constellation Crux) or a neighbour in the constellation of Centaurus.

1 Find a map of Crux and Centaurus, perhaps in a star atlas or a computer program that shows the stars (e.g. Google Sky or WikiSky). You’ll need to orient the map to match Figure 21.0.2 above.

2 Identify the brightest stars of Crux and Centaurus seen in Figure 21.0.2.

3 Select one of these stars and look it up in the Hipparcos online catalogue, which can be accessed on the companion website at Pearson Places <www.pearsonplaces.com.au>.

To do this, enter the name (e.g. Alpha Crucis) into the ‘Target Name’ field and then click on ‘Submit Query’. The database should respond with a single line of data about your star.

4 Can you interpret the data? Use the cursor to select each column heading to understand what each value means.

5 Record the following important values (and the units) from the catalogue for each star:• Trigonometricparallaxandthe

standard error in this value• Vmagnitude• ColourindexB–V

At the end of this Module, you should have a better understanding of many of these data values.

369

Figure 21.0.2 The stars of the Southern Cross (right) and the ‘pointers’ α (alpha) and β (beta) Centauri (left)

21 Eyes on the sky

370

Who needs a telescope? Looking at the sky with the naked eye involves an optical system with a maximum diameter of about 7 mm—the size of the pupil of your dark-adapted eye. The brightness of the image and the ability to resolve fine detail are both set by that size. Using a telescope increases the diameter of the light-collecting ‘aperture’ to perhaps 60 mm in a small backyard telescope, or 10 m in the largest optical telescopes currently in use. The aperture is the most important property of any telescope operating at any wavelength because it governs the ability to ‘see’ faint objects and resolve fine detail.

21.1 The first telescopesThe magnifying ability of lenses has been known since ancient times, but it seems that the combination of two lenses to form a practical optical telescope was first achieved by spectacle makers in the Netherlands in 1608. By mid-1609, Galileo Galilei (1564–1642), at the University of Padua, had built his first telescopes with magnification of 3×. He quickly progressed to instruments with increasing magnification and, for just a few months, his superior instruments gave him an unchallenged ability to observe the sky (Figure 21.1.1). This turning point in the history of science prompted the United Nations to declare 2009, the 400th anniversary of Galileo’s initial observations, as the International Year of Astronomy.

telescope, magnification, refracting telescope, field of view, focal length,

reflecting telescope, radio telescope, sensitivity, angular resolution,

diffraction, Airy disc, active optics, seeing, scintillation, adaptive optics,

interferometer, interferometry

Figure 21.1.1 Two of Galileo’s original telescopes, mounted for display. The longer telescope had an aperture of 26 mm, a magnification of 14× and was about 1 m long.

371

Astrophysics

Galileo’s influence was immense because, in March 1610, he quickly published a short record of his initial observations in Sidereus Nuncius (Sidereal Messenger). Among his major discoveries was the presence of mountains and valleys creating shadows on the Moon’s surface, which changed as the lighting of the surface changed. This rugged nature of the surface of the Moon contradicted Aristotle’s concept of heavenly perfection (Figure 21.1.2).

WhAt is mAgnificAtion?

The magnification of an optical telescope describes how much

bigger an object appears in the telescope compared to the naked-eye view. Technically, it is the ratio of the angular size of the object with and without the telescope. A pair of 7 × 50 binoculars, for example, has an angular magnification of 7 times (7×) and 50 mm diameter main lenses. The view through the binoculars is as if the object was only 1/7 th as far away.

Figure 21.1.2 One of Galileo’s drawings of the Moon compared with a photograph

He also saw four ‘stars’ moving back and forth relative to Jupiter in a pattern he recognised as orbital motion around Jupiter. The observation of these ‘Galilean’ satellites almost led to the discovery of the planet Neptune, more than two centuries before its actually discovery.

Galileo’s instruments were refracting telescopes that used a plano-convex lens (Figure 21.1.3) as the ‘objective lens’. The properties of the glass and the shape of this lens bend the light and bring it to a focus (Figure 21.1.4). The smaller ‘eyepiece lens’ then relays the light to your eye, which creates a new image on the retina at the back of the eye—and you ‘see’ the object!

Galileo’s telescopes used a plano-concave lens (Figure 21.1.3) as the eyepiece. This results in an image that is upright; but the telescope then has a very narrow field of view—the area of sky you can ‘see’ at any moment—a little like looking down a narrow pipe at the object.

In 1611, Johannes Kepler (1571–1630) pointed out that a plano-convex eyepiece lens would also work (Figure 21.1.4), although the image was inverted. Christoph Scheiner (1573–1650) popularised this form of ‘Keplerian’ refractor by pointing out that the field of view was larger than in the ‘Galilean’ telescope (as illustrated in Figure 21.1.5).

With the field-of-view problem reduced in the Keplerian (or ‘astronomical’) refractor, a race began for higher magnification. This was most easily achieved by increasing the focal length of the objective lens—the distance between the lens and the image it forms (see section 21.3 for more on calculating magnification). By the 1670s, long telescopes achieved magnifications of more than 100×, but at the cost of enormous size and resulting difficulty in mounting and operation.

Discuss Galileo’s use of the telescope to identify features of the Moon.

Figure 21.1.3 Early telescopes used simple lenses that were flat on one side and either concave or convex on the other. Modern single lenses are often curved on both sides.

plano-convex plano-concave

Eyes on the sky21

372

Another advantage of a long focal length objective lens was to minimise distortions inherent in the lens design. One of these was chromatic (colour) aberration caused by the fact that different colours of light come to a focus at slightly different distances from the lens. This causes a coloured halo around a white star image. It was Isaac Newton (1642–1727) who realised that this was caused by the properties of the glass itself. Combining different types of glass offered the possibility of partially cancelling the colour effect of one glass with the effect of the other. It was not until 1733 that such an ‘achromatic’ objective lens was first made. Today, any good-quality small telescope or set of binoculars has an achromatic objective lens and an eyepiece also composed of several lenses.

Despite these developments, and the construction in the late 19th century of some large refractors with apertures up to 1 m, the dominance of the refracting telescope in astronomy is long past. Again it was Newton who made the breakthrough by constructing a reflecting telescope in 1668, using a mirror rather than an objective lens to collect and focus the light (Figure 21.1.6). It offered better images, without a coloured halo, in a telescope that was much smaller than any comparable refractor.

Newton’s telescope used a polished metal primary mirror to focus the light, and a smaller, flat secondary mirror to divert the light out the side of the telescope tube to the eyepiece. In 1672, a variation of the Newtonian telescope was developed by Laurent Cassegrain (~1629–1693), who used a small hyperbolic secondary mirror to reflect light back through a hole in the primary mirror (Figure 21.1.7). By the mid-1700s reflecting telescopes with up to 15 cm

objective lens

objective lens

focus

to eye

to eye

eyepiece

eyepiece

focal length

focal length

starlight

starlight

a Keplerian telescope

b Galilean telescope

Newtonian telescope

primefocus

to eye

eyepiece

focal length

starlight

Cassegrain telescope

primefocus to

eye

eyepiece

focal length

starlight

flat secondarymirror

hyperbolicsecondary mirror

parabolicprimarymirror

parabolicprimarymirror

Figure 21.1.4 (a) Keplerian and (b) Galilean telescopes using an objective lens of the same focal length for easy comparison

Figure 21.1.6 A replica of Newton’s original reflecting telescope

Figure 21.1.7 Newtonian and Cassegrain telescopes using a primary mirror of the same focal length for easy comparison. In large telescopes, if no secondary mirror is in place, the prime focus position can be used.

Figure 21.1.5 These images, shown through (a) Keplerian and (b) Galilean telescopes at the same magnification, have very different fields-of-view.

a

b

373

Astrophysics

ChECkpOINT 21.11 Identify what sets the light-gathering ability and resolving power of an optical system.2 Outline key astronomical observations made by Galileo.3 Describe chromatic aberration and its cause.

apertures were available, and in 1789 William Herschel (1738–1822) built a reflector with a 124 cm aperture! This was the first ‘giant’ reflector—the forerunner of the 8 and 10 m aperture reflecting telescopes of today.

21.2 Looking upFor more than 200 years from the time of Galileo, all telescopes were optical telescopes designed to focus visible light onto the retina of the human eye. From about 1840, the use of photography revolutionised astronomy by allowing images of the sky to be permanently recorded. For the next century telescopes grew in size but remained fundamentally unchanged.

In 1933, Karl Jansky (1905–1950), a physicist working for Bell Telephone Laboratories in New Jersey, realised that part of the background ‘hiss’ heard in radio communication was coming from the sky. Janksy’s discovery was followed up in 1937 by Grote Reber (1911–2002), who built a reflecting telescope in his backyard in Chicago (Figure 21.2.1). This was a radio telescope, with a mirror made from sheet metal and a diameter of 9.5 m—almost four times the size of the largest optical reflecting telescope of the era. The basic optical principles of Reber’s radio telescope were exactly the same as those of an optical reflecting telescope, with the exception that the radio ‘light’ was detected by a radio receiver rather than by eye or photography.

Radio astronomy really began to develop after World War II, spurred by technical developments during the war years. This was possible because a wide range of radio wavelengths, as well as visible light, penetrate the Earth’s atmosphere without significant absorption and can therefore be observed by telescopes on the ground. Electromagnetic (EM) radiation of most other wavelengths is absorbed by the atmosphere before it reaches the ground (Figure 21.2.2).

At wavelengths longer than about 10 m, the Earth’s ionosphere (see in2 Physics @ Preliminary section 8.4) blocks radio energy from space. The radio ‘window’ opens at metre wavelengths and remains open down to millimetre wavelengths for telescopes on the ground. However, observing at these shortest radio wavelengths requires a high, dry location such as the site of the ALMA mm–sub-mm radio telescope, under construction in the Atacama Desert in Chile.

Discuss why some wavebands can be more easily detected from space.

Discuss the problems associated with ground-based astronomy in terms of resolution and absorption of radiation and atmospheric distortion.

Figure 21.2.1 Grote Reber’s radio telescope

Eyes on the sky21

374

21.3 The telescopic view Most telescopes operate on the same basic principles, no matter what

their operating wavelength. They use some type of mirror to collect the ‘light’ (EM radiation) and focus it onto an appropriate detector—often using the same prime focus or Cassegrain configurations illustrated in Figure 21.1.7. The combination of telescope and detector will set the field of view.

Magnification Perhaps surprisingly, the concept of magnification (sometimes called

‘power’) is not too important in most astronomical telescopes. It is easily calculated if the focal length of the objective lens or mirror, fo, and the focal length of the eyepiece, fe, are known. The magnification M is then given by:

M =fofe

Far infra-red (FIR) light at wavelengths of about 100 µm (0.1 mm) is absorbed by water vapour in the atmosphere. However, this blocking becomes patchy as wavelength decreases, with windows of less absorption at certain infra-red (IR) wavelengths between 1 and 10 µm. As a result, some infra-red astronomy can be done from mountaintop observatories or even aircraft. However, observing across the full IR range requires a telescope above the atmosphere—a role most recently and spectacularly filled by the 0.85 m aperture Spitzer Space Telescope.

At wavelengths shorter than IR (i.e. below 1 micron), the atmospheric window opens again for the narrow optical band between about 350 and 750 nm. It then quickly closes in the ultraviolet (UV) and the shorter wavelength, higher energy X-ray and γ-ray wavebands. All of these are now observed from space.

100%

50%

0%0.1nm

1nm

10nm

100nm

1μm

10μm

100μm

1mm

1cm

10cm

1 m 10 m 100 m 1 km

Wavelength

Abs

orpt

ion

by a

tmos

pher

e

FermiGalex Spitzer

Gemini

Parkes

Hubble

Figure 21.2.2 The Earth’s atmosphere stops light from many parts of the electromagnetic spectrum reaching the ground.

AntArctic Astronomy

It may surprise you to learn that the highest and driest sites

on Earth are in Antarctica. The altitude in the centre of the Antarctic continent and the extreme cold of the atmosphere make it a tempting site for an infra-red observatory. Australian astronomers have studied the conditions for several years and a design study for a 2.4 m aperture optical/IR telescope called PILOT has been conducted. However, a large Antarctic telescope is still in the future.

ChECkpOINT 21.21 Discuss how and why radio astronomy started.2 Construct a table listing the various wavebands of the electromagnetic spectrum and describe their ability

to penetrate the Earth’s atmosphere.

375

astrophysics

Worked example QuestionA small refracting telescope has an objective lens diameter D of 70 mm and focal length fo of 700 mm.

If this telescope has a typical ‘low power’ eyepiece focal length fe of 25 mm, what would be its magnification?

solution (i.e. 28×)m =

fo =700

= 28fe 25

Sensitivity The sensitivity of a telescope system describes its ability to ‘see’ faint objects.

Sensitivity is sometimes called ‘light-gathering power’. It depends on how much light the telescope collects and how much of that light is delivered to the detector.

The primary factor controlling light-gathering power is the diameter D of the telescope’s objective lens or mirror. Most telescopes have circular mirrors and so the collecting area is proportional to D2. A 10 m diameter telescope therefore collects 100 times the light of a 1 m diameter telescope. This is the primary reason for building bigger telescopes, although it is often not quite that simple, as illustrated by the example of the Anglo-Australian Telescope (AAT).

Worked example QuestionWhat is the collecting area of the 3.9 m AAT, and how does it compare with that of a 70 mm diameter refracting telescope?

solutionThe refracting telescope has D = 70 mm, so its collecting area can be calculated using:

Collecting area = p D 2

2

The AAT has D = 3.9 m; however, it also has an obstruction caused by the secondary mirror and its housing, as seen in Figure 21.3.1. In the f/8 Cassegrain configuration, this obstruction is about 20% of the collecting area. So the ratio of the collecting areas of the two telescopes is:

Collecting area AAT=

Collecting area 70 mm

≈ 2500

The AAT has 2500 times the light-gathering power of the 70 mm telescope.

Once the light has entered the telescope, it must be delivered to the detector and this may involve a series of lenses and mirrors. Each of these may reflect, absorb or scatter a little of the light and reduce the amount that reaches the detector. In the AAT there are just two mirrors between the sky and a detector placed at the Cassegrain focus, yet 10% of the light is lost.

Define the terms ‘resolution’ and ‘sensitivity’ of telescopes.

p 3900 2

× 0.82

p 70

2

2

Figure 21.3.1 The 3.9 m Anglo-Australian Telescope, showing the black 2-degree field (2dF) secondary mirror structure at the top of the telescope

Eyes on the sky21

376

Apart from the telescope itself, the overall sensitivity also depends on several other factors:• Atmospherictransmission—FortelescopesontheEarth’ssurface,the

proportion of the light that penetrates the atmosphere depends on wavelength, as illustrated in Figure 21.2.2.

• Detectorefficiency—Oftenmuchlessthan100%ofthelightstrikingthedetector is actually recorded.

• Observingtime—Unlikethehumaneye,mostdetectorsrecordmorelight if they are left exposed for longer times.

• Background—Theskyisnottrulydarkbecauseofnaturalsourcesandman-made pollution at optical and radio wavelengths.

Angular resolutionAngular resolution is often simply called resolution. Adding ‘angular’ reinforces the idea that we are concerned with an ability to discern (resolve) detail separated by very small angles in the sky. In a practical sense, angular resolution describes whether we can see two closely spaced objects—or do they look like one?

The limit to resolution is set by the wave-like properties of light. Whenever light encounters an obstacle, such as the aperture of a telescope, it doesn’t cast a sharp shadow. Instead diffraction caused by the edges of the obstacle alters the path of the light near the edge. For a telescope with a circular objective lens or mirror this creates a pattern of light known as an Airy disc (Figure 21.3.2) at the focus—not an image of a star. For a telescope of diameter D using light of wavelength λ, the angular size R of this pattern is given by the equation:

R (in arc seconds) = 1.22 λ × 206 265D

(Recall the arc second as a unit of angle from in2 Physics @ Preliminary section 13.4.)

The angular size of this pattern depends only on the wavelength of the light and the diameter of the telescope. For most telescopes, R is much bigger than the apparent size of even the largest star as seen from Earth.

If two stars are close together, their Airy discs may overlap so that it is impossible to see that there are two stars. They are ‘unresolved’. The stars can theoretically be ‘resolved’ and seen as two stars (Figure 21.3.3) if they are separated by at least


Resolution also affects images of larger objects such as the Moon or planets, since each point of the object acts like a distant star and is blurred into an Airy disc (Figure 21.3.3).

RD R2D

Figure 21.3.2 (a) A view of an Airy disc, and (b) the same view through an aperture with twice the diameter

Figure 21.3.3 A magnified view of an Airy pattern from an unresolved pair of stars (left), and a resolved pair (right)

a b

Figure 21.3.4 Simulated views of Saturn, showing the effect of resolution using (a) a 10 cm diameter telescope and (b) a 30 cm diameter telescope

a b

377

Astrophysics

Notice that the angular resolution of a telescope is expected to be ‘better’ (R is smaller and the image is sharper) for a big telescope. It is also better if the wavelength is small (e.g. much better for visible light than for radio waves). This is illustrated by the theoretical resolution values for 10 m diameter telescopes listed in Table 21.3.1. For comparison, the size of the Sun or Moon in the sky is around 1800 arc seconds, while a $1 Australian coin seen from a distance of 100 km would appear just 0.05 arc seconds across.

Table 21.3.1 Theoretical resolution of 10 m diameter telescopes at different wavelengths

Band Typical wavelengTh ResoluTion R foR d = 10 m(arc seconds)

Ultraviolet(UV) 100 nm 0.0025

Optical 500 nm 0.013

Near infra-red (IR) 2000 nm (2 µm) 0.050

Radio (millimetre) 1 mm 25

Radio (centimetre) 21 cm 5300

try this!The man in The moonWhen you look at the Moon, do you see the face of the ‘man in the Moon’?

We’re all limited by the angular resolution of our eyes. If the pupil of your eye is open to, say, 5 mm, the theoretical resolution is about 1/100 the size of the Moon in the sky, and not even quite that in practice. The result is that you can only see the broad details of the dark ‘mare’ and bright lunar highlands. Your imagination does the rest—and often in not the same way as someone else!

Figure 21.3.5 The face of the man in the Moon—in the southern hemisphere

Activity 21.1

prActicAL EXpEriEncEs


Identify data sources, plan, choose equipment or resources for, and perform an investigation to demonstrate why it is desirable for telescopes to have a large diameter objective lens or mirror in terms of both sensitivity and resolution.

ChECkpOINT 21.31 Calculate the magnification of a telescope that has an objective

lens with a focal length of 500 mm and an eyepiece with a focal length of 12.5 mm.

2 Define sensitivity of a telescope system.3 Define angular resolution.4 Describe an Airy disc.

21.4 sharpening the imageIn practice, the theoretical resolution of a telescope may not actually be achieved if the imaging performance of the telescope is not good enough. How good is good enough?

As a ‘rule of thumb’, the surface of a lens or mirror must be within one-eighth of a wavelength of its correct shape if it is going to produce sharp images. At optical wavelengths of about 500 nm, this means the surface must be smooth and the correct shape to within about 60 nm—about one-thousandth the thickness of human hair! This extraordinary precision is routinely achieved with optical polishing techniques. At radio wavelengths of 21 cm, one-eighth of a wavelength is 2.6 cm and this is easily achieved with metal plates and even wire mesh attached to a metal frame.

Active opticsThese tight tolerances may be impossible to achieve if the mirrors are so large that they bend under their own weight. Alternatively, the telescope itself may bend as it points in different directions. In the past, both effects could largely be overcome by building massive telescopes. The dish of the Parkes radio telescope (Figure 21.4.1) is 64 m across and weighs 300 tonnes, but also has a clever design to allow for flexure of the structure.

Eyes on the sky21

378

This approach has been abandoned in recent optical telescopes in favour of much lighter mirrors and structures that are controlled by an active optics system that keeps the telescope at optimum performance. For example, the mirrors of the 8.1 m diameter Gemini telescopes (Figure 21.4.2) are only 20 cm thick—much too thin to hold their shape. Instead, a system of 180 computer-controlled actuators push and pull on the back of the mirror to keep it in shape. The distortions are measured and corrections made about once a minute.

Outline methods by which the resolution and/or sensitivity of ground-based systems can be improved, including:• adaptiveoptics• activeoptics.

Figure 21.4.1 The 64 m Parkes radio telescope Figure 21.4.2 One of the 8 m Gemini telescopes, in which Australia owns a small share

SeeingFor all wavelength ranges, coping with the slowly changing distortions in the telescope itself is an issue. In contrast, the much faster effects of ‘seeing’ are problems restricted to ground-based optical and infra-red telescopes.

Seeing refers to the effects on images of tiny, rapidly changing temperature variations in the Earth’s atmosphere. These effects distort the path of the visible and, to a smaller extent, the infra-red light passing through the atmosphere so that even a perfect optical/IR telescope will not produce a sharp image. Images in a large telescope are distorted into a constantly moving and changing ‘speckle pattern’ that averages over time into a blurred ‘seeing disc’ (Figure 21.4.3). The same effects produce the familiar scintillation, or twinkling, of stars. Scintillation describes rapid changes in the brightness of a star, while seeing refers to blurring of a star image.

The turbulence that produces seeing is usually concentrated near the ground and in discrete layers in the atmosphere. It depends on local weather conditions and geography. The best locations for good seeing are typically tall mountains that project above the relatively smooth stream of air flowing across open oceans. Most of the world’s largest optical or IR telescopes are therefore sited on islands,

Discuss the problems associated with ground-based astronomy in terms of resolution and absorption of radiation and atmospheric distortion.

Activity 21.2



379

Astrophysics

Figure 21.4.3 Simulated images of a binary star seen through a large telescope in ‘good’ seeing—(a) without seeing effects, (b) the ‘speckle’ pattern seen in a 10 ms exposure time and (c) the ‘seeing disc’ seen in longer exposure times. The arrow indicates 1 arc second.

or mountains on the western edges of continents. Siding Spring Observatory in New South Wales, site of the Anglo-Australian Telescope, is relatively poorly sited in comparison.

The long-wavelength radio waves are unaffected by atmospheric turbulence and so a mountaintop site is unnecessary for most radio telescopes. Protection from artificial sources of radio waves is essential, and so many radio telescopes are in very remote sites, away from population centres.

Adaptive opticsObviously, the ultimate way to beat the seeing effects is to place a telescope in space, above the atmospheric turbulence. The 2.4 m diameter Hubble Space Telescope (HST) has spectacularly demonstrated the power of observing from space since its launch in 1990. However, it is incredibly expensive and the largest optical/IR telescopes are on mountain tops, not in space.

For at least 40 years, astronomers and engineers have been working towards ‘beating’ the seeing effects for ground-based telescopes by using adaptive optics. Adaptive optics is essentially active optics, working much faster—correcting hundreds of times a second to overcome the rapidly changing seeing effects.

An adaptive optics system (Figure 21.4.4) works by using a wavefront sensor to measure the image distortion of a target star. A tilting mirror is used to stop the image moving and a deformable mirror is used to sharpen the image. The result is a much sharper image, although still not perfect (Figure 21.4.6).

The target star may not be the real target of scientific interest, but it must be very close and bright enough for the adaptive optics system to work. A single system can measure and correct only a tiny patch of sky, since the blurring is different just a few arc seconds away. If there is not a bright enough adaptive optics target star close to the scientific target, then astronomers can provide one! A laser guide star is created by firing a high-power laser up from the telescope, causing the atmosphere to glow at about 90 km altitude. To the adaptive optics system it looks like a star that can be moved to where it is needed.

1” 1” 1”

natural guide star science target

layers of turbulencein the atmosphere

laser projection system

telescopesodium laser

distorted wavefrontsproducing a movingblurred image

deformablemirror

control signals fordeformable mirror

laser guide star createdin sodium layer at 90 km

tiltingmirror

real-timecontrol computer

wavefrontsensor

control signals fortilting mirror

moving, blurred image of guide star

smoothed wavefrontsproducing a sharp image

observinginstrument

sharp image of science target

Figure 21.4.4 Layout of an adaptive optics system, including the science target, a natural guide star and a laser guide star

a b c

Eyes on the sky21

380

Figure 21.4.6 The blurred image (left) of the star IW Tauri reveals a pair of stars separated by 0.3 arc second (right) when imaged using adaptive optics on the 5 m Hale Telescope.

ChECkpOINT 21.41 Identify the factors that limit the resolution of an optical telescope.2 Explain the physical problems that make active correction necessary.3 Describe what is meant by the term seeing.4 Outline the need for adaptive optics.5 Distinguish the differences between active and adaptive optics.

21.5 InterferometryThe goal of adaptive optics is to achieve the resolution limit set by the diameter of the telescope and the wavelength of the light. Given the long wavelengths of radio waves, it is impractical to build a single large radio telescope to achieve the resolution possible with even a very small optical telescope. The solution is to use two or more radio telescopes and link them together to form an interferometer. Provided the effective path to the detector via each telescope is the same to within a few wavelengths, interference will be seen. The interference pattern contains information about the image of the astronomical object.

The ability of an interferometer to resolve fine details is governed by the equation:


where D is now the distance between any two telescopes, not their individual diameters. The resolution of the interferometer is about the same as having a large telescope of diameter D—but only in the direction along the line between the telescopes. At right angles to that direction the resolution is the same as using only one telescope.

If the interferometer is formed from an array of several telescopes, then each pair provides good resolution in one direction. With enough telescopes, the array

sEEing is not bELiEving!

The water in a swimming pool provides a very clear illustration

of how light is affected by the medium through which it passes. In Figure 21.4.5 the surface of the water bends the light paths, concentrating the light in places and generally distorting the view of the bottom of the pool. A similar effect makes stars twinkle.

Another example is the ‘mirage’ seen above water or a road on a hot day, due to the effects of temperature variations in the air.

Figure 21.4.5 A distorted image of the bottom of a swimming pool, caused by ripples on the surface of the water

All large optical/IR telescopes now have some sort of adaptive optics system. The next generation of even larger telescopes will rely on multiple adaptive optics systems and laser guide stars.

Outline methods by which the resolution and/or sensitivity of ground-based systems can be improved, including:• adaptiveoptics• interferometry• activeoptics.

381

Astrophysics

can produce good resolution in many different directions—enough to build up an image of the target. Also, as the Earth rotates, these directions move relative to the target and the array can ‘synthesise’ an even better image.

An interferometer produces the resolution of a single large telescope, but with much less sensitivity, because only a small proportion of the light is collected by the array of telescopes (Figure 21.5.1).

Interferometry is fundamental to the operation of almost all large radio telescopes. The main interferometric array of radio telescopes in Australia is the Australia Telescope Compact Array (ATCA), located near Narrabri, NSW (Figure 21.5.2). It has six 22 m dishes spread over 6 km in a pattern that can be varied. However, it is sometimes used with other dishes spread around Australia or the world to form a much larger array.

Interferometry is also now being applied in optical/IR astronomy. The Sydney University Stellar Interferometer (Figure 21.5.3), also located at Narrabri, is one of the pioneering instruments in this field. The much shorter optical wavelengths make the tolerances much tighter on the optical components and their positions, and seeing effects are once again a major problem.

WhAt is intErfEromEtry?

Generally when you combine two beams of light from a source you

just get more light. If you split the light and then recombine those two beams you may get something different. Provided the two beams travel the same distance, they ‘interfere’ with each other in a way that depends on the path of the beams and the characteristics of the light source. An optical device used to do this is called an interferometer (see section 3.2).

individual telescopes

single telescopemirror of thesame resolution

Figure 21.5.1 The resolution of a large telescope can be achieved using an array of small telescopes. Each white line joins an interferometric pair of telescopes.

Figure 21.5.2 Some of the dishes of the Australia Telescope in a compact configuration Figure 21.5.3 The Sydney University Stellar Interferometer has an array of fixed stations along a 640 m baseline. Interference of the light occurs in the central building (upper centre of the image).ChECkpOINT 21.5

1 Outline how greater resolution is achieved in interferometry than with single telescopes.

2 Calculate the maximum theoretical resolution of the Australia Telescope Compact Array that has six 22 m dishes spread over 6 km.

3 Explain why the resolution of an optical interferometer differs from that of a radio interferometer.

Eyes on the sky21

382

21.6 Future telescopes The forefront ground-based optical/IR telescopes in the world today are 8 m and 10 m in diameter. Planning is now underway for the next generation of Extremely Large Telescopes (ELTs) with apertures up to 42 m! Two or three of these may be operational after 2016. Smaller telescopes still have a role of course, but even 4 m telescopes are only truly competitive if, like the AAT, they have excellent instrumentation.

Also by about 2016, radio astronomers hope to be commissioning their own new instrument—the Square Kilometre Array (SKA). This will actually be an interferometric array of radio telescopes spaced over 3000 km, with a collecting area of one square kilometre.

By then, the replacement for the 2.4 m diameter Hubble Space Telescope will be operational. The James Webb Space Telescope (JWST) will be a 6.5 m diameter IR telescope. Other smaller space telescopes will continue to observe the sky at other wavelengths inaccessible from the ground. Observations at many wavelengths are essential to understanding most astronomical objects (Figure 21.6.1).

Figure 21.6.1 (a) A composite image of the nearby active galaxy Centaurus-A, with (b) an X-ray image and (c) a radio image revealing much more than is seen in (d) the optical image they overlay.

ChECkpOINT 21.6Contrast the maximum theoretical resolution of a 42 m diameter ELT, the JWST and the SKA. (You will need to assume some appropriate wavelength values.)

a b

c

d


383

Astrophysics

ACTIVITY 21.2: A bETTER TELEsCOpEMeasure the amount of light collected by lenses of differing diameter. Also determine the size of the lens that can best separate (resolve) two close objects.

Discussion questions1 Determine the relationship between diameter of a lens and its light-

gathering ability.2 Explain how the diameter of the lens is related to resolution and explain

how this relationship was determined.

Identify data sources, plan, choose equipment or resources for, and perform an investigation to demonstrate why it is desirable for telescopes to have a large diameter objective lens or mirror in terms of both sensitivity and resolution.

chAptEr 21This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For more detailed instructions and advice, use the in2 Physics @ HSC Activity Manual.

ACTIVITY 21.1: MOUNTAINs ON ThE MOONUse an image of the Moon to make the same type of measurements that Galileo used to estimate the size of the mountains on the Moon.

Discussion questions1 State what assumptions are made in estimating the size of the mountains.2 Explain how advances in technology in astrophysics have allowed for more

accurate methods to determine the height of lunar mountains.

Figure 21.7.1 A portion of the Moon’s surface around the crater Plato

Eyes on the sky

384

21

phYsICALLY spEAkINgCopy and complete the following table with the knowledge that you have gained after studying this chapter.

concepT definiTion of concepT diagRam To illusTRaTe The concepT

Telescope

Angular resolution

Sensitivity

Active optics

Adaptive optics

Interferometry

chapter summary

review questions

• Themagnificationofanopticaltelescopedescribes how much bigger an object appears in the telescope compared with the naked-eye view.

• Refractingtelescopesuseanobjectivelenstocollect and focus light.

• Reflectingtelescopesuseamirrortocollectand focus light.

• ‘Fieldofview’describestheareaofskyanopticalinstrument can ‘see’ at any moment.

• Thefocallengthofalensormirroristhedistancebetween the lens or mirror and the image it forms of a distant object.

• Thebasicopticalprinciplesofmosttelescopesatallwavelengths are the same as for an optical reflecting telescope.

• Visiblelightandawiderangeofradiowavelengthspenetrate the atmosphere without significant absorption and can therefore be observed by telescopes on the ground. Most other wavelengths are absorbed by the atmosphere.

• Sensitivity(or‘light-gatheringpower’)ofatelescopesystem describes its ability to ‘see’ faint objects.

• Diffractioninatelescopewithacircularobjectivelensor mirror creates a pattern of light known as an Airy disc at the focus—not an image of a star.

• Angularresolutiondescribeswhetherwecandiscerntwo closely spaced objects as separate (resolved) or not (unresolved).

• Thesurfaceofalensormirrormustbewithinaboutone-eighth of a wavelength of its correct shape if it is going to produce sharp images.

• Activeopticssystemscontroltelescopemirrorsandstructures to keep the telescope at optimum performance.

• ‘Seeing’describestheblurringofimagescausedbytiny,rapidly changing temperature variations in the Earth’s atmosphere.

• Scintillation,ortwinkling,describesrapidchangesinthe brightness of a star caused by tiny, rapidly changing temperature variations in the Earth’s atmosphere.

• Adaptiveopticsmeasurestheimagedistortionofatarget star by using a wavefront sensor, and then corrects it hundreds of times a second, to overcome the rapidly changing seeing effects.

• Twoormoretelescopeslinkedappropriatelyformaninterferometer, provided the effective path to the detector via each telescope is the same to within a few wavelengths. The interference pattern contains information about the image of the astronomical object.

385

Astrophysics

REVIEWINg 1 Recall how the introduction of technology has influenced the discoveries

made in astronomy, with particular reference to Galileo and the telescope.

2 Describe some of the observations Galileo made with his telescope.

3 Name some telescopes that are used to view the universe using wavebands other than visible light, identifying the wavebands they use as well as where they are located.

4 Define the following terms and describe how they relate to a telescope:• collecting area• sensitivity• magnification

5 A small value for angular resolution is considered good. Explain what it means and how it can be achieved in a telescope.

6 Discuss the implications of ‘seeing’ in optical astronomy.

7 ‘Twinkle twinkle little star’ is a line from a popular nursery rhyme. Explain what physical phenomenon is being described in this phrase.

8 Outline how active optics works.

9 Compare and contrast active and adaptive optics.

10 Adaptive optics systems are complex because what they are attempting is difficult to achieve. Outline some of the factors that make adaptive optics difficult.

11 Justify the use of interferometry to increase resolution.

12 Assess the importance of telescopes observing at different wavelengths in gathering information about the universe.

sOLVINg pRObLEMs 13 Compare the sensitivity of a telescope with a diameter of 10 cm with one

of 20 cm.

14 The nearby spiral galaxy M31 in the constellation of Andromeda is approximately 2.5 million light-years from the Sun. What is the size (in light-years) of the smallest features in M31 that can theoretically be resolved by a 10 m telescope at the visible light wavelength of 550 nm?

15 What is the size (in light-years) of the smallest features in M31 that can theoretically be resolved by the 6 km long baseline of the Australia Telescope Compact Array operating at the radio wavelength of 21 cm?

Revie

w Questions

Eyes on the sky

386

21phYsICs FOCUs

bEAUTIFUL dEATh-sTAR COULd ThREATEN EARThThe Earth may be in the firing line when one of the sky’s most beautiful objects explodes, according to University of Sydney astronomer Peter Tuthill.

Dr Tuthill discovered the elegant rotating pinwheel system, named WR104, in the constellation Sagittarius. It includes a highly unstable star known as a Wolf-Rayet, widely regarded by astronomers as a ticking bomb—the last stop in a star’s life before a cataclysmic supernova explosion.

‘When it finally explodes as a supernova, it could emit an intense beam of gamma rays coming our way’, says Dr Tuthill.

New images of WR104 taken with the Keck Telescope in Hawaii by Dr Tuthill in 2008 show a glowing plume of hot dust and gas flung out into a whirling spiral as the two stars at the centre of the system orbit one another every 8 months.

Figure 21.7.2 Spiral pattern around the massive star WR104, as seen using masked aperture interferometry


387

Astrophysics

ButsomethingoddabouttheimagescaughttheattentionofDrTuthillandhisteam:‘ViewedfromEarth, the rotating tail appears to be laid out on the sky in an almost perfect spiral. It could only appear like that if we are looking nearly exactly down on the axis of the binary system.’

Dr Tuthill and his team worry this box-seat view might put us in the firing line when the system finally explodes. ‘Sometimes, supernovae like the one that will one day destroy WR104, focus their energy into a narrow beam of very destructive gamma-ray radiation along the axis of the system. If such a ‘gamma-ray burst’ happens, we really do not want Earth to be in the way,’ warns Dr Tuthill.

At only 8000 light-years distance, WR104 is just down the road in galactic terms, only one-third of the way to the centre of our Milky Way Galaxy. ‘Earlier research has suggested that a gamma-ray burst— if we are unfortunate enough to be caught in the beam—could be harmful to life on Earth out to these distances. Scientists have speculated that, eons ago, a gamma-ray burst from a distant star could explain mass extinctions seen in the fossil record,’ he said.

‘I used to appreciate this spiral just for its beautiful form, but now I can’t help a twinge of feeling that it is uncannily like looking down a rifle barrel.’

ButDrTuthillisnotpanickingjustyet.‘Therearestill plenty of uncertainties: the beam could pass harmlessly to the side if we are not exactly on the axis, and nobody is even sure if stars like WR104 are capable of producing a fully fledged gamma-ray burst in the first place.’

‘We probably have hundreds of thousands of years before it blows, so we have plenty of time to come up with some answers.’

1 What do the colours in the image (Figure 21.7.2)represent?

2 Do you believe that every detail you see in this image is real? Why or why not?

3 This image is ~0.5 arc seconds across. How does the resolution in this image compare with one you might expect from one of the Keck telescopes?

EXTENsION4 Peter Tuthill used a technique called ‘masked

aperture interferometry’ to obtain this image. It uses a single telescope instead of two or more. How does interferometry work on a single telescope?

5 How does the resolution of this technique compare with that possible using an interferometer such as the Sydney University Stellar Interferometer or the Australia Telescope Compact Array?

6 How much of a risk to Earth do you think is posed by the star WR104?

388

Measuring the stars22

astrometry, parallax, astronomical unit, light-year, parsec, spectroscopy,

spectrum, Fraunhofer lines, continuous spectrum, emission line spectrum,

absorption line spectrum, black body, black body curve, Planck curve,

spectroscope, prism, spectrograph, diffraction grating, spectral classes,

brightness, luminosity, Doppler effect, photometry, magnitudes,

apparent magnitude, absolute magnitude, distance modulus,

colour index, spectroscopic parallax

Fingerprinting the starsThe invention of the telescope led to a revolution in astronomy. The Moon and planets were found to be unique worlds, faint stars were revealed and new mysteries such as ‘spiral nebulae’ were discovered. However, astronomers still couldn’t answer the most basic questions: How far away are the stars? What are they?

The answer to the first question came in 1838 with the first successful measurement of the parallax of a star. The clue to the second question was already in place by then with Fraunhofer’s observation of spectral lines in light from the Sun. The understanding of these lines as fingerprints of the stars is the basis of modern astrophysics.

22.1 How far? Prior to the invention of the telescope, astronomy was the science of the positions and motions of the stars and planets. We now call this astrometry. The telescope made these measurements much more precise, but one of the prizes of astrometry remained elusive—a parallax measurement of the distance of a star.

Stellar parallax (see in2 Physics @ Preliminary Figure 13.3.2) is the apparent change in position of a nearby star relative to more distant stars. Over the period of a year, any nearby star will trace out a tiny ellipse in the sky, mimicking the Earth’s orbital motion (Figure 22.1.1). The closer the star, the larger the ellipse will be. The parallax angle p measures the angular length of half the major axis of the ellipse.

Parallax, as a surveying concept, was well known to ancient astronomers. It offers an unambiguous distance measurement based on simple geometry. Unfortunately, even the largest stellar parallax motion is less than 1 arc second—about the size a star appears because of seeing effects (see section 21.4). The first successful measurement was made by Friedrich Bessel (1784–1846) in 1838, using visual observation of the star 61 Cygni. Later in the 19th century photography replaced visual observations.

Define the terms parallax, parsec, light-year.

Explain how trigonometric parallax can be used to determine the distance to stars.

Figure 22.1.1 The parallax motion of a nearby star, showing the parallax angle p

pp

d

r

Earth now

Earth6 monthsfrom now

Sun

phototakennow

photo taken6 monthsfrom now

relativelynearby star

more distant stars

389

astrophysics

The distance d of the star and the parallax angle p are simply related by:

tan prd

=

Using the small angle approximation (sin p ≈ tan p ≈ p) since p is so small, and measuring p in arc seconds, r in astronomical units (AU) and d in parsecs (pc) results in an even simpler form for this equation:

pd

=1

Worked exampleQuestionProxima Centauri is the nearest star to Earth. It has a parallax p = 0.76887 arc seconds. What is its distance in parsecs, light-years and astronomical units?

solution

dp

= = = = = =1 10 76887

1.3006 4.2421 4 0133.

.pc ly 26 827 AU ××1016 m

Ground-based measurements of parallax are hindered by seeing effects, yet the uncertainty in the measured angles is, remarkably, only about 0.01 arc seconds. Realistically, the limit of ground-based parallax measurements with an acceptable uncertainty (of, say, 30%) is about 0.03 arc seconds, corresponding to a distance of 30 pc.

Between 1989 and 1993 Hipparcos (high precision parallax collecting satellite) measured the parallax of 118 000 stars to a precision of 0.001 arc seconds and more than 1 million stars to lower precision. This represents a factor of 10 increase in the distance range of parallax measurements to, say, 300 pc. However, compared with the distance of about 8000 pc to the centre of the Milky Way galaxy, the stars with accurate Hipparcos distances are still very local!

In about 2012 the astrometric boundaries should begin to expand even further, with the launch of the Gaia spacecraft. It will survey 1 billion stars with a typical precision of 0.00002 arc seconds, reaching out to the centre of the galaxy.

Discuss the limitations of trigonometric parallax measurements.

Units of distance

Parallax measurement is based on the size of the Earth’s orbit,

so a useful unit of distance is the astronomical unit (AU). This is approximately the average distance between the Earth and the Sun. 1 AU ≈ 1.4960 × 1011 mThe most familiar distance unit in astronomy is the light-year (ly). This is the distance that light covers in 1 year, travelling at 2.998 × 108 m s–1. 1 ly ≈ 63 241 AU ≈ 9.4605 × 1015 m. The parsec (pc) is an alternative unit that represents the distance of a hypothetical star that has a parallax angle of 1 arc second. 1 pc ≈ 206 265 AU ≈ 3.2616 ly ≈ 3.0857 × 1016 m

activity 22.1

practicaL eXperiences


activity 22.2



CHeCkPoint 22.11 Explain exactly what the parallax angle p measures.2 Outline the reasons for using satellites to measure parallax angles.

22.2 light is the key In his book The Positive Philosophy, the 19th century French philosopher August Compte (1798–1857) wrote about our knowledge of the stars and planets.

... We see how we may determine their forms, their distances, their bulk, and their motions, but we can never know anything of their chemical or mineralogical structure ...

Measuring the stars

390

22The distances in astronomy are an enormous barrier to learning about the stars and planets, but Compte was wrong in thinking we would never know about their composition. Astronomical spectroscopy is the study of the light from objects in the universe to reveal their composition and physical characteristics.

The first step had been made long before when Isaac Newton (1643–1727) realised that ‘white’ light is composed of component colours, revealed when the light is passed through a glass prism. In 1802, William Wollaston (1766–1828) looked more closely and noticed dark lines in the rainbow of colours that form the spectrum of the Sun. In 1814 Joseph von Fraunhofer (1787–1826) used a spectroscope to rediscover the lines and eventually catalogued over 570 of these Fraunhofer lines (Figure 22.2.1). By 1860, Gustav Kirchhoff (1824–1887) and Robert Bunsen (1811–1899) and others discovered that chemical elements produced bright spectral lines when heated and some of these lines matched the dark Fraunhofer lines. They deduced that the Fraunhofer lines were produced by absorption of light by these elements in the Sun, providing the key to understanding the composition of the stars.

Kirchhoff ’s empirical laws of spectrum analysis describe how to produce three of the key types of spectra we observe in astronomy:• Ahot,densegasproducesacontinuous spectrum.• Ahot,low-densitygasproducesanemission line spectrum.• Acontinuousspectrumsourceviewedthroughacool,low-densitygasproduces

an absorption line spectrum.

Continuous spectraWhen you heat an object such as a steel bar it begins to glow red and orange and then perhaps white before melting. A very low resolution spectrograph reveals that at any of these temperatures the bar is emitting a continuous range of wavelengths—a continuous spectrum (Figure 22.2.2).

Any relatively dense material—a solid, liquid or high density gas—will behave in the same way. They approximate a black body (see section 9.2), which is the idealised example of a hot object. A black body emits light at all wavelengths, with a distribution described by a black body curve (or a Planck curve) that depends only on temperature (Figure 22.2.6), not composition. In astronomy, the interior of a star is very close to a black body source.

The wavelength of peak emission of a black body curve λmax is given by Wien’s law (see in2 Physics @ Preliminary section 14.2):

λmax

.=

× −2 9 10 3

T

where T is the temperature on the kelvin scale. The change in λ max with temperature alters the balance of the visible colours and leads to a change in the overall colour of the object. Cool objects (T 4000 K) look red but emit most of their light in the infra-red. Peak emission from a hot object (T 10 000 K) is in the ultraviolet (UV) and the object looks blue-white to the eye.

The total energy emitted per second per unit area of the surface is represented by the area under the black body curve. This power increases very rapidly with temperature according to the Stefan–Boltzmann law:

P T= ×( )−5 67 10 8 4.

Describe the technology needed to measure astronomical spectra.

Figure 22.2.1 Fraunhofer’s spectral lines superimposed on the rainbow spectrum of the Sun

Account for the production of emission and absorption spectra and compare these with a continuous black body spectrum.

activity 22.3



Figure 22.2.2 A continuous spectrum

400 500 600 700

Wavelength (nm)

391

astrophysics

Figure 22.2.1 Fraunhofer’s spectral lines superimposed on the rainbow spectrum of the Sun

PHYsiCs FeAtuResPeCtRosCoPes AnD sPeCtRogRAPHs

A spectroscope is a device that disperses light into its component colours to be viewed by eye. Early

spectroscopes used a prism to refract the various colours to different angles, according to their wavelengths. Replacing the eye with a camera creates a spectrograph, the key instrument in astronomical spectroscopy.

The spectral resolution of a spectrograph defines how much detail can be seen in the spectrum, in the same way that angular resolution is used for images. Optical spectral resolution ranges from low (~0.5 nm) to ultra-high (~0.001 nm).

Anything beyond the lowest resolutions requires greater dispersion of the light than is possible with a prism. This can be achieved using a diffraction grating, often made by ruling very fine lines onto an optical surface. Interference between light beams diffracted by the lines results in a spectrum of the light striking the surface.

Most spectrographs use a slit to restrict the light beam. The light is dispersed in the direction perpendicular to the slit (Figure 22.2.3). This direction measures wavelength. Depending on the optical set-up, distance along the slit may measure distance across the target object. The resulting spectrum of a single point such as

a star can be shown as a coloured band (Figure 22.2.4a), but it is usually presented as an intensity profile versus wavelength (Figure 22.2.4b).

Spectrographs are the key instruments on all optical telescopes. At the Anglo-Australian Telescope, AAOmega (Figure 22.2.5) is a modern spectrograph, fed by optical fibres bringing light from multiple targets in the image.

slit

light fromtelescope

camera

focusingmirror

collimatingmirror

grating

Figure 22.2.3 A basic spectrograph

500 600 700Wavelength (nm)

Inte

nsit

y

Hα

HβHγ

measure intensityalong a slice

through the spectrum

a

b

Figure 22.2.4 A spectrum as (a) coloured light and (b) an intensity profile


Figure 22.2.5 Astronomer Rob Sharp examines the grating feeding a camera (the assembly in the right half of the image) in the AAOmega spectrograph.

Measuring the stars

392

22

activity 22.4



Worked examplesQuestionIn Figure 21.0.2, the two bright stars on the left are α and β Centauri—the pointers to the Southern Cross. α Centauri is actually a double star, and the brighter star has a surface temperature of ~5800 K. β Centauri is a triple system, with the brightest star having a surface temperature of ~23 000 K. Which star is which in the image?

solutionUsing Wien’s law for the two stars:

α Centauri

λ max =

≈

× −2.9 105800

500

3

nmThis is in the middle of the visible band, so we expect a yellow overall colour.

β Centauri

λ max =

≈

× −2.9 1023 000

130

3

nmThis is in the ultraviolet, so we expect a blue-white overall colour.

Clearly the left star in the image is α Centauri and the right star is β Centauri.

QuestionWhat is the relative power output of the two stars?

solutionUsing the Stefan–Boltzmann law for the two stars:

α Centauri

P = ×= ×

− − −

−

( . )( )

.

5 67 10 5800

6 4 10

8 2 4 4

7 2

Wm K

Wm

β Centauri

P = ×

= ×

− − −

−

( . )(

.

5 67 10 23000)

1 6 10

8 2 4 4

10 2

Wm K

Wm

So, each square metre of the surface of the hotter star β Centauri emits nearly 250 times more energy per second than the surface of α Centauri. The hotter star is also about 8 times larger and so has 64 times the surface area of the cooler star. Thus the luminosity of the hotter star is about 16 000 times the intensity of the cooler one.

Emission line spectra When you inject energy into a low-density gas, the spectrum is very

different. It shows bright emission lines at wavelengths characteristic of the elements present in the gas (Figure 22.2.7). Familiar examples include the spectra from gas discharge lamps and neon signs. Low-density gas clouds in interstellar space also glow with an emission line spectrum.

Figure 22.2.6 Black body curves for different temperatures show that the peak moves to shorter wavelengths with increasing temperature and the curves move up, indicating greater energy output at all wavelengths.

Wavelength (nm)

3000 K4500 K

6000 K

12000 K

Inte

nsit

y

7500 K

0 500 1000 1500 2000

Figure 22.2.7 An emission spectrum

447.1 471.3 492.1 501.5 587.5 667.8

Wavelength (nm)

393

astrophysics

The origin of emission lines lies in the structure of the atom (see Chapter 12 ‘From Rutherford to Bohr’). When an atom absorbs energy, an electron will make a transition to a higher energy state, provided the absorbed energy exactly matches the energy of the transition. It will often then quickly make one or more downward transitions back to lower energy states, emitting energy.

This process is described as the absorption or emission of photons. In making a downward transition from energy E2 to energy E1, an atom emits a photon with energy E related to its wavelength λ (or frequency f ) by the relation:

E E Ehc

hf= − = =2 1

λ

where Planck’s constant h = 6.626 × 10–34 J s and c = 3.00 × 108 m s–1 is the speed of light. These photon energies (and hence wavelengths or frequencies) correspond to the observed emission lines. The pattern of emission lines is a fingerprint for each element because it is matched to the energy levels within each atom of that element.

Absorption line spectraThe spectrum of the Sun and stars is a continuous spectrum, but crossed by dark lines, not bright ones (Figure 22.2.8). The origin of these absorption lines lies in the relatively cool gas overlying the hotter, denser gas deep in the star. The denser gas produces a continuous spectrum. Atoms in the overlying gas absorb light from the continuous spectrum, but only at the wavelengths matching differences in their energy levels. They then re-emit the light, but in all directions, not just outwards. The net effect is less light in the outward direction at those wavelengths, creating dark lines in the spectrum.

The spectral fingerprint of a gas is the same, whether seen in absorption or emission. In the Sun and stars, it is the outer layer of the star whose composition is imprinted on the spectrum as absorption lines (Figure 22.2.9).

The spectra of other astronomical objects depends on their physical conditions of temperature, density and composition. Table 22.2.1 gives some examples.

Where does bLack body Light coMe froM?

If light is emitted by atoms at very precise wavelengths, how is a

continuous spectrum produced? The answer lies in the density of the material emitting the light. If the density is low, the atoms are relatively far apart and they emit light at the discrete wavelengths expected. However, if the density is high, the atoms are closer together, and they influence one another and alter the energy levels of the atoms. The effect is to blur the lines into a continuous distribution.

Figure 22.2.8 An absorption spectrum

Figure 22.2.9 All three types of spectra are produced by the Sun: continuous spectrum from deeper layers, absorption lines from the photosphere and emission lines from the chromosphere and corona.

absorption spectrum

emission spectrum

continuous spectrum

diffuse gas(e.g. outer layers ofa star or a nebula)

black body source (e.g. deeper layersof star)

Describe how spectra can provide information on surface temperature, rotational and translational velocity, density and chemical composition of stars.

Identify the general types of spectra produced by stars, emission nebulae, galaxies and quasars.

Wavelength (nm)

Measuring the stars

394

22Table 22.2.1 Spectral characteristics of some astronomical objects

Object DescriptiOn spectrum exampleEmission nebulae

Regions of gas (mostly hydrogen and helium) that glow due to intense UV light from embedded young hot stars

Dominated by strong emission lines characteristic of the gas composition

Normal galaxies

Collections of billions of stars, gas and dust; light output is generally dominated by the mix of stars of various types

Absorption spectra, often a mix of bright blue-white types with many more numerous but fainter yellow stars

Quasars A type of galaxy with an active nucleus dominating the total energy output

A continuous spectrum with emission lines (that may be variable) suggesting fast-moving gas clouds

CHeCkPoint 22.21 Describe the Fraunhofer lines.2 Describe how emission lines are produced.3 Describe the relationship between the absorption lines and the emission lines of an element.

22.3 the stellar alphabetIn 1885 astronomers at Harvard College Observatory began to compile a photographic catalogue of spectra that became the Henry Draper Catalogue of Stellar Spectra. When the final extension was published in 1949, almost 360 000 stars had been classified.

Hγ

Hβ + [OIII] Hα

0.8

0.6

0.4

0.2

400 500 600 700 800

Rel

ativ

e fl

ux

λ (nm)

1

0.8

0.6

0.4

0.2

Rel

ativ

e fl

ux

λ (nm)

450 500 550 600 650

0.3

0.2

0.1

0400 500 600 700 800

Rel

ativ

e fl

ux

λ (nm)

395

astrophysics

Spectral classesThe Harvard classification scheme developed during the project remains in use today. It started as groups of spectra with similar spectral lines, given letters from A to N, with O, P and Q added for some unusual stars. Re-ordering and simplification of the sequence led to the current, seemingly random sequence of letters. The major spectral classes are:

O B A F G K M

(We can remember this odd sequence using various mnemonics, the best known being ‘Oh Be A Fine Girl [Guy] Kiss Me’). With additions of other classes, the sequence can be represented as:

W – O B A F G K M – L T The black body curves underlying the shape of the spectra clearly imply

a sequence in surface temperature from hot O stars to cool M stars. It was not until the 1920s that the understanding of electron energy levels in atoms progressed enough to understand how surface temperature also controlled the appearance of the spectral lines. Composition differences between stars are relatively small and are not the reason why spectra vary so much.

A sample of spectra from various spectral classes is shown in Figure 22.3.1. Some of the key features of each spectral class are described in Table 22.3.1, emphasising that the colour of a star, its spectral class and surface temperature are all closely related.

Each spectral class has been divided into 10 sub-classes. The Sun, for example, is a yellow G2 star with a surface temperature of ~5770 K, placing it at the hot end of spectral class G.

Luminosity classesWhen plotting a Hertzsprung–Russell (HR) diagram (see in2 Physics @ Preliminary section 15.3), we can use the spectral class on the horizontal axis, since it is closely related to the surface temperature (Figure 22.3.2).

Describe the key features of stellar spectra and describe how these are used to classify stars.

brightness and LUMinosity

Recall the definitions of brightness and luminosity

from in2 Physics @ Preliminary section 15.1. The brightness of a star is a measure of the energy received in a certain time per unit of collecting area (or power per unit area) W m–2. Its luminosity is the total power output of the star in watts. A star will look brighter when seen through a telescope. It obviously isn’t really more luminous; the telescope has a larger collecting area than your eye alone, making the star appear brighter.

Table 22.3.1 Characteristics of spectral classes of stars

spectral class effective temperature (K) cOlOur strength Of

hyDrOgen linesOther spectral

features% Of main sequence

stars

W >50 000 Blue Weak He, C, N emission lines Extremely rare

o 31 000–50 000 Blue Weak Ionised He+ lines, strong UV continuum 0.00003

B 10 000–31 000 Blue-white Medium Neutral He lines 0.1

A 7500–10 000 White Strong Ionised metal lines 0.6

F 6000–7500 White-yellow Medium Weak ionised Ca+ 3

g 5300–6000 Yellow Weak Ionised Ca+, metal lines 8

k 3800–5300 Orange Very weak Ca+, Fe, strong molecules, CH, CN 12

M 2100–3800 Red Very weak Molecular bands, e.g. TiO, neutral metals 76

l 1200–2100 Red Negligible Neutral metals, metal hydrides

Brown dwarf numbers uncertain

t <1200 Red Negligible Methane bands Brown dwarf numbers uncertain

Measuring the stars

396

22

TiO TiO TiO TiO

1200

1000

800

600

400

200

0350 400 450 500 550 600 650

Wavelength (nm)

05

09

B6A2

A6F0

F0K5

M2

M4

F7G2

G7

Rel

ativ

e fl

ux

Hδ

Hβ

Hγ

He

Na

Ca II

Hα

Figure 22.3.1 Examples of spectra representing stars of different spectral classes, with hot stars at the top and cool stars at the bottom. Spectral lines due to hydrogen and calcium and some molecular bands are identified.

The HR diagram clearly illustrates that stars with the same surface temperature may have very different luminosities. This can only imply that one star has more surface area than the other—one is a giant in comparison to the other (look again at the second worked example in section 22.2). These differences in luminosity, and hence size, are described by the luminosity classes listed in Table 22.3.2 and illustrated in Figure 22.3.2.

The Sun is a G2 V star on the main sequence of the HR diagram. A G-type Ib supergiant like δ Cephei (see section 23.4) has a similar spectrum but is about 2000 times more luminous than the Sun.

More from spectral linesAnother pointer to the size of stars can be found in their spectral lines. Supergiant stars have very sharp spectral lines because the atoms in the low-density outer layers are widely spaced. Dwarf stars like the Sun have higher densities in their outer layers, causing the atoms to interact more, slightly changing their energy levels and hence slightly broadening their spectral lines.

This is just one of many extra clues about stars hidden in the spectral lines. Many others are related to the Doppler effect (see in2 Physics @ Preliminary section 13.8) caused by motion of the gas emitting or absorbing the light. At velocities much less than the speed of light c, the Doppler shift ∆λ of a spectral line of wavelength λ is related to the velocity v along the line of sight (Figure 22.3.3) by:

Δλλ

=vc

The wavelength shifts lead to broadening of spectral lines that encode information about the temperature and motion of the gas.

Figure 22.3.2 An illustration of the distribution of stars in the HR diagram, emphasising that stars of different luminosity classes may have the same spectral type

30 000 10 000 7000 6000 4000Effective temperature (K)

10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 BO AO FO GO KO MO

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence (V)Subdwarf (VI)

Supergiants (I)

Giants (II, III)

Subgiants (IV)

Whitedwarfs

(VII)

Table 22.3.2 Luminosity classes

luminOsity class DescriptiOnIa Bright supergiant

Ib Supergiant

II Bright giant

III Giant

IV Subgiant

V Main sequence dwarf

VI Subdwarf

VII White dwarf

397

astrophysics

Worked exampleSince the first confirmed detection in 1992, more than 300 planets have been detected around other stars. Most have been found by looking for a shift in the star’s spectral lines due to the motion of the star caused by the orbiting planets.

QuestionJupiter’s motion causes the Sun to wobble around a small circle at 12 m s–1. What shift in wavelength does this represent at an observing wavelength of 500 nm?

solution ∆λ

500 1012

3 00 109

1

8 1×=

×−

−

−mms

ms.

The periodic shift in wavelength ∆λ = 2 × 10–14 m = 2 × 10–5 nm.

This periodic back and forth shift of all the spectral lines is very small and comparable to the broadening of spectral lines due to the density of gas in the Sun’s photosphere, but 1000 times less than the thermal broadening due to the motion of the hot gas atoms.

CHeCkPoint 22.31 Describe how the spectral classes came about.2 Describe what is meant by a luminosity class.3 Explain why dwarf stars have broader spectral lines than larger stars.4 Explain the origin of blue and red shifting of spectral lines in stars.

22.4 Measuring magnitudes Astronomical photometry is the measurement of the brightness of an astronomical source. In optical astronomy, this used to require a photoelectric photometer. This usually employed a photomultiplier tube that produced a pulse of electrons when struck by a photon from a target star. Photographic film could also be used, but is inefficient. Both have now been replaced in most situations by charge-coupled device (CCD) cameras, the same devices used for modern astronomical imaging and similar to the detectors in a modern digital camera. CCDs offer very high detection efficiency and can record many objects at once.

Sensible SI units for brightness are W m–2. However optical astronomy has inherited the traditional system of magnitudes (see in2 Physics @ Preliminary section 15.1) that is much harder to use!

The magnitude system is usually said to have originated with the Greek astronomer Hipparchus (~190–120 bce). He called the brightest stars magnitude 1 and the faintest stars seen by eye magnitude 6. In 1856 Norman Pogson (1829–1891) suggested formalising the system and making the magnitude scale a multiplying scale, so each step up in magnitude represents a factor of about 2.5 in brightness. More precisely, he defined a first magnitude star to be exactly 100 times the brightness of a sixth magnitude star (a difference of 5 magnitudes). Each magnitude is then a factor of 1001/5 ≈ 2.512. This behaviour is logarithmic and roughly mimics the behaviour of the eye.

Describe the advantages of photoelectric technologies over photographic methods for photometry.

Figure 22.3.3 The Doppler effect measures the radial velocity along the line of sight. This is generally only a component of the true velocity through space.

transverse velocity

(across line of sight)

velo

city

thr

ough

spac

e

radial velocity

(along line of sight)

star

line of sig

ht

Earth

Measuring the stars

398

22

activity 22.6



Mathematically, this can be expressed as:

m mI

IB AA

B

− =⎛

⎝⎜

⎞

⎠⎟2 50 10. log

or equivalently:I

IA

B

m m 5B A= −( )100

These equations show how a difference in magnitudes (mB – mA) is related to a ratio of brightnesses (or ‘intensities’) IA/ IB.

The system is difficult because it is logarithmic, not linear, and larger magnitudes represent fainter stars.

Apparent magnitude The apparent magnitude is a measure of how bright an object is in our

sky. The exact zero-point of the system must be defined carefully, but the Sun, Moon, some planets and even some stars are clearly brighter than zero magnitude. Their values are therefore negative. Many stars are invisible to the unaided eye and therefore have magnitudes much greater than 6 (Figure 22.4.1).

Worked examplesQuestionShow that a brightness ratio of 100 corresponds to a difference of five magnitudes.

solution I

IA

B

=100

m mB A− = ( ) = × =2 50 100 2 5 2 510. log .

QuestionThe two stars of the α Centauri binary system have apparent magnitudes of 0.01 and 1.34. α Centauri is believed to be a triple system, with the star Proxima Centauri much fainter at apparent magnitude 11.09 and some distance away on the sky. What is the relative brightness of these three stars?

solutionChoose mA = 0.01, mB = 1.34 and mC = 11.09

Then: (mB – mA) = 1.34 – 0.01 = 1.33, then II

A

B

= ≈100 3 41 33 5. .

(mC – mA) = 11.09 – 0.01 = 11.08, then II

A

C

= ≈100 2700011 08 5.

So the brightness ratios A : B : C are 11

3 41

27000:

.: .

Define absolute and apparent magnitude.

Figure 22.4.1 The brightness of some astronomical objects on the magnitude scale

–30

–25

–20

–15

–10

–5

0

5

10

15

20

25

30

App

aren

t m

agni

tude

(m

v)

Sun

full moon

Venus at brightest

Sirius

Polaris

naked eye limit

faintest objects seenby HST

399

astrophysics

Absolute magnitudeApparent magnitude doesn’t tell the whole story, since it does not indicate the true luminosity of a star. The magnitude system handles this by calculating how bright the stars would appear to be in our sky if all were moved to a standard distance of 10 pc (32.6 ly). To calculate this absolute magnitude we use the inverse square law (see in2 Physics @ Preliminary section 15.1), which can be expressed as:

I

I dA

A10

210

=⎛

⎝⎜

⎞

⎠⎟

where II

10

A is the ratio of the brightness (intensity) at its actual distance dA

(in parsecs) compared to its brightness at 10 parsecs. Using this expression in the equation defining magnitudes leads to:

m mI

I

d

10 1010

10

2 50

510

− =⎛

⎝⎜

⎞

⎠⎟

=⎛

⎝⎜

⎞

⎠⎟

AA

A

. log

log

The absolute magnitude is usually represented as M. The relationship between apparent magnitude m and absolute magnitude M for a star at distance d can then be written as:

m Md

− =⎛⎝⎜

⎞⎠⎟5

1010log

orM m

d= −

⎛⎝⎜

⎞⎠⎟5

1010log

The quantity m – M is often called the distance modulus.Absolute magnitude is a measure of the luminosity of a star. It is often used

as the vertical axis in an HR diagram.

Worked examplesQuestionAssuming the apparent magnitude of the Sun is –26.72, what is its absolute magnitude?

solutionThe distance of the Sun is 1 AU =

1206265 pc

M md

M

= −

− − =

510

26 72 51

2062650

10

10

log

. log = −31 57.

Therefore M = +4.85.

If the Sun were placed at a distance of 10 pc, it would appear as one of many relatively faint stars visible with the unaided eye.

Explain how the concept of magnitude can be used to determine the distance to a celestial object.

Measuring the stars

400

22QuestionThe brightest star in the sky is Sirius (α Canis Majoris) with m = –1.43 and M = +1.47. How far away is Sirius?

solution− − =

− =

1 43 1 47 510

2 905 10

10

10

. . log

.log

d

d

11010

2 90 5− =

. d

Thus d = 10 × 0.263 = 2.63 pc.

This makes Sirius one of the closest stars to Earth. Although it is more luminous than the Sun, it is the brightest star in our sky mainly because it is so close.

Solve problems and analyse information using: M m

d= −

510

log

and II

m mA

B

B A= −( )100 5

to calculate the absolute or apparent magnitude of stars using data and a reference star.

CHeCkPoint 22.41 Briefly outline the history of the magnitude system.2 Deduce (do not calculate) the intensity ratio between star A with m = +3 and star B with m = –2.3 Deduce (do not calculate) the absolute magnitude of a star that is d = 100 pc away and has a magnitude of m = +5.

22.5 Colour mattersAs we saw earlier, the colour of stars arises from the distribution of the wavelengths they emit. The black body curve (Figure 22.2.6) shows that hot stars emit most of their light at blue and UV wavelengths. Why is such a star usually described as ‘blue-white’ and not simply violet?

The answer lies in the response of the human eye to light. The spectral sensitivity of the eye peaks in the yellow-green part of the visual band and falls quickly at blue wavelengths. The overall colour impression is therefore a result of both the spectrum of the star and the spectral sensitivity of the eye.

Coloured filters are often used to define the spectral sensitivity of a detector. The importance of this is illustrated in Figure 22.5.1 in which stars of different colours have different relative brightness when observed through different coloured filters. Different observers using different filters may then measure different brightness for a single star. Who is ‘right’?

The solution to this problem is to agree on ‘standard’ filters to use when measuring brightness. The first standard was the Johnson UBV (ultraviolet, blue and visual) set. Magnitude values quoted previously were all ‘visual’ magnitudes measured through a V filter or converted to that standard. An apparent visual magnitude is often represented as simply V (although mV is better). An absolute visual magnitude would then be represented as MV.

Colour indexThe difference in brightness seen through different filters is a measure of the colour of a star. A blue-white star will appear brighter (with a lower magnitude value) through a filter that passes blue light than through a filter that passes only

activity 22.5



Figure 22.5.1 A star field imaged through (top) red and (bottom) blue filters. Note the change in relative brightness of many stars.

401

astrophysics

red light. Using these magnitude measurements, the concept of colour can be assigned a value. A colour index is the difference in the brightness of a star, in magnitudes, when measured through two different filters.

Many colour indices are possible, but the best known is the (B – V) colour index (or mB – mV), the difference between the blue and visual apparent magnitudes. The system is set up so that ‘white’ A0 stars have (B – V) = 0. The colour index of the Sun is +0.66. Blue-white stars have small negative colour index values, while orange-red stars have larger positive colour indices (Figure 22.5.2).

Worked exampleQuestionAssuming the colour index of the Sun is +0.66, what is the apparent B magnitude (or m B) of the Sun? What is the absolute B magnitude (M B) of the Sun?

solutionFrom an earlier worked example, V = m V = –26.76 and M V = +4.85.

Colour index = B – V 0.66 = B – (–26.76) Therefore B = –26.10

Similarly:

Colour index = M B – M V

0.66 = M B – 4.85

Therefore M B = 5.51

The fact that the blue magnitudes are greater than the visual magnitudes implies the Sun is fainter in the blue, as expected for a yellow star.

Because colour index quantifies the concept of colour, it is closely related to both surface temperature and spectral class. Any of these three may be used to plot an HR diagram (Figure 22.5.3), but colour index has the advantage of being a measured value.

Spectroscopic parallax Measuring the colour index of a star places it within a narrow strip on

the HR diagram. Studying details of the spectral lines identifies the star’s luminosity class within that strip. With these two measurements we have approximately located it on the diagram, without knowing its absolute magnitude (see Figure 22.5.3). However, reading the vertical scale now reveals absolute magnitude. Comparison of the absolute magnitude and the apparent magnitude will then yield a distance, using the equation given earlier.

This method of determining distance is called spectroscopic parallax. This is a poor name since parallax is not involved. This procedure is essentially identifying the type of star from the characteristics of its light. Then, knowing how bright such stars are, a distance can be calculated using the inverse square law.

Explain how two-colour values (i.e. colour index, B – V) are obtained and why they are useful.

Outline spectroscopic parallax.

Figure 22.5.2 Magnitudes and colour indices for some stars in and around the Southern Cross

α Centauri AV = –0.01B – V = +0.71

ϒ CrucisV = +1.59B – V = +1.60

δ CrucisV = +2.79B – V = –0.19

β CentauriV = +0.61B – V = –0.23 e Crucis

V = +3.59B – V = +1.39

α CrucisV = +0.77B – V = –0.24

β CrucisV = +1.25B – V = –0.24

Measuring the stars

402

22This is a powerful technique since the colour index

can be measured and the luminosity class can be estimated, provided there is enough light to obtain a good spectrum. Often there is no other way to know the distance to a star. However, the uncertainties are often large and the distances derived are only approximate.

Figure 22.5.3 Using spectroscopic parallax to estimate the distance of γ Crucis


α Crucisβ Crucisδ Crucis

ε Crucis

γ Crucisβ Centauri

α Centauri A

α Centauri BSun

10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 BO AO FO GO KO MO

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence (V)

Supergiants (I)

Giants (II, III)

Subgiants (IV)Subdwarf (VI)

Whitedwarfs

(VII)

CHeCkPoint 22.51 Outline the need for standard filters in measuring brightness.2 Interpret what a colour index of zero tells you about a star.3 Explain how knowing the colour index and luminosity class of a star can lead to finding its absolute magnitude

using the HR diagram.

Worked exampleQuestionOne of the stars identified in Figure 22.5.2 is γ Crucis, the brighter reddish star in the Southern Cross. Estimate its distance using spectroscopic parallax, given that it is a M3.5 III star with V = 1.59 and colour index +1.60.

solutionFrom the HR diagram in Figure 22.5.3, the absolute magnitude of γ Crucis is just above zero, say –0.5.

Then:

m Md

d

− =

− −( )=

510

1 59 0 5 510

2

10

10

log

. . log

..log

.

095 10

1010

10

2 09 5

=

=

d

d

Thus d = 10 × 2.62 ≈ 26 pc.

The distance measured by the Hipparcos satellite is 27.0 ± 0.5 pc, which shows remarkable agreement, considering the estimates involved in the spectroscopic parallax value.


403

astrophysics


ACtivitY 22.1: PARAllAxDetermine the angle of parallax of an object at the back of the classroom in order to determine how far away it is. Use this method to understand the use of parallax to determine the distance to nearby stars.Equipment: protractor, metre ruler.

Discussion questions1 Compare the method used in the classroom with the method used to

measure the distance to the stars. What differences are there between the methods?

2 Explain how the accuracy in results changes when the angle is increased.

ACtivitY 22.2: liMits oF PARAllAxResearch the limits of parallax measurement for ground-based and space-based telescopes. Use the template to help focus your search. Summarise your findings in a table.

Discussion questions1 What minimum angle of parallax can be measured using ground-based

telescopes before the uncertainty of the measurement is too great?2 Explain what other factors contribute to the minimum useful parallax

measured by ground-based telescopes.3 Determine the difference between this ground-based limit and the limit

from space-based telescopes.

ACtivitY 22.3: sPeCtRAObserve spectra of different objects with a spectroscope and determine if they are emission, absorption or continuous spectra.Equipment: spectroscope, fluorescent light, incandescent light.

Discussion questions1 Compare and contrast the different types of spectra.2 Identify an example of each type of spectra.

Solve problems and analyse information to calculate the distance to a star given its trigonometric parallax using: d

p= 1

Gather and process information to determine the relative limits to trigonometric parallax distance determinations using recent ground-based and space-based telescopes.

Perform a first-hand investigation to examine a variety of spectra produced by discharge tubes, reflected sunlight, or incandescent filaments.


404

22 Measuring the stars

ACtivitY 22.4: teMPeRAtuRe PReDiCtionExamine each of the given black body radiation curves of stars and determine their surface temperature.

Discussion questions1 State the relationship between peak wavelength and surface temperature

of a star.2 Outline the method that you followed in order to determine the

temperature of the star.

ACtivitY 22.5: MAgnituDe AnD ColouR inDexObserve a light through different filters to determine how brightness changes when only selected colours are viewed.Equipment: light meter, 2 red filters, 2 blue filters, light source.

Discussion questions1 Compare the brightness of a red star when viewed through a red filter with

that of the star when viewed through a blue filter.2 Compare this with viewing a blue star through the filters.3 Explain how the change in brightness when viewed through the filters

would affect the measured magnitude of each star.

ACtivitY 22.6: teCHnologiCAl ADvAnCesResearch some of the instruments and techniques used in astronomy that have aided in developing our understanding of celestial objects.

• Telescopes• Photomultipliers• CCDs• Opticalfibres• Thintelescopemirrorsandactiveoptics• Adaptiveoptics• Opticalinterferometry• IRarraydetectors• Radiotelescopes• Radiointerferometry• X-raymirrors• Scintillators• Balloon-bornetelescopes• Spacetelescopes

Discussion questions1 Outline what has been discovered or developed because of the use of the

item you have researched.2 Discuss what technological developments had to occur in order for this

piece of equipment to be used.

Perform an investigation to demonstrate the use of filters for photometric measurements.

Identify data sources, gather, process and present information to assess the impact of improvements in measurement technologies on our understanding of celestial objects.

Analyse information to predict the surface temperature of a star from its intensity/wavelength graph.

405

chapter summary astrophysics

• Stellarparallaxanglep is the apparent change in position of a nearby star at distance d, relative to more distant stars. It is caused by the orbital motion of the Earth. p (in arc seconds) and d (in parsecs) are related by:

pd

=1

• Fraunhoferlinesareproducedbytheabsorptionoflightby atoms in the outer layers of the Sun.

• Aspectroscopeorspectrographdisperseslightbywavelength, and enables the composition and physical characteristics of the source to be studied.

• Threetypesofspectraareobservedfromhotsources:– A hot, dense gas produces a continuous spectrum.– A hot, low-density gas produces an emission line

spectrum.– A continuous spectrum source viewed through a

cool, low-density gas produces an absorption line spectrum.

• Ablackbodyisanidealisedhotobjectthatemitslightat all wavelengths with a distribution described by a black body curve.

• Thewavelengthofpeakemissionofablackbodycurveλmax is given by Wien’s law:

λmax

.=

× −2 9 10 3

T• Thetotalenergyemittedpersecondperunitareaof

the surface of a black body is given by the Stefan–Boltzmann law: P T= ×( )−5 67 10 8 4.

• Thespectralfingerprintofanelementisthesame,whether seen in absorption or emission.

• ThemajorspectralclassesareOBAFGKM.• Thesurfacetemperatureofastarcontrolsthe

appearance of the spectral lines. Composition differences between stars are not the main reason why spectra vary so much.

• Differencesinluminosity(andhencesize)between stars of the same spectral class are described by the luminosity classes.

• TheDopplereffectiscausedbymotionofthegasemitting or absorbing the light. At velocities much less than the speed of light c, the Doppler shift ∆λ of a spectral line of wavelength λ is related to the velocity v along the line of sight by:

Δλλ

=vc

• Photometryisthemeasurementofthebrightnessofanastronomical source.

• Eachstellarmagnituderepresentsafactorofabout2.5in brightness. A difference of five magnitudes is equivalent to exactly 100 times in brightness.

• Apparentmagnitudeisameasureofhowbrightanobject appears in our sky.

• Absolutemagnitudedescribeshowbrightastarwouldappear in our sky if moved to a standard distance of 10 pc (32.6 ly).

• Theoverallcolourimpressionfromastarisaresultofboth the spectrum from the star and the spectral sensitivity of the detector.

• Acolourindexisthedifferenceinthebrightnessof a star in magnitudes, when measured through two different filters. The best known is the B – V colour index.

• Spectroscopicparallaxisadistancemeasurementmadeby identifying the type of star from the characteristics of its light and then comparing its apparent and absolute magnitudes to derive a distance.

review questionsPHYsiCAllY sPeAkingOn the right is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

Parallax Colour index Distance Colour

Spectroscopy Temperature PhotometryAbsorption spectra

Emission spectra

Magnitude LuminosityContinuous spectra

406

Measuring the stars22

RevieWing 1 Using a diagram, explain how parallax can be used to

determine the distance to a nearby star.

2 Explain why parallax measurements are so important in astronomy.

3 Outline the factors that set limits on the angle of parallax that can be measured from the ground.

4 Describe the significance of the Hipparcos and Gaia astrometry missions.

5 List four pieces of information about a star that can be obtained using spectroscopy.

6 Compare and contrast the three types of spectra described in this chapter.

7 Explain how emission spectra are created, in terms of energy levels of electrons in an atom.

8 Identify the spectral class of stars with:a strong hydrogen lines but weak calcium emissionb relatively weak hydrogen lines and few other

strong absorption linesc relatively weak hydrogen lines and broad

absorption bands.

9 Describe the major observational properties that are used to classify stars.

10 Relate temperature to the colour of stars.

11 Recall how the magnitude scale was developed and hence explain why the scale has negative values.

12 Justify the need for a colour index in describing a star.

solving PRoBleMs 13 Calculate the distance in metres of one kiloparsec

(1000 pc).

14 A star has an annual parallax of 0.017 arc seconds. Calculate the distance to the star in parsecs and light-years.

15 A star has an annual parallax of 0.029 ± 0.005 arc seconds. Calculate the range of possible distances (in parsecs) to this star suggested by this value.

16 The hydrogen lines in the visible part of the spectrum have wavelengths given by the following equation:

11 097 10

12

172 2λ

= ×( ) −⎛⎝⎜

⎞⎠⎟.

n

a Calculate these wavelengths for n = 3, 4, 5, 6, 7.b Identify the hydrogen lines in the spectrum of the

F-type star shown in Figure 22.6.1.

17 The Sun and the brightest star of the α Centauri system are very similar. Assuming the apparent magnitude of the Sun in –26.72 and that of α Centauri A is 0.01, calculate how much brighter the Sun appears in our sky than α Centauri A.

18 Star A has an apparent magnitude of +0.3 and a distance of 5 pc. Star B has an apparent magnitude of +4 and a distance of 100 pc. Assess which star has the greater luminosity.

19 A star has a B magnitude of +9.6 and a V magnitude of +8.0. Deduce the approximate spectral type of this star.

20 Betelgeuse is a spectral class M Ib supergiant star. It has an apparent visual magnitude of about 0.45 (but varies). Using the spectroscopic parallax technique, calculate its approximate distance from the Earth.

Figure 22.6.1 Spectrum of an F-type star

440

420

400

380

360

340

320

300

Flux

350 400 450 500 550 600 650

Wavelength (nm)

Revie

w Questions

407

23Stellar companions and variables

binary star, centre of mass, visual binary, proper motion,

astrometric binary, spectroscope binary, eclipsing binary, light curve,

cataclysmic variable, X-ray binary, mass–luminosity relationship,

variable star, extrinsic variables, intrinsic variables, non-periodic

variables, periodic variables, Cepheid variables, RR Lyrae variables,

period–luminosity relationship

Not quite typical The Sun is often said to be a typical star. It is a not-quite middle-aged G-type dwarf among billions of similar stars in the disc of the Milky Way galaxy. It leads a relatively quiet life, with little visible change apart from a cycle of small sunspots and occasional flares at the surface. In this it is unlike much larger and brighter G-type stars that fluctuate significantly in brightness over periods of days or weeks. The Sun itself will also become variable in its old age. Even today, the Sun is a little unusual in living alone, without a close stellar companion. This is not true for the majority of stars in the galaxy.

23.1 Binary starsUnlike the Sun, the majority of stars do not live alone in space. Most stars have one or more stellar companions, or are members of larger clusters of hundreds or thousands of stars.

The simplest possibility is a binary star system composed of two stars orbiting their common centre of mass. Each of the orbits is generally elliptical and this centre of mass point is the ‘balance point’ that always lies between the two stars (Figure 23.1.1). The balance is given by the equation:

m1r1 = m2r2where m1 and m2 are the two masses and r1 and r2 are the distance of each from the centre of mass. If we use r = r1 + r2 as the distance between the stars, then: m1r1 = m2(r – r1) = m2r – m2r1

Rearranging, we get: r1 = m2 rm1 + m2

If we make the two stars of equal mass, then m1 = m2 and therefore r1 is half of r — the balance point is half way between the two stars. If star 1 is more massive than star 2 (m1 > m2), then the balance point is closer to star 1.

Stellar companions and variables

408

23

Worked exampleQuestioNA planet orbiting the Sun is a similar situation to a binary star, but with a larger ratio between the two masses. Where is the centre of mass between the Sun and Jupiter?

soLutioNmSun = 1.99 × 1030 kg

mJupiter = 1.90 × 1027 kg ≈ 0.001 mSun

Use the equation:

rm

m m rSun

Jupiter

JupiterSun= +( )

Even without calculating exactly, we can see that, since Jupiter has only ~one-thousandth the mass of the Sun, rSun is only about one-thousandth of r. Knowing the size of the Sun (see in2 Physics @ Preliminary section 16.1) leads us to the conclusion that the balance point between the two is very close to the surface of the Sun. For smaller planets such as the Earth, the balance point is within the Sun. So, although we usually say that the planets orbit the Sun, this isn’t quite true.

Figure 23.1.1 The centre of mass of a binary system is the balance point between the two stars.

m2

m1

r2

r1

star2

focus 2focus 2

star1

centreof mass

orbit ofstar 1

orbit ofstar 2

Try ThiS!Make your own binary systeMSome of the principles of a binary star system are easily demonstrated with some balls and a rod.

Cut a hole through each ball so that it can be pushed onto the thin wooden or metal rod but not slide off easily. Rest the rod on your finger and slide it along until it is balanced horizontally. Tie a string to the balance point (the centre of mass) and suspend your ‘binary system’ by the string. Gently rotate the binary by giving one of the balls a push. Try a different ball and see how the centre of mass changes.

r1 r2

m2m1

Figure 23.1.2 The centre of mass of the binary system is the balance point.

The apparent size and shape of the orbit is interesting, but the real importance of binary stars is the possibility of determining their masses. Mass cannot be determined from spectroscopic information alone. We need to study the gravitational effect on another object, and a binary star system provides a laboratory in which to do this.

We can see how this works if we assume the orbit is circular and use equations from section 2.2. The centripetal force needed to maintain circular motion of star 1 around the centre of mass is provided by gravitational force between the two stars:

F F

Gm m

r

m v

r

gravity centripetal=

=1 22

1 12

1

Explain the importance of binary stars in determining stellar masses.

409

aSTrophySicS

We can include the period T of the orbit of star 1 by using vr

T112

=π

.This gives: Gm

r

r

T2

2

21

2

4=

π

Now we can insert the result for r1 derived earlier:

Gm

r T

m

m mr2

2

2

22

1 2

4=

+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

π

Rearranging this produces:

m mr

GT1 2

2 3

2

4+ =

π

orr

T

G m m3

21 2

24=

+( )π

This is a form of Kepler’s third law (section 2.2; see also in2 Physics @ Preliminary section 13.4) but involves the combined mass of both stars. It also requires r, the true distance between the stars, with allowance made for any tilt of the orbit relative to the plane of the sky. This means that the distance to the system must be known.

If r1, the distance from star 1 to the centre of mass can also be determined, then the masses m1 and m2 can be calculated individually by using a slight rearrangement of the earlier result for r1. The problem is that only rarely do we have enough information to do this.

Worked examplesQuestioNThe α Centauri system is 1.338 pc away. The A and B components orbit each other with a period of 79.92 years and an average distance of 23.7 AU. What is the total mass of the system?

soLutioN

m2 rm1 + m2

Try ThiS!binary statusIn binary or multiple star systems we generally call the brightest star the primary and assign it the letter A. The secondary star is B and any other stars in the system are C, D etc. Unlike the system in Figure 23.1.1, a binary system is often drawn centred on the primary, which is considered stationary. For example, in Figure 23.1.3 the system is centred on β Centauri A, with the very close companion β Centauri B in orbit around it. Notice that the orbit is elliptical, although not the same ellipse you would draw if the motions were represented relative to the centre of mass (as in Figure 23.1.1). However, even this elliptical orbit is only the apparent orbit on the ‘plane of the sky’. The ‘real’ elliptical orbit is tilted relative to the plane of the sky.Use m m

rGT1 2

2 3

2

4+ = πbut ensure the correct SI units are used.

m m1 2

2 12 3

11 9

4 3 55 10

6 672 10 2 522 10+ =

×( )×( ) ×( )−

π .

. .

m

s22

304 16 10

2 1

= ×≈

.

.

kg

solar masses

This is usually written with a special symbol for ‘solar mass’ as 2.1 M.

In fact the orbit is highly elliptical, but the result is still valid.

QuestioNMeasurements indicate that α Centauri A has an average distance of 11.2 AU from the system’s centre of mass. What is the mass of each star?


410

23soLutioN

Use rr

m

m m1 2

1 2

=+( )

1 68 103 55 10 4 16 10

12

122

30

.

. .

××

=×( )

mm kg

m

Therefore m2 = 1.97 × 1030 kg ≈ 0.99 M, leaving m1 = 2.19 × 1030 kg ≈ 1.1 M.

See the examples in Chapter 22 for further information on these two stars.

PHYsiCs FeAtuReβ CeNtAuRi

β Centauri has long been known as a visual binary with the two stars separated by more than 1 arc

second. Early observations with an interferometer suggested that the brighter star was itself a very close binary—much too close to be seen as separate through a conventional telescope.

Figure 23.1.3 shows the orbit of β Centauri B. Its orbit has been determined from high resolution observations using the Sydney University Stellar Interferometer (see section 21.5). The stars are very close, so angles on the sky are measured in milli-arc seconds (one-thousandth of an arc second). Combining this with spectroscopic data yields all the parameters of the system.

oRBit

Period: 357.00 ± 0.07 daysSemi-major axis: 25.30 ± 0.19 milli-arc secondsEccentricity: 0.821 ± 0.003 Inclination to the plane of the sky: 67.4 ± 0.3 degrees

DistANCe 102.3 ± 1.7 pc

PRimARY Mass: m1 = 9.1 ± 0.3 M

Absolute V magnitude: M1 = –3.85 ± 0.05

seCoNDARY Mass: m2 = 9.1 ± 0.3 M

Absolute V magnitude: M2 = –3.70 ± 0:05

The high accuracy of these results offers an important starting point for understanding the structure of the stars themselves.


CHeCkPoiNt 23.11 Draw a diagram of a binary system that shows the two stars, the centre of mass of the system and their elliptical

orbits around the centre of mass.2 Determine where the centre of mass is located between Jupiter and its moon Callisto.

MJupiter = 1.90 × 1027 kgMCallisto = 1.08 × 1023 kgAverage distance of Callisto from Jupiter = 1 883 000 km

N

E

–25

–20

–15

–10

–10 –5 0 5 10 15 20 25

–5

0

5

Pos

itio

n (m

illi-

arc

seco

nds)

Position (milli-arc seconds)

orbit of β Centauri B

β Centauri A

Figure 23.1.3 Observations and orbit of β Centauri B relative to β Centauri A

411

aSTrophySicS

23.2 Doubly differentAlthough most stars are in binary or multiple systems, they don’t all reveal themselves in the same way. We can classify the types of binaries according to how they are observed.

Visual binaries A visual binary (or multiple system) can be resolved into two (or more)

stars by a telescope under sufficiently good seeing conditions. Like α Centauri (Figure 23.2.1), many of the brighter stars are ‘doubles’ well known to amateur astronomers. A few are merely chance alignments of unrelated stars, but most are genuine binaries.

A simple calculation of angles will reveal that if a star is to be seen as a double in a telescope with angular resolution about 1 arc second, the stars must be at least tens of AU apart and the system must be relatively close. Such stars orbit each other slowly; but over many years the orbital motion may become apparent. More than 100 000 visual double stars are catalogued, but the orbits of only a few thousand are known.

Astrometric binariesSome stars are sufficiently close that their motion through space is apparent as motion across the sky, that is, as proper motion. A few of these stars reveal their binary nature by the wobbling of their paths across the sky (Figure 23.2.2). The centre of mass of such an astrometric binary follows a straight path, but the individual stars appear to wobble as they orbit.

Few binaries are discovered in this way, since it usually requires long-term observations of nearby stars. The Hipparcos astrometry mission (section 22.1) revealed many new examples, because its very high precision position measurements revealed the motions of stars much more quickly.

Spectroscope binariesMost binary systems are too distant to appear as either visual or astrometric binaries. However, the spectrum of an unresolved binary will have light contributed by both stars. As the stars orbit, one will typically have a component of its motion towards us, while the other is moving away. The light from the two stars will show small blue and red Doppler shifts that change as the stars orbit. Provided the stars are close in brightness, the result is a doubled-lined spectroscope binary in which each spectral line will appear doubled by these small shifts in wavelength (Figure 23.2.3). At other times, the motions of the stars may be entirely across the line of sight and show no Doppler shift and therefore display only a single set of lines.

If the stars are significantly different in brightness, the spectrum of the brighter star will flood the fainter one and only one set of moving spectral lines may be visible, forming a single-lined spectroscopic binary.

Describe binary stars in terms of the means of their detection: visual, eclipsing, spectroscopic and astrometric.

Figure 23.2.1 α Centauri A and B

Figure 23.2.2 Sirius is a visual and an astrometric binary.

Sirius B

Sirius A

centre of mass

1990

1980

1970

1960

1950

1940

1930

1920

1910

1900

a

B

B

B

B

A

A

A

A is moving towards the observer: lines blue-shiftedB is moving away from the observer: lines red-shifted

A and B are moving acrossthe observer’s line of sight:no Doppler shift

A is moving away from the observer: lines red-shiftedB is moving towards the observer: lines blue-shifted

A AAB B

A AB B AB

B

A & B A & B A & B

Figure 23.2.3 The changing pattern of spectral lines in a spectroscopic binary (very exaggerated)


412

23To be visible as a spectroscopic binary, the component of the orbital velocity

measured by the Doppler shift must be relatively large. This means that the stars must be close—much closer than most visual binaries, so they orbit quickly. Also, the orbit must be orientated so that the orbital motion has a component in the line of sight. This is not enough to really understand the system since, generally, the tilt of the orbit to the line of sight is unknown.

The brighter star of β Centauri is a singled-lined spectroscopic binary that has now been resolved by the SUSI interferometer (see Physics Feature p 410). This combination of spectroscopic and astrometric data is rare and valuable, since it allows all the parameters of the system to be derived, in particular the mass.

Eclipsing binaries If the stars of a binary system are close together and orientated so that the

orbital plane is close to edge-on, then it may be seen as an eclipsing binary. In these systems the stars regularly eclipse one another, periodically blocking out some of the light from the system. To be well enough aligned with our line of sight means that the stars are usually very close together, with orbital periods of just a few hours or days. They are therefore usually also spectroscopic binaries. They may be so close together that the stars are distorted by their mutual gravitational forces.

Figure 23.2.4 Algol (β Persei), a famous example of an eclipsing binary system, and its corresponding light curve

primaryeclipse

orbit ofsecondary

primaryeclipse

secondaryeclipse

primary

secondary1.61.82.02.22.42.62.83.03.23.43.6

Visu

al m

agni

tude

2.867321 days

activity 23.1

pracTicaL EXpEriENcES

activity Manual, Page 185

Figure 23.2.5 This artist’s impression of Circinus X-1 shows a binary system composed of an ordinary star losing material to the accretion disc around a neutron star.

The change in brightness is apparent in the light curve of the systems—a plot of the apparent magnitude of the system versus time (Figure 23.2.4) that repeats with every orbit. If the stars are different, the two eclipses will not be identical. A primary eclipse will result in a greater loss of light than the secondary eclipse, with the details determined by the tilt of the orbit, the relative size of the stars, their surface temperatures and even the structure of their atmospheres.

The importance of eclipsing binaries lies in the wealth of information we can glean from observations, including masses and distances.

Cataclysmic variables and X-ray binaries Some binaries are so close that they exchange

mass between the stars. When one of the stars is a compact stellar remnant such as a white dwarf, a neutron star or a black hole (see section 24.5), it may accrete gas from its companion. This gas releases gravitational potential energy, becomes very hot and emits high-energy radiation.

If the compact remnant is a white dwarf, its light output may vary dramatically as the system suffers one or more outbursts as a cataclysmic variable. If the remnant is a neutron star or black hole, it may be apparent as a source of X-rays from the infalling gas. It is then known as an X-ray binary (Figure 23.2.5).

413

aSTrophySicS

ThE maSS—LumiNoSiTy rELaTioNShip

One important product of our ability to determine the masses of some stars in

binary systems is the mass–luminosity relationship(Figure 23.2.6). This is a plot of the mass M of a main sequence star versus its luminosity L, which can be approximately fitted by the relationship L ∝ M 3.5. It indicates that the luminosity of a star increases very rapidly with its mass. As luminosity is based on the consumption of hydrogen fuel in the core of the star (see in2 Physics @ Preliminary section 15.4), this indicates that fuel consumption increases much more rapidly than fuel availability. The result is that high-mass stars on the main sequence have shorter lifetimes than low-mass stars.

1000 000

100000

10000

1000

100

10

1

0.1

0.01

0.0010.1 1 10 100

Mass (M/M )

Lum

inos

ity

(sol

ar lu

min

osit

ies)

L ∝ M3.5

Figure 23.2.6 The mass–luminosity relation for main sequence stars

CHeCkPoiNt 23.21 Recall the different types of binary systems and outline the characteristics of each type.2 Explain the significance of the mass–luminosity relationship.

23.3 Variable stars Eclipsing binaries and cataclysmic variables can also classified as variable stars. They vary in brightness because they are part of a close binary system. Other stars vary by themselves, whether part of a binary system or not, as a normal stage in their lives.

More than 30 000 variable stars have been catalogued, and many thousands more are suspected to be variable. In fact, all stars vary in brightness to some degree. The Sun varies by about 0.1% within an 11-year cycle because of the solar activity cycle (see in2 Physics @ Preliminary section 16.3). Also, rapid variations are associated with tiny oscillations in the Sun measured by helioseismology (see in2 Physics @ Preliminary section 16.2). These are both just part of the normal behaviour of a Sun-like star, and the Sun is not regarded as a variable star.

Other stars vary more in brightness and we track these changes with a light curve. There are many types of variable stars and the main groupings are shown in Figure 23.3.1.

Classify variable stars as either intrinsic or extrinsic and periodic or non-periodic.

Figure 23.3.1 Different types of variable star systems

GROUP CLASS TYPE

VARIABLESTARS

INTRINSICVARIABLE

EXTRINSICVARIABLE

PULSATINGSTARS

ERUPTIVE(cataclysmic)STARS

ECLIPSINGBINARIES

ROTATINGVARIABLES

Cepheids

RR Lyrae

RV Tauri

Long-periodvariables

SupernovaeNovaeDwarf novaeSymbiotic starsFlare starsR Coronae BorealisT Tauri stars

Type I Classical

Type II W Virginis

Mira type

Semiregular


414

23It divides variables stars into extrinsic variables and intrinsic variables. Extrinsic variables are perhaps less interesting, at least as variable stars, because their variation is due to a process external to the body of the star itself. Eclipsing binaries are the best example, although the group also includes stars that vary because of their rotation (e.g. the effect of ‘spots’ on their surfaces).

Intrinsic variables vary because of physical changes in the star or the stellar system. The HR diagram (see in2 Physics @ Preliminary section 15.3) summarises the properties of stars, and therefore different types of intrinsic variables occupy specific regions of the HR diagram (Figure 23.3.2).

Intrinsic variables can be split into non-periodic variables and periodic variables. As the names suggests, the difference is whether their variation repeats at reasonably regular intervals.

Non-periodic variables The non-periodic variables cover a wide range of

different types of stars that are physically very different, as illustrated by the systems described in Table 23.3.1.

Figure 23.3.2 The location of some types of variable stars on the HR diagram


10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence

Long-periodvariables

Miras

Semiregular

T Tauri stars

Flare stars

RR Lyrae

RV Tauristars

Type I

Classical Cepheids

Type II

W Virginis

Instability

strip

Table 23.3.1 Major classes of non-periodic variables

Variable type brightness change physical descriptionStars in binary systemsType I Supernovae ~20 magnitudes within hours, then

gradually fades over weeksAccretion of gas onto a white dwarf from its companion, leading to a runaway nuclear explosion

Novae Between 7 and 16 magnitudes, usually within a few days, then fades overs years to its initial brightness; some have been seen to brighten again

A close binary composed of a Sun-like star leaking gas onto a white dwarf that eventually will accumulate enough material to generate a surface nuclear explosion

Dwarf novae Between 2 and 5 magnitudes, repeating semi-regularly; different types behave differently

A close binary composed of a Sun-like star leaking gas onto a white dwarf; instability in the accretion of gas appears to produce the outbursts

Symbiotic stars Vary semi-regularly over a range of about 3 magnitudes

A close binary composed of a red giant and a white dwarf; outbursts from the red giant fall onto the white dwarf

Single stars (may be in a binary system)Type II supernovae ~20 magnitudes within hours, then

gradually fade over weeksCore collapse of a massive star blows off its outer layers

Flare stars Typically 1 or 2 magnitudes and fade within hours

Solar-like flares on the surface of faint red dwarfs contribute dramatically to the overall visible light output of the star

R Coronae Borealis Usually at maximum brightness but fade at irregular intervals by up to 9 magnitudes, returning to normal over months

Carbon-rich dust clouds obscure the surface of the yellow supergiant star

T Tauri Vary irregularly in brightness Very young stars with an accretion disc, still approaching the main sequence

415

aSTrophySicS

23.4 Cepheid variablesOf all the variable stars, the Cepheid variables are the most important, followed by their cousins, the RR Lyrae variables. Both serve as ‘standard candles’ in distance measurements, because their absolute magnitudes can be estimated from their pulsation properties.

The Cepheids are named for the prototype of the class, the northern naked-eye star δ (delta) Cephei. They vary regularly and have a characteristic light curve. As they oscillate in size they change in surface temperature and therefore spectral class (Figure 23.4.1).

Periodic variables Periodic variables show regular or semi-regular changes in brightness,

with periods ranging from hours to hundreds of days. The major types are shown in Table 23.3.2. Their brightness changes as the stars pulsate in size, surface temperature and colour.

The pulsation arises from a slight instability in the balance between the inward pull of gravity and the outward pressure of the gas and radiation (see in2 Physics @ Preliminary section 15.4). This instability only occurs in the outer layers of the stars and does not affect the energy production in the core. Conditions in pulsating stars are just right to allow the pulsations to continue instead of dying away, as they would in most stars. These conditions are found in stars in the ‘instability strip’ on the HR diagram, with another similar zone of ‘long-period variables’ among the red giants (Figure 23.3.2).

Table 23.3.2 Major classes of pulsating variables

Variable type brightness change physical descriptionCepheid Between 0.5 and 2 magnitudes with

periods from 1 to 70 days Luminous yellow supergiants

RR Lyrae Up to 2 magnitudes with periods of less than 1 day

Old giants stars with MV ≈ +0.6

RV Tauri Alternating deep and shallow minima with periods from 20 to 100 days

Yellow supergiants

Mira Between 2 and 10 magnitudes with periods from 80 to 1000 days

Red giants and supergiants

Semi-regular Up to 2 magnitudes with periods from 80 to 1000 days, but with irregularities

Red giants and supergiants

NamiNg of STarS

Star names and designations follow many different systems for

different types of stars. There are also many historical names and catalogues with designations still commonly in use. For example, the brightest star in the sky has many designations such as Sirius, α Canis Majoris, 9 Canis Majoris, HD 48915 and Hipparcos 32349 to name a few.Variable stars have one of the oddest systems. The first variable discovered in a constellation is called R, e.g. R Canis Majoris. Next comes, S, T, U … Z. Then what to do? We use RR, RS, RT … all the way to ZZ, and then AA, AB … to QZ. Finally, if we have more than 334 variables in a constellation, we start a sensible system with V335, V336 etc!Different classes of variable stars often take their name from the first one identified, for example R Coronae Borealis stars.

CHeCkPoiNt 23.31 Distinguish between extrinsic and intrinsic variable stars.2 Explain how the pulsation of periodic variables can tell us about

the structure of those stars.


416

23

In the early 20th century, Henrietta Leavitt (1868–1921) was studying Cepheids in the Magellanic Clouds, two satellite galaxies to the Milky Way. The Magellanic Clouds are relatively small compared with their distance, so all Cepheids in each Cloud are at essentially the same distance from us. The different apparent magnitudes of these stars therefore reflect true differences in luminosity (or absolute magnitude). With this simplification, Leavitt realised that Cepheids with longer periods were more luminous than Cepheids with shorter periods. This is the period–luminosity relationship for Cepheids (Figure 23.4.2).

It was discovered later that there are actually two classes of Cepheids. Type I (or classical) Cepheids are massive young stars crossing the instability strip as they evolve. Type II Cepheids (or W Virginis stars) are much older stars also crossing the instability strip.

The power of the period–luminosity relationship lies in the fact that period can be determined by following the brightness variation of any Cepheid we can see. Then its absolute magnitude can be estimated from the period–luminosity curve (Figure 23.4.2) and the distance determined by comparing the apparent and absolute magnitudes (see section 22.4). However, reality is always a bit more difficult; for example, interstellar dust can make the stars appear dimmer than they should.

Because they are supergiant stars, Cepheids can be identified in some nearby galaxies. As a result, Cepheid distances are a fundamental stepping stone in measuring much larger distances in the universe.

RR Lyrae variables are not as bright as Cepheids, but they are simpler since they all appear to have about the same luminosity (Figure 23.4.2). Once a star is recognised as an RR Lyrae variable, its distance can be determined by comparing its apparent and absolute magnitudes.

Explain the importance of the period–luminosity relationship for determining the distance of cepheids

RR Lyrae

Type I (Classical)Cepheids

Type II (W Virginis)Cepheids

104

103

102

1

0.5 1 3 5 10 30 50 100Period (days)

Lum

inos

ity

(L )

Figure 23.4.2 The period–luminosity relations for Cepheid and RR Lyrae variable stars. The green arrows indicate the luminosity relative to the Sun of a type I Cepheid that has a period of 10 days.

PERIOD

Bri

ghtn

ess

Time

Figure 23.4.1 Variations in the brightness, size and colour of a Cepheid variable during its pulsation

417

aSTrophySicS

CHeCkPoiNt 23.41 Recall why Cepheids and RR Lyrae variables are so important in astrophysics.2 Outline how Leavitt discovered the period–luminosity relationship.3 Describe the characteristics of the two types of Cepheids.

Worked examplesQuestioNζ (zeta) Geminorum is a Type I Cepheid variable star in the constellation of Gemini. It varies in brightness between 3.7 and 4.2 magnitudes every 10.2 days. Using the period–luminosity relation, estimate its absolute magnitude.

soLutioNUsing the arrows on Figure 23.4.2, a period of 10.2 days indicates a luminosity of approximately 3500L.

This can be converted to a magnitude difference using:

m mIIB A

A

B

− =

2 50 10. log

but working with absolute magnitudes. MV for the Sun is +4.85, so:

4.85 – MA = 2.50 log10 (3500)

and thus MA is –4.0.

QuestioNUsing ζ Geminorum’s average apparent magnitude, estimate its distance.

soLutioNIf we assume the average apparent magnitude is +3.95 then, using the relation between absolute and apparent magnitude:

3 95 4 0 51010. . log− −( )=

d

Thend10

107 95 5

= . and d ≈ 390 pc

This is close to the accepted value.

pracTicaL EXpEriENcES23 Stellar companions and variables

418

chapTEr 23This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

ACtiVitY 23.1: eCLiPsiNg BiNARies Go to the website given to observe the light curves of eclipsing binary stars. Record your observations and use them to interpret any light curves.

Discussion questions1 Identify the basic features of the light curve of an eclipsing binary.2 Sketch a light curve and explain the positions of the stars at specific points

of the curve.

Perform an investigation to model the light curves of eclipsing binaries using computer simulation.

chapter summary• Abinarystarsystemiscomposedoftwostarsorbiting

their common centre of mass.• Themajorimportanceofbinarystarsistheirusein

determining the masses of the stars.

• Kepler’sthirdlawintheformm mr

GT1 2

2 3

2

4+ =

π

involves the combined mass of both stars. With more information, the individual masses can be determined.

• Avisualbinary(ormultiplesystem)canberesolvedintotwo (or more) stars by a telescope under sufficiently good seeing conditions.

• Anastrometricbinaryrevealsitsbinarynaturebythewobbling of its path across the sky.

• Inadoubled-linedspectroscopebinaryeachspectrallinewill periodically appear doubled by the Doppler shift of the individual stars.

• Inasingle-linedspectroscopebinarythespectrumofthe brighter star will flood the spectrum of the fainter one and only one set of moving spectral lines will be visible.

• Ifthestarsofabinarysystemareclosetogetherandorientated so that the orbital plane is close to edge-on, it may be seen as an eclipsing binary in which the stars regularly eclipse one another.

• Thelightcurveofaneclipsingbinaryorothervariablestar is a plot of the apparent magnitude of the system versus time.

• Somebinariesarecloseenoughtoexchangemassbetween the stars, and become apparent as cataclysmic variables or X-ray binaries.

• Themass–luminosityrelationshipisaplotofthemassM of a main sequence star versus its luminosity L, which can be approximately fitted by the relationship L ∝ M 3.5.

• Variablestarsystemsvaryinbrightness.• Variationinextrinsicvariablesisduetoaprocess

external to the body of the star itself.• Variationinintrinsicvariablesisduetophysicalchanges

in the star or the stellar system.• Non-periodicvariablescoverawiderangeofdifferent

types of stars that are physically very different.• Periodicvariablesshowregularorsemi-regularchanges

in brightness due to pulsation.• CepheidvariablesandRRLyraevariablesserveas

‘standard candles’ in distance measurements.• Cepheidvariablesobeyaperiod–luminosityrelationship

that can be used to determine their distance.• RRLyraevariablesallappeartobeaboutthesame

luminosity, which can be used to determine their distance.

419

aSTrophySicSreview questions

ReViewiNg 1 Define what is meant by the term binary star.

2 Discuss why binary star systems are important in astronomy.

3 Draw a typical light curve that would result from an eclipsing binary system that consists of two identical stars.

4 Outline the characteristics that can be determined if two stars form a spectroscopic binary.

5 Explain why relatively few binary systems are discovered as astrometric binaries.

6 On the basis of the other types of binary systems described, propose an idea of why you might describe a system as an ‘interferometric binary’ system.

7 Define what is meant by the term variable star.

8 List examples of variable star types that would be called extrinsic or intrinsic.

9 List the properties of a star that vary in a pulsating variable star.

10 Construct a block diagram indicating the relationship between all the types of binary stars and variable stars described in this chapter.

11 Explain the properties that distinguish a Cepheid variable star.

12 Outline the importance of the periodic change in luminosity of Cepheid variables.

PHYsiCALLY sPeAkiNgComplete the passage below by filling in the missing words from this list:

Variable stars are stars that change ______________. This change can range from

0.001 to as much as 20 ______________ over ______________ of a fraction of a

second to years. There are a number of reasons why the brightness of variable

stars will change. ______________ swell and shrink due to internal forces, while a

star in an ______________ will dim when it is eclipsed by a faint ______________ and

then brighten when the occulting star moves out of the way.

Variable stars are classified as either ______________, when variability is caused

by physical changes such as ______________ or ______________ in the star or stellar

system, or ______________ when ______________ is caused by the ______________ of

one star by another or by the effects of stellar ______________.

brightness, companion, eclipse,

eclipsing binaries, eruption,

extrinsic, intrinsic, magnitudes,

periods, pulsating variables,

pulsation, rotation, variability

420

23 Stellar companions and variables

soLViNg PRoBLems 13 If a binary system consists of a 0.5 M red dwarf star and a 3 M red giant

star separated by 20 AU, calculate the location of the centre of mass.

14 Sketch the orbits around the centre of mass of the 0.5 M and 3 M stars in the previous question.

15 a Sirius, the brightest star in the sky is a binary composed of a main sequence star and a white dwarf. It is 2.631 pc away. The A and B components orbit each other with a period of 50.1 years and an average distance of 19.8 AU. Calculate the total mass of the system.

b Measurements indicate that Sirius A has an average distance of 6.5 AU from the system’s centre of mass. Calculate the mass of each star.

16 V Puppis is a very closely spaced eclipsing binary in the southern constellation of Puppis. Its light curve is shown in Figure 23.5.1. Estimate the period of the orbit and the amplitude of the eclipses, in magnitudes.

Figure 23.5.1 Light curve of V Puppis

4.2

4.4

4.6

4.8

5.0

–0.3 0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4

Time (days)

App

aren

t m

agni

tude

Solve problems and analyse information by applying: m m

rGT1 2

2 3

2

4+ = π

Revie

w Questions

17 β Doradus is a Type I Cepheid variable in the southern constellation of Dorado. It varies in magnitude from 3.46 to 4.08 in 9.94 days. Calculate its distance.

18 The star RR Lyrae is spectral class F5 with an average apparent magnitude of 7.1. Calculate its distance.

421

aSTrophySicS

PHYsiCs FoCus

Figure 23.5.2 Hubble images of the ‘light’ echo surrounding V838 Monocerotis

20 May 2002 2 September 2002

28 October 2002 17 December 2002

VARiABLes AND suPeRgiANtsby Jonathon Nally

For most of the 20th century, professionals and amateurs were separated by a wide gap in technology. Only the professionals could afford the large telescopes and complex support gear required for modern astronomy and astrophysics. That gap has now closed to some degree. Amateurs now have affordable access to sophisticated telescopes, electronic detectors, software, and all the other paraphernalia needed to contribute to many areas of astronomy.

But it’s great to see that there’s still a place for the good old human eye. And this is nowhere more evident than in the field of variable star research.

Some variable stars change brightness over a period of days, weeks, months or even years. Many change their brightness with perfect reliability, while others are totally unpredictable. This means that constant, long-term monitoring is needed, which is the sort of thing that professional astronomers really are unable to do—but it’s a job perfectly suited to amateurs.

Every night, all over the world, scores of dedicated variable star observers are outside, making observations and brightness readings of hundreds of variable stars. They tabulate their data and send them in to one of several umbrella organisations scattered around the world. The data is then made available to the professionals who, it cannot be stressed enough, rely so heavily on this work by the amateurs.

Usually, it takes many years of patient monitoring for enough data to be built up to make a difference in research programs. But sometimes things can happen much, much faster.

On the night of 6 January 2002, Aussie amateur astronomer Nick Brown spotted something unusual. A certain star was much brighter than it should have been—V838 Monocerotis had gone into outburst. It was a nova, and Nick was the first person to spot it.

He quickly reported it, and it came to the attention of a bunch of professional astronomers who were interested in just this particular kind of star. For the past year, this team has been studying V838, and

have found it to be very unusual indeed. It had expanded into a supergiant star over a period of just a few months, brightening three times before fading dramatically. At first they thought the fade was caused by obscuring dust, but now it turns out that the star has actually cooled down. In fact, it has become the coolest supergiant star ever found and all because of a single observation made by a dedicated amateur astronomer in Australia.

Source: Transcript from ABC Science Show, Radio National, 10 May 2003

1 Outline the advantages of an electronic detector over the human eye for variable star observation.

2 Describe why it takes years of observations of a variable star to build a useful data set. Is this true of all types of variables?

eXteNsioN3 V838 Monocerotis is now well known because of

images taken by the Hubble Space Telescope (Figure 23.5.2) showing a ‘light echo’ from the outburst. Define the term light echo.

4 Recount what has happened to V838 Monocerotis since the time of the radio broadcast in 2003.

5 Propose some other fields of astronomy in which you think amateur astronomers can make valuable contributions.

422

Birth, life and death24

interstellar medium, nebula, emission nebulae, dark nebulae,

reflection nebulae, giant molecular clouds, protostar, zero-age main

sequence, accretion disc, jets, proton–proton chain, carbon–nitrogen–oxygen

cycle, red giant, triple alpha process, helium flash, horizontal branch (HB),

asymptotic giant branch, planetary nebula, supernova, supernova remnant,

neutron star, black hole, pulsar

Stellar evolution: How do we know?Astronomers talk about the ‘evolution’ of a star, but not quite in the sense a biologist uses the word. Astronomers mean that an individual star transforms itself during its lifetime as it consumes its nuclear fuel. However, stars have also evolved over generations of birth and death. Stars born today are somewhat different from stars born closer to the time of the Big Bang.

How do we know about the evolution of the stars? It’s a little like biology in that we almost never see evolution happening in our lifetimes, but the evidence is there to be found. Some of the ancient stars are still around to see today.

24.1 The ISM Within the Milky Way galaxy, the space between the stars is not quite

empty. It is filled by a patchy medium of gas and dust called the interstellar medium (ISM). The medium is visible as a nebula (meaning cloud), when it interacts with starlight. In visible light images such as Figure 24.1.1, we see three types of nebulae:• brightemission nebulae in which the gas is energised by hot young stars to

emit an emission spectrum (see section 22.2)• dark nebulae in which the dust scatters starlight, reddening or completely

blocking our view of background stars• reflection nebulae in which we see the light scattered by the dust, especially

at blue wavelengths.

The importance of gas and dust in our galaxy is apparent by looking at its distribution in other similar galaxies (Figure 24.1.2), although it makes up only about 1% of the mass of all the stars.

Figure 24.1.1 The Trifid nebula, composed of an emission nebula (pink) and a reflection nebula (blue), crossed by dark dust lanes

423

astrophysics

The gas is about 70% hydrogen and 28% helium (if measuring mass), with traces of other elements and molecules. This is spread through several different components of the ISM with very different properties (Table 24.1). Of particular interest are the cold giant molecular clouds in which the gas is most dense, and simple molecules can form. Individual molecular clouds have masses up to millions of solar masses. They make up only about 1% of the volume of the ISM but contain 90% of the mass and are the sites of star formation.

Where massive stars have formed in a giant molecular cloud, their intense ultraviolet (UV) radiation eats into the cloud and creates an emission nebula. Elsewhere the embedded dust may be revealed by dark nebulae and reflection nebulae.

The dust is just a fraction of a millimetre in size and represents about 1% of the mass of the gas. Its characteristics vary with location, but it seems to consist of silicate or carbon grains, sometimes with a coating of various ices. At least some of it is formed in the cool outer atmospheres of red supergiant stars and blown outwards by the star’s stellar wind.

The ISM is part of a giant recycling system that includes the stars (Figure 24.1.3). Gas from the ISM forms stars, is processed by nuclear reactions inside the stars, and some is returned to the ISM when the largest stars age and die.

Figure 24.1.2 IR image of the galaxy M81, with the spiral arms traced by emission from dust

Figure 24.1.3 The galactic recycling system

nuclear fusionin stars

neutral atomclouds

ionised clouds molecular clouds

ISM

stellar outflowsand explosions

star formation

Table 24.1.1 Major components of the ISM, with the air around you included for comparison

Component temperature (K) Density (atoms per m3)Atmosphere at Earth’s surface 300 ~1025

Molecular clouds 20–50 109–1011

Neutral atomic clouds 50–150 106–109

Partly ionised intercloud medium 103–104 ~104

Highly ionised coronal gas 105–106 102–103

CHeCkpoInT 24.11 Recall the three types of nebulae and their characteristics.2 State the main components of ISM.3 Outline the significance of giant molecular clouds.

24.2 Star birth The first step in the recycling scheme is star formation. If a part of a giant

molecular cloud is sufficiently cool and dense, it can be pushed into gravitational collapse. Likely triggers include the shockwave from the explosion of a nearby star, or the ‘density waves’ that sweep through the disc of the galaxy.

Once the trigger has boosted the density sufficiently, gravity takes over and that piece of the cloud starts to collapse because of the gravitational forces between its own particles. Initially the gas falls freely inward and the density

Describe the processes involved in stellar formation.

Birth, life and death

424

24increases quickly at one or more centres to form cores with more slowly collapsing envelopes.

As each core collapses, the infalling gas releases gravitational potential energy (see in2 Physics @ Preliminary section 4.1). Some of this energy is converted to kinetic energy and heats the gas while the rest is radiated as infra-red light. As each core gets hotter, the gas pressure increases, slowing the collapse of the core while the surrounding gas continues to fall inwards. The collapsing fragment of the cloud is now a protostar, probably one of many, perhaps destined to form part of a binary system or even an entire cluster of stars. It is a luminous source of infra-red light tracing a path on the HR diagram (Figure 24.2.1), but buried deep in the larger cloud.

As the collapse continues, the core temperature rises, eventually passing 10 million K and initiating the nuclear fusion of hydrogen into helium (see in2 Physics @ Preliminary section 15.4)—the star ‘turns on’! The increased energy production slows and eventually stops the collapse of the core. Lower mass stars like the Sun go through a highly active phase as a T Tauri variable star, but eventually settle into a stable

Figure 24.2.1 Evolutionary tracks for protostars of various masses


10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence

104 years

105 years

106 years

107years

15M°

9M°5M

°3M

°

.

1M°

0.5M°

ZAMS

Discs anD jets

Surrounding the protostar is an accretion disc that forms naturally as the protostar collapses and

spins faster. Infalling material accumulates in the disc, but it is accompanied by jets of outflowing material—a situation often seen in astronomy. When the stars ‘turns on’, the system is swept clean by radiation from the star and may reveal a planetary system that formed within the disc.

protostar

jet

jet

accretion disc

outflow

inflow

Figure 24.2.2 The accretion disc and jet surrounding a protostar

equilibrium between the inward pull of gravity and the outward force of gas and radiation pressure. The star is ‘born’ on the zero-age main sequence (ZAMS) of the HR diagram, a line along the lower edge of the broader main sequence band on Figure 24.2.1.

The time taken to reach this stage is a function of the mass of the collapsing fragment of cloud. More massive protostars heat up more quickly and begin nuclear reactions sooner. Very low mass stars will reach the ZAMS as red dwarfs long after their higher mass counterparts have lived their entire lives and died! Objects with mass below about 0.08M form brown dwarfs and will never fuse hydrogen into helium.

Just how many stars form in each mass range is uncertain and undoubtedly varies. However, for every one 10 M star, there are typically about 300 stars like the Sun formed and many more M-type red dwarfs.

425

astrophysics

24.3 Stars in the prime of life A place on the ZAMS is a starting point for the main part of any star’s life.

That place, how long the star took to get there and everything that happens afterwards are determined almost entirely by the mass of the star. This arises because mass determines the gravitational forces within the star and therefore the pressure that must resist those forces if the star is to be stable. The pressure is generated by the high temperatures inside a star as a result of the energy produced by the nuclear reactions in its core. So the stability of the star depends on it having enough nuclear fuel available and processing it fast enough.

Stars change very slowly as they age, so they spend most of their lives on the main sequence, not far from the ZAMS, and that is where most stars are found in the HR diagram. How long they can maintain this peaceful existence depends on how much hydrogen fuel they start with and how quickly they consume it.

The mass–luminosity relationship for main sequence stars (section 23.2) tells us that high-mass stars spend their fuel reserves recklessly. Despite their larger supply, more massive stars have shorter lives on the main sequence before their fuel runs low and they need to evolve (Table 24.3.1). In comparison, the Sun is a G2 star with an expected main sequence lifetime of about 11 billion years.

Table 24.3.1 Properties of main sequence stars relative to the Sun

speCtral Class

mass (m)

luminosity (l)

surfaCe temperature (K)

raDius (r)

time on main sequenCe(million years)

O5 40 400 000 40 000 13 1.0

B0 15 13 000 28 000 4.9 11

A0 3.5 80 10 000 3.0 440

F0 1.7 6.4 7 500 1.5 3 000

G0 1.1 1.4 6 000 1.1 8 000

K0 0.8 0.46 5 000 0.9 17 000

M0 0.5 0.08 3 500 0.8 56 000

To understand why the high-mass stars are so rash with their fuel, compared to their more frugal lower mass cousins, we need to consider the two processes at work in main sequence stars converting hydrogen into helium: the proton–proton chain and the carbon–nitrogen–oxygen cycle (see in2 Physics @ Preliminary section 15.4). In both cases the net reaction is a combination of four hydrogen nuclei (protons) into one helium nucleus with the release of energy and some other light particles. Energy is available because the mass of the four hydrogen nuclei is more than that of one helium nucleus. The lost mass is converted to energy according to Einstein’s famous equation E = mc2.

Describe the types of nuclear reactions involved in Main Sequence and post-Main Sequence stars.

Discuss the synthesis of elements in stars by fusion.

CHeCkpoInT 24.21 Outline some of the events that trigger the birth of stars.2 Describe a protostar.3 Describe the ZAMS.4 Recall the forces that are in balance when a star is in equilibrium.


426

24

protonneutronpositronneutrinogamma ray

11H

11H

21H

32He

32He

42He

21H

11H

11H

11H

11H

11H

11H

12C recycled

11H

11H

11H

11HN13

7

C136

C126

C126

N157

N147

O158

He42

positronneutrinogamma ray

Figure 24.3.2 A series of reactions starting with carbon make up the carbon–nitrogen–oxygen cycle.

Figure 24.3.1 A series of reactions must occur to form helium in the proton–proton chain.

CNO energy production is negligible below about 13 million K, but dominates proton–proton energy production by 20 million K. The energy production rises dramatically with temperature (proportional to ~T 20!), with some other reaction sequences also possible.

The enormous increase in reaction rate produced by small changes in temperature, especially in the CNO cycle, explains why high-mass stars expend the reserves of hydrogen in their cores so quickly. It also explains the differences in later evolution of low- and high-mass stars. These different evolutionary paths are illustrated in Figure 24.3.3 and described in the following sections.

As it ages, a main sequence star builds up helium in its core. This dampens the reaction rate, leading to a loss of pressure in the core, which responds by contracting a little. This releases gravitational potential energy that raises the temperature and therefore boosts the reaction rate. As the reaction is so sensitive to temperature, the overall effect is to increase the energy output and inflate the whole star. This moves the star on the HR diagram and gives the main sequence some width (Figure 24.3.4).

In cooler stars like the Sun, the dominant reaction is the proton–proton chain (Figure 24.3.1). Several pathways are possible but, for core temperatures from 10 to 14 million K, the main reaction sequence is:

11

11

12

12

11

23

23

23

24 2

H H H e

H H He

He He He

+ → + +

+ → +

+ → + ×

+ ν

γ

111H

The reaction emits e+ (positrons, i.e. positive electrons), ν (neutrinos—elusive particles with very small mass) and γ-rays. At temperatures below 10 million K, this reaction barely works. At higher temperatures the energy production rises quickly (proportional to ~T 4).

In more massive main sequence stars, the core temperature is even higher and the carbon–nitrogen–oxygen (CNO) cycle takes over as the dominant reaction. This process also produces helium from hydrogen but uses carbon nuclei as a catalyst, as illustrated in Figure 24.3.2.

612

11

713

713

613

613

11

714

714

C H N

N C e

C H N

+ → +

→ + +

+ → +

+

γ

ν

γ

NN He O

O N e

N He C He

+ → +

→ + +

+ → +

+

11

815

815

715

715

11

612

24

γ

ν

427

astrophysics

Figure 24.3.3 A summary of the evolution of single stars of various masses. Gas is returned to the ISM as raw material for future stars.

Significant mass loss via stellar wind

Major mass loss via stellar wind

Major mass loss via stellar wind

Dramatic mass loss, including heavy elements via supernova remnant

Main sequence star H fusion in core via

CNO cycle

Red supergiant He fusion in core via triple-α process,

H fusion in shell

AGB red supergiant C fusion in core, He, H fusion

in shells, finally multiple shells of heavy element fusion

Black hole M ≥ 3M°

No fusion reactions

Type II supernovaproduces heaviest elements

Neutron star1.4M

° ≤ M ≤ 3M

°No fusion reactions

High mass 8M°

≤ M Low to medium mass 0.08M

° ≤ M ≤ 8M

°

Very low massM ≤ 0.08M

°

Main sequence star H fusion in core via p–p chain and CNO cycle

Brown dwarf No ongoing fusion

Small mass lossvia stellar wind

Larger mass lossvia stellar wind

Increasing mass lossvia stellar wind

White dwarf M ≤ 1.4M

°

No fusion reactions, slowly cooling

Planetary nebula

AGB red giant Possible C fusion in core, He fusion in shell, H fusion in shell

Red giantHe fusion in core via triple-α process, H fusion in shell

Significant mass loss via planetary nebula shell

ionised clouds molecularclouds

STAR FORMATION

neutral atom clouds

ISM

climate preDiction: getting hotter!

The Sun is 4.6 billion years into its main sequence lifetime. It is now about 5% bigger

and brighter and 200 K warmer at the surface than it was when it landed on the ZAMS. Over the remaining 6 billion years of its main sequence life, the Sun will gradually double in luminosity and get 25% larger. This is a small change on the HR diagram (Figure 24.3.4), but one that will dramatically alter the Earth, eventually boiling away its oceans.


10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence (V)

Supergiants (I)

Giants (II, III)

Subgiants (IV)

Whitedwarfs

(VII)

2 million years

30 million years

180 million years

640 million years

11 billion yearspresent sun4.6 billion years

56 billionyears

initial 10M°

initial 5M°

initial 3M°

.

initial 30M°.

initial 1M°

initial 0.5M°

ZAMS

CHeCkpoInT 24.31 Explain why the lifetime on the main sequence is different for stars of different mass. 2 Write the net nuclear reaction equations that summarise the proton–proton chain and CNO cycle.3 If the CNO cycle produces negligible energy at 13 million K, explain why it can dominate energy production at

20 million K.

Figure 24.3.4 The motion of a star on the HR diagram during its life on the main sequence


428

2424.4 Where to for the Sun?

Eventually a main sequence star runs very low on hydrogen fuel in its core. For a star like the Sun, there is no mixing of new gas into the core and no increase in the reaction rate can counteract the lack of fuel. The core collapse accelerates, raising the temperature as gravitational potential energy is released, and igniting hydrogen fusion in a shell around the helium core.

Computer modelling shows that the outer layers of the star respond by expanding outwards and cooling. The plotted position on the HR diagram rapidly moves to the right across the sub-giant range (Figure 24.4.1). The luminosity of the star then begins to grow and the outer layers inflate further as the star climbs the red giant branch. About 1.3 billion years after leaving the main sequence, the G class V main sequence Sun becomes a K or M class III giant, about 200 times its current size.

The star now has a dense core that is perhaps one-third its original size, but the grossly extended outer layers reach out to nearly envelope the orbit of the Earth. These outer layers are of very low density and only weakly held. As a main sequence star, the Sun looses mass at a rate of ~10–14 M per year as solar wind (see in2 Physics @ Preliminary section 16.4). As a red giant, the mass loss will be perhaps 107 times larger, and it will simply blow away a significant fraction of its mass.

In the core of the red giant the temperature eventually rises to about 100 million K and the fusion of helium into carbon begins via the triple alpha process:

24

24

48

48

24

612

He He Be

Be He C

+ → +

+ → +

γ

γ

Adding another 24He can produce 8

16O.For stars of less than about 2.6 M, the onset of helium fusion occurs as the

helium flash, since conditions in the cores of these stars cause the entire core to begin helium fusion almost at once. Slightly more massive stars begin helium fusion a little more sedately, because the physical state of the core is a little different.

We believe that less massive stars that begin life as red dwarfs probably never go through a red giant phase. They never attain the temperatures needed to fuse helium. It’s hard to be sure, as their time on the main sequence is much greater than the current age of the universe and so there are no evolved red dwarfs to look at.

In stars that become red giants, the new energy supply from helium fusion causes a reduction in the size of the star and the surface temperature begins to rise again. Stars fusing helium in their cores, with hydrogen fusion in a surrounding shell, appear as smaller yellow G- and K-type giants.

The Sun may spend 100 million years in this horizontal branch (HB) phase, but its evolution is accelerating. Helium fusion occurs much faster than the hydrogen reactions and, all too quickly, the helium in the core is consumed and the core must again begin to collapse and heat. The outer layers expand again and the star moves up the asymptotic giant branch (AGB) in the HR diagram. The star now has a core of carbon and oxygen surrounded by shells of helium and hydrogen fusion. In higher mass stars, other elements are formed by

Outline the key stages in a star’s life in terms of the physical processes involved.

429

astrophysics

nuclei capturing neutrons. The Mira-type long-period variable stars are AGB giants.

After a further 20 million years these shells become unstable and begin to rapidly switch on and off, because the helium reaction is spectacularly dependent on the temperature (proportional to ~T 40!). The star has been losing mass throughout the giant phases, but now the pulsations gently puff off the outer layers, exposing deeper layers of the star.

The intense UV light from the remains of the star lights up the ejected gas as a planetary nebula (Figure 24.4.2). The nebula returns gas to the ISM that has been enriched in carbon, nitrogen and oxygen by its passage through the nuclear fires of the star.

At the centre of the planetary nebula, the remains of the star contract and heat up to about 100 000 K on the surface, as the final fusion of hydrogen occurs in a shell around the carbon–oxygen core. The fusion quickly ends and the star, now a white dwarf, begins a long, slow cooling that will last for tens of billions of years. With surface temperatures of about 10 000 K, but a size comparable to that of the Earth, the white dwarfs are small and faint and appear near the bottom of the HR diagram (Figure 24.4.1).

The star has reached its stellar graveyard with only about half the mass it had on the main sequence and having only consumed about 12% of its original supply of hydrogen fuel. Half its mass has been returned to the ISM, somewhat enriched in heavier elements, as raw material for the next generation of stars.


10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence (V)

Supergiants (I)

Giants (II, III)

Subgiants (IV)

White dwarfs (VII)

initial 5M°

initial 1M°

initial 0.5M°

planetary nebulaAGB helium

ignition

heliumflash

RGB

AGB

HB

HB

He → CH → He

He → CH → He

H → He

H → He

H → He

RGB

no nuclearreactions

Figure 24.4.2 An example of planetary nebulae

Figure 24.4.1 Evolution of stars of low to medium mass after they leave the main sequence

Explain the concept of star death in relation to:• planetarynebula• whitedwarfs.

pHYSICS FeATUReWHen IS A STAR noT A STAR?

What is a star? A sensible definition would say that a star produces energy from nuclear reactions in its core and is stable because of

the balance between the inward pull of gravity and the outward forces of gas and radiation pressure. By that standard, a white dwarf is not a star!

A white dwarf is a glowing ember. There are no nuclear reactions to generate energy, despite a core temperature as high as 10 million K. The energy trapped within the white dwarf from its energetic past slowly leaks out through its relatively small surface area.

With as much as 1.4M of material in an object the size of the Earth, the density of a white dwarf is enormous—about 109 kg m–3— a million times the density of water. Gas pressure cannot support the white dwarf against the correspondingly enormous gravitational force. The balancing force comes from a new physical effect: degenerate electron pressure. Essentially, the electrons reach a state in which they refused to be packed more tightly.


430

24CHeCkpoInT 24.41 Outline the stages in the evolution of a solar mass star from main sequence star to white dwarf.2 Describe the processes by which a star loses sufficient mass to end its life as a white dwarf.3 Recall the characteristics of a white dwarf.

24.5 The fate of massive starsFor any star the basic problem is the same throughout its life. To be stable, it must generate enough energy to ensure that internal pressure balances the gravitational force. In a massive star, this means higher core temperatures to generate the energy required and a correspondingly shorter lifetime.

For stars that start life with more mass than the Sun, but still less than about 8 M (the middle of spectral class B), evolution is faster than for lower mass stars (Figure 24.4.1), but they have the same fate in store. They lose more mass via stellar winds and all end up below the 1.4 M limit for the mass of a stable white dwarf.

Stars starting life more than about 8 M evolve much more quickly and have a different ending (Figure 24.3.3). Their main sequence lifetimes are measured only in millions of years (Table 24.3.1). Even at this stage, their stellar winds lead to significant mass loss (perhaps 10–6 M per year for a 60 M star). Their evolutionary path is flatter on the HR diagram (Figure 24.5.1), indicating that their size changes dramatically but their luminosity does not. These stars become supergiants, perhaps 1000 times the size of the Sun. They cross back and forth across the instability strip, becoming Cepheid variables for part of their lives.

In their cores, the onset of helium burning is gradual, but as the temperature increases to more than 300 million K the fusion of carbon with helium to produce oxygen becomes the dominant reaction. When the temperature climbs above 500 million K, carbon nuclei can fuse together to produce sodium, neon and magnesium. All the carbon is consumed quickly and the core collapses further, pushing the temperature ever higher. Ultimately, fusion of silicon produces iron at core temperatures of 7 billion K.

The core of the supergiant resembles the layers of an onion in which an iron core is surrounded by shells in which silicon and sulfur, oxygen and carbon, helium and hydrogen are all undergoing fusion reactions (Figure 24.5.2). However this is a fleeting phase. A 25 M star will fuse hydrogen on the main sequence for about 7 million years. It then spends 500 000 years also consuming helium, 600 years consuming carbon, half a year consuming oxygen and just a day consuming silicon.

Synthesis of iron is the limit of normal fusion processes, since any further reaction costs the star energy rather than generating it, accelerating its collapse. Within seconds a

Describe the types of nuclear reactions involved in Main Sequence and post-Main Sequence stars.

Discuss the synthesis of elements in stars by fusion.


10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence (V)

Giants (II, III)

Subgiants (IV)

Whitedwarfs

(VII)

Supergiants (I)initial 10M

°

heliumignition

supernova

HB

AGB

RGB

He → CH → He

H → He

Figure 24.5.1 Evolution of high-mass stars after they leave the main sequence

431

astrophysics

portion of the core the size of the Earth collapses to just a few kilometres across, achieving densities of 1017 kg m–3. The speed of collapse causes the core to bypass the electron degeneracy pressure that supports a white dwarf. At 1012 K, degenerate neutron pressure finally halts the collapse at the centre; however, the surrounding layers hit the core and bounce back, blowing the star apart in a supernova explosion. In a few seconds 1046 J of energy is released, 99% of it as elusive neutrinos. Just 1% of the energy appears as visible light, but this is sufficient for it to outshine all other stars in the galaxy for a few days!

During the explosion, iron nuclei are ripped apart and a large number of neutrons are released. These neutrons can be captured by existing heavy nuclei to produce heavier nuclei than could be produced by any process earlier in the star’s history.

The blast wave and several solar masses of gas ejected by the supernova explosion speed outward at several thousand km s–1, sweeping up gas blown away earlier by the stellar wind (Figure 24.5.3). The result is a supernova remnant (Figure 24.5.4), glowing across the full range of wavelengths and carrying gas enriched in a full range of heavy elements back into the ISM. After perhaps 100 000 years, the expanding remnant fades and merges into the ISM, but it may have already triggered a new burst of star formation nearby.

Back at the site of the explosion, some of material very quickly falls back onto the core of degenerate neutrons. Computer simulations remain very uncertain, but they indicate that stars with initial masses of between 8 M and roughly 25 M will produce a core of less than 3 M. The core will then be stable as a neutron star. Much rarer stars with initial masses of between roughly 25 M and 40 M will have more material falling back. The core mass will exceed the 3 M limit of stability for a neutron star. No known force will stop the collapse and the result is the formation of a black hole. Computer simulations suggest that even more massive stars may form a black hole before an explosion can occur!

Explain the concept of star death in relation to:• supernovae• neutronstars/pulsars• blackholes.

H shellHe shell

C, O shellSi, S shell

coreregion

supergiant star

Fe core

Figure 24.5.2 The shell structure in the core of a high-mass star just before it explodes

Figure 24.5.4 The 14 light-year diameter remnant from Kepler’s 1609 supernova is shown in this image assembled from images taken by the Hubble, Spitzer and Chandra space telescopes.

Figure 24.5.3 The expanding remnant from supernova 1987A brightens in its radio emission as the outrushing gas crashes into the surrounding medium.

1994.4 2000.9

2005.2 2008.3


432

24

neUTRon STARS: pUlSARS

A neutron star packs a mass about that of the Sun into an object

typically only 10 km across. As the core shrinks to this size, the rotation of the neutron star increases to a rate that is typically dozens of times a second! The acceleration due to gravity at the surface is about 1012 m s–2 (compared with 9.8 m s–2 at the surface of the Earth) and its magnetic field near the surface is about 108 T!

Neutron stars were first proposed in the 1930s, but no-one expected to be able to see such tiny objects. In 1967 radio emission was detected from a rapidly pulsing source that was quickly recognised to be a rotating neutron star—a pulsar.

Pulsars produce two beams of radiation from near their magnetic poles. As these poles are usually not aligned with the rotation axis, these beams sweep around the sky as the pulsar rotates rapidly (Figure 24.5.5). As it spins, the beams may sweep across the Earth, to be seen as a pulse. The pulses arrive at a

very precise rate that reduces only very slowly as the pulsar loses energy.

A few pulsars have been observed at visible, X-ray and γ-ray wavelengths, but most of the 1700 known pulsars have been discovered by radio telescopes. Many of these, including the first binary pulsar system, have been found by the 64 m Parkes radio telescope (Figure 21.4.1).

pHYSICS FeATURe

BlACk HoleS

Black holes are even more bizarre than neutron stars. With no known force able to resist the

collapse, all the mass is concentrated at a point and surrounded by an intense gravitational field.

They are ‘black’ because not even light can escape once it has crossed a theoretical boundary called the ‘event horizon’. At the event horizon, the escape velocity (see section 1.3) is the speed of light.

Black holes have many intriguing physical properties, both within the event horizon and outside it. Many of these do not directly affect their behaviour as astronomical objects, as their results cannot be observed. Others enable us to detect the hole, because of the effect on surrounding matter, despite the tiny size and ‘blackness’ of the hole.

A black hole in a close binary system may draw gas off its companion. The gas will be heated as it falls onto an accretion disc (something like that around a protostar) around the black hole. The gas emits high-energy radiation at UV, X-ray and γ-ray wavelengths.

Stellar mass black holes are thought to range in mass from 3M to about 15M. A supermassive black hole of 3 to 4 million M is be believed to mark the centre of the Milky Way galaxy, and even larger black holes are thought to be present in some distant galaxies.

Figure 24.5.6 An artist’s impression of an accretion disc around a black hole with jets of emission

magnetic axisrotation axis

possibleline ofsight toEarth

beamedradiation

magnetic field

beamedradiation

neutron star

Figure 24.5.5 The ‘lighthouse’ model sweeps beams of radiation around the sky as the pulsar rotates.

433

astrophysics

CHeCkpoInT 24.51 Outline what happens in a supernova explosion.2 If iron is the heaviest element made through normal fusion reactions, explain how heavier elements are made.3 Outline the properties of a neutron star.

24.6 How do we know?How do we know that the picture of stellar evolution presented in this chapter is correct, when almost every step takes far longer than a human lifetime?

Our understanding is based on computer modelling of the structure of stars. This takes the physical laws we know and carefully applies them to the situation inside stars. Comparison of our models with reality gives us confidence that our understanding of stellar structure and evolution is largely correct.

A key set of evidence lies in the HR diagrams of star clusters. Open (or galactic) star clusters typically contain a few hundred relatively young stars in a loose grouping about 10 light-years across (Figure 24.6.1). In contrast, a globular star cluster usually contains hundreds of thousands of old stars in a sphere about 100 light-years across (Figure 24.6.2).

Stars in an open cluster are expected to have formed from a single cloud with the same initial composition. We therefore expect a well-defined ZAMS line to mark the birth of the cluster stars on an HR diagram. Plotting that diagram is made easier since all the stars are at essentially the same distance from us, so the apparent magnitudes of the stars reflect their true relative luminosity.

Explain how the age of a globular cluster can be determined from its zero-age main sequence plot for a HR diagram.

Figure 24.6.1 The open star cluster M44 Figure 24.6.2 The globular star cluster M92


434

24

ZAMS

12b

14

16

18

20

220 0.4 0.8 1.2 1.6 2.0

V

–2

0

2

4

6

B – VB – V

V

4.0

6.0

8.0

10.0

12.0

14.0

0.0 0.4 0.8 1.2 1.6 2.0

ZAMS

a

Figure 24.6.3 Colour index versus apparent magnitude for stars in the (a) M44 cluster and (b) M92 cluster

pHYSICS FeATUReA FInAl THoUgHT

Simon Newcomb (1835–1909) was a Canadian-American astronomer and mathematician who,

in the late 19th century, said: ‘We are probably nearing the limit of all we can know about astronomy.’

One hundred and twenty years later his comments seem a bit premature, since much of our modern

understanding of the stars was unknown to Newcomb. We should take this as a warning not to think that our impressive knowledge today is complete. There is much we don’t understand and probably even more we don’t know anything about. That’s what makes astrophysics so exciting!

CHeCkpoInT 24.61 Explain how the HR diagrams of star clusters support our understanding of stellar evolution.2 If black holes emit no light, explain how we can observe them.

activity 24.1

practical eXperiences


Figure 24.6.3a is an HR diagram for the open cluster M44. The line on the diagram estimates the ZAMS. Most of the stars are clearly still on the main sequence. However, stars at the high-mass end are starting to evolve away from the main sequence, and a few more-evolved supergiants are apparently also present. This is consistent with our expectations that stars move off the main sequence to the right on the HR diagram and massive stars evolve most quickly.

Figure 24.6.3b is a rather different HR diagram for the globular cluster M92. The ‘turn-off point’ is much lower down the main sequence, and a well-developed giant branch is apparent. The turn-off point is an indicator of the age of the cluster; the further down the main sequence the turn-off point is, the older the cluster. The M44 cluster is estimated to be about 600 million years old, while the M92 cluster was formed early in the universe—about 13 billion years ago.

practical eXperiences

435

astrophysics

Giants (III)

106

105

104

103

102

101

100

10–1

10–2

10–3

10–4

White dwarfs

Dwarfs (V)

Supergiants (I)

1000R°100R

°

1R°

0.1R°

0.01R°

0.001R°

10R°

Lum

inos

ity

(L°)

40000 20000 10000 5000 2300

Red dwarfs

Blue giants

Red giants

Temperature (K)

Figure 24.7.1 Simulation of HR diagram


ACTIvITY 24.1: STAR ClUSTeRS on THe HR dIAgRAMRevise what you learnt in the Module 4 ‘The Cosmic Engine’ of in2 Physics @ Preliminary by briefly reviewing the properties of different types of stars.

Discussion questions1 Explain the differences in properties of open clusters and globular clusters.2 Describe how these differences are revealed on an HR diagram.

Present information by plotting Hertzsprung–Russell diagrams for: • nearbyorbrighteststars• starsinayoungopencluster• starsinaglobularcluster.

436

24 Birth, life and death chapter summary

review questions

• Thespacebetweenthestarsisfilledbythegasanddustof the interstellar medium (ISM).

• Thegasofemissionnebulaeisenergisedbyhotyoungstars to emit an emission spectrum.

• Thedustofdarknebulaescattersstarlight,reddeningorcompletely blocking our view.

• Thelightofreflectionnebulaeisscatteredbydust,especially at blue wavelengths.

• Giantmolecularcloudsareregionswherethegasismostdense and simple molecules can form. They are the sites of star formation.

• GasfromtheISMformsstars,isprocessedinsidethestars, and some is then returned to the ISM.

• Giantmolecularcloudscanbepushedintogravitationalcollapse by ‘triggers’ such as supernovae.

• Protostarswarmupasinfallinggasreleasesgravitationalpotential energy and raises the central temperature until nuclear fusion of hydrogen into helium begins.

• Surroundingtheprotostarisanaccretiondiscandjetsof outflowing material.

• Starsare‘born’ontothezero-agemainsequence(ZAMS).• ThelocationofastarontheZAMS,thetimetoget

there and its subsequent evolution are almost entirely determined by the mass of the star.

• Theproton–protonchainandthecarbon–nitrogen–oxygen cycle both combine four hydrogen nuclei into one helium nucleus, with the release of energy and some other light particles.

• Whenamainsequencestarrunsverylowonhydrogenfuel in its core, its core shrinks and is surrounded by a shell of hydrogen fusion, while the outer layers inflate to create a red giant.

• Whenthetemperatureofthecoreofaredgiantisgreater than 100 million K, the fusion of helium into

carbon begins via the triple alpha process, explosively in some stars as the helium flash.

• Astarmovesuptheasymptoticgiantbranch(AGB)when it has a core of carbon and oxygen surrounded by shells of helium and hydrogen fusion.

• Aplanetarynebulaispuffedoffbyalow-tomedium-mass star and returns gas to the ISM.

• Awhitedwarfcoolsslowlyfortensofbillionsofyears,supported by electron degeneracy pressure.

• Starsstartinglifewithmorethanabout8 M evolve much more quickly and have strong stellar winds, leading to significant mass loss.

• Thecoreofaredsupergiantbrieflyresemblesthelayersof an onion in which an iron core is surrounded by shells of silicon and sulfur, oxygen and carbon, and helium and hydrogen fusion.

• Corecollapseandbouncebackbythesurroundinglayers blows a massive star apart in a supernova explosion.

• Theblastwaveandejectedgasfromasupernovasweepup other gas to form a supernova remnant.

• Starswithinitialmassesofbetween8M and roughly 25 M will produce a neutron star that packs a mass of up to 3 M into a rapidly spinning, highly magnetised object typically about 10 km across.

• Pulsarsareneutronstarsthatproducebeamsofradiationfrom near their magnetic poles that sweep across the Earth to be seen as very precisely timed pulses.

• Starswithinitialmassesgreaterthanroughly25 M will produce a black hole with mass greater than 3 M.

• Blackholesare‘black’becausenotevenlightcanescape.They are detected because of their effect on the surrounding matter.

• The‘turn-offpoint’onthemainsequenceisanindicator of the age of a star cluster.

pHYSICAllY SpeAkIngMatch each term to the best definition.

term DefinitionPlanetary nebula Pre-main sequence stage of a star

Supernova Final state of a low mass star

Giant molecular cloud Relatively gentle ejection of the outer layers of a star

Protostar Location in space of the ‘birth’ of a star

ZAMS Possible final state of a high mass star

White dwarf Location on the HR diagram of the ‘birth’ of a star

Neutron star Explosive ejection of the outer layers of a star

437

astrophysicschapter summaryRevIeWIng 1 Stars A and B form in the same giant molecular

cloud. Star A is 1M, star B is 5M.a Predict which star will reach the main sequence

first.b Predict which wavelength bands you would best

use to observe the early stages of the formation of the stars.

2 Why do stars need core temperatures in excess of 10 million K before fusion commences?

3 Explain how the mass of a star determines where it sits on the main sequence.

4 Explain how the mass of a star determines its lifetime on the main sequence.

5 A 1M star goes through several stages in its life, many of which are listed here but are out of order. Construct a list of these terms in the correct time sequence:

planetary nebula, white dwarf, asymptotic giant branch, giant molecular cloud, red giant branch, protostar, helium flash, main sequence, T Tauri star

6 Construct a table of the major nuclear reactions in the life of the Sun. In the table, list:a the temperature needed for the process to occurb the stage in the Sun’s life at which it occurs.

7 Recall the nuclear reactions that occur during

different stages of the life of a 5 M star.

8 Explain how a white dwarf can produce light with no nuclear fusion in the core.

9 Outline the properties of a neutron star.

10 Outline the properties of a black hole.

Present information by plotting on an HR diagram the pathways of stars of 1, 5 and 10 solar masses during their life cycle.

11 Construct a flow chart that maps the evolution of stars

of masses 1, 5 and 10M. Add as much detail as possible.

12 On HR diagrams, construct the complete evolutionary

path from birth to death of stars of 1, 5 and 10M.

13 Estimate the time associated with the different stages

in the lifetime of stars of 1, 5 and 10M. Construct a timeline to scale of the life of each star.

14 Compare the HR diagrams of an open cluster and a globular cluster.

Analyse information from an HR diagram and use available evidence to determine the characteristics of a star and its evolutionary stage.

15 For each of the stars marked on the HR diagram in Figure 24.7.2, identify the spectral class and luminosity class and describe its future.

SolvIng pRoBleMS 16 The Sun is mainly powered by nuclear reactions of

the proton–proton cycle. Each reaction contributes 4.2 × 10–12 J to the solar luminosity of 3.84 × 1026 W.a Calculate how many hydrogen atoms fuse to form

helium every second to produce this luminosity.b If the mass of the proton is 1.67 × 10–27 kg,

calculate how much mass this represents.c Assuming 70% of the Sun’s initial mass of

1.99 × 1030 kg was hydrogen, calculate how many years it will take to convert all the hydrogen to helium.

d Contrast this result to the expected lifetime of the Sun. Explain any discrepancy.


Supergiants (I)

Giants (II, III)

Subgiants (IV)

White dwarfs

(VII)

10–4

10–1

10–2

10–3

1

104

103

102

10

105–8

–10

–6

–4

–2

0

2

4

6

8

10

12

14

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence (V)

E

B

AC

D

Figure 24.7.2

Revie

w Questions

438

multiple choice(1 mark each) 1 Which of the following telescopes must be used

above Earth’s atmosphere to observe astronomical objects?A OpticalB RadioC X-rayD Infra-red

2 The star Canopus (α Carinae) has an apparent visual magnitude of –0.7 and the star Deneb (α Cygni) has an apparent visual magnitude of +1.3. Which of the following statements about Canopus is true?A 2.0 times fainter than DenebB 2.0 times brighter than DenebC 6.3 times fainter than DenebD 6.3 times brighter than Deneb

3 The spectra of two stars indicate that they are of the same spectral type, but star A has relatively broad absorption lines while star B has very narrow lines. What does this tell us about the two stars?A Star A is more massive than star B.B Star A has a more dense atmosphere than star B.C Star A is older than star B.D Star A is moving away from us faster than star B.

4 Consider two single stars with different masses, both on the zero-age main sequence. What properties would you expect for the more massive star, relative to the less massive star?A Lower luminosity and shorter lifetimeB Lower luminosity and longer lifetimeC Higher luminosity and longer lifetimeD Higher luminosity and shorter lifetime

5 What properties would you expect to be approximately common between stars within an open star cluster?A AgeB LuminosityC Age and initial chemical compositionD Age, initial chemical composition and luminosity

short response 6 Describe the problems associated with viewing

astronomical objects through ground-based optical telescopes. (2 marks)

7 Explain why the limit of accurate parallax measurement from Earth is different to that from space. (2 marks)

8 The apparent magnitude of star A is 2.4 while that of star B is 0.3. a Calculate the brightness ratio of star A to star B.b Explain which star is brighter. (2 marks)

9 Outline how to calculate the distance to a Cepheid variable star. (2 marks)

10 Describe the evolution of a 1 M star. Make sure to include the various stages reached and nuclear reactions taking place. (2 marks)

11 Explain why the HR diagram of a globular cluster is different from the HR diagram of an open cluster. (2 marks)

6 The review contains questions that address the key concepts developed in this module and will assist you to prepare for the HSC Physics examination. Please note that the questions on the HSC examination that address the option modules are different in structure and format from those for the core modules. Past exam papers can be found on the Board of Studies NSW website.

439

astrophysics

extended response12 Outline an experiment you undertook in class to

observe emission, absorption and continuous spectra. Describe the differences you observed between the spectra, and identify an object that produces each type of spectra. (3 marks)

13 For each of the stars marked on the HR diagram in Figure 24.8.1, list the properties and probable stage of evolution. (3 marks)

Figure 24.8.1


10–4

10–1

10–2

10–3

1

104

103

102

10

105–10

–8

–6

–4

–2

0

2

4

6

8

10

12

14

O5 B0 A0 F0 G0 K0 M0

Abs

olut

e m

agni

tude

(M

v)

Lum

inos

ity

com

pare

d to

Sun

Spectral class

Colour index

Main sequence

E

B F

A

C

D

–0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0

7 skills

Figure 25.0.1 It is important to fine tune your skills in order to get the most out of your experiment.

440

In the preliminary text you were introduced to many skills that are needed to get through Physics. In the HSC year it is important to further develop these skills, and to make sure you know how to use the skills without thinking twice. You will now be able to fine tune and add to what you have learnt.

One of the most important skills for a physicist is being unambiguous, because the mathematical nature of physics demands that everything be defined strictly and clearly. It’s also important as a scientist to have the skill of communicating clearly the results of research to other scientists.

The HSC year is about clearly getting across what you have learnt. It is important not to be ambiguous, so you must learn how to interpret questions and construct your answers. Throughout this module we will focus on the skills you need to make sure what you have learnt is conveyed clearly.

Context

441

InquIry ACtIvIty

the Bungee jumper

Let’s say you are responsible for getting the bungee cord length (lo) correct for the next jumper.

The elasticity (k) of the cord is always known. For simplicity, assume that the energy conversion is close to 100% efficient. You are given the chance to take any measurements of the jumper.

How could you calculate the length of the cord you need to attach to the jumper if the bridge stands 100 m above the river below and the jumper is to stop between 10 and 100 cm from the water for maximum thrill. Not too close for safety, but not too far for thrill.

HintsA sketch of the situation (Figure 25.0.3) has been given to you to help you along the way.1 List the types of energy that is available at the top of the jump

(this will include gravitational U = mgh, elastic potential energy EPE = EPE = 12

2kxkx 2

and kinetic energy EK = EPE = 12

2kxmv 2) and the values of things that are known.

2 Using the law of conservation of energy, write an equation that includes all the types of energies you have listed.

3 hi is the maximum height that the cord can stretch to. It is made of the original length of the cord plus the stretched length, i.e. hi = lo + x.

4 Rearrange the equation to solve for the bungee cord length (lo).

Think about how to do this and then have a go at doing it in a scaled-down experiment.

What will you need to know, or take measurements of? How accurate will you need to be?

Set up your experiment to test it.

questIons1 List any assumptions that you have made in your calculations.2 Determine how it is possible for you to test your prediction accurately.3 List problems involved with your experiment and how it is possible to

reduce them.

Figure 25.0.2 A bungee jumper

Figure 25.0.3 A sketch of the situation

442

skills stage 2 25

prefix, systematic error, random error, accuracy, precision

skills introductionSkills are a very important component of the Year 12 syllabus. In Year 11 you started learning them and in Year 12 you will develop them further. The skills that are important to develop include planning and carrying out investigations, researching and communicating what you have found, and problem solving. Most of these skills are developed while you learn the material in the modules, but there are a few specific concepts that will be discussed separately in this chapter, to aid with your development.

25.1 metric prefixesIn the preliminary course, you were taught to express numbers using scientific notation. This is a way of representing a large or small number in an unambiguous, compact way. To add to this, we will introduce the idea of prefixes. Although you may not know the term prefix, you are probably very familiar with some of these in everyday life, for example kilogram and millimetre and centimetre. They are usually printed on the back of exercise books.

Prefixes help reduce the written size of the unit of measurement. A list of common prefixes (and some less common ones) is given in Table 25.1.1. For example, the basic SI unit of distance is the metre (m). A centimetre (cm) is then 1/100th of a metre. The prefixes that are commonly used in your course have an asterisk placed next to them. You are expected to know the values of these, as they will be used in questions and their value will not be given in any exam you undertake, so it is in your best interests to become familiar with them.

443

skills

Sometimes you will need to convert from one prefix to another. The easiest way to carry out the conversions is to follow the steps set out in the worked example below.

Worked examplequestIonConvert 5.49 mm to µm.

solutIonStep 1: Express the number in scientific notation.

5.49 × 10–3 m

Step 2: Look at the prefix that you need to convert the number to. Express the prefix in scientific notation.

micro = 10–6

Step 3: Look at the difference between the two.

10–3 m to 10–6 m

To change the exponent to 10–6, the decimal place must be moved 3 places to the right.

5.490 × 10–3

5490 × 10–6 = 5490 µm

25.2 numerical calculationsIn terms of exam performance, exam technique can be just as important as understanding the physics that you have been studying. If you cannot decipher the question and communicate your answer clearly, you cannot show that you understand physics. Chapter 26 ‘Revisiting the BOS key terms’ will go through the technique of answering questions that require a written response. Here we will look at numerical answers.

When attempting any question involving numerical calculations, follow this series of steps:1 Highlight numerical values within the question.2 Identify the unknown that must be determined.3 Write the relevant equation.

In each of steps 1–3, write the item down at the top of your answer, then:4 Show all working.5 Write your answer with units.

CheCkpoInt 25.11 Express the following masses using prefixes in terms of the

specified unit.a 0.45 g in mg b 345 g in kg c 5 600 000 g in Mg

2 Convert 4.5 km to Mm. (Note the difference between Mm and mm.)3 Convert 0.54 nm to km.

Table 25.1.1 Common metric prefixes

Number Prefix Symbol10–1 deci- d

10–2 centi- c *

10–3 milli- m *

10–6 micro- µ *

10–9 nano- n *

10–12 pico- p

10–15 femto- f

10–18 atto- a

10–21 zepto- z

10–24 yocto- y

101 deka- da

102 hecto- h

103 kilo- k *

106 mega- M *

109 giga- G

1012 tera- T

1015 peta- P

1018 exa- E

1021 zeta- Z

1024 yotta- Y

skills stage 2

444

25Worked example questIonA driver of a car travelling at 16.0 m s–1 slams on the brakes when he sees a ball on the road in front of him. The ball is 25 m away at the time the brakes are applied. Will the car stop in time, given that the deceleration is –2.5 m s–2?

solutIonStep 1: The data have been highlighted in the question.

Step 2: List them below.

Assume the direction of initial velocity is positive.

u = +16.0 m s–1

v = 0

a = –2.5 m s–2

s = +25 m

The unknown is actual displacement s taken to stop.

Even though our list shows that there is a displacement, this is not the actual displacement of the car, but the initial displacement of the ball from the car.

s = ?

Step 3: v 2 = u 2 + 2as

Step 4: 0 = (16.0)2 + 2(–2.5)s

s = 256/5

s = 51.2 m

Step 5: This is more than 25 m, therefore the car will not stop in time.

Recall from the preliminary course that the answer to a calculation needs to be expressed correctly in order to reflect the accuracy of the measurements.

First you must look at the calculation itself. The calculation should be completed with all the available digits in each value. When multiplication or division is used, the answer should be expressed to the least number of significant figures used in the values in the question. With addition and subtraction, the answer is expressed to the lowest number of decimal places within the question.

Remember that a calculator can ‘invent’ accuracy. Just because the calculator gives you an answer with 10 digits doesn’t mean that the answer has become that accurate. Refer back to the question and reduce your answer accordingly.

Worked example questIonCorrectly express the answer from the worked example above.

solutIonThe equation has multiplication and division in it, therefore we will need to look at the answer in terms of significant figures.

The least number of significant figures in the question is 2.

Therefore the answer to the calculation is 51 m.

445

skills

CheCkpoInt 25.21 An aeroplane has touched down on the runway at 290 km h–1. The braking ability allows it to decelerate at a rate of

–11 m s–2. The runway is 3.9 km long. Will the aeroplane stop in time? 2 A satellite needs to be launched into a circular near-Earth orbit. At what height above the Earth’s surface would the

satellite need to be placed in order to maintain a period of 90 minutes?

25.3 sourcing experimental errorsThere are errors involved in all measurements. The magnitude of the error and how many errors are present can be defined by the equipment you choose to use in an experiment as well as the method you use. So how can you minimise the effect of errors and make sure your experiment is both valid and reliable?

A valid experiment is one that actually tests what the aim has stated. This sounds obvious but it can be easy to stray from the task at hand.

A reliable experiment is one that can be repeated each time with the same results, no matter who is carrying it out. How do you achieve this?• Makeclearstatementsinyourmethod.• Expressclearlytheneedforrepetitionandaverages.• Usethemostaccurateequipmentavailable.

It is also important to understand the difference between systematic and random uncertainities (often simply called ‘errors’).

Systematic error is something within the experiment that causes the readings to be consistently high or low; they are always the same magnitude and sign no matter how often the experiment is repeated. This includes things such as errors in calibration, or a repetitive mistake in experimental technique. Systematic error can be minimised by careful experimental techniques and proper understanding of the experimental apparatus.

Random error is the unpredictable fluctuation of results, due to rapidly changing external influences, limitations of the readability or sloppiness of an instrument or method. If a measurement is taken repeatedly, it would be noted that the values would be readings scattered around a number. Although random error cannot be eliminated, it can be reduced by good experimental practice and by repeating the measurements many times and taking an average.

Two words closely related to these ideas are accuracy and precision. An accurate measurement is one with small systematic error, while a precise measurement is one with small random error (see Figure 25.3.1).

It is not always clear if an uncertainty (error) is systematic or random. Some causes could be considered partly systematic and random. Consider the effect on the result and how best it fits in the definitions before applying the rule blindly.

Remember to record your values to the accuracy of your instruments. If you have equipment that is accurate to 1/1000th of a unit, record it in its entirety. Remember to place any measured zeros at the end if appropriate (e.g. 7.600 m). If you don’t, you have lost that accuracy and use of the equipment is pointless. Then perform any calculations to the full number of figures available in the data and only reduce the number of figures expressed in the answer at the end.

Figure 25.3.1 Accuracy and precision

inaccurate and imprecise

precise but inaccurate

accurate but imprecise

accurate and precise

skills stage 2

446

25You may be asked in an exam to express how you made sure that your

experiment was valid and reliable, so make sure you do this when you are doing the experiment (refer to Chapter 17 ‘Physics skills’ in in2 Physics @ Preliminary).

25.4 presenting research for an examDuring the course of your study you were required to gather a lot of data from primary and secondary sources. It is important to know how much of this you need in order to accurately answer questions based on this material.

In the activity manual you have found templates to help guide your search so that you had information that covered the content to the right depth. The discussion questions in the student book also help guide you in reaching the depth that is required. If you can answer these from your research you are on the right track.

Once you have extracted the information, practise questions on the topic from past exam papers so you can get a feel for what can be asked.

The biggest mistake you can make is storytelling. Remember to answer the question, don’t just regurgitate everything that you have researched. How to answer questions as they are asked will be looked at in more detail in Chapter 26 ‘Revisiting the BOS key terms’.

CheCkpoInt 25.31 You are required to carry out an experiment that determines the reaction time of your lab partner. Hold a ruler just

above your partner’s fingers (see Figure 25.3.2). Your partner must catch the ruler after you drop it without warning. Record how far the ruler has dropped by the time it is caught. Use your knowledge of physics to determine the reaction time of your lab partner.

Repeat the process five times at an interval of 30 s. Does your partner become better?

After a day or so, do a similar experiment, this time using a data-logger sensor. Does your partner become better?

Figure 25.3.2 Your partner must catch the ruler after you drop it without warning.

2 Determine if the experiment is reliable.3 Is the experiment valid?4 How can the experiment be made more reliable and valid?

CheCkpoInt 25.4During your study of Module 1 ‘Space’, you were required to research a scientist in relation to their contribution to space exploration. Use this information to answer the following question: How has your scientist influenced those who have come after him/her?

447

skills

CheCkpoInt 25.5For your chosen scientist, state the research carried out and explain how the research relates to the Physics course.

25.5 Australian scientistOne skill that you need to undertake in Physics is the study of an Australian scientist. In this student book there is information about Australian scientists. Pick one that interests you to study in detail. Things that you are looking to know about will include their name, the area of physics in which their current research is being carried out, what the research is, where they are currently working and what area of your study relates to this scientist’s work.

25.6 linearising a formulaWhen looking for a relationship between two variables it is usually simpler to deal with a linear relationship. What happens if the relationship between two variables in a formula is not a straight line? It is not necessarily a problem, as sometimes we can rearrange the formula in order to graph the two variables so that they relate to each other in a linear way. Confused? Let’s look at the steps to follow with an example.

Worked examplequestIonA student is looking at how the velocity v of a ball relates to its displacement s as it falls through the air when dropped. To do this, the student drops a ball and records the start velocity, the displacement it goes through and the end velocity.

What does the student need to graph in order to get a linear relationship?

solutIonThe formula that we can use to relate these two variables is:

v 2 = u 2 + 2as

From this equation, we can see that there is not a linear relationship between v and s, but there is a linear relationship between v 2 and s. Graphing v 2 against s would produce a straight line.

Although it is a good skill to know, linearising is less important today when Excel(andsimilarapplications)canfitotherfunctionssuchaspowerlawsdirectly(a power law with power 2 in this example). The fit may not be identical because the least squares fitting will be minimising slightly different values.

CheCkpoInt 25.61 Determine the linear relationship between length and period from the formula T

Lg

= 2π .

2 What variables would need to be placed on the axes of a graph if a straight line was to result?

448

Revisiting the BOS key terms26Verbs in actionIn the Preliminary course, you were introduced to the idea of grouping the BOS key terms in order to help you answer questions.

Below is a table with the BOS key terms in the allocated groups.From working through the examples in the Preliminary course,

you should be very familiar with how to formalise your answers. In the HSC year you will not have as much of an opportunity to fill

in a table and analyse your answer as completely as you have up to now. Because you are now familiar with the structure of the answers, we will introduce the short examination technique to help you. If you are still not confident with structuring your answers, continue to work as outlined in in2 Physics @ Preliminary.

Table 26.1.1 BOS key terms in their allocated groups

Group 1: KnowledGe Group 2: Comprehension Group 3: AppliCAtionDefine

Extract

Identify

Outline

Recall

Recount

Account

Clarify

Compare

Contrast

Describe

Discuss

Distinguish

Extrapolate

Interpret

Predict

Apply

Calculate

Classify

Demonstrate

Examine

Group 4: AnAlysis Group 5: synthesis Group 6: evAluAtionAnalyse

Explain

Investigate

Construct

Propose

Summarise

Synthesise

Appreciate

Assess

Deduce

Evaluate

Justify

Recommend

449

SKILLS

26.1 Steps to answering questions1 Highlight the verb.2 Highlight the main topics that make the question.3 Recall the group the verb falls into and determine the depth of information

needed.– Group 1: List information.– Group 2: Interpret details of topic.– Group 3: Apply knowledge.– Group 4: Look at relationships and organisation.– Group 5: Put together information to decipher.– Group 6: Make judgements and draw conclusions based on fact.

4 Record in point form in the margin what needs to be stated—include diagrams. This should be a brief outline to organise your ideas so the order is correct and you don’t go off topic as you progress in your writing.

5 Elaborate on each point in the space provided—refer to your diagrams.

The best way to make sure you can write a good answer is to practise writing good answers.

What follows is a set of worked examples for each of the verb groups.

Group 1: Knowledge

Worked example 1QueStionDefine orbital velocity with reference to satellites.

noteSDefine falls into group 1, therefore this requires you to list information.

Orbital velocity: tangential velocity of satellite when moving in an orbit around the Earth

The speed of the satellite is given by the following formula:

vGmr

=

This shows that the velocity is affected by the mass of the central object—the Earth—and the radius from the centre of the object.

SolutionOrbital velocity of an object is the tangential velocity of an object that is in orbital motion around another object.

If the orbit is circular, the satellite must have a velocity that is determined by the formula:

vGmr

=

The speed of the satellite will be determined by the radius from the centre of the Earth as well as the mass of the Earth. The velocity will vary if the orbit is eliptical.

Group 2: Comprehension

Worked example 2QueStionDuring an experiment carried out in class, students had two current-carrying wires, one placed near an electronic balance, the other on it. They noticed that the readings fluctuated as they changed the current moving through one of the wires. Discuss how these readings changed in relation to the current.

noteSDiscuss falls into group 2, therefore this requires you to list information and interpret details of topic.

– A current-carrying wire has a magnetic field around it.

– Two current-carrying wires near each other will attract or repel, depending on the direction of the current.

– One wire is on the balance; balance will register the weight of it.

– If the wires are repelling, the reading also registers the extra force.

– If the size of the current or the direction of the current is changed, the reading will change by the amount the force is changing.

– Force is determined by:

Fl

kI Id

= 1 2

Revisiting the BOSkey terms

450

26SolutionTwo current-carrying wires will either attract or repel. If the current is travelling in the same direction in both wires, the wires will attract. If the current is in different directions, then they will repel. The size of the force is determined by:

Fl

kI Id

= 1 2

As one wire is resting on the balance, the balance is registering both the weight of the wire and the force created by the interaction of the wires.

Force is proportional to the amount of current. If there is a repulsive force originally, then the registered reading will decrease as the current is decreased and increase as the current is increased. If there is an attractive force originally, then the registered reading will decrease as the current is increased and increase as the current is decreased.

If the direction of the current is reversed, then the force will change from attractive to repulsive or vice versa, with a change in the readings as mentioned above.

Group 3: Application

Worked example 3QueStionCalculate the radius of orbit for a geostationary satellite.

noteSCalculate falls into group 3, therefore this requires you to list information and interpret details of topic and apply knowledge.

– Geostationary = 24 h, orbit = T

– It is in orbit around the Earth, therefore M = 6.0 × 1024 kg.

– Equation that applies is

r

T

GM3

2 24=

πSolution

r

T

GM3

2 24=

π

r = ?

M = 6.0 × 1024 kg

T = 24 × 60 × 60 = 86 400 s

G = 6.67 × 10–11

r 3

2

11 24

286 400

6 67 10 6 0 10

4= × × ×−. .

π

r = 6 67 10 6 0 10

4

11 24 2

23

. .× × × ×−

π86 400

= 4.2 × 107 m

≈ 42 000 km

Group 4: Analysis

Worked example 4QueStionExplain how the debate over the apparent inconsistency in behaviour of cathode rays was finalised.

noteSExplain falls into group 4, therefore this requires you to list information and interpret details of topic and apply knowledge and consider relationships and organisation.

– Inconsistent behaviour refers to cathode rays displaying both wave and particle properties.

– English versus German research.

– German findings supported wave theory as electric field did not cause deflection.

– Hertz showed that cathode rays could pass through thin layers of metal such as gold and silver.

– Resolved when JJ Thomson could deflect the ray with an electric field.

SolutionBoth Hertz and Crookes observed properties of the cathode rays that showed properties of both waves and particles. The main reason for Hertz believing that the ray was a wave was from two experiments: one showed that when an electric field was applied there was no evidence of the ray moving, and the second showed that the rays could pass through thin metal. This resembled the behaviour of light, which was known to be a wave.

It was not until JJ Thomson performed the same experiment again that the debate was put to rest. Thomson produced a near-perfect vacuum, eliminating any trace atoms within the tube, and coated the end of the tube with a fluorescent screen. It was noted that when the electric field was applied the ray was bent in the direction in which a negative particle would move.

451

SKILLS

Group 5: Synthesis

Worked example 5QueStionPropose an experimental method to be able to determine Planck’s constant.

noteS:Propose falls into group 5, therefore this requires you to list information and interpret details of topic and apply knowledge and consider relationships and organisation and put together information to decipher.

– E = hf

– W = qV so measuring V will allow you to get a measure of E.

– f can be varied by using coloured filters.

– Set up apparatus as shown.

– Measure the stopping voltage for each frequency.

– Produce a graph of V against f.

– Convert V to E by multiplying by 1.6 × 10–19.

– Determine the gradient of the graph. This is equal to h.

Solution• Setuptheequipmentasshown.(Drawadiagramlike

Figure 9.3.1.)

• Measurethestoppingvoltageforeachfrequencyoflight.(Stoppingvoltageisthevoltagetostopcurrentflowinginthe circuit.)

• ProduceagraphofE against f. To determine E multiply V by 1.6 × 10–19.(Drawagraphlikethatshownbelow.)

• Thegradientofthegraphish.

Group 6: Evaluation

Worked example 6QueStionJustify the extensive safety precautions that are evident on the Space Shuttle in order to protect the astronauts.

noteSJustify falls into group 6, therefore this requires you to list information and interpret details of topic and apply knowledge and consider relationships and organisation and put together information to decipher and make judgement and draw conclusion based on fact.

– Heat shields: heat energy not allowed to travel into the Shuttle

– Lying down on take off: improves tolerance of g-forces

– Parachutes: remaining kinetic energy is removed on landing

– Entry angle

– Nose cone: rounded to change kinetic to heat energy efficiently

SolutionThe Space Shuttle was the first reusable space transport. It was designed to fly many missions and be reused quickly while protecting those inside.

On take off, the g-forces that are applied to the astronauts are massive. As they need to be able to withstand the g-force, the astronauts lie down during take off, so the blood does not rush away from their vital organs and they remain conscious throughout the launch.

Most of the precautions are taken for the re-entry. On the approach the astronauts must aim the shuttle to come in at an angle of between 1° and 2°, in order to avoid burning up in the atmosphere or skipping off the atmosphere. The burning is due to friction. A small amount of this has to occur, and so heat shields are used to minimise the effects on the astronauts. The underside of the surface of the Shuttle is covered with tiles that can be superheated and dissipate the energy quickly. This avoids extensive heat inside and converts the massive amounts of KE to heat quickly, slowing the Shuttle.

The nose cone of the Shuttle is fairly blunt. This also aids in the conversion of KE to heat and therefore helps to slow the shuttle. A pointed nose would be more likely to melt.

There is a limit to how much energy can be converted to heat safely, so the final means of slowing down is the use of parachutes. These parachutes are deployed on landing and allow the Shuttle to slow and stop safely.

The main purpose of the Shuttle is to transport humans safely. These precautions must be taken in order for the occupants to survive.

EKEK = hf – φ

–φ

slope = h

Frequencythresholdfrequency

452

Numerical answers

These are selected numerical answers only. A complete set of answers can be found in the Teacher Resource.

Module 1 SpaceChapter 1Checkpoints: 1.1 1 ah = 0, av = 9.8 m s–2 down 2 θ = 45°,

θ = 90° 3 drag 4 both at the same time. 6 see Table 1.1.1 1.2 1 F = Gm1m2)/d 2 3 1/d 2

1.3 2 EP = (–Gm1m2)/r 3 0

Review questions: 3 2.5 s 5 a 2Fi b 4Fi c Fi /4 8 11.2 km s–1 10 a 1.38 s b 4.37 m above ground c 18.0 m (right) d 14.3 m s–1, 25º below horizon 11 a 25.3 m s–1, 0.883 s b 17.9 m s–1, 1.25 s 13 5.35 m s–1 14 260.5 N, 89.99º from x-axis 15 7.0 m 17 5.30 × 108 J 18 10 300 m s–1 19 a 42 100 m s–1

Chapter 2 Checkpoints: 2.1 1 Sergey Korolyov, Wernher von Braun

6 a g-force = 1 b g-force = 0 c g-force = 0 2.2 3 ac = v 2/R 8 True 9 a = r 10 Mercury 2.3 1 circle, ellipse, hyperbola and parabola 2 hyperbola 2.4 2 hyperbolic 3 they are equal 4 ∆vmax = 2VP 2.5 2 false

Review questions: 15 –6.25 × 107 J, 2.12 × 1012 m 16 a 10 690 kg s–1 b 5.2 m s–2, 12.1 m s–2 17 a w = mg b N = 0 c ac = g d v > 7.00 m s–1 e 0 19 32.7 h–1

20 16.8 days 22 a 4.22 × 107 m b 2.66 × 107 m

23

Type of orbiT Ke Two-body pe Me =

Ke + pe−GmM

2a

Geostationary 9.47 × 109 J –1.887 × 1010 J –9.40 × 109 J –9.44 × 109 J

Circular semi-synchronous

1.504 × 1010 J –2.99 × 1010 J –1.49 × 1010 J –1.50 × 1010 J

24 hyperbolic, no 26 a 2.22 × 1011 J b 2.45 × 1011 J

Chapter 3 Review questions: 1 D 13 1.73 m s–2 16 a 80 kg b 121 kg

17 0.781c, 5.60 years 18 1.64 × 10–13 J 19 6.91 MeV 20 6.3 × 1013 J 21 v = 0.994c

Module 1 reviewMultiple choice: 1 B 2 D 3 A 4 A 5 C

Short response: 7 –2.36 × 1010 J 8 0.9679c 10 greater by 1.08 × 10–8 kg ≈ 11 µg at equator

Module 2 Motors and GeneratorsChapter 4 Checkpoints: 4.3 1 at 90° to each other

2 F ∝ B, F ∝ I, F ∝ l, F ∝ sin θReview questions: 13 4.3 Ω 17 3 N 18 b gradient = B/F = 2.47

c 1/gradient = 0.4, Il = 0.4, 100%

Chapter 5 There are no numerical answers for this chapter.

Chapter 6 Review questions: 16 1380 Nm 17 5.4 × 10–4 Nm

18 b 2.7 c nBA cos θ (= nBA when 0 = 0º) d 30 T

Chapter 7 Checkpoints: 7.2 4 1200 V

Review questions: 22 3600 V 23 8%

24

Coils in priMary

priMary volTage (v)

Coils in seCondary

seCondary volTage (v)

sTep-up or sTep-down

100 6 200 12 Step-up

320 240 66.7 50 Step-down

50 000 393 30 500 240 Step-down25

CurrenT in priMary Coil (a)

volTage in priMary Coil (v)

CurrenT in seCondary Coil (a)

volTage in seCondary Coil (v)

Turns raTio

5 6 0.125 240 0.025

0.1 240 2 12 20

0.5 200 0.1 1000 0.2

Module 2 reviewMultiple choice: 1 A 2 D 3 A 4 C 5 B

Extended questions: 10 a up b F = BIl sin θ = 5 × 10–4 N 11 b 2 × 10–4 N c 2.5 × 10–3 Nm

Module 3 From Ideas to ImplementationChapter 8 Checkpoints: 8.2 2 0.86 N C–1 8.3 2 9.6 × 10–19 N

Review questions: 12 2 × 103 N C–1 13 0.3 J 14 a 100q J b 1.0 × 103 N c q × 103 N d 1.6 × 10–16 N 15 EK = q × E × s 16 1.76 × 1014 m s–2 17 1.88 × 106 m s–1 18 5.1 × 10–14 N 19 a 0.22 m b 3.4 × 10–12 N 20 9.09 × 10–31 kg.

Chapter 9 Review questions: 19 3.0 × 10–19 J 20 b 0.25 × 1015 Hz

c 1.66 × 10–19 J f 6.4 × 10–34 J s

Chapter 10 Review questions: 14 a 3.06 × 10–19 J b 3.27 × 1015

15 1.09 × 10–6 m 16 5.94 × 10–20 J = 0.37 eV

Chapter 11 Review questions: 19 2.53 × 10–6 m 20 a 75 K, 90 K

Module 3 reviewMultiple choice: 1 C 2 C 3 A 4 B 5 B

Extended response: 12 a 4.5 × 1014 Hz b 3.0 × 10–19 J

Numerical answers

453

Numerical answers

Module 4 Quanta to QuarksChapter 12 Checkpoints: 12.4 5 6.6 × 10–7 m 6 Balmer series

Review questions: 13 a 9.61 × 10–19 J b 1.45 × 1015 Hz c 2.07 × 10–7 m 14 a –0.85 eV b 12.75 eV c 3.08 × 1015 Hz d 9.73 × 10–8 m e Lyman series 15 a 1.22 × 10–7 m c n = 2 and n = 1 d 1.22 × 10–7 m or 122 nm e Lyman series 17 a n = 1 to n = 5 b n = 5 to n = 1 c n = 3, n = 4, n = 5 to n = 2 d n = 5 to n = 4 18 b 9.38 × 10–8 m or 94 nm c ni = 4 and nf = 2 d 4.86 × 10–7 m or 486 nm 19 a ni = 6 c Balmer series 20 c 2.09 × 10–18 J d 9.50 × 10–8 m

Chapter 13 Checkpoints: 13.2 2 2.4 × 10–9 m 3 6.56 × 10–25 kg m s–1

13.5 2 1.93 × 10–10 m

Review questions: 12 b 2.4 × 10–11 m 13 a 7.3 × 10–4 m s–1 14 a i 1.2 × 10–9 m ii 6.6 × 10–13 m 15 2.87 × 105 m s–1 16 a i 4 × 10–36 m ii 3 × 10–34 m iii 3 × 10–38 m 17 a + infinity b no 18 a 1.1 m

Chapter 14 Checkpoints: 14.2 1 a factor of ~ 1035 greater

14.6 2 5.2 × 10–27 kg 3 280 MeV

Review questions: 14 a Fe = 231 N b Fg = 5.6 × 10–34 N

15 0.034348 amu 16 a 28.4 MeV b 0.03049 amu c 4.0026 amu 17 c 8.9 MeV per nucleon

18 a 39102Y (yttrium) b Uranium 7.6 MeV/nucleon,

iodine 8.5 MeV/nucleon, yttrium 8.6 MeV/nucleon

19 a a = 01

2965

2966CuCun →+ b 1123

1022

+10Na Ne e v+ → + 20 17.406 MeV


Module 4 reviewMultiple choice: 1 C 2 C 3 B 4 D 5 A

Short response: 1 b 1.03 × 10–7 m or 103 nm

Module 5 Medical PhysicsChapter 16Checkpoints: 16.4 3 40%

Review questions: 22 8.16 × 10 6 rayl 23 1039.7 kg m–3 24 0.034%


Chapter 18 Review questions: 9 26.7° 10 a 43.98° b 67.5°

11 n2 = 1.33, water

Chapter 19 Review questions:

17 a U-238 → Th-234: α Th-234 → Pa-234: β Pa-234 → U-234: β U-234 → Th-230: α Th-230 → Ra-226: α Ra-226 → Rn-222: α Rn-222 → Po-218: α Po-218 → Pb-214: α, Po-218 → At-218: β At-218 → Bi-214: α Pb-214 → Bi-214: β Bi-214 → Tl-210: α, Bi-214 → Po-214: β TI-210 → Pb-210: β Po-214 → Pb-210: α Pb-210 → Bi-210: β Bi-210 → Po-210: β Po-210 → Pb-206: α18 a 3 years

Chapter 20 Review questions: 18 6 19 1.41 × 10–26 J 20 2.8 × 1010 Hz

Module 5 reviewMultiple choice: 1 C 2 B 3 B 4 B 5 B

Module 6 AstrophysicsChapter 21Checkpoints: 21.3 2 40× 21.5 2 ≈ 0.1 arc second

Review questions: 13 ¼ 14 ~0.2 ly 17 ~90 ly

Chapter 22Review questions: 13 ≈ 3.0857 × 1019 m 14 59 pc, 190 ly

15 59 pc, 34 5 6 65 1

. ..

+−

pc

16 a 656.3 nm, 486.2 nm, 434.1 nm, 410.2 nm, 397.0 nm 17 ~5 × 1010 20 d ≈ 200 pc

Chapter 23Review questions: 13 3 AU from the centre of the red giant

15 a 6.16 × 1030 kg (~3MSun) b m1 = 4.14 × 1030 kg ≈ 2.1 MSun, m2 = 2.02 × 1030 kg ≈ 1.0 MSun 16 T = 1.5 days; amplitude (main eclipse) = 0.6 magnitudes, (secondary eclipse) = 0.5 magnitudes 17 360 pc or ~1170 ly 18 d ≈ 220 pc or ~720 ly

Chapter 24Review questions: 16 a 3.66 × 1038 b 6.11 × 1011 kg

c 72 billion years d ~6 times the expected lifetime of the Sun

Module 6 reviewMultiple choice: 1 C 2 D 3 B 4 D 5 C

Short response: 8 a 1.4 b star B

Module 7 SkillsChapter 25 Checkpoints: 25.1 1 a 450 mg (milligrams) b 0.345 kg

(kilograms) c 5.6 Mg (megagrams) 2 0.0045 Mm 3 0.00000000000054 km 25.2 2 300 km above surface

25.6 1 T

g L2

2

π

= 2 s and t 2


454

Numerical answers

Glossary2D real-time scan a scan in which a convex array transducer

launches a rapid series of beams along different directions (sector scan) to build up a fan-like two-dimensional image of B-mode scans in real time

3D ultrasound a rapid series of sector scans are performed (oriented over a range of angles) to yield a three-dimensional image

4D ultrasound 3D ultrasound images are acquired so rapidly that a series of three-dimensional images can be displayed as a movie

ablation excess heat is carried away from a heat shield by material in the heat shield melting or vaporising

absolute magnitude the apparent brightness of an astronomical object (usually a star) as it would appear in our sky if it were moved to a standard distance of 10 pc (32.6 ly), measured in magnitudes and often represented as M

absorption line spectrum a spectrum with dark emission lines at wavelengths characteristic of the elements present in the light source, usually produced in astrophysical situations by a source of a continuous spectrum source viewed through a cool, low-density gas; most stars produce an absorption line spectrum

acceptor energy level an extra energy level is created in the energy gap near the valence band of a semiconductor as a result of the introduction of acceptor impurities

acceptor impurities produce a deficiency of electrons in the crystal lattice of a semiconductor

accretion disc a disc of material around a compact object (e.g. a protostar or a black hole) formed from infalling gas

accuracy an accurate measurement is one with small systematic error

acoustic impedance is a measure of how easily a medium oscillates in response to a sound. It is given by the formula Z = ρv where ρ is density of the medium and v is the velocity of sound in the medium

active optics a system to detect and overcome the slowly changing effects of gravity and temperature that distort the mirror or structure of a telescope and degrade the image quality

adaptive optics a system to detect and overcome the rapidly changing effects of turbulence in the Earth’s atmosphere (seeing) that are apparent as motion and blurring of the image produced by a telescope

aether drag the hypothesised tendency for aether to be trapped and dragged along by mountains and valleys on the surface of the rotating, orbiting Earth

aether wind the movement of the hypothetical aether relative to the surface of the rotating, orbiting Earth; also called ‘aether drift’.

Airy disc the diffraction pattern caused by light passing through a circular aperture such as the objective lens or mirror of a telescope

alpha decay a form of radioactive decay in which an atomic nucleus emits an alpha particle

alpha particle a highly energetic helium nucleus (two protons plus two neutrons) produced by radioactive decay

alternating current (AC) electric current that changes direction periodically

A-mode scan an ultrasound scan in which the wavefront is launched in a single direction, yielding a one-dimensional oscilloscope plot of echo intensity versus depth data

angular resolution resolution that describes the ability to resolve details separated by very small angles in the sky

anode in physics, the positive electrode. (Caution: The definition in chemistry is more complicated.)

anode glow small luminous region in a discharge tube adjacent to the anode

anti-nodes the points on a standing wave where the amplitude is maximum

antiparallel when one vector points in the opposite direction to another

antiparticle a particle with ‘opposite’ properties to another particle; for example, a positron is the antiparticle of an electron

aphelion in a satellite’s orbit around the Sun, the farthest position from the Sun

apoapsis in a two-body system, the farthest position of a satellite from the central mass

apogee in a satellite’s orbit around the Sun, the farthest position from the Earth

apparent magnitude the apparent brightness of an astronomical object as it appears in our sky, measured in magnitudes and often represented as m

armature the part of a motor or generator that contains the main current-carrying coils or windings

Aston dark space the non-luminous region nearest the cathode in a discharge tube

astrometric binary stellar system that reveals its binary nature by the wobbling of its path across the sky

astrometry the measurement of the positions and motions of astronomical objects

astronomical unit approximately the average distance between the Earth and the Sun: 1 AU ≈ 1.4960 × 1011 m

asymptotic giant branch the portion of the evolutionary path of a red giant in which the star has a core of carbon and oxygen surrounded by shells of helium and hydrogen fusion

atomic mass is the mass of an atom made up from the total mass of the protons, neutrons and electrons

atomic mass unit (amu) it is equal to the mass of one twelfth of an atom of Carbon-12, which is approximately the mass of a proton or neutron

atomic number the number of protons in a nucleus

atomic pile the name given to a type of nuclear fission reactor that was constructed using a pile of graphite bricks

back emf an emf that is produced in accordance with Faraday’s law within the coils of a motor and opposes supply emf

455

Glossary

ballistic trajectory the trajectory of a projectile subject only to gravity and air resistance

ballistics the science of projectile motion

Balmer series a set of hydrogen spectral lines containing all the visible spectral lines that are produced when an electron transitions to the 2nd energy level or orbital (n = 2)

band gap same as energy gap

baryon a family of particles that are made up of three quarks (protons and neutrons are both baryons)

base part of a bipolar transistor

BCS theory microscopic theory of superconductivity developed Bardeen, Cooper and Schrieffer

beam splitter a partially silvered mirror that allows a fraction of a light beam to pass through and another fraction to be reflected

beta decay a form of radioactive decay in which an atomic nucleus emits a beta particle

beta particle an electron ejected from the nucleus of a radioactive nuclide

binary star a system of two stars orbiting their common centre of mass

biopsy a procedure in which a tissue sample is obtained for medical tests

bipolar transistor a semiconductor device used to finely control the flow of electric current

black body an idealised example of a hot object that produces a continuous spectrum accurately described by a black body curve; an object that absorbs or transmits all the wavelengths of the electromagnetic spectrum

black body curve (or Planck curve) the distribution of light versus wavelength produced by a black body, an idealised example of a hot object; the shape, peak wavelength and intensity depend simply on the temperature of the object

black hole a remnant of a supernova explosion with a mass greater than about three solar masses, and a gravitational field so strong that even light cannot escape

blanking the blocking of the electron beam, in a cathode ray oscilloscope, on its way back to the left of the screen so that it does not retrace over itself

B-mode scan or ‘brightness mode scan’; similar method to an A-mode scan except that the echo intensity versus depth data are represented as a one-dimensional line of pixels, with pixel intensity proportional to echo intensity, yielding a one-dimensional slice of an image

bone density or bone mineral density, is related to the mass per unit volume of bone and depends on its mineral (calcium) content. The lower the calcium content, the lower the bone density and so the weaker the bone

bone scan a diagnostic imaging technique in which a patient is given technetium-99m methylene diphosphonate, which accumulates preferentially in excessively active bone and can indicate the presence of cancer or other diseases

boson a family of particles within the Standard Model and includes photons, gluons and the proposed Higgs particle

Bragg law mathematical relationship between wavelength, slit spacing and wavelength for constructive interference of electromagnetic radiation from a diffraction grating

Bremsstrahlung ‘braking radiation’; X-rays with a broad wavelength range that result from the conversion of kinetic energy of rapidly braking high-speed electrons directly into X-ray photons

brightness is a measure of the energy received in a certain time per unit of collecting area of a source (e.g. a star)—or power per unit area—W m–2

carbon–nitrogen–oxygen cycle the chain of reactions that dominates the conversion of hydrogen into helium in the hot cores of stars of higher mass than the Sun

CAT (computed axial tomography) a form of imaging in which multiple X-ray radiographs (‘projections’) are collected at different angles through a patient, from which a computer reconstructs a 3D stack of slices of the patient or ‘tomogram’; also known as CT

cataclysmic variable a binary system in which one star is so close to its white dwarf companion that it pours mass onto the white dwarf, causing the system’s light output to vary dramatically as the system suffers one or more outbursts

cathode in physics, the negative electrode (Caution: The definition in chemistry is more complicated.)

cathode glow first luminous region near the cathode in a discharge tube

cathode ray tube glass tube that contains two electrodes (an anode and cathode) that produce electron beams

cathode rays collimated beams of electrons in an evacuated vessel

central body a large mass around which other much smaller masses orbit, via gravitational attraction

centre of mass the ‘balance point’ that always lies between the two stars about which each of them can be considered to orbit

Cepheid variables periodic variables that vary between 0.5 and 2 magnitudes with periods from 1 to 70 days. They are yellow supergiants that are important distance indicators to nearby galaxies

characteristic X-rays X-rays of sharply defined wavelengths that are characteristic of each individual chemical element and are given off when a target is bombarded by accelerated electrons, resulting from rearrangements of electrons in the inner shells of target atoms

cladding in a clad optical fibre, the lower index coating that protects the core from damage and ensures total internal reflection

classical theory electromagnetic wave theory used to calculate the mathematical relationship for the black body radiation curve

closed or stable orbit an orbit that repeats indefinitely. Its two- body mechanical energy is negative

coherence length the nominal separation between Cooper pairs

coherent fibre bundle a bundle of optical fibres in which the order of fibres is the same at each end, ensuring that an image projected on one end is transmitted accurately to the other

coils loops of current-carrying wire in motors, generators and transformers

collector part of a bipolar transistor

collimate to make a beam of radiation more focused or parallel by using opaque barriers to remove components of the radiation travelling at the wrong angle

456

Glossary

collimator any kind of cylindrical, conical or planar barrier used to collimate radiation that can’t be focused using mirrors or lenses

colour Doppler imaging a sonogram employing the Doppler effect in which the colour of the pixel represents the velocity of the tissue being observed

colour index the difference in the brightness of a star in magnitudes when measured through two different filters, indicating the colour of the star. The B – V or mB – mV colour index is most commonly used

commutator brushes conducting contacts that connect the commutator to the external circuit in an electric motor

commutator slip-ring a cylindrical metal contact in a simple AC motor or generator that provides a continuous connection between the coils and the external circuit

commutator split-ring a segmented cylindrical metal contact in a simple DC motor or generator that reverses the connection between the coils and the external circuit

conduction band range of energy levels of free electrons in a solid

conduction level energy a valence electron must acquire to become free and contribute to electrical conduction

constructive interference two identical waves combine to produce a wave of greater amplitude when their crests overlap

continuous spectrum a spectrum with light of all wavelengths, usually produced in astrophysical situations by a hot, dense gas

contrast agent substance that when injected into a patient increases the contrast for target organs or tissues in medical imaging techniques such as X-ray, CAT or MRI

control rods in a nuclear fission reactor the control rods absorb neutrons and are adjusted so that the chain reaction proceeds at a constant rate

controlled nuclear reaction the conditions for a controlled nuclear chain reaction are that the available neutrons which cause the fission are regulated, resulting in a constant rate of nuclear reactions

conventional current the conventional direction assigned to a current is the direction of motion of positive particles, opposite to the actual flow of electrons

convex array transducer a piezoelectric transducer with a convex front face; the most commonly used shape of transducer for medical ultrasound imaging

coolant a substance, commonly water, used to transfer heat energy away from the core of a nuclear reactor

Cooper pair the temporary binding (or coupling) of two electrons in a superconductor mediated by a lattice deformation (phonon)

core in a clad optical fibre, the higher index central fibre that carries the light

core in a nuclear fission reactor the core houses the fuel rods, control rods, a coolant system and a moderator material; it is where fission occurs

cosmic rays extremely energetic particles that originate from outer space

critical angle the smallest angle of incidence (approaching a boundary with a lower refractive index material) that will allow some of the incident light to pass through the boundary instead of being reflected by it

critical field magnetic field that can destroy the superconducting state

critical mass the minimum mass of nuclear fuel which is required to produce sufficient neutrons to cause a sustained fission reaction

critical temperature characteristic temperature below which superconductivity occurs

CRO cathode ray oscilloscope

Crookes dark space the second non-luminous region near the cathode of a discharge tube separated from the Aston dark space by the cathode glow

crystal a three-dimensional regular arrangement of atoms in a solid

current (I) the net flow of charges through a region per unit time in amperes

current-carrying conductor a conducting material (typically copper wire) through which an electric current is flowing.

current-carrying loop a circular coil of wire through which an electric current is flowing

cut-off frequency the frequency of electromagnetic radiation below which no electrons are emitted from a photocathode

cyclotron is a type of particle accelerator that accelerates particles in a spiral trajectory. For medical purposes, the cyclotron produces accelerated protons that can convert non-radioactive nuclei into short-lived radioisotopes, usually β– emitters

cyclotron motion circular motion of a charged particle at right angles to the direction of the magnetic field

dark nebulae a portion of the interstellar medium where the dust scatters starlight, reddening or completely blocking our view of background stars

decay series a map or sequence of nuclear transformations from one element or isotope to another

density mass of a material per unit volume

depletion region the region in the centre of a p–n junction that is depleted of charge due to the recombination of electrons with holes

destructive interference two identical waves will cancel to produce a resulting wave of zero amplitude when the crest of one wave coincides with the trough of the other

diffraction the tendency of wavefronts to bend around the edges of small obstacles, instead of casting a sharp shadow

diffraction grating an optical surface, often reflective, that uses fine lines ruled onto its surface to disperse visible light into its component colours using diffraction and interference

diffraction pattern interference pattern from a diffraction grating

diffusion movement of free particles from places of high to low concentration

diode a device that allows current to travel in only one direction

direct current (DC) an electric current that travels in only one direction along a conductor

discharge tubes glass tubes in which there is conduction as a result of ionisation of the contained gas initiated by electron motion between imbedded electrodes

distance modulus the difference (m – M) between the apparent magnitude m of an object and its absolute magnitude M; related to its distance

457

Glossary

donor energy level an extra energy level created in the energy gap just below the conduction band of a semiconductor as a result of the introduction of donor impurities

donor impurities produce unbonded electrons in a semiconductor

doping the introduction of impurities into the crystal lattice of a solid

Doppler effect in which the frequency (perceived by an observer) of any kind of wave (including sound and light) is changed by the motion of the wave source (or a reflecting surface). Increased frequency means the observer and source are approaching; decreased frequency means they are moving apart

Doppler ultrasound a form of sonogram in which detection of a change in reflected ultrasound frequency (due to the Doppler effect) is used to deduce the velocity of a tissue, usually blood

drag the resistive force exerted on an object moving through any fluid (for example air or water). In air, it is also known as air resistance

drain one of the three components of a field effect transistor

dual X-ray absorptiometry imaging technique that compares the X-ray absorption by a specimen using X-rays of two different energies, making it more sensitive to differences in bone density than a single energy X-ray

dwarf novae a small outburst, repeating semi-regularly, produced by instabilities in the flow of gas from a Sun-like star onto a close white dwarf companion; classified as a non-periodic intrinsic variable star

dynodes electrodes in a photomultiplier tube that are used to increase the number of electrons emitted from a photocathode

eccentricity a measure of how far the position of the centre of a satellite’s orbit lies from the central mass. A measure of how elliptical an orbit is; a circle has zero eccentricity

echocardiography a form of colour Doppler ultrasound used to diagnose conditions of the heart

eclipsing binary binary systems in which the stars are close together and orientated so that the orbital plane is close to edge-on, causing the stars to regularly eclipse one another, periodically blocking of some of the light from the system

eddy current circulating current in a conducting material caused by a changing magnetic field

eddy current braking a braking system that utilises eddy currents to slow a moving object

effectively weightless the effect of being in free-fall or orbit. The only significant force acting on a body is gravity. Because there are no other opposing forces acting, the body experiences no compression or tension and so behaves like a truly weightless object in an inertial reference frame

electromagnet a magnet in which the magnetic field is produced by the flow of electric current

electromagnetic induction the generation of an electric current by a changing magnetic field

electromagnetic wave consists of oscillating electric and magnetic fields at right angles to each other and travelling at the speed of light

electron volt an atomic-sized energy unit; 1 eV is the energy gained by an electron when accelerated through a potential of one volt (1 eV = 1.6 × 10–19 J)

electronics method of finely controlling the flow of electrical current

ellipse a ‘compressed’ or ‘stretched’ circle

emf a measure of the strength of a source of electrical potential energy that produces a separation of charge; typically used to describe the energy per unit charge that produces a potential difference in an open circuit

emission line spectrum a spectrum that shows bright emission lines at wavelengths characteristic of the elements present in the light source; usually produced in astrophysical situations by a hot, low-density gas

emission nebulae a portion of the interstellar medium where the gas is energised by hot young stars to emit an emission spectrum

emitter part of a bipolar transistor

endoscope a device employing optical fibres and/or small cameras to collect images of internal organs with minimal surgical invasion

endoscopy procedures involving an endoscope

energy gap range of forbidden energy levels between the valence and conduction bands; energy is needed to destroy the superconducting state

escape velocity the minimum velocity required for a projectile to permanently escape the gravitational field of an astronomical body

exclusion principle the Pauli exclusion principle was formulated by Pauli in 1925 and states that no two fermions can possess the same quantum numbers in the same system; explains the energy states and orbitals in the atom

exhaust velocity the velocity of the exhaust gas as it leaves the rocket nozzle

extrinsic semiconductor one in which conduction is dominated by donor or acceptor impurities

extrinsic variables variable stars whose variation in brightness is due to a process external to the body of the star itself, for example eclipsing binaries or stars that vary because of their rotation

Faraday dark space a non-luminous region in a discharge tube between the negative glow and the positive column

Faraday’s law the induced emf in a coil is proportional to the product of the number of turns and the rate at which the magnetic field changes within the turns

fictitious forces apparent forces used for convenience to ‘explain’ apparent changes in velocity when a system is observed from within a non-inertial (accelerating) frame of reference

field of view the area of object (e.g. the sky) you can see at any moment with, for example, a telescope

field-effect transistor uses an electric field to control the flow of current through a conducting channel in a semiconductor

filter a slab or sheet of material designed to selectively remove certain wavelengths, frequencies or energies from a beam of radiation

fission the splitting of an atomic nucleus into two or more fragments

flare star a faint red dwarf that flares in visible brightness because of solar-like flares on the surface; classified as a non-periodic intrinsic variable star

458

Glossary

fluorescence the phenomenon by which a material absorbs photons of one energy (frequency) and immediately re-emits photons of lower energy (frequency)

fluoroscope a traditional apparatus that produces real-time moving X-ray images by exposing a fluorescent screen to a broad beam of X-rays passing through a patient or specimen. Modern digital versions use electronic detectors in place of the screen

flux leakage loss of magnetic flux as it passes from the primary coil through the secondary coil, resulting in loss of induction in the secondary coil and, hence, loss of energy

flux pinning the stopping of the motion of vortices through a type II superconductor by crystal defects and boundaries

focal length the distance between a lens or mirror and the image it forms of a distant object

forbidden energy gap see energy gap

force (F) a push, pull or twist measured in newtons (N)

forward bias connecting the positive and negative terminals of a power supply to the p and n sides respectively of a p–n junction

fuel rod tubes filled with nuclear fuel and located in the centre of the reactor known as the core

functional MRI a form of MRI in which the contrast agent is designed to concentrate in parts of the brain that are active during certain activities, making it possible to identify the functions of different regions of the brain

fusion the joining of two nuclei to form a new nucleus

Galilean transformation the formula for transforming velocities relative to a different frame of reference; vB (rel. to A) = vB – vA

galvanometer a device used to measure the relative strength and direction of electric current

gamma camera an array of collimated gamma detectors for imaging patients who have been given diagnostic gamma-emitting radiopharmaceutical

gamma decay decay involving emission of a gamma ray

gamma photons high-energy photons mostly emitted in nuclear reactions and positron emission tomography

gamma rays a form of electromagnetic radiation (photons) with wavelength <~10 pm. This definition overlaps with X-rays, so this term is usually reserved for photons produced by atomic nuclei

gasoline US term for petroleum or ‘petrol’

gate one of the three components of a field effect transistor

Geissler tubes sealed tubes that contain gas at low pressure and used to produce an electric discharge

general relativity Einstein’s relativity theory (see special relativity) generalised to include non-inertial reference frames; it was the replacement for Newton’s laws of gravity

generator a device that transforms kinetic energy into electrical potential energy

geostationary a circular, geosynchronous orbit directly over the equator. A satellite in such an orbit will appear stationary to an observer on Earth’s surface; also called a ‘Clarke orbit’

geosynchronous an orbit with the semimajor axis chosen so the orbital period equals the rotational period of Earth

g-force an overall measure of the magnitude of internal forces and reactions within bodies resulting from the combined effects of

gravity and acceleration. It is the ratio of the ‘effective weight’ to true weight on Earth. g-force is measured in units of Earth gravity g

giant molecular clouds a portion of the interstellar medium where the gas is most dense; simple molecules can form and star formation occurs

gimbal a pivoted frame that allows rotation around more than one axis; common example is the mount that supports a gyroscope

gradient coils coils in an MRI machine that apply gradients in the magnetic field across a patient so that variations in the detected Larmor frequency of relaxing nuclei can be used to obtain the position information needed to create a tomogram

gravimeter a device for measuring the local value of the gravitational field (or equivalently, the gravitational acceleration

gravitational field g a vector representing the direction and strength of gravity at a point in space. The magnitude g of the vector equals the acceleration due to gravity at that point

gravitational potential energy potential energy associated with a gravitational field

gravitationally bound a body is locked in an indefinite orbit around a central body it is said to be gravitationally bound. Its two-body mechanical energy is negative and its speed is never greater than the escape velocity

gravity assist using the gravitational influence of a planet to transfer some of the planet’s momentum (and kinetic energy) to a satellite that has been deliberately steered to pass close to it

hadron a particle comprising of quarks. There are two types: mesons, which contain two quarks (comprising a quark and anti-quark pair), and baryons, which contain three quarks

half-life the time taken for half of the nuclei of a radioisotope sample to decay

heat shield a layer of material that protects the vulnerable surfaces of a re-entering spacecraft from air resistance-induced heating using the principles of thermal insulation, radiative cooling and/or ablation

helium flash the explosive ignition of helium burning in red giant stars with masses comparable to that of the Sun

hertz the SI unit of frequency (Hz)

hole a vacancy left behind by the removal of an electron from the valence band in a solid

horizontal branch (HB) stage of stellar evolution that immediately follows the red giant branch in stars whose masses are similar to the Sun’s. Horizontal branch stars are powered by helium fusion in the core and by hydrogen fusion in a shell surrounding the core.

hyperbola the shape of a graph of y = k/x

hypersonic usually defined as speeds five times the speed of sound in air

impedance matching any material or device of intermediate impedance, connecting a material of higher impedance and one of lower impedance in order to reduce reflections, for example gel smeared between an ultrasound transducer and human skin; also known as acoustic coupling

impulse change in momentum

induction cooktop an appliance for cooking which uses rapidly changing magnetic fields to cause resistive heating in metallic cookware

459

Glossary

induction motor type of AC motor in which the power is induced in the rotating device by means of electromagnetic induction

inertial frame of reference a non-accelerating frame of reference

insulator a material that resists the flow of electric current

integrated circuit a thin wafer of semiconductor on which many elements (transistors, resistors, capacitors) are combined to form a useful circuit

interferometer an instrument that performs highly sensitive measurements (usually of distance or time) by exploiting light interference

interferometry splitting the light from a source (e.g. by observing a star with two telescopes) and then combining the beams may produce ‘interference’ between the beams, provided they have travelled the same distance. The interference effect depends on the path of the beams and the characteristics of the light source

interstellar medium (ISM) a patchy medium of gas and dust in the space between the stars

intrinsic semiconductor one in which conduction is dominated by the creation of electrons and holes as a result of ambient thermal energy

intrinsic variables variable stars whose variation in brightness is due to physical changes in the star or the stellar system, for example pulsating stars

invariant any quantity or relationship that remains unchanged in moving from one inertial frame of reference (or one position) to another

inverse square law a mathematical formula describing any physical quantity (e.g. light intensity or strength of gravity) that decreases in magnitude proportionally to the distance squared

ionising radiation radiation energetic enough to generate ions

irradiation a process in which a substance is exposed to a form of radiation

isotope atoms of an element which have nuclei with the same number of protons but different numbers of neutrons

jet a stream of material emitted from the polar regions of an accretion disc around a compact object (e.g. a protostar or a black hole)

Kepler’s laws the three laws that govern the motion of planets around the Sun

Larmor frequency the frequency of precession of the axes of protons aligned by a magnetic field

launch angle the angle between the horizontal and the initial velocity vector of a projectile

law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another

law of universal gravitation Newton’s theory of gravitation which was assumed to apply to all massive objects in the universe

length contraction the phenomenon in which the length of an object (e.g. a ruler) moving relative to an observer is seen to decrease

Lenz’s law an induced current in a closed conducting loop will appear in such a direction that it opposes the change that produced it

lepton a family of elementary particles (e.g. electrons, muons and neutrinos) in the Standard Model; do not participate in the strong interaction

lift upward force on an object (moving through air) due to resulting pressure differences above and below the object. This keeps an aeroplane in the air

lift-off the launch of a rocket, also known as take-off

light curve a plot of the changing apparent magnitude of a variable star system versus time

lightning protector wires placed above transmission lines to significantly reduce the chance of lightning striking the transmission lines

light-year the distance that light covers in one year, travelling at the speed of light: 1 ly ≈ 63 241 AU ≈ 9.4605 × 1015 m

linear accelerator a type of particle accelerator that accelerates particles in a straight line

linear array transducer a piezoelectric transducer with a flat front face; it has been largely replaced by the convex array transducer for medical ultrasound imaging

longitudinal relaxation time constant T1 a measure of the time taken for protons to return to their normal ratio of parallel and antiparallel configurations relative to the field, after the initial RF pulse

Lorentz factor the factor by which the time between ticks on a clock moving relative to the observer is observed to increase, or the factor by which the length of a ruler moving relative to the observer is observed to decrease

Los Alamos a US national research laboratory set up during World War II to develop the atomic bomb

low Earth orbit orbits around Earth at an altitude lying between (roughly) 160 and 2000 km above the Earth’s surface

luminiferous aether a hypothetical (non-existent) medium once supposed to be the medium required to carry light waves

luminosity is the total power output of a source (e.g. a star) in watts

luminosity classes describe differences in luminosity and hence size between stars of the same spectral class

magnetic field a region of space that can exert a force on a compass needle

magnetic field strength (B) is a measure of the magnetic field strength per square metre measured in tesla (T) or equivalently in webers per square metre (Wb m–2)

magnetic flux (ΦB) is a measure of the amount of magnetic field passing through a given area in webers

magnetic flux density see magnetic field strength

magnetic hysteresis method of heat production in induction cookware in which a magnetic field is applied to and then removed from a magnetic material and a second field in the opposite direction is applied

magnetic moment a measure of the strength of the magnetic field around a magnetic object

magnetic resonance imaging (MRI) a tomographic imaging technique that detects RF radiation given off by precessing nuclei (usually protons) aligned in a strong magnetic field

magnification a concept most commonly used with optical devices such as telescopes to describe how much bigger an object appears compared to the view with the unaided eye. Technically, it’s the ratio of the angular size of the object with and without the telescope

460

Glossary

magnitudes a measure of the brightness of a star in which 5 magnitudes represents a factor of 100 in brightness; brighter objects have smaller (or negative) magnitudes

Manhattan Project the code name for the US secret atomic bomb project during World War II

mass number the total number of nucleons in a nucleus

mass-energy many physicists dislike the definition of ‘relativistic mass’ and prefer not to distinguish mass and energy, but rather lump them together as a single quantity called ‘mass-energy’

massive possessing mass

mass–luminosity relationship a relationship between the mass M of a main sequence star and its luminosity L that can be approximately fitted by the relationship L ∝ M3.5

Maxwell’s equations four equations (in their modern versions) that summarise all possible phenomena of electromagnetism

mechanical medium any material (e.g. air or water) through which a mechanical wave can travel

medium Earth orbit orbit higher than a low Earth orbit but lower than a geostationary orbit

meson a particle belonging to the hadron family; it is comprised of one quark and one antiquark

Michelson–Morley experiment the historical experiment designed to measure the effect of aether wind on the speed of light

microprocessor the main component of a computer that is at the centre of all computer operations

Mira variable star a pulsating variable red giant or supergiant star with a period of between 80 and 1000 days

M-mode scan a rapid series of B-mode scans displayed side by side to represent echo intensity versus depth as a function of time; used to detect organ motion

moderator a material used in a nuclear fission reactor to slow down neutrons and so improve their chance of being captured by a nucleus

MOSFET metal–oxide–semiconductor field-effect transistor

motor a device that converts electrical potential energy into kinetic energy

motor effect the force experienced by a current-carrying conductor in a magnetic field

muon a lepton; it is the heavier cousin of the electron in the Standard Model

mutual annihilation the process in which two antiparticles meet (e.g. electrons and positrons), their opposite properties cancelling, leaving only their rest mass-energy, which is released in the form of two gamma rays

nebula the portion of the interstellar medium where interaction with starlight reveals the gas and dust

negative glow the luminous region in a discharge tube next to the Crookes dark space and before the Faraday dark space

neutron uncharged subatomic particle

neutron scattering the neutron can be fired into sample materials and analysis of the resulting interactions can determine the motion, spacing, magnetic structure and inner structure of many materials

neutron star a remnant of a supernova explosion with a mass about that of the Sun in an object typically only 10 km across

nodes the null points on a standing wave

non-coherent fibre bundle a bundle of optical fibres for which the order of fibres is not same at both ends, making it unsuitable for imaging but suitable for remote illumination

non-periodic variables intrinsic variable stars whose variation in brightness repeats at irregular intervals

novae a smaller outburst than a supernova, produced when a Sun-like star leaks enough gas onto a close white dwarf companion to generate a surface nuclear explosion; classified as a non-periodic intrinsic variable star

n-type semiconductor a doped semiconductor that results in extra unbonded electrons in the crystal lattice

nuclear reactor a device in which a controlled nuclear chain reactions occurs at a sustained and steady rate. The energy can be used to produce electricity and the neutrons emitted can be used for industrial and medical purposes

nucleon the name given to protons and neutrons when they are present in a nucleus

nuclides the name given to a specific isotope of an element

null result when a well-designed, carefully performed experiment fails to observe an expected effect

optical fibre a thin, highly transparent (usually silica) fibre that conducts light with little loss because of total internal reflection at the outer boundary

optical telescope a telescope designed to image optical (visible) light

orbit the path an electron takes around the nucleus in classical atomic theory

orbital decay decrease in the orbital radius as orbital kinetic energy is converted into thermal energy, e.g. by drag

orbital velocity the tangential velocity of a satellite in orbit

oscilloscope a device used to measure the variation of voltage in time across an electrical component

osteoporosis a condition of weakened bone due to low bone density

parabola the shape of the graph of a quadratic equation

paraffin a common name for a hydrocarbon wax

parallax the apparent change in position of a nearby object relative to a more distant background caused by a change in viewing position, such as the apparent change in position of nearby stars caused by the Earth’s orbital motion

parallel when one vector points in the same direction as another

parsec the distance of a hypothetical star that has a parallax angle of 1 arc second: 1 pc ≈ 206 265 AU ≈ 3.2616 ly ≈ 3.0857 × 1016 m

particle accelerator a device for accelerating charged particles

path length distance travelled by a wave

payload the cargo, instruments or passengers delivered into space by a vehicle

periapsis in a two-body system, the position of closest approach of a satellite to the central mass

perigee in a satellite’s orbit around the Earth, the position of closest approach

461

Glossary

perihelion in a satellite’s orbit around the Sun, the position of closest approach

periodic variables intrinsic variable stars whose variation in brightness repeats with a regular period

period–luminosity relationship a relationship between the period of variation of a Cepheid variable star and its luminosity L that allows the luminosity and hence the distance of the star to be estimated

perpendicular area the area of a wire loop perpendicular to the magnetic field lines passing through the loop

PET positron emission tomography

phonons lattice vibrations with discrete energy; analogous to photons

photocathode an electrode that emits electrons in an evacuated vessel when struck by electromagnetic radiation

photocell a device that converts light to an electrical signal

photoelectric effect electron emission from the surface of metals when irradiated with electromagnetic radiation, mostly visible or ultraviolet

photoelectrons the electrons emitted from an electrode in an evacuated vessel when struck by electromagnetic radiation

photometry the measurement of the brightness of a light source

photomultiplier tube an evacuated tube with a photocathode, anode and dynodes used to convert light to an electrical signal

photons radiant electromagnetic energy consisting of concentrated bundles of energy; a ‘particle’ of light

photovoltaic cells semiconductor devices used to convert light into electricity

piezoelectric effect the phenomenon in which some materials produce a voltage when squeezed and conversely become slightly compressed (or expanded) when a suitable voltage is applied

piezoelectric transducer a transducer (often used to generate ultrasound) composed of a material (such as PZT) that exhibits the piezoelectric effect

pion the lightest meson

pixel abbreviation of ‘picture element’; the smallest dot represented in a digital image

Planck curve or black body curve is the distribution of light versus wavelength produced by a black body, an idealised example of a hot object. The shape, peak wavelength and intensity depend simply on the temperature of the object

Planck’s constant a fundamental constant h = 6.63 × 10–34 J s

planetary nebula a short-lived, small cloud of gas expelled by a low mass star to reveal its core as a white dwarf star

plate part of a thermionic device used to collect the electron current

polarisation the direction of the electric field in an electromagnetic wave

positive column largest luminous region in a discharge tube and the most prominent feature of the discharge; situated near the anode

positron an antimatter electron

positron decay subset of beta decay involving emission specifically of a positron

positron emission tomography a form of imaging in which a biologically accumulating substance tagged with a positron

emitting isotope is used to identify the position of diseased tissue. A tomogram is built up by deducing the original positions of the positrons by detecting pairs of gamma rays that result from mutual annihilation with surrounding electrons

postulate a fundamental principle assumed to be true but for which there is no direct proof, which forms the basis upon which a theory can be built

potential difference a measure of the difference in electrical potential energy between two points in a circuit. This quantity is measured in volts and is often replaced by the term voltage

power station an industrial facility for the generation of electric power

precession the conical motion of the axis of rotation that results when an external torque is applied to a spinning object

precision a precise measurement is one with a small random error

prefix mathematical word that can be added in place of scientific notation

principle of relativity physical laws remain invariant in all inertial frames of reference

prism a wedge-shaped glass block that disperses visible light into its component colours using refraction within the glass

projectile an object (e.g. a cannonball) projected through open space

propellant material (usually combustible) used to generate exhaust through a rocket nozzle, hence producing thrust

proper length the length of an object as measured by an observer stationary relative to that object

proper mass see rest mass

proper motion the apparent motion of a star across the sky due to its the transverse motion through space

proper time the time interval between two events that take place in the same position as measured by an observer stationary relative to those events

proton–proton chain the chain of reactions that dominates the conversion of hydrogen into helium in the relatively ‘cool’ cores of lower mass stars like the Sun

protostar the hot core of a collapsing fragment of a gas cloud, perhaps destined to form a star

p-type semiconductor a doped semiconductor that results in a deficiency of electrons in the crystal lattice

pulsar a neutron star that is visible because beams of radiation produced near its magnetic poles sweep across the Earth to be seen as a pulse as the pulsar rotates

PZT lead-zirconate-titanate, a material commonly used in piezoelectric transducers

quanta the emission or absorption of energy in discrete units

quantum mechanics a set of principles that describe physical reality at the atomic level of matter (molecules and atoms) and the subatomic (electrons, protons and even smaller particles), including the simultaneous wave-like and particle-like behaviour of matter and radiation (wave–particle duality)

quantum number a set of numbers used to describe quantities in a quantum system

quark a set of particles developed to explain the properties of a family of particles called hadrons

462

Glossary

R Coronae Borealis star a yellow supergiant star that fades significantly at irregular intervals as carbon-rich dust clouds obscure the surface; classified as a non-periodic intrinsic variable star

radiation particles or waves that propagate outwards from a source

radiation shielding protects people and the environment against excessive radiation and prolongs the working life of a nuclear reactor facility

radio frequency (RF) electromagnetic radiation frequencies in the range (roughly) 3 kHz to 300 GHz

radio telescope a telescope designed to collect radio wavelengths of electromagnetic radiation

radioactive any nucleus that undergoes radioactive decay

radioactive decay the process whereby certain unstable atomic nuclei approach a more stable state by releasing alpha, beta or gamma rays

radiograph the oldest form of X-ray image; produced by exposing film using a broad beam of X-rays passing through a patient or specimen

radiographer a technician who takes medical images of patients

radioisotopes atoms that have an unstable ratio of protons to neutrons and will decay via alpha or beta decay to attain a more stable configuration; may also emit gamma radiation

radiopharmaceuticals radioisotopes incorporated into compounds that are used in medicine and can be classified into diagnostic and therapeutic

radiotherapy a medical procedure in which radiation emitted from a radioactive source is directed at an area of diseased tissue

random error an uncertainty in a measurement governed by random statistical fluctuations

range the maximum horizontal displacement of a projectile

reaction device any device (such as a rocket) that is driven along by the reaction force from material being expelled

real-time image when image data are collected and processed so rapidly that the delay between an event and its displayed image is negligible

receiver apparatus used to detect an electromagnetic wave

red giant a star that has evolved off the main sequence and grown in size and luminosity, having largely exhausted the hydrogen fuel in its core

re-entry the process of bringing a spacecraft back through the atmosphere

reflecting telescope a telescope that uses a mirror as the objective element to gather the light (not just visible light). The mirror reflects the light and brings it to a focus

reflection nebulae a portion of the interstellar medium in which we see the light scattered by the dust, especially at blue wavelengths

refracting telescope a telescope that uses a lens as the objective element to gather the light (almost always visible light). The lens refracts the light and brings it to a focus

refraction bending of the path of a wave as it passes from one medium to the next at an angle with respect to the surface (interface) of these media

relativistic mass the phenomenon in which the mass of an object moving relative to the observer is appears to increase

relaxation the process of a previously aligned system (e.g. nuclear spin) to gradually become misaligned

resistive heating heating that occurs in a conductor when energy is transferred from the moving charges of an electric current to the atoms of the conductor

resonance the tendency for one object possessing a natural frequency of oscillation to oscillate strongly in the presence of an external source at that same frequency

rest mass the mass of an object as measured by an observer who is stationary relative to that object

reverse bias connecting the negative and positive terminals of a power supply to the p and n sides of a p–n junction respectively

RF transceiver coils the coils that both transmit the RF pulse and receive the RF echo of relaxing nuclei in MRI

rotor coils the central rotating component of a motor or generator consisting of wire coils wound around a laminated iron frame attached to an axle or shaft

RR Lyrae variables pulsating variable stars, commonly found in globular clusters, that vary up to 2 magnitudes with periods less than 1 day

RV Tauri star a pulsating variable yellow supergiant star with a period of between 20 and 100 days

Rydberg’s constant constant with a value of 1.097 × 107 m–1 used in the calculation of hydrogen spectral line wavelengths

satellite any object in orbit under the gravitational influence of a much larger body

sawtooth the waveform of the potential difference across the horizontal plates inside the tube of a cathode ray oscilloscope

scintillation the rapidly changing effects of tiny, rapidly changing temperature variations in the Earth’s atmosphere that are apparent as rapid changes in the brightness of a star; commonly known as ‘twinkling’

scintillator any substances that produces light flashes when struck by ionising radiation

sector scan fan-shaped ultrasound beam that emanates from a convex array transducer. The wavefronts of ultrasound are launched at different angles at separate times, to avoid simultaneously detecting confusing echoes from different directions

seeing rapidly changing effects of tiny, rapidly changing temperature variations in the Earth’s atmosphere that are apparent as motion and blurring of the image produced by a telescope

semiconductors materials used for the manufacture of modern day electronic components

semimajor axis the distance between the apoapsis of an orbit and the centre of the orbit

semi-regular variable star a pulsating variable red giant or supergiant star with an irregular period of between 80 and 1000 or more days

semi-synchronous an orbit with the semimajor axis chosen so the orbital period equals half the rotational period of Earth

sensitivity (light-gathering power) describes the ability of a telescope system to ‘see’ faint objects; depends on how much light the telescope collects and how much of that light is delivered to the detector

463

Glossary

shaded pole induction motor an induction motor in which four small copper shading rings are inserted into the stator on each side of the rotor on opposite poles. The currents induced in these shading rings act to delay the magnetic flux passing through the rotor, producing an asymmetric magnetic field

shadow mask a metal sheet with an array of holes that is placed behind the phosphor screen of a CRT television screen. Each hole guides the three beams to their respective coloured phosphor as the beams move horizontally and vertically

shock wave the wavefront of sharp pressure increase that builds up in front of an object moving through a gas at speeds faster than the speed of sound in the gas

simultaneity the state of being simultaneous—two or more phenomena taking place at the same time as seen by an observer

single-phase AC refers to the distribution of AC electric power using a system in which only a single voltage oscillation is supplied to the user; used for household power supply

slingshot effect see gravity assist

solid state physics branch of physics that includes the study of properties of solid materials

sonogram an image created using ultrasound

source one of the three components of a field-effect transistor

spacetime a 4D set of coordinate axes consisting of the usual x, y and z axes of space plus an extra axis representing time. The equations of special relativity and Maxwell’s equations are expressed in these four dimensions

special relativity Einstein’s theory applied to inertial reference frames in which the speed of light is assumed to be invariant and the principle of relativity is assumed to apply to all physical laws. It expands on Galileo and Newton’s laws of mechanics

spectral classes groups of stars with similar spectral lines, indicating similar surface temperatures

spectrograph an instrument that disperses light into its component wavelengths to be recorded by a detector

spectrometer an instrument that measures the intensity of electromagnetic radiation for a range of wavelengths

spectroscope an instrument that disperses visible light into its component colours to be viewed by eye

spectroscope binary stellar systems that reveal their binary nature by the motion of spectral lines in the spectrum

spectroscopic parallax a poor name for a stellar distance estimated by identifying the type of star from the characteristics of its light

spectroscopy the study of the light from objects to reveal their composition and physical characteristics

spectrum the pattern that results when light from a source is spread into its component wavelengths (colours if visible light)

spin the property of subatomic particles that gives them a magnetic moment. It is (very roughly) the microscopic equivalent of the spin of a spinning top

squirrel cage rotor the rotor of an AC induction motor containing parallel conducting bars around its circumference

Standard Model the current, scientifically accepted model to describe the nature of matter

standing wave a wave that remains in a constant position. It consists of two equivalent waves overlapping and travelling in

opposite directions so that the maximum amplitude and null points on the wave do not move in space

stationary state a definite energy; in atomic theory relates to a stable orbital

stator the stationary part of a motor or generator surrounding the circumference of the rotor

step-down transformer a transformer that produces an output voltage that is less than the input voltage

step-up transformer a transformer that produces an output voltage that is greater than the input voltage

stopping potential the potential difference required to stop electrons from leaving the surface of a photocathode

strong force the nuclear force holding the nucleus together

substation a subsidiary station of an electrical distribution network where voltage is either stepped up or down using transformers

superconductivity total disappearance of electrical resistance in a material when it is cooled to below a certain temperature

supernova a cataclysmic explosion of a star, producing a massive outburst of light that fades over many weeks. They are sometimes classified as non-periodic intrinsic variable stars. There are two major types: type I—accretion of gas onto a white dwarf from its companion leads to a runaway nuclear explosion; type II—collapse of the core of a massive star

supernova remnant the cloud of gas that was expelled by a star and made to glow by a supernova explosion

supersonic speeds greater than the speed of sound in a gas

supply emf the emf applied to a circuit

symbiotic star a close binary composed of red giant and white dwarf stars, with irregular outbursts from the red giant falling onto the white dwarf; classified as a non-periodic intrinsic variable star

synchrotron a type of particle accelerator

systematic error consistent error that is in every measurement

T Tauri star a very young star with an accretion disc that varies irregularly in brightness as it approaches the main sequence; classified as a non-periodic intrinsic variable star

telescope a device that collects electromagnetic radiation (‘light’) and focuses it to create an image on a detector that is brighter and with greater spatial resolution than could be achieved using the detector alone

test mass a small mass used (in practice or in theory) to measure the gravitational field. Its mass should be so small as to make a negligible contribution to the gravitational field being measured

thermionic device a cathode ray tube device used to control the flow of electrons

three-phase AC three circuit conductors carry three alternating currents (of the same frequency) which reach their instantaneous peak values at different times, resulting in constant power transfer over each cycle of the current; used by high power or industrial machines

thrust the reaction force exerted back onto a rocket by the exhaust gas it expels; this reaction is responsible for driving the rocket

time dilation the phenomenon in which the time between ticks on a clock moving relative to the observer is observed to increase

time of flight the time elapsed between the launch of a projectile and when it hits a barrier

464

Glossary

timebase the time (x) axis on an oscilloscope controlled by the timebase dial

tomogram a 3D image in the form of a stack of 2D slices produced by any form of tomography

tomography any imaging technique that produces tomograms

torque the turning effect (or turning moment) of a force

total internal reflection the complete reflection that occurs when light in a higher index material meets a boundary with a lower index material at an angle of incidence larger than the ‘critical angle’

trace the plot of voltage against time displayed on the screen of an oscilloscope

trajectory the path of a projectile in flight

transducer any device that converts energy from one form to another

transformer a device that alters the voltage and current of AC electricity

transistor a semiconductor device used to finely control the flow of electrical current in a circuit

transition the movement between two orbitals or energy states

transmission tower typically a steel tower used to support wires for the long distance transmission of electricity

transmitter apparatus used to generate an electromagnetic wave

transmutation the process of changing one element into another

transverse relaxation time constant T2 a measure of the time taken for the magnetic component perpendicular (transverse) to the external field to return to zero, after the initial RF pulse

triode a thermionic device used to finely control the flow of electric current

triple alpha process the chain of reactions that describes the conversion of helium into carbon in the hot core of a red giant star

twin paradox a thought experiment that explores the symmetry of time dilation by supposing one identical twin takes a journey in a spacecraft travelling at high speed while the other stays on Earth

type I superconductor one in which the internal magnetic field remains zero until a critical applied magnetic field strength is reached, at which a sudden transition to the normal state occurs

type II superconductor one that has two critical magnetic field strengths and the superconductivity can be maintained up to the upper critical field, but there is partial penetration of the field into the superconductor at between these fields

ultraviolet catastrophe the erroneous prediction by classical theory that the intensity of radiation from a black body will increase towards infinity at the ultraviolet end of the electromagnetic spectrum

uncertainty principle the Heisenberg uncertainty principle states that there is a limit to how precisely you can measure pairs of quantities such as position–momentum, and energy–time and that both cannot be known to arbitrary precision

uncontrolled nuclear reaction chain reaction occurs when the production of neutrons goes unchecked and the fission reactions increase at an accelerating rate

universal gravitational constant G the constant G = 6.67 × 10–11 N m2 kg–2 that appears in Newton’s equation for gravitational force; it is assumed to be constant throughout the universe

valence band range of energy levels of electrons bound to an atom

valence electron bound outer electron of an atom

valence level the energy of a valence electron while being bound to its atom

variable star a star or star system that appears to vary in brightness

vertical and horizontal components two vectors parallel to the vertical and horizontal directions respectively that add up to the vector being analysed

visual binary a binary star system that can be seen as two stars by a telescope under sufficiently good seeing conditions

voltage the SI unit for potential difference; voltage is also commonly used to replace the term potential difference

vortices swirls of electrical current surrounding a normally conducting region that are embedded in a type II superconductor

voxel volume element; the smallest part of a three-dimensional image (or tomogram)

wave function a solution to Schrodinger’s wave equation; the square of this function provides you with a probability density that allows you to predict the likelihood of finding a particle

wave mechanics commonly used interpretation of quantum mechanics

wavefronts a line or surface joining all points of equal phase in a wave, e.g. the circular crest of a single ripple in a pond

white dwarf the core of a low mass star revealed when the outer layers are gently blown away, and visible as a faint ‘star’ slowly cooling off over tens of billions of years

work function energy required to just remove the electrons from the surface of a metal

X-ray binary a binary system in which one star is so close to its neutron star or black hole companion that it pours mass onto the companion and emits X-rays from the infalling gas

X-rays a form of electromagnetic radiation (photons) with wavelength 10 nm

Zeeman effect the splitting of spectral lines in the presence of a magnetic field

zero-age main sequence (ZAMS) the line on the Hertzsprung–Russell diagram where collapse stops and a newborn star is powered entirely by nuclear reactions in its core

465

Index

IndexA-mode ultrasound scans 312ablating materials 49absolute magnitude, of stars 399absorption spectra, lines in 232, 393AC electric motors 121–4

synchronous 130–1AC generators

compared with DC 135simple 131–3

AC generators and transformers, affect on society 146–7

AC induction motorsactivity 126single-phase 124three-phase 122–3

AC powergeneration and delivery 142–4losses during transmission and

distribution 144–6accelerated particle beams, uses of 288acceleration due to gravity, activities 20–1acceptor energy level 193accuracy in experiments 445acoustic coupling 312acoustic impedance (Z) 310–12active optics in telescope mirrors 377–8adaptive optics in telescopes 379–80aether drag 64aether model for light transmission 61–2agricultural radioisotopes 284air resistance (drag) 7Airy disc 376Algol (β Persei) eclipsing binary star system

412alpha decay 264, 341alpha (α) particles 341alternating current (AC) 85

electric motors 121–4amplification of currents using triodes 198Anderson, Carl, and cosmic rays 286angle (θ), in motor effect 93Ångström, Anders, and hydrogen spectrum

233angular resolution of telescopes 376–7anode glow 158ANSTO (Australian Nuclear Science and

Technology Organisation)Bragg Institute at 272Echidna neutron diffractometer 285National Medical Cyclotron at Royal

Prince Alfred Hospital 282OPAL reactor facility at Lucas Heights

272, 282, 345radiopharmaceutical production at 353

antennae, radio 175–6apparent magnitude of stars 398apparent weight 31–2artefact standards of mass and length 79Aston dark space in discharge tubes 158astrometric binary stars 411

astrometry 388–9astronauts, forces on during take-off 31–5astronomical unit (AU) of distance 389asymptotic giant branch 428ATLAS particle detector in Sydney 290–1atom

Bohr’s postulates for model of 235–6Rutherford’s model 228–30

atomic bombs 279atomic mass, and the neutron 261atomic mass number (A) 262atomic mass unit (amu) 266atomic number (Z) 262atomic piles 270atomic spectra

for hydrogen 232–5for larger atoms 239

atoms, historical understanding 229Australia Telescope Compact Array (ATCA),

for interferometry 381Avogadro project (CSIRO) 79

B-mode ultrasound scans 313back emf in DC electric motors 120ballistic trajectories 3Balmer, Johann

Bohr’s explanation of series 236–8emission spectrum series for hydrogen

233band gap (forbidden energy gap) 189barium emission spectrum 239barium meal X-ray procedure 325baryons 292BCS (Bardeen, Cooper and Schrieffer) theory

215, 217Bessel, Friedrich, and stellar parallax motion

388beta (β) particles 341beta decay 260, 341–2

positron production 342beta (minus) and (plus) decay 264β Centauri A and B orbits 410bias, forward and reverse 194–5binary stars 407–10

β Centauri 410types of 411–12

binding energy of nucleus 268–9bipolar transistors 198–9black body absorbers and emitters 178black body radiation 178–9black holes 431, 432blood flow and Doppler effect 315–16Bohr, Nils

explanation of Balmer series for hydrogen 236–8

postulates for atomic model 235–6and Rutherford’s atomic model 230

bone-density measurement using ultrasound 315

bone scans with gamma camera 346–7

BOS key terms 448–51Bragg, William Henry, particle properties of

X-rays 208, 248Bragg, William Lawrence (Sir), law of 208,

248Bragg Institute at ANSTO OPAL reactor

facility 272brain, MRI scans of 360, 362brain function 303braking using eddy currents 112–13Bremsstrahlung 323brightness of stars see magnitudesbrushless DC motors 115bungee jumping 441Bunsen, Robert, and spectral lines 390

calcium emission spectrum 239carbon–nitrogen–oxygen cycle 425–6Cassegrain, Laurent, telescope 372CAT (computed axial tomography) 321CAT scans

benefits over other methods 329process and image construction 327–8

CAT X-ray images 326–9cataclysmic variables 412cathode glow 158cathode ray oscilloscope (CRO) 167–8cathode ray tubes, history of 156–7cathode rays

nature of 157, 160see also electrons

Cavendish, Henry, measuring density of Earth 25

CDs and DVDsas diffraction gratings 206as spectrometers 155

Centaurus-A galaxy images 382centripetal force 36Cepheid variable stars 415–17Chadwick, James, discovers neutron 260–1chain reaction game 227chain reactions 270–1characteristic X-rays 323charge to mass ratio of electron 166charged particles

in electric fields 160–3forces on in magnetic fields 89, 164–5

circular orbits 41–2, 57classical theory approach to black body

radiation 178–9closed orbits 41coherence length between Cooper pairs 215,

217coherent bundles of optical fibres 335–6coil, torque on 116–17Colour Doppler imaging 315colour index of stars 400–1compact fluorescent lights 173Compte, August, knowledge of stellar objects

389–90

466

Index

computed axial tomography see CATconduction bands 189–90conduction level 189conic sections as orbits 42conservation of energy, law of 105continuous spectra 390, 392contrast agents in X-ray imaging 325control rods in nuclear reactors 281conventional current 85convex array transducers 308–9coolant in nuclear reactors 281Coolidge, William, X-ray tube 320Coolidge X-ray tubes 321–2Cooper pairs 215, 217core losses, in transformers 140core of nuclear reactor 280Cormack, Allan McLeod, and tomography

321cosmic rays 286critical angle in optical fibres 334critical field strengths (Bc1 and Bc2) 213critical mass 271critical temperature (Tc) for superconductors

211Crookes, William (Sir), and cathode rays 157Crookes dark space, in discharge tubes 158Crookes magnetic deflection tube 90crystal structure, and X-ray diffraction

gratings 208–9crystal structure of metals, and electrical

conductivity 209–10crystals, plane spacing (d) in 209Curie, Jacques, and piezoelectric effect 308Curie, Pierre, and piezoelectric effect 308current amplification by triode 198current-carrying loop (coil), torque on

116–17current (I) in motor effect 93currents, electric 84–8cut-off frequency (f0) in photoelectricity 182cyclotron motion 164cyclotrons 287

producing radioisotopes in 282, 345

dark nebulae 422Davisson, Clinton, and electron waves 251Davy, Humphrey (Sir), mentor of Faraday

100DC electric motors 114–20

back emf in 120DC generators

compared with AC 135simple 133–4

de Broglie, Louis (Prince)hypothesis confirmed 251–2matter wave equation 248–50

De Forest, Lee, invents triode 198decay series for uranium-238 265deceleration during space shuttle re-entry 48degenerate electron pressure 429depletion regions 194diffraction

of waves 250–1of X-rays 207

diffraction angle (θ) 209diffraction gratings 206–7, 391diffusion of charges, in semiconductors 194diodes (p–n junctions) 193–5direct current (DC) 85

electric motors 114–20transmission at high voltage 146

discharge tubes, structure of 158–9Discovery space shuttle, re-entry problems 49distance modulus for stars 399, 400donor energy level 192doping of semiconductors 191–3Doppler shift (Δλ) of spectral lines 396Doppler ultrasound imaging 306

for blood flow 315–16drag (air resistance) 7dual X-ray absorptiometry (DXA or DEXA)

315dynodes 184

Earth, weighing of 25Echidna (high-resolution neutron powder

diffractometer) at OPAL 285echocardiography 316eclipsing binary stars 412eddy currents 106–8

braking in trains and roller-coasters 112–13

in induction cooking 108losses in transmission lines due to 145resistive heating due to 140

Edison, Thomasand DC electricity 141and incandescent lamps 198

Einstein, Alfredand photoelectric effect 182–3principle of relativity 58–9proposal of photons 179, 181special theory of relativity 64–8

electric current 84–8control of 197–200

electric field lines 161electric field strength (E) 160–1

between parallel plates 161–3electric motors

activity 126alternating current 121–4characteristics of different types 125compared with generators 135direct current 114–20home-made 83universal 122

electric power (P)generation and delivery 142–4losses during transmission and

distribution 144–6and resistance 86in transformers 138

electrical conduction and energy bands 189–90

electrical conductivity of metals, and crystal structure 209–10

electrical resistance see resistance, electricalelectricity supply network 143

electromagnetic induction 100–3without relative motion 103see also induction cooking; induction

motorselectromagnetic wave emitter energy 180electromagnetic wave theory, Maxwell’s

174–5electromagnetism, exploring 136electron capture 265electron orbits, as standing waves 252–3electron spin 217electron volt (eV) 190, 343electrons

charge to mass ratio 166de Broglie’s hypothesis of wave nature

248–52electrons (cathode rays) 165–6electrostatic particle accelerators 286elliptical orbits

deductions from perturbations of 40of Halley’s Comet 39Kepler’s laws for 37–9properties 37

emf (ε) 85, 102in DC electric motors 120

emission nebulae 422emission spectra

for emission nebulae, normal galaxies and quasars 394

lines in 232, 392–3endoscopy 333, 335

medical uses 336–7energy, equivalence with mass 72–4energy band diagrams 189–90energy gap 189–90, 217environment, effect of widespread electricity

generation 147equal areas, Kepler’s second law of 38–9escape velocity (ve) 18–19Esnault-Pelterie, Robert (REP), space trip

calculations 27evolution of stars 425–7excitation of hydrogen proton by RF pulses

357exclusion principle, Pauli’s 255experimental errors 445–6extremely large telescopes (ELTs) 382extrinsic and intrinsic semiconductors 193extrinsic variable stars 414

Faraday, Michaeldark space in discharge tubes 158law of 100–3

Fermi, Enricoand controlled nuclear reactions 270and transuranic elements 269

Fermilab Tevatron 288fictitious forces 59, 60field-effect transistors (FETs) 200fission, nuclear 269fission reactors 280–1Fitzgerald, George, and aether 63Fleming, John Ambrose, builds first diode

198

467

Index

flight time of projectile 6fluorescence, in discharge tubes 156–7fluorescent lights 173flux leakage, in transformers 139flux pinning, in magnetic levitation 214focal length of lens 371–2food irradiation 284forbidden energy gap (band gap) 189force carriers 292formulae, linearising 447forward bias in p–n junctions 194–5forward-biased LEDs 1954D ultrasound 314frames of reference, inertial and non-inertial

58–9Fraunhofer, Joseph von, lines in spectrum of

Sun 390Friedman, Jerome, and quarks 292Fritsch, Otto, and radioactive isotopes 269fuel rods in nuclear reactors 280functional MRI images 362fundamental physical property standards of

mass and length 79fusion bombs 280

g-force 31–4galactic recycling system 423galaxies (normal), emission spectra for 394Galilean transformation 5Galileo Galilei 4–5

and telescope 368, 370–2Galileo space probe, use of gravity assist 45galvanometers 119gamma camera 346–7gamma photons 184gamma radiation 265gamma ray emitters 342–3γ Crucis, distance of 402gastroscopy see endoscopyGeiger, Johannes, and model of atom 229–30Geissler, Heinrich, and vacuum pump for

discharge tubes 156–7Geissler tubes 156Gell-Mann, Murray, and quarks 292generators, electricity 130–5

compared with motors 135comparing AC and DC 135

geographic poles 88geostationary satellites 43geosynchronous satellites 43Germer, Lester, and electron waves 251giant molecular clouds 423Glashow, Sheldon, and electro-weak theory

292global positioning system (GPS)

receivers 67satellites 43–4

globular star clusters 433–4Goddard, Robert, US rocket physicist 27gradient coils in MRI scanners 361gravitational fields

variations in 14–15weight in 13–14

gravitational potential energy (GPE) 16–19

gravity 10–15activity 20effect on orbits 35–40

gravity assist (slingshot effect) 44–5

hadrons, properties 294Hahn, Otto

aand radioactive isotopes 269and beta decay 265

half-life of radioisotopes 282, 343Halley, Edmund, comet of 39Harriot, Thomas, and telescope 368heart imaging by echocardiography 316heat shields 49heating during space shuttle re-entry 48–50Heaviside, Oliver, and Maxwell’s equations

61Heisenberg, Werner, uncertainty principle

of 254–5helium flash 428Herschel, William, and reflector telescope

373Hertz, Heinrich

and cathode rays 157discovers photoelectric effect 182measures speed of radio waves 61, 175–7verifies electromagnetic wave theory

174–5Hertzsprung–Russell (HR) diagram 395–6

activity on star clusters 435and distance of γ Crucis 402life of star on main sequence 427showing evolutionary tracks for protostars

424showing variable stars 414for star clusters 434

Higgs particle 291high-mass stars, evolution of 430–1high-temperature superconductors 216high-voltage DC (HVDC) transmission 146Hipparchus, and star magnitudes 397Hohmann transfer orbits 57holes in valence bands 191hollow structures, X-ray images of 325homopolar electric motor 83Hooke, Robert, law of 15Hounsfield, Godfrey, and tomography 321,

327hydrogen atom

Balmer series for 233emission spectrum 239spectra and energy levels of 232–5

hydrogen bomb 280hydrogen proton

as energy carrier 285precession of 356spin in magnetic field 355–6subjected to RF pulses 357–9

hyperbolic orbits 41–2hypersonic flight 48–9

impedance matching 312induction cooking 108

see also electromagnetic induction

induction motorssingle-phase AC 124see also electromagnetic induction

industrial radioisotopes 283inertial frames of reference 58–9inertial and non-inertial frames of reference

activity 75infra-red light 197insulators

energy band diagrams for 189–90in high-voltage AC power transmission

145–6integrated circuits (ICs) 200interference, constructive and destructive 205interferometry

in Michelson–Morley experiment 62–4in radio telescopes 380–1

International Thermonuclear Experimental Reactor (ITER), superconductors in 218

interstellar medium (ISM) 422–3intrinsic variable stars 414inverse square law 11isotopes

of atoms 262–3, 341radioactive decay of 340–3

James Webb space telescope (JWST) 382Jansky, Carl, and background radiation 373Joliot-Curie, Irene, and radioactive isotopes

269junction transistors 198–9

Kamerlingh Onnes, Heike, liquefies helium 211

Kelvin, Lord (William Thomson), and temperature scale 210

Kendall, Henry, and quarks 292Kepler, Johannes

laws of planetary motion 35–9telescope 371–2third law 409

Kirchhoff, Gustav, and spectral lines 390Korolyov, Sergey (aka Sergei Korolev), USSR

space program leader 28Kronig, Ralph, and electron spin 240

Lagrange points 29landing of spacecraft 50–1Large Hadron Collider (LHC) at CERN

288, 290–1Larmor frequency 356laser diodes 195Laue, Max von, X-ray diffraction patterns

207–9launch angle of projectile 6law of universal gravitation 11

in finding new planets 40in predicting small deviations in planetary

orbits 39Lawrence, Ernest, and linear accelerator 287length contraction 69–70length (l ) in motor effect 93Lenz, Heinrich

application to DC motors 121law of 104–5

468

Index

leptons 292–4levitated magnets 213–14Lewis, Gilbert, names photons 231light

aether theory of transmission 61–2infra-red 197intensity and wavelength 178speed of 65as wave and particles 179wave properties of 206–7

light-emitting diodes (LEDs) 195light-year (ly) distance unit 389lighting, fluorescent and neon 173lightning protection 147linear array transducers 309linear electric motors 129linear generator principle 133linear particle accelerators 286–7linearising formulae 447Livingston, Stanley, and linear accelerator

287longitudinal relaxation time constant (T1)

358Lorentz, Hendrik

and aether 63factor (γ) 67

loudspeaker-making activity 126loudspeakers 91, 93low Earth orbits (LEOs) 43luminiferous aether 61luminosity classes 395–6Lyman, Theodore, spectral series for

hydrogen atom 233–4

M-mode ultrasound scans 313Mach, Ernst (1838–1916), and ratio of speed

of sound 48magnetic field strength (B)

as magnetic flux density 101in motor effect 92

magnetic fieldsand electric currents 87forces on charged particles in 89

magnetic flux (ΦB) 101–3magnetic hysteresis losses

in induction cookware 108in transformers 140

magnetic levitation (maglev) 213–14trains 129, 218–19

magnetic moment 355magnetic poles 88magnetic resonance imaging (MRI) 217–18

applications 362behaviour of hydrogen proton in 357see also MRI scanners

magnetism, and spin of atomic particles 354–5

magnification of telescopes 371, 374–5magnitudes of stars 397–400main sequence stars, properties relative to

Sun 425mammograms 325Manhattan Project 279–80Mariner 10, use of gravity assist by 45Marsden, Ernest, and model of atom 229–30

massequivalence to energy 72–4relativistic 71

mass defect 267mass–energy 73mass–luminosity relationship 413massive stars, fate of 430–1matter, crystal structure of 204–5matter waves, de Broglie’s equation 248–50Maxwell, James Clerk

electromagnetic wave theory 174–5equations of 61–2

medical radioisotopes 283medium Earth orbits (MEOs) 43Meissner effect 212Meitner, Lise

and beta decay 265and radioactive isotopes 269

mesons 292metal-oxide-semiconductor field-effect

transistor (MOSFET) 200metals, crystal structure and electrical

conductivity 209–10methamphetamines, brain tissue loss in users

302metric prefixes 442–3Michell, John, weighing the Earth 25Michelson, Albert, and speed of light 62Michelson–Morley experiment 62–4

results interpretation activity 75–6microscopes, optical to electron 259Minkowski, Hermann, and spacetime 72moderators in nuclear reactor 281molybdenum-99 production by ANSTO 353momentum conservation, in slingshot effect

44Moore’s law 223Morley, Edward, and speed of light 62Moseley, Henry 247

and model of atom 230motion, components of 5motor effect (F) 90–3

activity 97MRI scanners

operation of 360–1see also magnetic resonance imaging

(MRI)muons 286

n-type semiconductors 192National Medical Cyclotron at Royal Prince

Alfred Hospital, Sydney, radioisotope production at 282, 345, 353

nebulae, types 422negative glow 158neon lights 138, 173neutrinos, Pauli’s proposal 265–6neutron, discovery of 260–1neutron diffractometers, ANSTO’s Echidna

285neutron scattering, applications of 272, 278neutron stars 431, 432neutrons, number (N) in atomic nucleus

262–3

Newcomb, Simon, on future of astronomy 434

Newton, Isaaccomposition of light 390law of universal gravitation 10–11telescope of 372

non-coherent bundles of optical fibres 335–6non-elliptical orbits 41–2non-inertial frames of reference 59non-periodic variable stars 414nuclear chain reactions, controlled and

uncontrolled 270–1nuclear fission 269nuclear fission reactors, components 280–1nuclear fusion energy generation 218nuclear reactors, producing radioisotopes in

345nucleons, gravitational attraction of 261–2numerical calculation skills 443–4

Oberth, Herman, space flight and travel 27Ohm’s law, and electrical resistance 86O’Neill, Gerard, US physicist 28–9OPAL nuclear research reactor

Bragg Institute at 272Echidna at 285radioisotope production at 282, 345

open orbits 41–2open star clusters 433–4optical fibres 334

in endoscopy 333, 335–7orbital decay 46–7orbital velocity 36

of Earth 35orbits

elliptical 37–9manoeuvres between 57of planets 35–40of satellites 43–4types 41–2

osteoporosis detection using ultrasound 315

p–n junctions (diodes) 193–5as solar cells 196

p-type semiconductors 192parabolic orbits 41–2parabolic trajectories 3–7parallax angle (p) 388–9parallax measurement of star distance 388–9parallel wires, forces between 93–6parsec (pc) distance unit 389particle accelerators

and radioisotopes 282types 286–9

particle detectors 289path length of waves 205Pauli, Wolfgang

exclusion principle 255and neutrino 266

pendulum, oscillation period T 13period–luminosity relationship, for Cepheid

stars 416periodic variable stars 415periods, Kepler’s third law of 38

469

Index

Perrin, Jean, and cathode rays 160PET see positron emission tomographyphonons 210photocathode and photoelectrons 182photocell 184photoelectric effect

applications 184–5discovery 182–3

photometry, astronomical 397photomultiplier tube 184photon energy (E) 393photons, Einstein’s proposal 179, 181photovoltaic cells (PVs) 196piezoelectric effect and transducers 308–9Planck, Max

and black body radiation curve 179constant (h) 181, 231–2and electromagnetic wave emitter energy

180planetary nebula 429planets, finding new 40Plücker, Julius, and vacuum discharge tubes

156–7‘plum pudding’ model of atom 229plutonium-239 atomic bomb 279Pogson, Norman, and star magnitude scale

397Poincaré, Jules Henri, and speed of light 65polarised electromagnetic waves 177poles, geographic and magnetic 88positive column 158positron decay 342positron emission tomography (PET) 184–5,

283, 347–9potential difference 86power, electric see electric power (P)power generation, transmission and storage

using superconductors 219power-line transmission structures 145–6precession in rotation of magnetic moment

356precision in experiments 445prefixes, metric 442–3presenting research 446–7pressurised water reactor 281principle of relativity, Einstein’s 58–9projectile motion 4–10

activity 20proper length 69proper mass (rest mass) 71proper time 67proton–proton chain 425–6protons, electrostatic repulsion of 261–2protostars, evolution of 424pulsars 432

quanta of energy 179quantum computers 223quantum mechanics, Solvay Conference

(1927) 247–8quantum number (n) 231

of wavelengths in Bohr orbit 253quantum physics 226

quarks 292–4quasars, emission spectra for 394

racing magnets 107radiation shielding, in nuclear reactors 281radio antennae 175–7radio frequency (RF) waves 356radio telescopes 372–3radio transmitters and receivers 174–5radio waves, speed of 175–7radioactive decay of isotopes 340–3radioactive tracers 282radiographs, conventional 324radioisotopes

decay of 340–3half-life of 343production of 345range of uses 282–4

radiopharmaceuticals 283choice of 344–5

radiotherapy 283random error 445Reber, Grote, and radio telescope 373rectifiers 195re-entry into Earth’s atmosphere 46–50

safe corridors 47–8reflecting telescopes 372–3reflection nebulae 422refracting telescopes 371–2regenerative braking 134relativistic mass and speed 71relativity, Einstein’s principle of 58–9relaxation of hydrogen protons after cessation

of RF pulse 358–9relaxation time constants (T1 and T2) of

body tissues 359–60research, presenting 446–7resistance, electrical

from crystal structure 210losses in transmission lines due to 144–5and Ohm’s law 86and power 86

resistive heating (Q)in induction cooktops 108in transformers 140

resonance 358rest mass (proper mass) 71reverse bias in p–n junctions 195RF transceiver coils in MRI scanners 361right-hand grip rule 878right-hand palm (or push) rule 89, 90–1,

123rocket engines and stages 30–1rocketry

history of 26–7, 29researchers in 20th century 27–9

rocketsforces during take-off 334thrust on 30

roller-coasters, eddy current braking in 112–13

Röntgen, Wilhelm Conrad, discovers X-rays 207

RR Lyrae variable stars 415–16on HR diagram 414

Rutherford, Ernest, atomic model of 228–30Rutherford–Bohr model of atom 236

limitations of 239–40Rydberg, Johannes, constant (R) of 233

Salam, Abdus, and electro-weak theory 292satellites, orbits of 43–4Savitch, Pavle, and radioactive isotopes 269sawtooth waveform on CRO 167–8scans

MRI 217–18, 360–1, 362ultrasound 312–14

Schrodinger, Erwinand de Broglie’s hypothesis 250wave function of 253–4

seeing in telescopes 378–9Segré chart for uranium-238 decay series 264semi-synchronous satellites 43semiconductor devices 193–6semiconductors, explanation 190–3sensitivity of telescopes 375–6shaded-pole AC induction motors 124shock waves from projectiles 49silicon doping 192simple DC motors, operation of 117–19simultaneity, relativity of 65–6single-emission computed tomography

(SPECT) 347single-phase alternating current 121

induction motors 124sinusoidal waveform (trace) on CRO screen

168slingshot effect (gravity assist) 44–5solar cells 196solid-state devices, compared with

thermionic devices 199solid state physics 190Solvay conference (1927) participants 248sonograms 306Space Shuttle launch, g-force during 34Space Shuttle re-entry and landing 48–51spacecraft, launching 26–35spacetime interval 72special relativity, Einstein’s theory 64–8spectra, continuous 390, 392spectral classes 395spectral lines

fine and hyperfine 240relative intensity of 239and size of stars 396–7

spectrographs 391spectrometers and black body radiation 180spectroscope binary stars 411–12spectroscopes, CDs and DVDs as 155spectroscopes (spectrometers) 232

and spectrographs 391spectroscopic parallax, for star distance

401–2spectroscopy, astronomical 390spin of atomic particles, and magnetism

354–5spin of electrons 217springs, behaviour of 15Square Kilometre Array (SKA) telescope 382

470

Index

squirrel-cage rotors 123stable orbits 41Standard Model of matter 292–4standards for mass and length 79standing waves 175–6

as electron orbits 252–3Stanford linear accelerator 287star catalogues 369star clusters

activity 435age of 433–4

starsageing of 425–7binary 407–10birth of 423–4colour index 400–1composition of 389–94distance modulus 399–400evolution 425–7evolution of low to medium mass 429evolution of massive 430–1magnitude measurement 397–9naming 415size from spectral lines 396spectroscopic parallax 401–2variable 413–17

step-down and step-up transformers 137Stoney, George Johnston, names electron 166Strassman, Fritz, and radioactive isotopes 269strong nuclear force 262strontium emission spectrum 239substations 142–3Sun

future of 428–9 life on main sequence of HR diagram

427superconducting quantum interference

device (SQUID) 218superconductors

applications 217–19critical temperature (Tc) 211discovery of 211high-temperature 216Meissner effect in 212type-I and type-II 212–13

supergiant stars 421supernova 431supersonic flight 48–9supply emf in DC electric motors 120surface-conduction electron-emitter (SCE)

display TV sets (SED-TVs) 173synchronous AC motors 130–1synchrotrons 288systematic error 445Szilard, Leo, and chain reactions 270

tangential velocity at rocket launch 34–5Taylor, Richard, and quarks 292technetium-99m, generation and use 346telescopes

angular resolution 376–7early optical 370–3of the future 382interferometry in 380–1

magnification 374methods of sharpening images 377–80radio 372–3sensitivity 375–6

televisionCRT tubes used in 168–9LCD, plasma and SED-TVs 173

Tesla, Nicola, and AC motors and generators 141

thermionic devices, compared with solid-state devices 199

thermionic diodes and triodes (valves) 198thermonuclear bombs 280Thomson, George P, and electron beam

diffraction 252Thomson, Joseph John 160

discovery of electron 165–6Thomson, William (Lord Kelvin), and

temperature scale 210three-phase AC induction motors 122–3three-phase alternating current 121–23D real-time ultrasound images 314thrust on rocket (FT) 30time dilation 66–7time of flight of projectiles 6torch without batteries 133torque (τ) on a rotating coil 115–17total internal reflection, in optical fibres 334trains, eddy current braking in 112–13trajectories 3

components 5–7transducers

piezoelectric 308–9for ultrasound 305

transformersefficiency and design 137–40in electricity distribution 144–5in the home 144principles 136–7

transistors 198–9transmission towers for high-voltage AC

power 145–6transmutation, artificial and natural 263–5transverse magnetisation (Bxy) 359transverse relaxation time constant (T2) 359triodes as current amplifiers 198triple alpha process 428Tsiolovsky, Konstantin, rocket equation 27twin paradox 682D real-time ultrasound scans 313–142D slice image construction from CAT

scanning 327–8type-I and type-II superconductors 212–13

ultrasound 304–5reflection at tissue boundaries 310–12

ultrasound imagingand acoustic impedance 310–11limitations 307–8in obstetrics 307principles 305–7types of scans 312–16using piezoelectric transducers 308–9

ultrasound scans, types 312–14

ultraviolet catastrophe 179, 180uncertainty principle, of Heisenberg 254–5uniform circular motion, activity 52universal electric motors 122universal gravitation

Newton’s law of 10–11see also law of universal gravitation

universal gravitational constant G 11uranium-235 atomic bomb 279

V838 Monocerotis, supergiant star 421valence bands 189–90

holes in 191valence electrons 189valence level 189Vallebona, Alessandro, and tomography 321valves (thermionic devices) 198Van de Graaff accelerator 288variable stars 413–15

observed changes in 421vascular structures, X-ray images of 325verbs, use of 448–51visual binary stars 411voltage (V) 86von Braun, Werner, Moon mission leader 28vortex states, in magnetic levitation 214Voyager 1 and 2 space missions 46

wave function, Schrodinger’s 254wave interference 205–7wave mechanics of Schrodinger 250wave nature of electrons, de Broglie’s

hypothesis 248–52wave properties of light 206–7wavelength of X-rays (λe) 209waves, path length 205weak nuclear force 266weight in gravitational fields 13–14weightlessness effect 31Weinberg, Steven, and electro-weak theory

292Westinghouse, George, and AC electricity

141white dwarfs 429Wollaston, William, and lines in Sun

spectrum 390work and gravitational potential energy

16–17WR 104 star system 386–7

X-ray binary stars 412X-ray detector technology 326X-ray diffraction by crystals 207X-ray images

production 324–5types 320–1

X-ray tubes 321–2X-rays, types and properties 322–3

Zeeman, Pieter, effect of 240zero-age main sequence (ZAMS) 424Zweig, George, and quarks 292

471

Acknowledgements

We would like to thank the following for permission to reproduce photographs, texts and illustrations. The following abbreviations are used in this list: t = top, b = bottom, c = centre, l = left, r = right.

AIP Emilio Segre Visual Archives: p. 216.

Alamy Pty Ltd: p. 259r.

Andrew Dunn: p. 372b.

ANSTO – Australian Nuclear Science & Technology Organisation: p. 345/Vanessa Peterson: pp. 278, 285.

Australia Telescope National Facility: pp. 378l, 431l.

Barnaby Norris: pp. 368, 375, 391.

BioMed Central: p. 313.

Brian James: p. 96.

CartoonStock: p. 299.

CERN: pp. 226, 288, 289, 293.

Corbis: pp. 61, 303.

CSIRO Publishing: p. 188.

Damian Peach: p. 376.

Dave McKinnon: p. 411.

David Malin Images: pp. 369, 400, 401.

Dreamstime: pp. 14, 314b, 324l, 352tl, 356, 360.

Peter Tuthill and WM Keck Observatory: p. 386.

D. Smyth: p. 381l.

Fourmilab: p. 11.

Fundamental Photographs, NYC: p. 251l.

Gemini Observatory: p. 378r.

Greg Konkel: p. 383.

Harald Hess: p. 214.

Physics Stage 6 Syllabus © Board of Studies NSW for and on behalf of the Crown in the right of the State of New South Wales, 2007: pp. x–xxiii, Formulae and Data sheet: p. 473, Periodic Table of the Elements: p. 474. The Board of Studies does not endorse model answers prepared by or for the Publisher and accompanying the material. The Office of the Board of Studies takes no responsibility for errors in the reproduction of the material supplied by the Office of the Board of Studies to the Publisher.

Images.com: p. 2.

Institut International de Physique Solvay: p. 248.

Intel: p. 223.

iStockphoto: pp. 138, 139, 142, 178, 217, 307, 326, 338, 348br, 350tl, 380b, 441.

ITER: pp. 218.

Jason Lee: p. 290.

Jim Mosher: p. 372t.

John Rowe Animation: p. 412.

Lorojon Pty Ltd: p. 197b.

NASA: pp. 17, 26, 28t, 28b, 33, 36, 49a,b,d, 50, 60, 72, 382, 421, 423, 429, 431r.

NRAO/AUI/NSF: p. 373.

PA/Jeff Stanger: pp. 83, 90, 115, 118, 124, 132, 133, 144r /John O’Byrne: p. 381r.

Palomar Observatory, Caltech: p. 380t.

Photolibrary Pty Ltd: pp. 3, 12, 13, 15, 27, 28c, 91, 114, 119, 122, 127, 141, 146, 154b, 155, 167, 176, 180, 185, 197t, 204, 207, 210, 213, 227, 228, 235, 246, 249, 251c, 251r, 252, 254, 255, 260, 271, 306, 308, 309, 314t, 314c, 315, 319, 324r, 325tr, 328, 332r, 333, 337, 348tl, 349, 352tr, 362, 367, 390b, 392, 422, 433, 440.

472

Sebastian Terfloth: p. 112.

The Picture Source: p. 7.

Radiation Oncology Department of the Prince of Wales Hospital, Randwick NSW: p. 366.

Rod Nave: p. 241.

Reproduced with permission from the Ministero per i Beni e le Attività Culturali, Italy/Biblioteca Nazionale Central, Firenze: p. 4.

Sebastian Egner: p. 379.

Shutterstock: pp. 59, 82, 86, 121, 129, 144l, 145, 154t, 259l, 325tl, 332l, 365l, 377.

Smithsonian National Air and Space Museum: p. 49 (Soyuz).

U.C.L.A./Dr. Paul Thompson: p. 302.

University of Arizona/Tim Hunter: p. 324c.

UBC & Vancouver Coastal Health Research Institute/Department of Radiology: pp. 347, 352br.

Every effort has been made to trace and acknowledge copyright. However, should any infringement have occurred, the publishers tender their apologies and invite copyright owners to contact them.

Acknowledgements

473

FORMULAE SHEET cOnTinUEdF qvB= sinθ

AV

V=0

out

in

EVd

= V

VRR

= −out

in

f

i

E hf=

c f= λ

Z v= ρ

I

I

Z Z

Z Z

r =−[ ]+[ ]

2 12

2 12

0

dATA SHEETCharge on electron, qe –1.602 × 10–19 C

Mass of electron, me 9.109 × 10–31 kg

Mass of neutron, mn 1.675 × 10–27 kg

Mass of proton, mp 1.673 × 10–27 kg

Speed of sound in air 340 m s–1

Earth’s gravitational acceleration, g 9.8 m s–2

Speed of light, c 3.00 × 108 m s–1

Magnetic force constant, k = μπ0

22.0 × 10–7 N A–2

Universal gravitational constant, G 6.67 × 10–11 N m2 kg–2

Mass of Earth 6.0 × 1024 kg

Planck constant, h 6.626 × 10–34 J s

Rydberg constant, R (hydrogen) 1.097 × 107 m–1

Atomic mass unit, u 1.661 × 10–27 kg

931.5 MeV/c2

1 eV 1.602 × 10–19 J

Density of water, ρ 1.00 × 103 kg m–3

Specific heat capacity of water 4.18 × 103 J kg–1 K–1

FORMULAE SHEETv = fλ

E Gm m

rp = − 1 2

Id

12

F mg=

vv

ir=1

2

sinsin

v ux x=2 2

EFq= v u at= +

RVI

= v u a yy y y= +2 2 2 Δ

P VI= x u tx=Δ

VIt=Energy y u t a ty y= + 12

2Δ

vrt

=avΔΔ

r

T

GM=3

2 24π

avt

av u

t= = −

therefore vavaΔΔ F

Gm m

d= 1 2

2

F ma=Σ E mc= 2

Fmv

r=2

l lv

cv = −

2

210

E mvk = 12

2t

t

v

c

v =

−2

21

0

W Fs=m

m

v

c

v =

−2

21

0

p mv=

Ft=Impulse d p= 1

F

lk

I I

d= 1 2 M m

d= − 510

log

F BIl= sinθ I

IA

B

m mB A=−( )

1005

Fd=τm m

r

GT+ = 4

1 2

2 3

2

π

nBIA= cosτ θR

n nf i

= −1 1 12 2λ

V

V

n

np

s

p

s=

hmv=λ

Formulae and data sheets

474

PERI

ODIC

TABL

E OF

THE E

LEME

NTS

KE

Y

Ato

mic

Num

ber

79

A

u S

ymbo

l of

ele

men

t

Ato

mic

Wei

ght

19

7.0

G

old

Nam

e of

ele

men

t

1

H

1.0

08

H

ydro

gen

2

He

4.0

03

H

eliu

m

3

Li

6.9

41

Li

thiu

m

4

Be

9.0

12

B

eryl

lium

5

B

10

.81

B

oron

6

C

12

.01

C

arbo

n

7

N

14

.01

N

itro

gen

8

O

16

.00

O

xyge

n

9

F 1

9.0

0

Fluo

rine

10

N

e 2

0.1

8

Neo

n

11

N

a 2

2.9

9S

odiu

m

12

M

g 2

4.3

1

Mag

nesi

um

13

A

l 2

6.9

8

Alu

min

ium

14

S

i 2

8.0

9

Sili

con

15

P

3

0.9

7

Pho

spho

rus

16

S

3

2.0

7

Sul

fur

17

C

l 3

5.4

5

Chl

orin

e

18

A

r 3

9.9

5

Arg

on

19

K

3

9.1

0

Pot

assi

um

20

C

a 4

0.0

8

Cal

cium

21

S

c 4

4.9

6

Sca

ndiu

m

22

Ti

4

7.8

7

Tita

nium

23

V

50

.94

Va

nadi

um

24

C

r 5

2.0

0

Chr

omiu

m

25

M

n 5

4.9

4

Man

gane

se

26

Fe

5

5.8

5

Iron

27

C

o 5

8.9

3

Cob

alt

28

N

i 5

8.6

9

Nic

kel

29

C

u 6

3.5

5

Cop

per

30

Zn

6

5.4

1

Zinc

31

G

a 6

9.7

2

Gal

lium

32

G

e 7

2.6

4

Ger

man

ium

33

A

s 7

4.9

2

Ars

enic

34

S

e 7

8.9

6

Sel

eniu

m

35

B

r 7

9.9

0

Bro

min

e

36

K

r 8

3.8

0

Kry

pton

37

3

8

39

4

0

41

4

2

43

4

4

45

4

6

47

4

8

49

5

0

51

5

2

53

5

4

Rb

Sr

Y Zr

N

b M

o Tc

R

u R

h P

d A

g C

d In

S

n S

b Te

I

Xe

85

.47

8

7.6

2

88

.91

9

1.2

2

92

.91

9

5.9

4

[97

.91

] 1

01

.1

10

2.9

1

06

.4

10

7.9

1

12

.4

11

4.8

1

18

.7

12

1.8

1

27

.6

12

6.9

1

31

.3

Rub

idiu

m

Str

onti

um

Yttr

ium

Zi

rcon

ium

N

iobi

um

Mol

ybde

num

Te

chne

tium

R

uthe

nium

R

hodi

um

Palla

dium

S

ilver

C

adm

ium

In

dium

Ti

n A

ntim

ony

Tellu

rium

Io

dine

Xe

non

55

5

6

57

–71

7

2

73

7

4

75

7

6

77

7

8

79

8

0

81

8

2

83

8

4

85

8

6

Cs

Ba

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

13

2.9

1

37

.3

17

8.5

1

80

.9

18

3.8

1

86

.2

19

0.2

1

92

.2

19

5.1

1

97

.0

20

0.6

2

04

.4

20

7.2

2

09

.0

[20

9.0

] [2

10

.0]

[22

2.0

] C

aesi

um

Bar

ium

La

ntha

noid

s H

afni

um

Tant

alum

Tu

ngst

en

Rhe

nium

O

smiu

m

Irid

ium

P

lati

num

G

old

Mer

cury

Th

alliu

m

Lead

B

ism

uth

Pol

oniu

m

Ast

atin

e R

adon

87

8

8

89

–10

3

10

4

10

5

10

6

10

7

10

8

10

9

11

0

11

1

Fr

Ra

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

[22

3]

[22

6]

[26

1]

[26

2]

[26

6]

[26

4]

[27

7]

[26

8]

[27

1]

[27

2]

Fran

cium

R

adiu

m

Act

inoi

ds

Rut

herf

ordi

um

Dub

nium

S

eabo

rgiu

m

Boh

rium

H

assi

um

Mei

tner

ium

D

arm

stad

tium

Roe

ntge

nium

Lant

hano

ids

57

La

1

38

.9

Lant

hanu

m

58

C

e 1

40

.1

Cer

ium

59

P

r 1

40

.9

Pra

seod

ymiu

m

60

N

d 1

44

.2

Neo

dym

ium

61

P

m

[14

5]

Pro

met

hium

62

S

m

15

0.4

S

amar

ium

63

E

u 1

52

.0

Eur

opiu

m

64

G

d 1

57

.3

Gad

olin

ium

65

Tb

1

58

.9

Terb

ium

66

D

y 1

62

.5

Dys

pros

ium

67

H

o 1

64

.9

Hol

miu

m

68

E

r 1

67

.3

Erb

ium

69

Tm

1

68

.9

Thul

ium

70

Yb

1

73

.0

Ytte

rbiu

m

71

Lu

1

75

.0

Lute

tium

Act

inoi

ds

89

A

c [2

27

] A

ctin

ium

90

Th

2

32

.0

Thor

ium

91

Pa

23

1.0

P

rota

ctin

ium

92

U

2

38

.0

Ura

nium

93

N

p [2

37

] N

eptu

nium

94

P

u [2

44

] P

luto

nium

95

A

m

[24

3]

Am

eric

ium

96

C

m

[24

7]

Cur

ium

97

B

k [2

47

] B

erke

lium

98

C

f [2

51

] C

alif

orni

um

99

E

s [2

52

] E

inst

eini

um

100

Fm

[25

7]

Ferm

ium

101

Md

[25

8]

Men

dele

vium

102

No

[25

9]

Nob

eliu

m

103

Lr

[26

2]

Law

renc

ium

For

elem

ents

tha

t ha

ve n

o st

able

or

long

-liv

ed n

uclid

es,

the

mas

s nu

mbe

r of

the

nuc

lide

wit

h th

e lo

nges

t co

nfir

med

hal

f-lif

e is

list

ed b

etw

een

squa

re b

rack

ets.

The

Inte

rnat

iona

l Uni

on o

f P

ure

and

App

lied

Che

mis

try

Per

iodi

c Ta

ble

of t

he E

lem

ents

(O

ctob

er 2

00

5 v

ersi

on)

is t

he p

rinc

ipal

sou

rce

of d

ata.

Som

e da

ta m

ay h

ave

been

mod

ifie

d.

In 2 Physics

Documents

Transcript of In 2 Physics