Modern Physics Part II : Introduction to Quantum physicsCh.# 402 weeks (8 lectures)

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الدكتور المادة : مدرس1431 – 1430

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1- Planck’s Quantum Hypothesis; Blackbody Radiation

Any object in thermal equilibrium at any temperature is constantly emitting and absorbing radiation. However, not all materials are equally capable of absorbing and emitting radiation in different parts of the spectrum (i.e. in all frequency 0 to infinity). According to Kirchhoff’s law, any object which is a good absorber of radiation of a particular wavelength is also a good emitter of radiation of the same wavelength. A body which absorbs radiation of ALL wavelengths is called BLACKBODY.

Lecture # 1Ch. 40

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What is a blackbody?What is a blackbody? An object that absorbs all incident radiation, An object that absorbs all incident radiation,

i.e.i.e. no reflection. no reflection.

At thermal equilibrium with its surroundings At thermal equilibrium with its surroundings radiates as much energy as it absorbs, i.e. radiates as much energy as it absorbs, i.e.

blackbody is perfect emitter as well as absorberblackbody is perfect emitter as well as absorber

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The spectrum of blackbody radiation has been measured; it is found that the frequency of peak intensity increases linearly with temperature.

•The radiation is absorbed in the walls of the cavity•This causes a heating of the cavity walls•Atoms in the walls of the cavity will vibrate at frequencies characteristic of the temperature of the walls•These atoms then re-radiate the energy at this new characteristic frequency

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• This figure shows blackbody radiation curves for three different temperatures. Note that frequency increases to the left.

Wien’s displacement law

Plots characteristics:1. The distribution of frequencies is a function of the temperature of the blackbody.2. The total amount of radiation emitted increases with increasing temperature. 3. The position of the peak maximum shifts toward higher frequencies with increasing equilibrium temperature4. The wavelength at the peak of the curve is inversely proportional to the absolute temperature .

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peak vs Temperature

peak = 2.9 x 10-3 m

T(Kelvin)T

3100K(body temp)

2.9 x 10-3 m3100

=9x10-6m

58000K(Sun’s surface)

2.9 x 10-3 m58000 =0.5x10-6m

infrared light

visible light

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• This spectrum could not be reproduced using 19th-century physics. A solution was proposed by Max Planck in 1900:The energy of atomic oscillations within atoms cannot have an arbitrary value; it is related to the frequency:

The constant h is now called Planck’s constant.

Planck found the value of his constant by fitting blackbody curves:

Planck’s proposal was that the energy of an oscillation had to be an integral multiple of hf. This is called the quantization of energy.

The birth of the quantum theory

Basic Laws of Radiation

1) All objects emit radiant energy.

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Basic Laws of Radiation

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects.

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Basic Laws of Radiation

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects. The amount of energy radiated is proportional to the temperature of the object.

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Basic Laws of Radiation

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects. The amount of energy radiated is proportional to the temperature of the object raised to the fourth power.

This is the Stefan Boltzmann Law

F = T4

F = flux of energy (W/m2)T = temperature (K) = 5.67 x 10-8 W/m2K4 (a constant)

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All objects emit radiation whose total intensity is proportional to the fourth power of their temperature. This is called thermal radiation; a blackbody is one that emits thermal radiation only.

Basic Laws of Radiation

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects (per unit area). The amount of energy radiated is proportional to the temperature of the object.

3) The hotter the object, the shorter the wavelength () of emitted energy.

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Basic Laws of Radiation

1) All objects emit radiant energy.

2) Hotter objects emit more energy than colder objects (per unit area). The amount of energy radiated is proportional to the temperature of the object.

3) The hotter the object, the shorter the wavelength () of emitted energy.

This is Wien’s Law

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Stefan Boltzmann Law.

F = T4

F = flux of energy (W/m2)T = temperature (K) = 5.67 x 10-8 W/m2K4 (a constant)

Wien’s Law

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We can use these equations to calculate properties of energy radiating from the Sun and the Earth.

6,000 K 300 K

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T

(K)

max

(m)

region in spectrum

F

(W/m2)

Sun6000

Earth300

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T

(K)

max

(m)

region in spectrum

F

(W/m2)

Sun60000.5

Earth30010

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Electromagnetic Spectrum

(m)

1000 100 10 1 0.1 0.01

ultravioletvisiblelightinfraredmicrowaves x-rays

HighEnergy

LowEnergy

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T

(K)

max

(m)

region in spectrum

F

(W/m2)

Sun60000.5Visible(yellow?)

Earth30010infrared

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• Blue light from the Sun is removed from the beam by Rayleigh scattering, so the Sun appears yellow when viewed from Earth’s surface even though its radiation peaks in the green

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T

(K)

max

(m)

region in spectrum

F

(W/m2)

Sun60000.5Visible(green)

Earth30010infrared

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Stefan Boltzman Law.

F = T4

F = flux of energy (W/m2)T = temperature (K) = 5.67 x 10-8 W/m2K4 (a constant)

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T

(K)

max

(m)

region in spectru

m

F

(W/m2)

Sun60000.5Visible(green)

7 x 107

Earth30010infrared

460

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1. The human eye is most sensitive to 560-nm light. What is the temperature of a black body that would radiate most intensely at this wavelength?

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9

2.898 10 m K5.18 10 K

560 10 mT

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2. (a) Lightning produces a maximum air temperature on the order of 104 K, whereas (b) a nuclear explosion produces a temperature on the order of 107 K. Use Wien’s displacement law to find the order of magnitude of the wavelength of the thermally produced photons radiated with greatest intensity by each of these sources. Name the part of the electromagnetic spectrum where you would expect each to radiate most strongly.

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max 4

2.898 10 m K 2.898 10 m K~ ~10 m

10 KT

ultraviolet

310

max 7

2.898 10 m K~ ~10 m

10 K

ray

(a)

(b)

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3- The Photoelectric Effect

What Is The Photoelectric Effect?when light illuminates a piece of metal, the light will kick off electrons from the metal’s surface and these electrons can be detected as a change in the electric charge of the metal or as an electric current, emitted electrons are called PHOTOELCTRONs. Hence the name: photo for light and electric for the current. The explanation behind this simple phenomenon opened the door to revolutionary modern physics concepts regarding the composition of light, quantum mechanics, and what is now referred to as the “wave-particle duality” of nature.

Lecture # 2Ch. 40

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What is the Photoelectric Effect?What is the Photoelectric Effect?

A photon with energy hf strikes an electron and ejects it from the metal. E=hf = K.E. + w :-

w = work to remove electron from metal

K.E. = kinetic energy of ejected electron.

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Most commonly observed phenomena with light can be explained by waves. But the photoelectric effect suggested a particle nature for light.

The remarkable fact that the ejection energy was independent of the total energy of illumination showed that the interaction must be like that of a particle which gave all of its energy to the electron! This fit in well with Planck's hypothesis that:light in the blackbody radiation experiment could exist only in discrete bundles with energy

hE

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1- if v is kept constant, the photoelectric current increases with increasing intensity I of the incident radiation.

3- The emission of photoelectrons takes place ONLY if the frequency of the incident radiation is equal to or greater than a certain minimum frequency vs called the THRESHOLD FREQUENCY . vs is different for different surfaces.

2- Photoelectrons are emitted within less than 10-9 sec after the surface is illuminated by light, i.e. NO TIME LAG BETWEEN ILLUMINATION AND EMISSION.

We can obtain the following results experimentally

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4-The maximum kinetic energy, Kmax, of photoelectrons is independent of the intensity I of the incident light.

The stopping potential is the same for light of three different intensities but of

the same frequency

I3˃ I2 ˃ I1

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5- The maximum kinetic energy, Kmax, of photoelectrons depend on the frequency of the incident radiation.

The stopping potential is different for different v

even though I is the same

vs1˃ vs2 ˃ vs3

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6- There is a linear relation between Kmax, and v.

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Explanation of Classical “Problems”

• The effect is not observed below a certain cutoff frequency since the photon energy must be greater than or equal to the work function– Without this, electrons are not emitted, regardless of the intensity of the light

• The maximum Kmax depends only on the frequency and the work function, not on the intensity

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Einstein’s “quantum theory” Explanation

• A tiny packet of light energy, called a photon, would be emitted when a quantized oscillator jumped from one energy level to the next lower one– Extended Planck’s idea of quantization to electromagnetic radiation

• The photon’s energy would be E = hƒ

• Each photon can give all its energy to an electron in the metal

• The maximum kinetic energy of the liberated photoelectron is Kmax = hƒ – Φ• Φ is called the work function of the metal

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Photoelectric effect (Albert Einstein)

• The photoelectric effect is a phenomenon where electrons are ejected from the surface of a metal (or some other material) when light shines on it

• shine light rays --> current• increase intensity(total energy) --> current goes up• That makes sense.• The strange thing about this phenomenon, is that a

certain minimum frequency of light is required before any current is detected.

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Verification of Einstein’s Theory

• Experimental observations of a linear relationship between KE and frequency confirm Einstein’s theory

• The x-intercept is the cutoff frequency

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Cutoff Wavelength

• The cutoff wavelength is related to the work function

• Wavelengths greater than C incident on a material with a work function don’t result in the emission of photoelectrons

c

hc

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Example:Light of wavelength λ=5893 A° (589.3 nm)is incident on a potassium surface. The stopping potential for the emitted electrons (i.e. the maximum energy of the photoelectron) is 0.36 volt. Calculate: A) the work function, and B) the threshold frequency (or cutoff wavelength).

A)

The work function is ,

B)The cutoff wavelength is given by the relation

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4- Application of the Photoelectric Effect

The most obvious example is probably solar energy, which is produced by photovoltaic cells. These are made of semi-conducting material which produce electricity when exposed to sunlight. An everyday example is a solar powered calculator and a more exotic application would be solar power satellites that orbit around the earth.

- The photo-cell

The photo-cell is the most ranging of applications of the photoelectric effect. It is most commonly found in solar panels. it works on the basic principle of light striking the cathode which causes the emission of electrons, which in turn produces a current.

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5- Compton effect Arthur H. Compton observed the scattering of x-rays from electrons in a carbon target and found scattered x-rays with a longer wavelength than those incident upon the target. The shift of the wavelength increased with scattering angle according to the Compton formula (Compton Shift Equation):

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Compton explained and modeled the data by assuming a particle (photon) nature for light and applying conservation of energy and conservation of momentum to the collision between the photon and the electron.

The scattered photon has lower energy and therefore a longer wavelength according to the Planck relationship.

Q: In the Compton shift equation, show that the maximum shift is given as ,

where is the Compton wavelength

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Derivation of the Compton shift equation:

Energy conservation:

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0 'cm

hccm

hcee

Momentum conservation:

sinsin'

0 :direction

coscos'

:direction 0

vmh

y

vmhh

x

e

e

Three equations, three unknowns (', v and ) eliminating v and relation between ', 0 and :

)cos1(' 0 cm

h

e

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)cos1(''

2cos1'

2

'2

'cos

'2

')2(

)1(

)2( '

2'

'

2'

'

''

)1( cos'

2'

sin'

cos'

sin'

sin sinsin'

0

cos'

cos coscos'

000

2

0

2

00

222

0

0

2

0

2

0

2

2

0

2

222

2

2

022222

22

2

2

0

22

0

0

222

0

2

22

0

200

cm

hhhcm

h

hhcm

hhhhh

hhcm

hhvm

hchccm

hchcvcm

cmvcmhc

cmhc

cmvcmcm

cmhc

cmhc

cmhc

cmhc

hhhvm

hhhvm

hvmvm

h

hhvmvm

hh

ee

e

eeee

eee

eee

eeee

e

e

ee

ee

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7. Calculate the energy, in electron volts, of a photon whose frequency is 3.10 GHz, and determine the corresponding wavelength for this photon and state the classification of the electromagnetic spectrum

34 9 1 519

1.00 eV6.626 10 J s 3.10 10 s 1.28 10 eV

1.60 10 JE hf

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9

8

6

3.00 10 m s9.68 10 m 9.68 cm, radio wave

3.10 10 Hz

3.00 10 m s6.52 m, radio wave

46.0 10 Hz

cf

cf

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22. X-rays having an energy of 300 keV undergo Compton scattering from a target. The scattered rays are detected at 37.0° relative to the incident rays. Find (a) the Compton shift at this angle, (b) the energy of the scattered x-ray, and (c) the energy of the recoiling electron.

1 cose

hmc

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1331 8

6.626 101 cos37.0 4.88 10 m

9.11 10 3.00 10

00

hcE

34 8

3 19

0

6.626 10 3.00 10 m s300 10 eV 1.60 10 J eV

120 4.14 10 m 12

0 4.63 10 m

34 814

12

6.626 10 J s 3.00 10 m s4.30 10 J 268 keV

4.63 10 m

hcE

0 300 keV 268.5 keV 31.5 keVeK E E

:

:

a:

c:

b:

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An X-ray photon with =60 nm is scattered over 1500 by a target electron.(a) Find the change of its wavelength.(b) Find the angle between the directions of motion of the recoil electron

and the incident photon.(c) Find the energy of the recoil electron.

)a( ' 1 coseC

0 3cos150 0.866

2

12 121 cos 2.43 10 1 0.866 4.53 10eC m m

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)c(

momentum conservation:)b(

cos 'cosep p p along x axis:

along y axis: sin 'sinep p

cos 'cosep p p

sin'sin sin'tan'cos ' coscos

'

hcp

hc hcp p

0tan 0.258 14.4

/p h

' / 'p h

ep

x

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An X-ray photon with wavelength 0.8nm is scattered by an electron at rest. After the scattering the electron recoils with a speed equal to 1.4106m/s (non-relativistic case).NOTE: IN THE FOLLOWING, DO THE CALCULATIONS TO AT LEAST 4-DECIMAL PLACE ACCURACY

(a) Calculate the energy of the scattered photon in units eV (use conservation of energy).(b) Calculate the Compton shift in the photon’s wavelength, in meters.(c) Calculate the angle through which the photon was scattered.

(a)

231 6

19

9.1 10 1.4 10 /5.57

2 1.602 10 /

kg m sK eV

J eV

34 8

10 19

6.626 10 3 10 /1551

8 10 1.602 10 /

J s m shceV

m J eV

1551 5.6 1545.4'

hceV eV eV

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15 8

9'

4.136 10 3 10 /' 0.8029 10

1545.4ph

eV s m shcm

E eV

(b)

(c)

9 12' 0.0029 10 2.9 10m m

' 1 coseC

2.9cos 1 1 0.2

2.4eC

0102

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1- Photons and Electromagnetic waves1.1. Wave-Particle Duality: Light (Photon)

Most commonly observed phenomena with light can be explained by waves. But the photoelectric effect and the Compton scattering suggested a particle nature for light having energy hf and momentum h/λ.

How can light be considered a photon when we know it is a wave?Is light wave or particle?

Both are correct, but depend on the phenomenon being observed. And we must accept both models. The particle model and the wave model of light complement each other

Lecture # 3Ch. 40

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PhenomenonCan be explained in terms of

waves.Can be explained in terms of

particles.

Then electrons have a dual natures.

Reflection

Refraction

Interference

Diffraction

Polarization

Photoelectric effect

Compton scattering

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1.2. The wave properties of particles

Louis de Broglie postulated that because photons have both wave and particle characteristics, perhaps all forms of matter have both properties.

•The de Broglie wavelength of a particle is .

and the frequency of a particle is

So according to Louis de Broglie, every object of mass m and momentum p = mv has wave properties, with a de Broglie wavelength given by :

mv

h

p

h

Note: p = mv for v << c , and p = mv for any speed v, where = (1-v2/c2)-1/2.

p

h

h

Ef

de Broglie hypothesis the dual nature of matter:

• Particle nature (concepts): p and E.

• Wave nature (concepts): λ and ƒ.

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Q: Compare the de Broglie wavelengths of (a) an electron traveling at a speed that is one-hundredth the speed of light and (b) a baseball of mass 0.145 kg having a speed of 26.8 m/s (60 mph).

mv

hλ

Electronme = 9.11 × 10-31 kgv = c/100 = 3.00 × 106 m/s

Baseballm = 0.145 kgv = 26.8 m/s

Electronme = 9.11 × 10-31 kg v = 3.00 × 106 m/s

Baseballm = 0.145 kg v = 26.8 m/s

s

m10x3.00kg10x9.11

sJ10x6.63λ

631

34

2.43 × 10-10 m

s

m26.8kg0.145

sJ10x6.63λ

34

1.71 × 10-34 m

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protonelectron λλ

electronelectronelectronλ

vm

h

protonprotonprotonλ

vm

h

electronelectronvm

h

protonprotonvm

h

Q: The proton is heavier than the electron. How would the speed of these particles compare if their corresponding de Broglie wavelengths were about equal?

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electron

proton

proton

electron

m

m

v

v

electronelectronvm protonprotonvm

kg 10 x 9.10938

kg 10 x 1.6726231-

-27

proton

electron v

v

310 x 5 1.8361proton

electron

v

v

proton. the of that times 1800 than

more is electron the ofvelocity The

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1.3. Properties of Waves

•Wavelength (λ) is the distance between identical points on successive waves.

•Amplitude is the vertical distance from the midline of a wave to the peak or trough.

•Frequency (f) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).•The speed (V) of the wave = λ x f• Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.

• Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiationλ x f = c

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Example : A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region?

λ = 3.00 x 108 m/s / 6.0 x 104 Hz λ = 5.0 x 1012 nm

λ x f = c λ = c/f

Radio wave

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2.1 The double-slit experiment revisited

Electron diffraction experiment:• Parallel mono-energetic electron

beams are used.• Small slit widths compared to the

electron wavelength.• A typical wave interference pattern

is observed.

A minimum occurs at . / where2

,2

sin phpd

hd

Closing one slit alternatively: no interference pattern can be observed.

Conclusions:• An electron interacts with both slits simultaneously. It passes trough both

slits.• If an attempt is made to determine experimentally which slit the electron

goes through, the act of measuring destroys the interference pattern.

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Closing one slitNo interference patternThe result is the sum of the individual slit

The double-slit experiment shows the dual nature of the electron: the electrons are detected as a particle at a localized spot at some instant of time, but the probability of arrival at that spot is determined by finding the intensity of two interfering waves

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2.2 The Uncertainty Principle

In 1927, Werner Heisenberg showed that the position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision.

There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time

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If a measurement of the position of a particle is made with uncertainty Δx and a simultaneous measurement of its x component of momentum is made with uncertainty Δpx, the product of the two uncertainties can never be smaller than h/4π

Heisenberg uncertainty

principle

That means , IT IS PHYSICALY IMPOSSIBLE TO MEASURE SIMULTANEOUSLY THE EXACT POSTION

AND EXACT MOMENTUM OF A PARTICLE

The uncertainty dose not arise from imperfections in particle measuring instruments, rather arises from the quantum

structure of mater

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From de Broglie relation p=h/λ, we can calculate the momentum precisely, since any region along a single-wavelength is the same, and because we can not decide or identified the position of the particle that produced the wave, this mean we have infinite uncertainty in the position of the particle.

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HomeworkP. 1315, Ch. 40, Problems: #33, 34.

33. Calculate the de Broglie wavelength for a proton moving with a speed of 1.00 × 106 m/s.

34. Calculate the de Broglie wavelength for an electron that has kinetic energy (a) 50.0 eV and (b) 50.0 keV.

Q: What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?

Q: Calculate the de Broglie wavelength for an electron (me = 9.11x10-31

Kg) moving at 1x107 m/s

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67

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