Section 12.1: Introducing Quantum Theory Section 12.1 ...

39
Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.1-1 Section 12.1: Introducing Quantum Theory Section 12.1 Questions, page 619 1. In classical physics, energy can be transferred by collisions between particles and larger objects, including near-collisions when objects interact at a distance (e.g., charged particles, astronomical objects) and by waves, including water waves, sound, and light. 2. Classical particles differ from classical waves in that a particle occupies a small region of space with some mass density that abruptly changes outside the particle. In contrast, a wave spreads out over a large region and oscillates. Also, for transferring energy, particles tend to transfer energy in discrete parcels (lumps), one at a time, whereas waves continuously transfer energy and can interfere destructively. 3. (a) The double-slit experiment with electrons shows that the energy arrives in discrete lumps, which indicates that the electrons are particles. (b) The double-slit experiment with electrons also shows destructive interference, which indicates that the electrons are waves. 4. In the early twentieth century, classical physics could describe nearly everything physical. However, there were a few puzzling things related to microscopic phenomena, particularly the behaviour of atoms and electrons, that disagreed with predictions of classical physics. Atoms and electrons behave differently than large objects such as baseballs, rockets, and planets. In particular, microscopic objects can exhibit wave-like behaviour. Quantum theory was developed to describe both the wave-like and particle-like behaviour of microscopic things such as atoms and electrons. 5. The electron double-slit experiment showed that electrons can interfere like waves, which is a non-classical behaviour. The result could not be explained by classical physics and thus gave a major push to the development of quantum theory. 6. Quantum theory is more complete than Maxwell’s theory because it can describe both the wave-nature and the particle-nature of light. Maxwell’s theory describes only the wave-nature of light. 7. Golf balls hit toward a wall with two narrow slits would leave a distribution of marks on a wall behind the slits. The distribution would show two dense areas directly behind each slit. Many balls would not even hit the slits, but assuming that a high fraction of them did, the distribution would look like the one below. 8. (a) The distribution of baseballs behind the two open slits would look similar to (b). (b) The distribution of electrons behind the two open slits would look similar to (a). (The slits would have to be narrower in accordance with the smaller size of the electrons.)

Transcript of Section 12.1: Introducing Quantum Theory Section 12.1 ...

Page 1: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.1-1

Section 12.1: Introducing Quantum Theory Section 12.1 Questions, page 619 1. In classical physics, energy can be transferred by collisions between particles and larger objects, including near-collisions when objects interact at a distance (e.g., charged particles, astronomical objects) and by waves, including water waves, sound, and light. 2. Classical particles differ from classical waves in that a particle occupies a small region of space with some mass density that abruptly changes outside the particle. In contrast, a wave spreads out over a large region and oscillates. Also, for transferring energy, particles tend to transfer energy in discrete parcels (lumps), one at a time, whereas waves continuously transfer energy and can interfere destructively. 3. (a) The double-slit experiment with electrons shows that the energy arrives in discrete lumps, which indicates that the electrons are particles. (b) The double-slit experiment with electrons also shows destructive interference, which indicates that the electrons are waves. 4. In the early twentieth century, classical physics could describe nearly everything physical. However, there were a few puzzling things related to microscopic phenomena, particularly the behaviour of atoms and electrons, that disagreed with predictions of classical physics. Atoms and electrons behave differently than large objects such as baseballs, rockets, and planets. In particular, microscopic objects can exhibit wave-like behaviour. Quantum theory was developed to describe both the wave-like and particle-like behaviour of microscopic things such as atoms and electrons. 5. The electron double-slit experiment showed that electrons can interfere like waves, which is a non-classical behaviour. The result could not be explained by classical physics and thus gave a major push to the development of quantum theory. 6. Quantum theory is more complete than Maxwell’s theory because it can describe both the wave-nature and the particle-nature of light. Maxwell’s theory describes only the wave-nature of light. 7. Golf balls hit toward a wall with two narrow slits would leave a distribution of marks on a wall behind the slits. The distribution would show two dense areas directly behind each slit. Many balls would not even hit the slits, but assuming that a high fraction of them did, the distribution would look like the one below.

8. (a) The distribution of baseballs behind the two open slits would look similar to (b). (b) The distribution of electrons behind the two open slits would look similar to (a). (The slits would have to be narrower in accordance with the smaller size of the electrons.)

Page 2: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.1-2

9. Some of our intuitions gained through our experiences in the macroscopic world of large objects do not apply to the quantum world because objects in the quantum world exhibit both wave-like and particle-like behaviour. Things in the macroworld exhibit only wave-like or particle-like behaviour.

Page 3: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-1

Section 12.2: Photons and the Quantum Theory of Light Tutorial 1 Practice, page 624 1. Given: W = 4.60 !10"19 J; h = 6.63!10"34 J # s Required:

Ephoton

Analysis: Ephoton =W

Solution:

Ephoton =W

Ephoton = 4.60 !10"19 J

Statement: The lowest photon energy that can cause emission of electrons from calcium is 194.60 10 J−× , the work function of calcium. 2. Given: !0 = 268 nm = 2.68"10#7 m; h = 6.63"10#34 J $s; c = 3.0!108 m/s Required: the work function, W Analysis: The lowest photon energy must satisfy the equation

Ephoton =W . Use

Einstein’s equation for the photon energy, written in terms of the wavelength, and substitute the threshold wavelength λ0.

photon0

hcEλ

=

Solution:

Ephoton =hc!0

= (6.63"10#34 J $ s ) (3.0"108 m/s )(2.68 "10–7 m )

Ephoton = 7.42"10#19 J

Statement: The minimum photon energy to release an electron from the material is 7.42!10"19 J , which is much closer to silver’s value ( 7.43!10"19 J ) than to lead’s ( 6.81!10"19 J ). So, the material is silver. Tutorial 2 Practice, page 626 1. Given: ! = 450 nm = 4.50"10–7 m; h = 6.63"10#34 J $s Required:

pphoton , the momentum of the photon

Analysis: Use the relation pphoton =

h!

.

Solution:

pphoton =h!

= 6.63"10#34 J $s4.50"10–7 m

pphoton = 1.5"10#27 kg $m / s

Statement: The photon’s momentum is 1.5!10"27 kg #m / s .

Page 4: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-2

2. Given: ! = 630 nm = 6.30 "10–7 m; h = 6.63"10#34 J $s; c = 3.0"108 m/s Required:

Ephoton , the energy of the photon

Analysis: Use the relation Ephoton =

hc!0

.

Solution:

Ephoton =hc!0

= (6.63"10#34 J $ s )(3.0"108 m/s )6.30"10–7 m

Ephoton = 3.2"10#19 J

Statement: The photon’s energy is 193.2 10 J−× . 3. Given: 14 8

photon 2.2 10 J; 3.0 10 m/sE c−= × = ×

Required: pphoton , the momentum of the gamma photon

Analysis: Ephoton = hf and

pphoton =

hfc

, so pphoton =

Ephoton

c.

Solution:

pphoton =Ephoton

c

= 2.2!10"14 J3.0!108 m/s

pphoton = 7.3!10"23 kg #m / s

Statement: The momentum of the gamma ray is 7.3!10"23 kg #m / s . Tutorial 3 Practice, page 629 1. Given: T = 5100 K Required: !

Analysis: Use Wien’s law, !max =

2.90 "10#3 m $KT

.

Solution:

!max =2.90 "10#3 m $K

T

= 2.90 "10#3 m $ K5100 K

!max = 5.7 "10#7 m

Statement: The maximum wavelength of the 5100 K blackbody is 75.7 10 m−× , which gives a yellow colour.

Page 5: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-3

2. Given: !max = 510 nm = 5.10 "10#7 m Required: T Analysis: Use Wien’s law, rearranged to obtain T.

!max =2.90 "10#3 m $K

T

T = 2.90 "10#3 m $K!max

Solution:

T = 2.90 !10"3 m #K$max

= 2.90 !10"3 m #K5.10 !10"7 m

T = 5700 K

Statement: The temperature of the aqua-coloured blackbody is 5700 K. 3. Given: T = 37 ˚C = 310 K Required: !max

Analysis: Use Wien’s law, !max =

2.90 "10#3 m $KT

.

Solution:

!max =2.90"10#3 m $K

T

= 2.90"10#3 m $ K3.1"102 K

!max = 9.4"10#6 m

Statement: The wavelength of maximum intensity for a human body is 9.4×10–6 m, or 9.4 µm. This is infrared light, which can be imaged with an infrared camera, but not the human eye. Research This: Exploring Photonics, page 630 Answers may vary. Students will choose a technology or process that uses the particle nature of light. The sample answers for A to C below are for the photomultiplier tube: A. A photomultiplier tube relies upon the photoelectric effect, and thus relies upon the particle nature of light. The tube also relies upon secondary emission, another process that is similar to the photoelectric effect but occurs for massive, charged particles such as electrons. Photomultiplier tubes consist of a photocathode and a series of electrodes, called dynodes, in an evacuated glass enclosure. The electrodes are each maintained at a more positive potential. A photon that strikes the photoemissive cathode releases an electron from the photocathode (material with very low work function). The released electron gets focused and accelerated by a magnetic electron lens and strikes the first dynode. The collision releases several electrons in a process called secondary emission. The electrons strike the next dynode, releasing even more electrons, and the process cascades along many more dynodes. This cascading effect creates 105 to 107 electrons for each photon hitting the first cathode, depending on the number of dynodes and the

Page 6: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-4

accelerating voltage. The resulting avalanche of electrons hits the output (anode), where the amplified signal can be measured. B. The photomultiplier tube counts photons, which would not be possible to interpret without the quantum theory of light. C. The avalanche photodiode is the solid-state equivalent to the photomultiplier tube. D. As of 2009, there were over 5000 Canadian companies in the Canadian Photonics Consortium, and these companies employed nearly 300 000 people. Section 12.2 Questions, page 631 1. Given: W = 5.0 eV = 8.0 ! 10–19 J; h = 6.63!10"34 J # s Required: f0, the minimum photon frequency that can eject an electron Analysis: Use Einstein’s relation for the photon energy,

Ephoton = hf0 , rearranged for f0,

and then substitute the photon energy with the work function, Ephoton =W .

Ephoton = hf0

f0 =Ephoton

h

Now, substitute W for Ephoton .

f0 =

Wh

Solution:

f0 =Wh

= 8.0!10"19 J6.63!10"34 J #s

f0 = 1.2!1015 Hz

Statement: The minimum frequency to eject an electron is 1.2!1015 Hz , which is ultraviolet, according to Table 3. 2. (a) The lower-energy material, cesium, is a better choice than aluminum for the metal piece in the photocell because it will capture all the photons that the aluminum will plus those with energy between 1.95 eV and 4.20 eV. Judging from Table 3, we would probably need a work function less than half the 5.0 eV of the aluminum to capture the lower-frequency visible light. (b) Given: W = 1.95 eV = 3.12 ! 10–19 J; h = 6.63!10"34 J # s Required: f0, the minimum photon frequency that can eject an electron Analysis: Use Einstein’s relation for the photon energy,

Ephoton = hf0 , rearranged for f0,

and then substitute the photon energy with the work function, Ephoton =W .

Ephoton = hf0

f0 =Ephoton

h

Now, substitute W for Ephoton .

Page 7: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-5

f0 =

Wh

Solution:

f0 =Wh

= 3.12!10"19 J6.63!10"34 J #s

f0 = 4.71!1014 Hz

Statement: The lowest photon frequency that can be measured with the photocell is 4.71!1014 Hz . (c) A frequency of 4.71!1014 Hz is near the lower boundary of the visible range of the electromagnetic spectrum ( 4.3!1014 Hz ). 3. The higher-intensity red light will not be able to eject electrons according to the photoelectric effect. The ability of light to eject electrons depends only on the frequency, not the intensity. So, if the low-intensity red light could not eject electrons, the higher-intensity red light cannot either. 4. (a) Given: f = 100 MHz = 1.00!108 Hz; h = 6.63!10"34 J #s; c = 3.0!108 m/s Required: the energy, E, and momentum, p, of the photon

Analysis: Combine the equations E = hf and p = hf

c to obtain

p = E

c. Since we can

use E to calculate p, first calculate E. Solution:

E = hf

= (6.63!10"34 J # s )(1.00!108 Hz )E = 6.63!10"26 J (two extra digits carried)

p = Ec

= 6.63!10"26 J3.0!108 m/s

p = 2!10"34 kg #m / s

Statement: The energy of an FM radio station with a frequency of 100 MHz is 7 !10"26 J , and the momentum is 2!10"34 kg #m / s . (b) Given: ! = 633 nm = 6.33"10–7 m; h = 6.63"10#34 J $s; c = 3.0"108 m/s Required: the energy, E, and momentum, p, of the photon

Analysis: Combine the equations E = hf and ! =

cf

to obtain E = hc

! , and use

p = E

c

from part (a). Since we can use E to calculate p, first calculate E.

Page 8: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-6

Solution:

E = hc!

= (6.63"10#34 J $ s )(3.0"108 m/s)6.33"10–7 m

= 3.1422"10#19 J (two extra digits carried)

E = 3.14"10#19 J

p = Ec

= 3.1422!10"19 J3.0!108 m/s

p = 1.05!10"27 kg #m / s

Statement: The energy of a red light with a wavelength of 633 nm is 3.14!10"19 J and the momentum is 1.05!10"27 kg #m / s . (c) Given: ! = 0.070 nm = 7.0 "10–11 m; h = 6.63"10#34 J $s; c = 3.0"108 m/s Required: the energy, E, and momentum, p, of the photon Analysis: The same as in part (b) above.

Solution:

E = hc!

= (6.63"10#34 J $ s )(3.0"108 m/s)7.0 "10–11 m

= 2.841"10#15 J (two extra digits carried)

E = 2.8"10#15 J

p = Ec

= 2.841!10"15 J3.0!108 m/s

p = 9.5!10"24 kg #m / s

Statement: The energy of an X-ray photon of wavelength 0.070 nm is 152.8 10 J−× , and the momentum is 9.5!10"24 kg #m / s . 5. An X-ray photon has greater energy than an ultraviolet photon because it has a greater frequency. 6. (a) There are 1.60 × 10–19 J in one electron-volt, so we multiply the 13.6 eV by 1.60 × 10–19 J to get 2.176 × 10–18 J (one extra digit carried). There are 2.18 × 10–18 J in 13.6 eV.

Page 9: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.2-7

(b) Given: E = 2.176×10–18 J; h = 6.63!10"34 J # s; c = 3.0 !108 m/s Required: ; f λ Analysis: Use Einstein’s relation for the energy of a photon, E = hf , rearranged to

isolate f: f = E

h . Then, use the relation between wavelength and frequency,

! = c

f, to

calculate λ.

Solution:

f = Eh

= 2.176!10"18 J6.63!10"34 J #s

= 3.2821!1015 Hz (two extra digits carried)f = 3.28!1015 Hz

! = cf

= 3.0"108 m/s3.2821"1015 Hz

! = 9.14"10#8 m

Statement: The frequency of the highest-energy photons emitted by a hydrogen atom is

3.28 !1015 s"1 , and the wavelength is 9.14!10"8 m , which correspond to ultraviolet photons. 7. The working of a solar cell relies on quantum mechanics through the photoelectric effect and the quantum theory of semiconductors. The photoelectric effect removes an electron from a special region within the semiconductor wafer, and the electron flows through a circuit, producing electrical power. The special region inside the semiconductor is similar to that in an LED. One side has been made to have electron charge carriers, and the other side to have missing electrons, which act as “holes” that carry positive charge according to the quantum theory of semiconductors. At the boundary, the electrons and holes combine, forming the special region called the depletion region. The depletion region does not have many charge carriers. The combining of the electrons and holes creates an electric field and thus a voltage. To generate power, the depletion region needs charges that can flow. These charges are produced by the photovoltaic effect, which is similar to the photoelectric effect; briefly, an incident photo travels into the depletion region and causes an electron to move up to a higher energy level in which it can conduct. The electron then gets pushed by the electric field through the circuit, generating electricity.

Page 10: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.3-1

Section 12.3: Wave Properties of Classical Particles Tutorial 1 Practice, page 634 1. Given: p = 1.8 !10"25 kg #m/s; h = 6.63!10"34 J # s Required: λ

Analysis: Use the de Broglie relation, hp

λ = .

Solution:

! = hp

=6.63"10#34 kg $m 2 /s

1.8 "10#25 kg $ m/s

! = 3.7 "10#9 m

Statement: The de Broglie wavelength of the electron is 3.7 !10"9 m , or 3.7 nm. 2. Given: m = 1.7 !10"27 kg; v = 3.4 !105 m/s; h = 6.63!10"34 J # s Required: λ Analysis: The speed of the proton is much less than light speed, so we can use the

classical momentum p = mv. Thus, the de Broglie relation, hp

λ = , becomes hmv

λ = .

Solution:

! = hmv

=6.63"10#34 kg $m 2 /s

(1.7 "10#27 kg )(3.4 "105 m/s )

! = 1.1"10#12 m

Statement: The proton’s de Broglie wavelength is 121.1 10 m−× . 3. Given: m = 140 g = 1.4 !10–1 kg; v = 140 km/h; h = 6.63!10"34 J # s Required: λ Analysis: The speed of the proton is much less than light speed, so we can use the

classical momentum p = mv. Thus, the de Broglie relation, hp

λ = , becomes hmv

λ = .

First, convert kilometres per hour to metres per second.

140

kmh

! 103 m1 km

! 1 h3.6 !103 s

= 3.89 !101 m/s (one extra digit carried)

Solution:

! = hmv

= 6.63"10#34 J(1.40 "10–1 kg)(3.89 "101 m/s)

! = 1.2 "10#34 m

Statement: The de Broglie wavelength of the baseball is 341.2 10 m−× .

Page 11: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.3-2

4. The de Broglie wavelength of the baseball is 19 orders of magnitude smaller than the diameter of a proton; therefore, we could never expect to see any wave-like behavior of a macroscopic object like a baseball. Research This: Exploring Quantum Computers, page 638 Answers may vary. Sample answers: A. Quantum computers differ fundamentally from digital computers in the basic unit of information. For a digital computer, the basic unit is the bit, an element that can be in only one of two states, a “0” and a “1”. For a quantum computer, the basic unit is a quantum bit or “qubit.” A qubit can be in any superposition of two states, just like the electron trapped in a box (Figure 4 on page 636 of the Student Book) can be in a superposition of state 1 and state 2. Moreover, reading the state of a qubit is much different than reading the state of a bit. The reading of the state destroys the quantum superposition. B. Several problems presently stand in the way of building practical quantum computers. One is the difficulty of making a computer with many qubits. Another problem is the fragility of the quantum superposition state; it is relatively easy to disturb the system, so that the superposition state gets destroyed. Another difficulty is finding a way to easily read the qubits. C. A quantum computer’s design should allow it to perform very quickly at some computations that are very difficult for digital computers, so some possible applications of quantum computing include the factoring of large numbers, database searching, and the simulation of quantum mechanical systems. Section 12.3 Questions, page 639 1. Given: m = 9.11!10"31 kg; # = 150 nm = 1.5!10–7 m; h = 6.63!10"34 J $ s Required: speed of the electron, v Analysis: Notice that the wavelength here is larger than that in the solution to Sample Problem 1 of Tutorial 1, and in that case the electron’s speed is much less than that of light. A larger wavelength means that the speed is slower, so use the classical momentum in the de Broglie relation and solve for v.

! = h

mv, so v = h

m!.

Solution:

v = hm!

= 6.63"10#34 J $ s(9.11"10#31 kg)(1.5"10–7 m)

v = 4.9 "103 m/s

Statement: The electron’s speed is 4.9 ! 103 m/s, a non-relativistic speed. 2. Given: mproton/melectron = 1800; λproton = λelectron Required: Eproton/Eelectron Analysis: Assume that the two particles are non-relativistic (otherwise, we would need to know if the energy is the total energy or just the kinetic energy). In addition to using the

Page 12: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.3-3

de Broglie relation, use the classical relation between E and v, as well as that between p and v.

!electron =

hpelectron

and !proton =h

pproton

But, as λproton = λelectron, then it follows that pproton = pelectron. The classical kinetic energy can be written in terms of the classical momentum mv.

2proton proton proton

2proton proton

proton

2proton

protonproton

12( )12

12

E m v

m vm

pE

m

=

=

=

The same relation holds for the electron. 2electron

electronelectron

12pEm

=

Thus, 2electron

electron electron2protonproton

proton

2electron protonelectron

2proton electron proton

1212

pE m

pEm

p mEE m p

=

=

Solution: 2electron protonelectron

2proton electron proton

proton

electron

electron

proton

18001

p mEE m p

mm

EE

=

=

=

Statement: When the proton’s wavelength equals that of the electron, then they both have the same momentum. And when they have the same momentum and have non-relativistic speeds, then the ratio of their classical kinetic energies is 1800:1, with the electron having the higher energy because it is lighter.

Page 13: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.3-4

3. (a) Given: m = 1000.0 kg; v = 100.0 km/h; h = 6.63!10"34 J # s Required: λ

Analysis: Use the de Broglie relation, assuming non-relativistic speed: hmv

λ = .

First, convert kilometres per hour to metres per second.

100.0

kmh

! 103 m1 km

! 1 h3.6!103 s

= 2.778!101 m/s (one extra digit carried)

Solution:

! = hmv

= 6.63"10#34 J $s(1000.0 kg)(2.778"101 m/s)

! = 2.39"10#38 m

Statement: The de Broglie wavelength of the car travelling at 100.0 km/h is 2.39!10"38 m . (b) Given: m = 1000.0 kg; v = 10.0!103 km/h = 1.0!104 km/h; h = 6.63!10"34 J #s Required: λ Analysis: The car’s speed, though fast, is still non-relativistic (~2800 m/s), so we can use

the same relation as in (a), hmv

λ = . First, convert kilometres per hour to metres per

second.

1.0!104

kmh

! 103 m1 km

! 1 h3.6!103 s

= 2.778!103 m/s (one extra digit carried)

Solution:

! = hmv

= 6.63"10#34 J(1000.0 kg)(2.778"103 m/s)

! = 2.39"10#40 m

Statement: The de Broglie wavelength of the car travelling at 1.0!104 km/h is

2.39!10"40 m . (c) The de Broglie wavelength of the car at rest is undefined. As the speed decreases, the wavelength increases, and at zero speed, the wavelength blows up. (This result seems impossible, because we always see parked cars as solid objects and not spread out. However, consider how small the speed needs to be for the wavelength of the car to exceed 1 µm. The observer would have to establish that the speed of the car was less than about 6×10–31 m/s. Such a determination would be impossible, so we do not see parked cars spread out like a wave.) 4. In classical physics, particles occupy a definite position in space, and we can calculate exactly how a particle’s position changes with time. Moreover, we can determine both the particle’s position and the particle’s velocity at each instant of time with arbitrary precision. In quantum mechanics, we do not know what happens to the particle between

Page 14: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.3-5

measurements. Moreover, a measurement cannot determine the particle’s position and velocity with arbitrary precision. Instead, quantum mechanics gives us the probabilities for obtaining various outcomes of the measurement. 5. Answers may vary. Sample answers: (a) An example of experimental evidence for wave-like properties of matter is the Davisson–Germer experiment with electrons diffracting from a crystal. Other experiments have shown diffraction of larger particles. (b) An example of experimental evidence for particle-like properties of electromagnetic radiation is the early experiments by Heinrich Hertz on the photoelectric effect. (Other experiments include that of the photovoltaic effect (e.g., solar cells)). 6. I think wave functions are real. Wave functions cannot be observed directly, so one might conclude that they are not real. However, we can say the same thing about atoms, and yet atoms seem to be quite real; we can touch objects and we can feel the wind. Similarly, we can infer the existence of wave functions through their influence on measurements. For example, the probability distribution of electrons striking the wall behind a pair of slits is a result of the electron’s wave function. 7. Presently, all interpretations of quantum mechanics are consistent with the same observable results that we measure and experience. Yet quantum mechanics describes things that we cannot observe directly, such as the wave function. This indeterminacy of various aspects of quantum mechanics makes it possible for several views to be consistent with what we observe. Thus, different interpretations of quantum mechanics exist. I think the Copenhagen interpretation is most likely because I am comfortable with the idea that there are things we simply cannot know. I do not like the pilot-wave interpretation, as it seems to imply that future events are predetermined. Future events might be predetermined, but I am not comfortable with the idea. Similarly, I do not like the many-worlds interpretation because it is hard for me to picture the universe continually splitting in two. The collapse interpretation is not so objectionable, but I prefer the Copenhagen interpretation. 8. According to the Heisenberg uncertainty principle, one cannot take exact measurements of an electron (or any other object) when it is at rest. If the electron is at rest, then Δp = 0 and Δx, the uncertainty in the electron’s position, blows up. Thus, we could not determine where the electron was. 9. Willard Boyle earned a PhD in physics from McGill University. He worked at Bell Labs in New Jersey, then left for a job providing NASA with technological support for the Apollo space program, and then returned to Bell Labs in 1964, where he worked on developing electronic devices, including the charge-coupled device. Charge-coupled devices (CCDs) are designed around the photoelectric effect and the quantum mechanics of semiconductors. Other physical aspects of their operation are the motion of charges under an applied voltage, and for CCDs used for imaging, their operation depends on optics. They were originally designed to be used in several applications, including use as a memory device and shift register, but their most common application is for imaging. They were immediately useful in astronomy because, as imaging sensors, they could detect far fainter objects than those detected using film. They are now used in nearly all digital cameras.

Page 15: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.5-1

Section 12.5: Raymond Laflamme and Quantum Information Theory Section 12.5 Questions, page 643 Answers may vary. Sample answers: 1. Raymond Laflamme studies quantum information theory, the theoretical concepts behind the workings of a quantum computer. One contribution he has made is his devising of a mathematical framework for error-correcting codes for quantum computing. Error-correcting codes are particularly important for quantum computers because of their greater sensitivity to noise and other disruptions. His research group now (2011) has the world’s largest quantum processor, and he is developing methods to protect quantum computers from the type of quantum noise called decoherence. 2. Laflamme’s work in developing error-correcting codes will help prevent errors in quantum computers. The errors arise from noise in the circuits and interference from outside. Without such codes, quantum computers would not be reliable. 3. Quantum information theory describes how quantum information processing can become completely secure. Thus, this type of processing may greatly improve cryptography. In particular, the method can allow one to send secure information across the Internet without the usual concerns about someone intercepting the information. If such information gets read by anyone except the intended receiver, the information will become fundamentally changed. 4. Noise and decoherence degrade information transmission. 5. Laflamme’s research will affect society through the impact that quantum computers and quantum cryptography will have on society. Quantum computers hold much promise in their potential ability to greatly increase the speed of computation. Quantum cryptography should one day make shopping and banking over the Internet far more secure, possibly allowing us to do things we cannot presently do. 6. The primary difference between conventional digital computers and quantum computers is the way they store information. The conventional computer stores information in a bit that can have one of two possible states. A quantum computer stores information in quantum bits, or qubits, that exist in an arbitrary superposition of two quantum states. Because of this ability, a quantum computer processes information in multiple quantum states simultaneously. 7. In addition to the topics discussed here, Laflamme also studies experimental quantum information processing, quantum cryptography using satellite communications, the physical systems for quantum information processing, and the simulation of quantum systems using quantum computing.

Page 16: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.6-1

Section 12.6: The Standard Model of Elementary Particles Tutorial 1 Practice, page 647 1. The orbits in Figure 5 have energy levels 4 and 7. So, energy levels 5 and 6 are allowed between these values.

2. The first three energy levels of a Bohr atom are sketched below.

Section 12.6 Questions, page 653 1. The Rutherford planetary model of the atom is inconsistent with classical physics because classical electrodynamics shows that an accelerating charged particle radiates energy. If an atom had orbiting electrons, as in the planetary model, then classically they would radiate and fall into the nucleus. Such a collapse would occur quickly. As we know that atoms are stable (otherwise, how could we exist?), the planetary model is inconsistent with classical physics. 2. The progression of atomic models involved the Rutherford model, the Bohr model, the later quantum model (Schrödinger and others), and the standard model. Rutherford’s model was classical and did not work well, as explained in Question 1 above. The Bohr model, proposed by Niels Bohr and later strengthened by Louis de Broglie, was the first quantum model. It explained the hydrogen atom very well, but could not explain atoms with many electrons. Later quantum models, based on the Schrödinger equation, did well at explaining the behaviour of larger atoms and molecules, but did not explain the nucleus. The standard model is a complete model of the entire atom, describing not only the electrons, but also the nucleus.

Page 17: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.6-2

3. (a) Sketches of the constituent parts of the proton and neutron are below. The smaller circles with “u” and “d” are the quarks, and the wiggly lines connecting them represent the gluons, the force-mediating particles that hold the nucleus together.

(b) Answers may vary. Sample answer: I decided to change one of the d quarks in the neutron to a c quark.

(c) In looking up the above particle in a list of baryons, I found it labelled the “charmed lambda.” Because the charm quark has the same charge as the up quark, the charmed lambda has a charge of +1. It is unstable, with a lifetime of about 2.00 × 10–13 s, and it is more than twice as heavy as the proton, at 2286.46 MeV/c2. 4. Bosons have an important role in the standard model because they mediate the forces between particles. For example, the photon is the boson that mediates the electrodynamic force. In addition, the Higgs boson gives particles their mass. 5. (a) Fermions and bosons together are the building blocks of matter. The fermions include the leptons and quarks and have can have charge. The bosons mediate the forces and provide mass. (b) Mesons and baryons are hadrons, in that they are composed of quarks. Whereas the baryons have three quarks and include the particles of the nucleus, the proton and neutron, the mesons have just two quarks and are all unstable. (c) Leptons and hadrons are both particles, but leptons do not contain quarks. Leptons are the family of particles that include the electron, its neutrino, and its antiparticle. Unlike the leptons, hadrons are composite particles, being made out of quarks and including mesons and baryons. 6. The standard model describes how matter is composed and how interactions occur through the electromagnetic force and both the weak and the strong nuclear force. The model describes matter as being made up of quarks and leptons, with their forces mediated by bosons. However, the standard model does not describe gravity.

Page 18: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12.6-3

7. The only significant missing piece in the standard model is the Higgs boson. The standard model says the Higgs exists, but it has never been detected. So, the discovery of the Higgs boson would help to verify the standard model and help us understand the origin of mass. 8. Scientist may think that there must be a theory of everything for a few reasons. One, there is the general idea that the universe can be understood. Without a theory of everything, we do not understand why the microscopic realm is described by one theory, the standard model, whereas the large-scale structure is described by another theory, general relativity. In addition, the trend in science is to unify. First Newton unified the mechanics on Earth with the motions of the planets. Later Maxwell unified electricity and magnetism. And more recently, the standard model unified electromagnetism with the nuclear forces. Next on the list is to unify all forces in one grand theory. With a grand theory of everything, scientists may better understand the origin of the universe as well as many other mysteries, and also discover new phenomena. 9. Newton unified physics in the sense that the same gravitational force that objects on Earth respond to also acts on celestial objects. Moreover, Newton proposed that all matter attracts other matter gravitationally; that is, gravity is a universal force between all masses. 10. Answer may vary. Various pieces of the standard model have come from many scientists. Students will research the history of the standard model and choose a scientist who made an important contribution to the standard model. Scientists fitting this description would have to include many Nobel Prize winners—Murray Gell-Mann, Jerome I. Friedman, Henry W. Kendall, Richard E. Taylor, Yoichiro Nambu, Makoto Kobayashi, Toshihide Maskawa, Emilio Gino Segrè, Owen Chamberlain, Sin-Itiro Tomonaga, Julian Schwinger, Richard P. Feynman, Sheldon Lee Glashow, Abdus Salam, Steven Weinberg, Gerardus ’t Hooft, Martinus J.G. Veltman, David Politzer, David Gross, Frank Wilczek, Burton Richter, Samuel Ting, Martin Perl, Frederick Reines, Leon M. Lederman, Melvin Schwartz, Jack Steinberger, Chen Ning Yang, Tsung-Dao (T.D.) Lee, Carlo Rubbia, and Simon Van der Meer—and many others, including Gerson Goldhaber, Francois Pierre, James Bjorken, John Iliopoulos, Luciano Maiani, Donald Perkins, Charles Prescott, Harald Fritzsch, George Zweig, O.W. Greenberg, M.Y. Han, Harald Fritzsch, and Chien-Shiung Wu.

Page 19: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-1

Chapter 12 Self-Quiz, page 659 1. (c) 2. (c) 3. (b) 4. (c) 5. (a) 6. (a) 7. (a) 8. False. Newton’s laws do not work well on the atomic scale. 9. True 10. False. The work function, W, is measured in joules or electron-volts. 11. True 12. True 13. False. Max Planck explained the problems of blackbody radiation by assuming that the energy in the cavity must be emitted in discrete quanta. 14. True 15. True 16. True 17. False. When Rutherford first set up his gold foil experiment, he did not expect to see many alpha particles rebounding off the gold foil. 18. False. Electrons and positrons are examples of leptons.

Page 20: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-2

Chapter 12 Review, pages 660–665 Knowledge 1. (d) 2. (a) 3. (d) 4. (b) 5. (a) 6. (a) 7. (a) 8. (a) 9. (b) 10. (b) 11. (d) 12. (c) 13. (d) 14. True 15. False. When two classical particles approach each other, one particle cannot pass through the other. 16. False. If the frequency of a light source doubles, then the energy of each photon from that light source will double. 17. True 18. False. Planck’s constant is independent of the metal that is used. 19. False. The photoelectric effect is an experiment that demonstrates the particle nature of light. 20. False. Planck’s constant has units of energy-time. 21. True 22. False. As the temperature of an object increases, the wavelength of maximum radiated intensity will decrease. 23. True 24. True 25. False. Superconducting quantum interference devices interpret brain wave signals and convert them into instructions for operating artificial limbs. 26. True 27. False. Two up quarks and one down quark make up a proton. 28. True 29. False. Particles in the lepton family are fermions, not bosons. 30. (a) (v) (b) (iv) (c) (vi) (d) (i) (e) (iii) (f) (ii)

Page 21: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-3

31. The double-slit experiment can be used to show that matter has wave-like properties, just like it is used to show the wave nature of light. To show wave-like properties of a particle, one sends a particle beam through the slits instead of light. The particles must have a de Broglie wavelength λ of about the width of the slit or larger to show the interference effect. So, slow-moving, not high-energy, particles should be used. 32. (a) The essential features of the photoelectric effect are an incident source of light, a material surface, and an ejected electron. They are related as sketched below. Light of frequency f strikes a surface (typically metal) of work function W, and if the frequency of light is such that hf > W, an electron gets ejected with speed v.

(b) Light with a higher intensity does not provide more kinetic energy to the electrons because increasing the light intensity just increases the number of photons hitting the surface; if the photon energy is less than W, no electron gets ejected, regardless of the intensity. This is because the photon collisions are single events, and thus the individual photon energy must itself be large enough to eject the electron. (c) The kinetic energy of the ejected electron is less than the photon’s energy because of conservation of energy. The electron is attached to the surface with energy W, and thus the initial energy of the system equals hf – W. The final energy must be the same, and thus must be less than hf by the amount W. Understanding 33. (a) Money is quantized just like energy. We can spend money in units of a cent, but not a fraction of a cent. All things bought and sold have a price in units of a cent or multiples of a cent, such as a dollar. (b) The quantization of money is analogous to the quantization in quantum physics. Money gets exchanged in units of cents. Similarly, energy is exchanged via light in units of photons. In the photoelectric effect, photons of energy less than W do not remove an electron from the metal, no matter how many photons strike the surface. Similarly, if we go to a store that sells widgets for amount W, then only customers with money exceeding W can remove the widget from the store, regardless of the number of customers entering the store. (c) Answers may vary. Sample answer: Many other things we experience are quantized. Brick houses are built in units of bricks. School progress is measured in units of credits. Warehouses are stocked in units of items. Bread is used in units of slices. Time periods are given in units of years, days, or hours. Digital clocks give time in units of minutes or seconds. Companies hire work in units of people. 34. The main difference between classical and quantum mechanics is the quantization of energy. In classical mechanics, light of frequency f can have any energy, whereas in quantum mechanics, the energy of light must be in multiples of hf. 35. (a) The double-slit experiment with subatomic particles shows the pattern on the screen building up point by point, showing that the impact on the screen is particle-like.

Page 22: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-4

(b) The experiment shows the pattern on the screen to have fringes, showing that the particle beam has interference, which is characteristic of a wave. (c) The experiment with a double slit shows the same type of result with photons as with massive particles. The particles (massless, as with photons, or massive, as with other particles) strike the screen in discrete lumps like particles yet interfere like waves. Thus, they exhibit wave–particle duality. 36. (a) The relative intensity of light from a blackbody varies with wavelength as sketched below.

(b) Classical physics predicts that the intensity will increase without limit as the frequency increases, so on the graph of intensity versus wavelength, λ = 0 would be an asymptote where the intensity increases very rapidly. (c) The problem of the blackbody intensity increasing infinitely was solved by Max Planck by assuming that the energy modes (later called photons) in the blackbody cavity were quantized in units proportional to a constant h. By doing this, he showed that the relative intensity vanished at the highest frequencies, and the relative intensity fit the data exactly when h equalled a constant. The constant was later named Planck’s constant. 37. Answers may vary. Sample answer: The timeline below shows three pieces of evidence that support the particle model of light.

38. The frequency of the light determines whether or not photoemission will occur. The reason is that the frequency of light determines the energy of the photon, and photoemission occurs through single-photon collisions with the atoms. Thus, each photon must have sufficient energy to knock the electron out.

Page 23: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-5

39. An electron of energy 1 eV differs from a photon of the same energy in several ways. The table below shows four quantities that differ: charge, mass, momentum, and wavelength.

Particle Energy, E

(eV) Charge

(C) Mass, m

(kg) Momentum, p

(J·s/m) Wavelength, λλ

(m)

photon 1 eV 0 0 E / c = 3.3!10"28 h / p = 2.0 !10"6

electron 1 eV !1.6 "10!19 9.11!10"31 2mE = 4.3!10"25 h / p = 1.5!10"9

The electron has greater charge, mass, and momentum, whereas the photon has greater wavelength. 40. An experiment showing electron diffraction from a single slit of various widths supports the Heisenberg uncertainty principle in the following way: When the slit is made smaller, the uncertainty in the electron’s position at the slit Δx decreases but the uncertainty in the electron’s momentum along the slit-width direction Δp increases. We know that this momentum uncertainty increases because the position where the electron strikes the screen is spread out. Thus, the product of ΔxΔp cannot be made arbitrarily small. 41. The Davisson–Germer experiment in which electrons diffracted like a wave from a crystal supports the wave model of matter. 42. An optical microscope cannot magnify an image as much as an electron microscope because the wavelength of visible light is much larger than that of an electron of the same energy. The wavelength of an electron of the same energy as a photon is 1000 times smaller. A smaller wavelength can resolve smaller distances and thus have greater magnification. 43. The chapter discusses four interpretations of quantum mechanics. Applied to the double-slit experiment, they are as follows: 1) The collapse interpretation says that the particle leaves the source as a particle, spreads out as a wave and goes through the slit, and then collapses back into a particle when it is measured at the screen. 2) The pilot wave says that the particle remains a particle, but is guided by a “pilot wave” that follows the laws of quantum mechanics. The pilot wave diffracts like a wave through the slit, but the particle is detected at the screen. 3) According to the many-worlds interpretation, an electron goes through both slits, but in our universe, only one electron reaches the screen; the other electron is in another universe. 4) In the Copenhagen interpretations, the electron’s whereabouts between the measurement points cannot be determined, so it is meaningless to ask about it.

Page 24: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-6

44. Quantum cryptology differs from traditional cryptology. In quantum cryptology, any attempt to intercept a message will likely be detected because the message is sent via polarized photons. The sender and receiver know the polarization, but a potential eavesdropper would not. In quantum mechanics, the act of measuring a system will alter the system. Thus, the attempt at eavesdropping on the message will alter the message, and the alteration can be detected. 45. The main theoretical problem with the planetary model of the atom was that it would continuously lose energy through radiation. The atom would thus be unstable, in contradiction to observations. 46. The development of our understanding of the atom involved the atomic models of Rutherford and Bohr. Rutherford tested an earlier model of the atom in which there was no nucleus. His experiment sent a beam of high-speed, positively charged particles at a piece of gold foil and found greater deflection than expected. He proposed that the atom had a small nucleus around which the electrons orbited like planets around the Sun. But the model had the theoretical problem of predicting unstable atoms (as described in Question 45 above). The Bohr model, proposed by Niels Bohr, and later strengthened by Louis de Broglie, was the first quantum model. His model had the electrons orbiting with discrete energy levels that did not radiate and thus were stable. Photons could radiate if they dropped down to lower energy levels, or the atom could absorb radiation by going to a higher level. It explained the experiments done on the hydrogen atom very well, but did poorly at explaining atoms with many electrons. 47. The three fundamental forces covered in the standard model are the electromagnetic and the two nuclear forces. They are mediated by the bosons listed in the table below.

Fundamental force Boson electromagnetic force photon weak nuclear force W+, W–, and Z strong nuclear force gluon 48. The standard model has two types of particles, fermions and bosons. The bosons mediate the forces between the fermion particles. The fermions have two sub-types: leptons and hadrons. The leptons include the electron family (electron, muon, tau, plus their neutrinos and antiparticles) and quarks. Only the quarks make up composite particles called hadrons. The hadrons with two quarks are called mesons, and the hadrons with three quarks are called baryons. The most familiar baryons are protons and neutrons, the nuclear particles in atomic nuclei.

Page 25: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-7

Analysis and Application 49. Given: !max = 625 nm = 6.25 × 10–7 m Required: T Analysis: Use Wien’s law, rearranged to find T.

!max =2.90 "10#3 m $K

T

T = 2.90 "10#3 m $K!max

Solution:

T = 2.90!10"3 m #K$max

= 2.90!10"3 m #K6.25!10"7 m

T = 4640 K

Statement: The orange blackbody has a temperature of 4640 K. 50. (a) The temperature 37.0º C equals 3.10 × 102 K. (b) Given: T = 3.10 × 102 K Required: λmax Analysis: Use Wien’s law:

!max =

2.90 "10#3 m $KT

Solution:

!max =2.90 "10#3 m $K

T

= 2.90 "10#3 m $ K3.10 "102 K

!max = 9.35"10#6 m

Statement: is 69.35 10 m−× or 9.35 µm.

51. Given: 2; ; ; cE mc p mv E hf fλ

= = = =

Required: relation found by Compton, photonhpλ

=

Analysis: From 2E mc= , assume that a photon of energy E has an “effective mass” of

2

Emc

= . So (at non-relativistic speeds) the momentum of such a mass would be mv, or

2

Ep vc

= . But from Einstein, E = hf for a photon. Putting this relation into the expression

for p gives 2

hfp vc

= .

Page 26: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-8

Setting v = c and then using cfλ

= gives

2

hfp cch cchp

λ

λ

=

=

=

Solution: hpλ

=

Statement: Using Einstein’s relation for the rest mass energy and photon energy,

Compton could derive de Broglie’s relation between momentum and wavelength, hpλ

= .

52. Given: Two data points are given: one for light of wavelength λ, the other with

wavelength 2λ . The actual wavelength is not important. But notice that this corresponds

to a data point using light of frequency f and frequency 2f. We will use frequency instead of wavelength to avoid a conversion step in the photoelectric equation. From E = hf – W, we have the two results: 1) 1.0 eV2) 3.0 eV 2

hf Whf W

= −= −

The above are our givens. Required: W Analysis: Subtract twice the first equation from the second equation.

3.0 eV = 2hf !W2(1.0 eV) = 2hf ! 2W

Subtract the bottom equation from the top, and simplify. 3.0 eV 2.0 eV= 2 2 ( 2 )

1.0 eVhf hf W WW

− − − − −=

Solution: 1.0 eVW = Statement: The work function W equals 1.0 eV. 53. The first method determines the work function directly and is sketched in Figure 2 of Section 12.2. The metal plate is set parallel a short distance from a second plate. Both plates are in a vacuum and connected by a voltage source to make a circuit. Turn up the voltage until a current crosses the gap between the plates. Multiply this minimum voltage by the charge of an electron to get the work function. The second method uses the photoelectric effect. You can use the same two plates, but spaced far enough apart that light can shine on one of them. Then, increase the frequency of the light until a current crosses the gap. Multiply this minimum frequency by Planck’s constant to get the work function.

Page 27: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-9

54. (a) Given: ! = 3.20"10#7 m; W = 1.95 eV = 3.12 "10–19 J; c = 3.0"108 m/s;

h = 6.63!10"34 J #s Required: E Analysis: Use the photoelectric effect equation and convert frequency to wavelength.

E hf WcE h Wλ

=

=

Solution:

E = h c!

–W

= (6.63"10#34 J $ s )3.0"108 m/s3.20"10#7 m

%&'

()*

– 3.12 "10–19 J

= 3.096"10#19 J (two extra digits carried)

E = 3.10"10#19 J

Statement: The kinetic energy of the ejected electron is 3.10!10"19 J . (b) Given: E = 3.096!10"19 J Required: voltage V needed to stop the electron from reaching the plate Analysis: An electron crossing a voltage V would acquire a kinetic energy of eV.

Thus, EVe

= .

Solution:

V = Ee

= 3.096!10"19 J1.60!10"19 C

V = 1.93 V

Statement: An applied voltage of 1.93 V would stop the ejected electrons. 55. Given: E = 1.7 eV = 2.72!10–19 J ; W = 4.52 eV = 7.23!10–19 J ; c = 3.0!108 m/s;

h = 6.63!10"34 J #s Required: λ Analysis: Use the photoelectric effect equation, rewritten to obtain λ :

E hf WcE h W

cE W h

chE W

λ

λ

λ

=

=

+ =

=+

Page 28: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-10

Solution:

! = h cE +W

= (6.63"10#34 J $ s ) 3.0"108 m/s2.72"10–19 J + 7.23"10–19 J

%&'

()*

! = 2.0"10#7 m

Statement: The wavelength of the light is 2.0 ! 10–7 m. 56. (a) According to the graph in Figure 3, the photosynthesis process is using the red and the blue-violet parts of the spectrum. So, assuming that most of the absorption of light is due to photosynthesis, then very little green gets absorbed by the plant. If the light does not get absorbed, then the light gets scattered back into our eyes, making the plant appear green. (b) Pure green light produces a very low reaction rate because the plot shows little photosynthesis at wavelengths corresponding to green. (c) Given: T = 5300 K Required: λmax

Analysis: Use Wien’s law, !max =

2.90 "10#3 m $KT

.

Solution:

!max =2.90"10#3 m $K

T

= 2.90"10#3 m $ K5300 K

!max = 5.5"10#7 m

Statement: The light from an incandescent bulb at 5300 K produces light of maximum wavelength 5.5!10"7 m or 550 nm, which lies in the middle of the green, where the rate of photosynthesis is lowest. The amount of photosynthesis will depend on how much off-peak intensity lies in the red and the blue-violet. 57. (a) Given: Vmin = 5.65 V; 191.60 10 Ce −= × Required: W Analysis: To get the work function W, multiply the minimum voltage V by the electric charge e.

minW eV= If we use the units of eV, then the work function is numerically the voltage. Solution:

W = eVmin

= (1.60 !10"19 C)(5.65 V)W = 9.04 !10"19 J

Statement: In electron-volts, the work function is 5.65 V. In joules, the work function is 9.04!10"19 J . (b) According to Table 1, the metal with this work function is platinum.

Page 29: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-11

58. (a)

Given: WCu = 8.17 !10"19 J; WPb = 6.81!10"19 J; h = 6.63!10"34 J # s Required: f0, the threshold frequencies for Cu and Pb Analysis: Use the photoelectric effect equation, rewritten to obtain f when E = 0.

0

0

0

–0E hf W

hf Whf W

Wfh

== −=

=

Solution: 0Wfh

=

For copper

f0 =8.17 !10"19 J

6.63!10"34 J # sf0 = 1.23!1015 Hz

For lead

f0 =6.81!10"19 J

6.63!10"34 J # sf0 = 1.03!1015 Hz

Statement: The frequency for copper is slightly higher than that for lead. The values should be close to the values deduced from the graph. (b) The lines in the graph have the same slope because they arise from the equation

–E hf W= , which has the slope of h, a universal constant. 59. (a) Given: W = 7.47 !10"19 J = 4.67 eV; h = 6.63!10"34 J #s Required: f0, the threshold frequency for Fe

Analysis: 0Wfh

=

Page 30: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-12

Solution:

f0 =Wh

= 7.47 !10"19 J6.63!10"34 J # s

f0 = 1.13!1015 Hz

Statement: The minimum frequency of light to eject electrons from iron is 151.13 10 Hz× .

(b) Given: 157.9 10 Hzf = × Required: energy of electron ejected from iron, E Analysis: Use the photoelectric effect relation, –E hf W= . Solution:

E = hf – W

= (6.63!10"34 J # s )(7.9 !1015 s"1 ) – (7.47 !10"19 J)E = 4.5!10"18 J

= (4.5!10"18 J) 1eV1.60 !10"19 J

$%&

'()

E = 28 eV

Statement: The ejected electron has a kinetic energy of 4.5 ! 10–18 J or 28 eV. 60. (a) Given: W = 8.17 !10"19 J ; E = 2.1 eV = 3.36!10–19 J; h = 6.63!10"34 J #s Required: frequency of the light, f Analysis: Use the photoelectric effect equation, rewritten to obtain f.

–E hf WE W hf

E Wfh

=+ =

+=

Solution:

f = E +Wh

= 3.36!10"19 J +8.17 !10"19 J 6.63!10"34 J #s

f = 1.7 !1015 Hz

Statement: The frequency of the incident light is 1.7 ! 1015 Hz. (b) Light of frequency 1.7 ! 1015 Hz is ultraviolet. 61. Given: ! = 580 nm = 5.80 "10#7 m; h = 6.63"10#34 J $ s Required: momentum of the photon, p

Analysis: Use the relation hpλ

= .

Page 31: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-13

Solution:

p = h!

= 6.63"10#34 J $ s5.80 "10#7 m

p = 1.1"10#27 kg $m/s

Statement: The photon’s momentum is 1.1!10"27 kg #m/s . 62. (a) Given: p = 2.03!10"24 kg #m/s; h = 6.63!10"34 J #s Required: the photon’s wavelength, λ

Analysis: hp

hp

λ

λ

=

=

Solution:

! = hp

= 6.63"10#34 J $s2.03"10#24 kg $m/s

= 3.2660 "10#10 m (two extra digits carried)

! = 3.27 "10#10 m

Statement: The X-ray wavelength is 3.27 !10"10 m or 0.327 nm. (b) Given: ! = 3.2660 "10#10 m; c = 3.0"108 m/s Required: the photon’s frequency, f

Analysis: cfλ

=

Solution:

f = c!

= 3.0"108 m/s3.2660"10#10 m

f = 9.19"1017 Hz

Statement: The frequency is 9.19!1017 Hz . 63. Given: E= 6.40 eV = 1.024!10–18 J; c = 3.0!108 m/s Required: momentum of the photon, p

Analysis: Use the relation Epc

= .

Page 32: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-14

Solution:

p = Ec

= 1.024!10–18 J3.0!108 m/s

p = 3.41!10"27 kg #m/s

Statement: The momentum of the photon is 3.41!10"27 kg #m/s . 64. Although we do not directly experience quantum mechanical effects in our everyday lives, we use many electronic devices that rely upon quantum mechanics to operate. Moreover, the processes we observe in our everyday lives, including the very act of seeing, rely upon quantum mechanics. 65. To detect diffraction effects, the slit width should be less than or equal to the wavelength of the impinging wave. So, we will assume a maximum slit width equal to the electron’s wavelength and use de Broglie’s expression for the wavelength. As 4 eV is a low energy for an electron, we will use the classical expression for the momentum. Given: E = 4 eV = 6.4 ! 10–19 J; m = 9.11!10"31 kg; h = 6.63!10"34 J # s Required: λ Analysis: First, find the electron’s momentum using p = 2mE , then use de Broglie’s

relation, ! = h

p.

Solution:

p = 2mE

= 2(9.11!10"31 kg)(6.4!10"19 J)

p = 1.08!10"24 kg #m/s (two extra digits carried)

! = hp

= 6.63"10#34 J $ s1.08 "10#24 kg $m/s

! = 6 "10#10 m

Statement: The slit width should be at most 6 !10"10 m or 0.6 nm. 66. Given: m = 9.11!10"31 kg; v = 5.80 !106 m/s; h = 6.63!10"34 J # s Required: the de Broglie wavelength, λ Analysis: As the speed of the electron is much less than c, use the classical momentum p in de Broglie’s relation.

hphmv

λ

λ

=

=

Page 33: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-15

Solution:

! = hp

= 6.63"10#34 J $ s(9.11"10#31 kg)(5.80 "106 m/s)

! = 1.25"10#10 m

Statement: The electron’s wavelength is 1.25!10"10 m or 0.125 nm. 67. (a) Given: ! = 0.220 nm = 2.20 "10–8 m; m = 9.11"10#31 kg; h = 6.63"10#34 J $ s Required: speed of the electrons, v Analysis: As the speed of the electron is much less than c, use the classical momentum p in de Broglie’s relation, but rewrite to determine v.

! = hp

! = hmv

v = hm!

Solution:

v = hm!

= 6.63"10#34 J $ s(9.11"10#31 kg)(2.20 "10–8 m)

v = 3.308 "106 m/s (one extra digit carried)

Statement: The electron’s speed is 63.31 10 m/s× . (This is much less than c, which justifies our use of classical momentum.) (b) Given: 63.308 10 m/sv = × ; 319.11 10 kgm −= × Required: the electron’s kinetic energy, E

Analysis: As the speed is non-relativistic, use the classical kinetic energy, 212

E mv= .

Solution:

E = 12

mv2

= 12

(9.11!10"31 kg)(3.308 !106 m/s)2

E = 4.984 !10"18 J (one extra digit carried)

= 4.984 !10"18 J1 eV

1.60 !10"19 J#$%

&'(

E = 31.2 eV

Statement: The electron’s kinetic energy is 184.98 10 J−× or 31.2 eV. (c) Given: 184.984 10 JE −= × ; 191.60 10 Ce −= × Required: voltage, V, to accelerate the electrons Analysis: The energy equals the electronic charge times the voltage.

Page 34: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-16

E eVEVe

=

=

Solution:

V = Ee

= 4.984 !10"18 J1.60 !10"19 C

V = 31.2 V

Statement: The voltage needed to accelerate the electrons is 31.2 V. 68. Given: 31 27

e p e p e9.11 10 kg; 1.67 10 kg; 0.025 ; m m v c v v− −= × = × = =

Required: !e / !p

Analysis: Use the de Broglie relation for both electron and proton and simplify. Because the speeds are much less than c, we can use the classical momentum (but the result is the same even if we use the relativistic momentum).

ee e

pp p

e e e

p

p p

hm vhm vhm vhm v

λ

λ

λλ

=

=

=

Now we can cancel the hs and both vs (because both vs are equal). pe

p e

mm

λλ

=

Solution:

!e

!p

=mp

me

=1.67 "10#27 kg9.11"10#31 kg

!e

!p

= 1800

Statement: The ratio of the electron to the proton wavelength equals the inverse ratio of their masses, which is 1800, so the ratio of the de Broglie wavelengths of a proton and of an electron both moving at 0.025c is 1 : 1800.

Page 35: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-17

69. Answers may vary. Sample answer: In the early 1900s, experiments involving the bombardment of atoms and nuclei by various particles showed that nuclei were made of protons and neutrons. For a short time, electrons, protons, and neutrons were thought to be the fundamental particles of all matter. However, this conjecture was soon disproved by cosmic ray studies, collision experiments, and nuclear decay experiments by various researchers. The research showed that protons and neutrons were made of smaller particles called quarks. Protons, neutrons, and their anti-particles belong to the hadron group. It is assumed hadrons are divisible. Electron and positrons belong to the lepton group. It is assumed that leptons are elementary particles that are not divisible. With the discovery of these new particles, the standard model of elementary particles became the current theory of fundamental particles and the forces that interact with them. Particle physics research is readily applied to the nuclear power technology. Understanding how nuclear fission works and the energy created is due to the theories of particle physics. Evaluation 70. Answers may vary. Sample answer: I think Niels Bohr made the largest contribution to quantum mechanics for several reasons. One, his atomic model, in accurately explaining the spectrum from hydrogen, gave a huge impetus to the development of quantum mechanics. He also did much other work on various technical results in quantum mechanics. Two, he developed the correspondence principle, the idea that the results of quantum mechanics should approach those of classical physics as the quantum numbers become large. Three, his idea of complementarity proposed that quantum mechanics allows complementary ways of viewing a system, so that we can view the double-slit experiment as exhibiting wave-like or particle-like behaviour. Four, Bohr contributed greatly to the ideas behind the Copenhagen interpretation. Finally, he nurtured many younger scientists who would later go on to make significant contributions to quantum mechanics. 71. A bright light bulb produces a different spectrum of radiation than the Sun, and thus our skin responds to the two spectra differently. The Sun is much hotter than a light bulb, and thus from Wien’s law, produces more high-energy, high-frequency radiation. The higher-energy radiation can damage the skin when lower-energy radiation does not, even when the higher-energy radiation has much lower intensities. The reason is related to the photoelectric effect. The higher-energy ultraviolet (UV) radiation may dislodge an electron from some tissue material, as it does for a metal in the photoelectric effect, and this process damages the tissue. Such damage will not occur for low-energy radiation. 72. (a) The completed table is below.

Wavelength (nm) Frequency of light (Hz) Kinetic energy of ejected electrons (J) 600 5.0 !1014 0 520 5.8 !1014 0 440 6.8 !1014

4.6 !10–20

360 8.3!1014 1.4 !10–19

280 1.0 !1015 2.9 !10–19

200 1.5!1015 5.7 !10–19

Page 36: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-18

Given: ! = 600 nm = 6.0 "10–7 m; c = 3.0 "108 m/s Required: frequency, f Analysis: Use the general relation for waves, applied to light and written to obtain f. c f

cf

λ

λ

=

=

Solution:

f = c!

= 3.0 "108 m/s6.0 "10–7 m

f = 5.0 "1014 Hz

Statement: This frequency value is listed in the table above, along with those for the other wavelengths. (b)

f = c!

= 3.0"108 m/s440 nm

1"109 nm1 m

#

$%

&

'(

f = 6.818"1014 Hz

(c) From inspection of the plot, the x-intercept is approximately 6.1!1014 Hz . (d) From inspection of the plot, the slope is approximately 6.4 !10"34 J/Hz or

6.4 !10"34 J # s . (e) At the lowest frequencies, the photons do not have enough energy to eject an electron. 73. The term “mediate” is appropriate as a description of what bosons do in particle physics. A mediator stands between and communicates between two parties. In this sense, the bosons mediate the force because they lie between the particles and convey the one particle’s position to the other particle, which tells the other particle how to respond and vice versa.

Page 37: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-19

Reflect on Your Learning Answers may vary. Sample answers: 74. I found the wave–particle duality most interesting because it seems much different from other physics, which seems more “rigid” and one-sided. Also the photoelectric effect was nice to learn about because it helped me to better understand how light interacts with matter. I would like to learn more about these topics in more advanced classes. 75. I would use examples and analogies, not abstract concepts. For example, I would talk about the two-slit experiment and the photoelectric effect. I would compare the electron in a box to a macroscopic ball in a box. For the standard model, I would talk about how boson particles mediate a force, but first I would like to learn how a particle causes an attractive force. 76. I do not need to consider quantum mechanical effects in navigating through my daily life. But I sometimes reflect on how quantum mechanics can seem so strange yet allow researchers and engineers to devise wonderful new devices as well as older devices that I use regularly. For example, I use computers regularly, and the workings of a microchip rely upon quantum mechanical principles. Research 77. (a) In 1927, Clinton Davisson and Lester Germer confirmed de Broglie’s matter–wave hypothesis by showing that a beam of electrons will scatter from a crystal surface and produce a diffraction pattern, just like X-rays of the same wavelength. This confirmed the wave-like nature of electrons. The diffraction effect occurs because the surface of a crystal has a periodic arrangement of atoms, much like a diffraction grating except with a much smaller spacing. (b) The setup of Davisson and Germer involved a low-energy electron beam that was aimed at a nickel surface, the entire setup being in a vacuum so the electrons would not get scattered by gas molecules. They also had a detector that collected the scattered electrons over a small angular region. They could change the relative angle between the incident and scattered electrons, the angle between the surface and incident beam, and the electron-beam energy. In this way, they could establish that the electrons diffracted from the surface like waves. 78. (a) In the 1920s, Edwin Hubble discovered that the other galaxies in the universe were moving away from us, as well as from each other, and the speed increased with their distance. (b) In 1965, Arno Penzias and Robert Wilson discovered the uniform microwave background radiation from space. The wavelength indicated a thermal background to the universe of 3.5 K. They made the discovery inadvertently while testing their microwave antenna for communication applications. (c) Both the motion of the galaxies discovered by Hubble and the microwave background radiation discovered by Penzias and Wilson support the big-bang hypothesis of the origin of the universe. In an expanding universe, all galaxies would be moving away from each other, like dots on the surface of an inflating balloon, and the average temperature of the universe would decrease, like the temperature of an expanding gas.

Page 38: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-20

(d) Physicists have suggested that only quarks, leptons, and bosons existed at the origin of the universe. As the universe expanded and cooled, hadrons formed, then small nuclei such as hydrogen and helium, and then light atoms. 79. Answers may vary. Sample answers: (a) Experiments and ideas about the photoelectric effect came from men such as Heinrich Hertz, Aleksandr Stoletov, Joseph (J. J.) Thomson, Philipp Lenard, and Max Planck. (b) Solar photovoltaic cells operate on the principle of the photovoltaic effect, which is similar to the photoelectric effect. In the latter, the electrons are dislodged from the surface, whereas in the photovoltaic effect, the electrons move to a higher energy level in the material. In the solar cell, the material is a semiconductor layer that has been “doped” to have excess electrons on one side and missing electrons (“holes”) on the other side. This doping provides a way for the photovoltaic-generated electrons to move one way in a circuit, thus allowing the solar cell to produce electricity. Although photovoltaic cells, even solar cells, preceded Einstein’s work on the photoelectric effect, modern solar cells rely on the principle of the quantum theory of semiconductors, which grew out of Einstein’s early impetus to quantum theory via his theory of the photoelectric effect. 80. Answers may vary. Sample answers: (a) Gravity posed a problem for Einstein in two senses of the word. One, he wondered why gravitational mass was the same as inertial mass. It seemed to be a coincidence in physics. He argued that it was not a coincidence, that instead they were the same thing. This led to the idea that gravity produced curvature in space. And this curvature led to the second “problem” with gravity: the mathematics of curved space-time was very difficult. (b) In trying to develop a theory that unifies gravity with the standard model, physicists have particular difficulty with time and quantum uncertainty. Time is absolute in quantum mechanics, but relative in general relativity. With quantum uncertainty in position, or space, the relativity of time can lead to predictions that massive objects like black holes can appear everywhere. 81. Answers may vary. Sample answers: (a) When one tries to separate the quarks in a proton or neutron, the required force becomes so strong that new particles are created, that is, quark–anti-quark pairs, that produce other quark composites such as mesons and other hadrons. The strong nuclear force becomes stronger at larger separations instead of weaker, as in the more familiar forces of electromagnetism and gravity. (b) When protons or neutrons are collided at high energy, the constituent quarks start to separate and then produce new quark–anti-quark pairs as described in (a) above. (c) One cannot separate two quarks for the same reason as described in (a) above. 82. Answers may vary. Sample answers: (a) Richard Feynman contributed to our understanding of quantum mechanics in various ways. One, he developed a path-integral approach to quantum mechanics that is useful for some problems. Two, in parallel to the work of Julian Schwinger and Shin-ichiro Tomonaga, he developed the theory of quantum electrodynamics. And, other than solving other problems in quantum mechanics, he produced a series of lectures and popular books that contained topics in quantum mechanics. (b) A Feynman diagram is useful to physicists because it conveys in a simple picture a complex mathematical operation.

Page 39: Section 12.1: Introducing Quantum Theory Section 12.1 ...

Copyright © 2012 Nelson Education Ltd. Chapter 12: Quantum Mechanics 12-21

83. (a) A CCD works on principles similar to the solar cell (see Question 84b), but with special circuitry to collect and transport the photo-generated electrons in an orderly fashion. (b) CCDs have been useful to astronomers because they can detect lower levels of light than film and have less noise. 84. (a) Russel Ohl made the first semiconductor type of solar cell in 1946. They were first used on satellites in 1958. (b) The essential part of a solar cell is a special region inside the semiconductor (usually silicon) that is similar to that in an LED. One side has been made to have electron charge carriers, and the other side to have missing electrons, which act as “holes” that carry positive charge according to the quantum theory of semiconductors. At the boundary, the electrons and holes combine, forming the special region called the depletion region. The depletion region does not have many charge carriers, but the combining of the electrons and holes creates an electric field and thus a voltage. But to generate power, the depletion region needs charges that can flow. These charges are produced by the photovoltaic effect, which is similar to the photoelectric effect; briefly, an incident photon travels into the depletion region and causes an electron to move up to a higher energy level in which it can conduct. The electron then gets pushed by the electric field through the circuit, generating electricity. (c) An array consists of many cells. (d) Other physics concepts at work in a solar cell are in the fields of optics and electricity. 85. (a) The radiation emitted from a small hole in a larger cavity nearly approximates a perfect blackbody. Black velvet cloth is also a good blackbody. (b) Blackbody radiation is useful for measuring the temperature of objects remotely. For example, the ear thermometer is a useful application. Astronomers rely on blackbody radiation to estimate the temperature of distant objects. Blackbody radiation is also a cooling mechanism in which a body cools by radiating heat to the surroundings. 86. (a) Synchrotron radiation in general is radiation from charged particles travelling in a circular path. Travelling in a circular path means that the particles are continuously accelerating inward, and any accelerating charge must radiate according to classical electrodynamics. The name “synchrotron radiation” arises because the particle accelerators typically used for this radiation are synchrotrons in which the particles are highly relativistic. The radiation is usually X-rays. (b) Electrons have several advantages for generating synchrotron radiation. Electrons are easy to produce and electrons lose a greater fraction of their kinetic energy during bending acceleration than larger particles such as protons. (c) Synchrotron radiation has many applications. Some general uses include photolithography, tomography, and spectroscopy.