walker4 ism ch30

34
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 30 – 1 Chapter 30: Quantum Physics Answers to Even-Numbered Conceptual Questions 2. If energy is quantized, as suggested by Planck, the amount of energy for even a single high-frequency photon can be arbitrarily large. The finite energy in a blackbody simply cant produce such high-frequency photons, and therefore the infinite energy implied by the ultraviolet catastrophecannot occur. In classical physics, any amount of energy can be in the form of high-frequency lightthe energy does not have to be supplied in discrete, large lumps as in Plancks theory. Therefore, classical physics implies that all frequencies of light have the same amount of energy, no matter how high the frequency. This is what leads to the catastrophe.4. Plancks theory of blackbody radiation implies a one-to-one relationship between the absolute temperature of a blackbody and the frequency of light at the peak of its radiated energy spectrum. This relationship is given by Wiens displacement law (equation 30-1). Therefore, by measuring the peak in the radiated energy from a star, we can tell its temperature. In broad terms, a blue star is very hot, a red star much less so, and a yellowish star like our Sun is intermediate in temperature. 6. A monochromatic source of light meansliterallythat it emits light of a single color. This means that all the photons emitted by the source have the same frequency, and hence they also have the same energy. 8. (a) A photon from a green light source always has less energy than a photon from a blue light source. (b) A photon from a green light source always has more energy than a photon from a red light source. The reason for these results is that the energy of a photon depends linearly on the frequency of light; that is, E = hƒ. 10. Classically, it should be possible to eject electrons with light of any frequencyall that is required is to increase the intensity of the beam of light sufficiently. The fact that this is not the case means that the classical picture is incorrect. In addition, the fact that there is a lowest frequency that will eject electrons implies that the energy of the photon is proportional to its frequency, in agreement with E = hƒ. 12. Yes. An electron and a proton have the same de Broglie wavelength ( ) hp λ = if they have the same momentum. Solutions to Problems and Conceptual Exercises 1. Picture the Problem: The blackbody spectrum of blackbody A peaks at a longer wavelength than that of blackbody B. Strategy: Use Wien’s Displacement Law (equation 30-1) to answer the conceptual question. Solution: 1. (a) Because blackbody A has the longer wavelength it also has the lower frequency. We know from equation 30–1, however, that the peak frequency of a blackbody spectrum is proportional to the absolute temperature of the blackbody. Therefore, the temperature of blackbody A is lower than the temperature of blackbody B. 2. (b) The best explanation is II. Blackbody B has the higher temperature because an increase in temperature means an increase in frequency, which corresponds to a decrease in wavelength. Statement I is false. Insight: Statement I is false because increasing the temperature of a blackbody shifts the peak emission to greater frequency, not greater wavelength. 2. Picture the Problem: The star Betelgeuse emits a peak frequency that is proportional to its surface temperature. Strategy: Solve Wien’s Displacement Law (equation 30-1) for the temperature of Betelgeuse. Solution: Solve equation 30-1 for the temperature: ( ) 10 1 1 peak 14 10 1 1 5.88 10 s K 1.82 10 Hz 3100 K 5.88 10 s K f T T = × × = = × Insight: Betelgeuse, like all red-giant stars, has a cooler surface temperature than our own Sun. The first printing of the third edition text had incorrectly listed the peak frequency as 3.09×10 14 Hz, which would correspond to an incorrect surface temperature of 5260 K, much closer to the 5800 K surface temperature of our Sun.

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Ch 30

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Page 1: Walker4 Ism Ch30

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 1

Chapter 30: Quantum Physics

Answers to Even-Numbered Conceptual Questions 2. If energy is quantized, as suggested by Planck, the amount of energy for even a single high-frequency

photon can be arbitrarily large. The finite energy in a blackbody simply can’t produce such high-frequency photons, and therefore the infinite energy implied by the “ultraviolet catastrophe” cannot occur. In classical physics, any amount of energy can be in the form of high-frequency light—the energy does not have to be supplied in discrete, large lumps as in Planck’s theory. Therefore, classical physics implies that all frequencies of light have the same amount of energy, no matter how high the frequency. This is what leads to the “catastrophe.”

4. Planck’s theory of blackbody radiation implies a one-to-one relationship between the absolute temperature of a blackbody and the frequency of light at the peak of its radiated energy spectrum. This relationship is given by Wien’s displacement law (equation 30-1). Therefore, by measuring the peak in the radiated energy from a star, we can tell its temperature. In broad terms, a blue star is very hot, a red star much less so, and a yellowish star like our Sun is intermediate in temperature.

6. A monochromatic source of light means—literally—that it emits light of a single color. This means that all the photons emitted by the source have the same frequency, and hence they also have the same energy.

8. (a) A photon from a green light source always has less energy than a photon from a blue light source. (b) A photon from a green light source always has more energy than a photon from a red light source. The reason for these results is that the energy of a photon depends linearly on the frequency of light; that is, E = hƒ.

10. Classically, it should be possible to eject electrons with light of any frequency—all that is required is to increase the intensity of the beam of light sufficiently. The fact that this is not the case means that the classical picture is incorrect. In addition, the fact that there is a lowest frequency that will eject electrons implies that the energy of the photon is proportional to its frequency, in agreement with E = hƒ.

12. Yes. An electron and a proton have the same de Broglie wavelength ( )h pλ = if they have the same momentum.

Solutions to Problems and Conceptual Exercises 1. Picture the Problem: The blackbody spectrum of blackbody A peaks at a longer wavelength than that of blackbody B.

Strategy: Use Wien’s Displacement Law (equation 30-1) to answer the conceptual question.

Solution: 1. (a) Because blackbody A has the longer wavelength it also has the lower frequency. We know from equation 30–1, however, that the peak frequency of a blackbody spectrum is proportional to the absolute temperature of the blackbody. Therefore, the temperature of blackbody A is lower than the temperature of blackbody B.

2. (b) The best explanation is II. Blackbody B has the higher temperature because an increase in temperature means an increase in frequency, which corresponds to a decrease in wavelength. Statement I is false.

Insight: Statement I is false because increasing the temperature of a blackbody shifts the peak emission to greater frequency, not greater wavelength.

2. Picture the Problem: The star Betelgeuse emits a peak frequency that is proportional to its surface temperature.

Strategy: Solve Wien’s Displacement Law (equation 30-1) for the temperature of Betelgeuse.

Solution: Solve equation 30-1 for the temperature: ( )10 1 1peak

14

10 1 1

5.88 10 s K

1.82 10 Hz 3100 K5.88 10 s K

f T

T

− −

− −

= × ⋅

×= =

× ⋅

Insight: Betelgeuse, like all red-giant stars, has a cooler surface temperature than our own Sun. The first printing of the third edition text had incorrectly listed the peak frequency as 3.09×1014 Hz, which would correspond to an incorrect surface temperature of 5260 K, much closer to the 5800 K surface temperature of our Sun.

Page 2: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 2

3. Picture the Problem: The human body can be considered a blackbody radiator with a surface temperature of 95°F. The

peak radiation frequency and wavelength are functions of that temperature.

Strategy: Solve Wien’s Displacement Law (equation 30-1) for the peak frequency. Remember to convert the Fahren-heit temperature to Kelvin. Use equation 14-1 to calculate the wavelength, where the wave speed is the speed of light.

Solution: 1. Calculate the peak frequency: ( )( ) ( )

10 1 1peak

10 1 1 59

13peak

5.88 10 s K

5.88 10 s K 95 32 273.15 K

1.81 10 Hz

f T

f

− −

− −

= × ⋅

= × ⋅ − +⎡ ⎤⎣ ⎦

= ×

2. Calculate the wavelength:

8

13

3.00 10 m/s 16.6 m1.812 10 Hz

cf

λ μ×= = =

×

Insight: This wavelength falls in the infrared portion of the electromagnetic spectrum (see Chapter 25).

4. Picture the Problem: The initial Big Bang can be considered as a blackbody radiator whose peak frequency is a function of its temperature.

Strategy: Solve Wien’s Displacement Law (equation 30-1) for the peak frequency. Use equation 14-1 to calculate the wavelength, where the wave speed is the speed of light.

Solution: 1. (a) Calculate the peak frequency: ( )( )( )

10 1 1peak

10 1 1 11

5.88 10 s K

5.88 10 s K 2.7 K 1.6 10 Hz

f T− −

− −

= × ⋅

= × ⋅ = ×

2. (b) Calculate the wavelength:

8

11

3.00 10 m/s 1.9 mm1.6 10 Hz

cf

λ ×= = =

×

Insight: This wavelength falls in the microwave portion of the electromagnetic spectrum (see Chapter 25).

5. Picture the Problem: The Sun is a blackbody radiator with a surface temperature of 5800 K.

Strategy: Solve Wien’s Displacement Law (equation 30-1) for the peak frequency. Solution: Write the frequency using equation 30-1: ( )

( )( )

10 1 1peak

10 1 1 14

5.88 10 s K

5.88 10 s K 5800 K 3.4 10 Hz

f T− −

− −

= × ⋅

= × ⋅ = ×

Insight: This peak corresponds to a wavelength of 880 nm and falls in the infrared portion of the electromagnetic spectrum (see Chapter 25). We feel this infrared light from the Sun as warmth.

6. Picture the Problem: A blackbody emits radiation according to Wien’s Displacement Law.

Strategy: Use equation 30-1 to calculate the ratio of the peak frequencies for temperatures of 20.0 K and 40.0 K. Repeat for the Kelvin temperatures that correspond to 20.0°C and 40.0°C.

Solution: 1. (a) Calculate the ratio of peak frequencies for 20.0 K and 40.0 K:

( )( )

10 1 120peak,40

10 1 1peak,20 10

5.88 10 s K 40.0 K 2.0020.0 K5.88 10 s K

Tff T

− −

− −

× ⋅= = =

× ⋅

2. (b) Calculate the ratio of peak frequencies for 20.0°C and 40.0°C:

( )( )

10 1 140peak,40

10 1 1peak,20 20

5.88 10 s K 273.15 40.0 C 1.068273.15 20.0 C5.88 10 s K

Tff T

− −

− −

× ⋅ + °= = =

+ °× ⋅

Insight: Doubling the temperature in Kelvin actually doubles the average kinetic energy of the molecules. However, in this case doubling the temperature in Celsius only corresponds to a slight (6.8%) increase in the Kelvin temperature.

Page 3: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 3

7. Picture the Problem: A negative image of the double star Albireo is shown at right. When

viewed through a telescope, the upper-left star A is a warm golden color and the lower-right star B is a brilliant blue.

Strategy: According to Wien’s Displacement Law, a hotter star will emit light with a higher peak frequency. Because blue has a higher frequency than yellow, the blue star will be the hotter star. Use Wien’s Displacement Law (equation 30-1) to calculate the ratio of the peak frequencies from these two stars.

Solution: 1. (a) Star B is the blue star. A higher temperature corresponds to a higher frequency, and blue yellow .f f>

2. (b) Calculate the ratio of the peak frequencies:

( )( )

10 1 1A

10 1 1B

5.88 10 s K 4700 K 0.3613,000 K5.88 10 s K

A

B

Tff T

− −

− −

× ⋅= = =

× ⋅

Insight: Star A’s peak frequency is in the infrared spectrum (see Chapter 25), so it has a higher intensity in the red-yellow part of the visible spectrum and it appears golden. Star B’s peak frequency is in the ultraviolet spectrum, so it has a higher intensity in the blue end of the spectrum and it appears blue.

8. Picture the Problem: An incandescent light bulb operates at a cooler temperature than a halogen bulb. Therefore, the

blackbody radiation from the two light bulbs will have different peak frequencies. Strategy: According to Wien’s Displacement Law, a hotter bulb will emit light with a higher peak frequency. Because

the halogen bulb is hotter, it will emit a higher peak frequency than the incandescent bulb. Use Wien’s Displacement Law (equation 30-1) to calculate both of the peak frequencies and calculate their ratio.

Solution: 1. (a) Because peak ,f T∝ the halogen bulb has the higher peak frequency.

2. (b) Calculate the ratio of the peak frequencies:

( )( )

10 1 1halhal

10 1 1std std

5.88 10 s K 3400 K 1.22900 K5.88 10 s K

Tff T

− −

− −

× ⋅= = =

× ⋅

3. (c) Calculate the peak frequencies of the two bulbs: ( )( )( )( )

10 1 1 14hal

10 1 1 14std

5.88 10 s K 3400 K 2.0 10 Hz

5.88 10 s K 2900 K 1.7 10 Hz

f

f

− −

− −

= × ⋅ = ×

= × ⋅ = ×

4. The halogen bulb produces a peak frequency that is closer to 145.5 10 Hz× than the standard incandescent bulb. Insight: Because the halogen bulb produces a peak frequency closer to the sensitive frequencies of the human eye, more

of the radiation from the halogen bulb is visible, making it appear brighter for the same power output. 9. Picture the Problem: The tungsten filament in a standard light bulb can be considered a blackbody radiator.

Strategy: Use Wien’s Displacement Law (equation 30-1) to find the peak frequency of the radiation from the tungsten filament.

Solution: 1. (a) Solve Eq. 30-1 for the peak frequency:

( )( ) ( )

10 1 1peak

10 1 1 14

5.88 10 s K

5.88 10 s K 2850 K 1.68 10 Hz

f T− −

− −

= × ⋅

= × ⋅ = ×

2. (b) Because the peak frequency is that of infrared electromagnetic radiation, the light bulb radiates more energy in theinfrared than the visible part of the spectrum.

Insight: This radiation in the infrared spectrum is what makes the light from a standard light bulb feel warm. By contrast, a fluorescent light bulb emits most of its radiation in the ultraviolet and visible spectrum, and therefore the light from the fluorescent bulb does not feel warm. Fluorescent lamps are also much more efficient at converting electrical energy into visible light, because most of the light from a standard light bulb is in the infrared.

Page 4: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 4

10. Picture the Problem: The image shows two oxygen atoms oscillating

back and forth, similar to a mass on a spring. The image also shows the evenly spaced energy levels of the oscillation.

Strategy: Use equation 13-11 to calculate the period of oscillation for the two atoms. Calculate the frequency, using equation 13-1, from the inverse of the period. Multiply the period by Planck’s constant to calculate the spacing of the energy levels.

Solution: 1. (a) Set the frequency equal to the inverse of the period:

1326

2

1 1 1 1215 N/m 4.792 10 Hz2 2 1.340 10 kg

mTk

kfT m

π

π π −

=

= = = = ××

2. (b) Multiply the frequency by Planck’s constant: 34 13 20(6.63 10 J s)(4.792 10 Hz) 3.18 10 JE hf − −= = × ⋅ × = ×

Insight: Oxygen molecules in the atmosphere absorb light waves with a frequency 134.792 10 Hz× , and the energy from the light is converted into the oscillations of the atoms.

11. Picture the Problem: A source of red light, a source of green light, and a source of blue light each produce beams of

light with the same power.

Strategy: Use the concepts of wavelength, frequency, and photon energy to answer the conceptual questions. Note that these sources have the same power; that is, they give off the same amount of energy per time. In addition, note that red photons have less energy than green photons, and green photons have less energy than blue photons.

Solution: 1. (a) The frequency of light increases as you go from red to green to blue, which means that the wavelength decreases. Therefore, the ranking in order of increasing wavelength is blue < green < red.

2. (b) The ranking in order of increasing frequency is red < green < blue.

3. (c) More red photons must be emitted to give the same energy as blue photons. Therefore, the ranking in order of increasing number of photons emitted per second is blue < green < red.

Insight: If each source were to emit the same number of photons per second, the ranking of the power emitted would be red < green < blue.

12. Picture the Problem: A source of red light has a higher wattage than a source of green light.

Strategy: Use the expression for photon energy (equation 30-4) to answer the conceptual question.

Solution: 1. (a) The energy of photons emitted by the red source is less than the energy of photons emitted by the green source, regardless of the wattage of the source. This conclusion follows from the relation E = h f and the fact that green light has a higher frequency than red light.

2. (b) The best explanation is II. The red-source photons have less energy than the green-source photons because they have a lower frequency. The wattage of the source doesn’t matter. Statements I and III are each false.

Insight: A higher wattage source in general produces more photons than does a lower wattage source, but the energy of each photon is determined by the frequency, not the power of the source.

13. Picture the Problem: A source of yellow light has a higher wattage than a source of blue light.

Strategy: Use the expression for photon energy (equation 30-4) to answer the conceptual question.

Solution: 1. (a) The number of photons emitted per time by the yellow source is greater than the number of photons emitted per second by the blue source. There are two reasons for this. First, its photons have less energy than the photons from the blue source, because red light has a lower frequency than blue light and E = h f. Therefore, even if the two sources had the same power, the yellow source would emit more photons per second. Second, the yellow source has the greater power, requiring even more photons per second.

Page 5: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 5

2. (b) The best explanation is I. The yellow source emits more photons per second, because (i) it emits more energy per second than the blue source, and (ii) its photons have less energy than those of the blue source. Statements II and III are each false.

Insight: The number of photons emitted per second might be equal if you compared a low wattage yellow source with a higher wattage blue source of light.

14. Picture the Problem: Light of a particular wavelength does not eject electrons from the surface of a given metal.

Strategy: Use the concepts that describe the photoelectric effect to answer the question.

Solution: 1. (a) To eject electrons from this metal requires photons with greater energy, which means that the frequency of the light must be increased. If the frequency of the light is increased, its wavelength must be decreased.

2. (b) The best explanation is II. The energy of a photon is proportional to its frequency; that is, inversely proportional to its wavelength. To increase the energy of the photons so they can eject electrons, one must decrease their wavelength. Statement I erroneously asserts that increasing the wavelength will increase the energy.

Insight: Increasing the intensity of the first light source will have no effect, because that would only direct more photons to the metal surface, none of which have sufficient energy to eject an electron.

15. Picture the Problem: Light of a certain wavelength and intensity does not eject electrons from the surface of a metal.

Strategy: Use the concepts that describe the photoelectric effect to answer the question.

Solution: No. If the intensity of the light is increased there will be more photons striking the surface per second. Each photon, however, will still have too little energy to eject an electron.

Insight: Electrons could only be ejected if the wavelength of the light were decreased until the photon energies are larger than the work function of the metal.

16. Picture the Problem: UV photons have an energy of 196.5 10 J.−× This energy corresponds to a specific frequency and

wavelength.

Strategy: Solve equation 30-4 for the frequency of the photon. Insert the frequency into equation 14-1 to calculate the wavelength.

Solution: 1. Calculate the frequency:

1914

34

6.5 10 J 9.8 10 Hz6.63 10 J s

Efh

×= = = ×

× ⋅

2. Calculate the wavelength from the frequency:

8

14

3.00 10 m/s 310 nm 0.31 m9.80 10 Hz

cf

λ μ×= = = =

×

Insight: As expected, this frequency and wavelength lie in the UV spectrum (see Chapter 25). 17. Picture the Problem: The signal from a radio station can be considered as the transmission of many photons of the

same frequency. The power output is equal to the number of photons emitted per second multiplied by the energy of each photon.

Strategy: Divide the emitted power by the energy of each photon (equation 30-4) in order to calculate the rate of photon emission.

Solution: Calculate the rate of photon emission:

( )( )3

3234 3

270 10 W 4.6 10 photons/s6.63 10 J s 880 10 Hz

P PE hf −

×= = = ×

× ⋅ ×

Insight: Note that a radio station that broadcasts with the same power output, but at a higher frequency, would emit fewer photons per second. For example, a station broadcasting 270 kW at 1400 kHz would emit 322.9 10 photons/s.×

Page 6: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 6

18. Picture the Problem: A helium atom is ionized when it absorbs the energy from a 50.4-nm photon.

Strategy: Set the energy of the photon (from equations 30-4 and 14-1) equal to the ionization potential of helium.

Solution: Replace the frequency in equation 30-4 with the speed of light divided by the wavelength:

( )( )34 818

9

6.63 10 J s 3.00 10 m/s3.95 10 J

50.4 10 mhcEλ

−−

× ⋅ ×= = = ×

×

Insight: Photons with wavelengths longer than 50.4 nm cannot completely ionize the helium atom because they have too little energy. When photons with wavelengths that are shorter than 50.4 nm interact with the helium atom, they will both ionize the atom and give extra kinetic energy to the emitted electrons.

19. Picture the Problem: The light from a flashlight can be considered as the emission of many photons of the same

frequency. The power output is equal to the number of photons emitted per second multiplied by the energy of each photon.

Strategy: Divide the emitted power by the energy of each photon, given by equation 30-4, to calculate the rate of photon emission.

Solution: Calculate the rate of photon emission: ( )( )

1834 14

2.5 W 7.3 10 photons/s6.63 10 J s 5.2 10 Hz

P PE hf −= = = ×

× ⋅ ×

Insight: A flashlight emitting photons of a lower frequency would need to emit more photons per second to achieve the same power output. For instance, if the frequency were 144.6 10 Hz× , then 8.2×1018 photons/s would need to be emitted in order to produce a power of 2.5 W.

20. Picture the Problem: As photons interact with silver, some of the energy of the photon frees an electron from the silver

atom. The remainder of the energy becomes kinetic energy of the electron.

Strategy: Solve equation 30-7 for the work function of silver.

Solution: Calculate the work function:

( )( )max 0

34 14 190 max

19

6.63 10 J s 9.95 10 Hz 0.180 10 J

= 6.42 10 J

K hf W

W hf K − −

= −

= − = × ⋅ × − ×

×

Insight: In order for a photon to eject an electron from silver, it must have a frequency greater than 149.68 10 Hz.× At this frequency, all of the energy of the photon goes into the work function and the ejected electron has no kinetic energy.

21. Picture the Problem: As photons interact with gold, some of the energy of the photon frees an electron from the gold.

The remainder of the energy becomes kinetic energy of the electron.

Strategy: Solve equation 30-7 for the frequency of light. The work function is given in units of electron volts and must be converted into joules.

Solution: 1. Solve eq. 30-7 for the frequency:

max 0

max 0

K hf WK W

fh

= −+

=

2. Insert the given data:

( ) ( )( )19 1915

34

6.48 10 J 4.58 eV 1.60 10 J eV2.08 10 Hz

6.63 10 J sf

− −

× + ×= = ×

× ⋅

Insight: If this same photon were to interact with silver (which has a work function of 4.0 eV), the ejected electron would have a greater kinetic energy.

Page 7: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 7

22. Picture the Problem: Each photon contains a quantized amount of energy, determined by the photon wavelength. The

total energy of 2.5 J is obtained by the addition of many photons.

Strategy: Calculate the energy of a single photon from its wavelength using equations 30-4 and 14-1. Divide the total energy by the energy of the single photon to calculate the number of photons.

Solution: 1. (a) Use equations 30-4 and 14-1 to write the energy of a single photon:

cE hf hλ

= =

2. Divide the total energy by the energy of a single 350-nm photon:

( )( )( )( )

918total total

34 8photon

350 10 m 2.5 J4.4 10 photons

6.63 10 J s 3.00 10 m/sE E

nE hc

λ−

×= = = = ×

× ⋅ ×

3. (b) Divide the total energy by the energy of a 750-nm photon:

( )( )( )( )

918

34 8

750 10 m 2.5 J9.4 10 photons

6.63 10 J s 3.00 10 m/sn

×= = ×

× ⋅ ×

Insight: More red photons than UV photons are required in order to produce the same amount of energy. 23. Picture the Problem: A monochromatic light bulb emits many photons per second. The number of photons per second

multiplied by the energy per photon is the power radiated by the light bulb.

Strategy: Divide the power output by the energy per photon (from equations 30-4 and 14-1) to obtain the rate of photon emission in all directions. The fraction of these photons that enters your eye is the ratio of the area of your pupil to the surface area of a sphere with radius 15 m. Use this fraction to find the number of photons entering your eye per second.

Solution: 1. (a) Calculate the energy per photon:

cE hf hλ

= =

2. Divide the power by the energy per photon to find the rate of emission:

( )( )( )( )

920

34 8

650 10 m 45 W1.5 10 photons/s

6.63 10 J s 3.00 10 m/sP PnE hc

λ−

×= = = = ×

× ⋅ ×

3. (b) Multiply the photon rate by the ratio of the areas of your pupil and the 15-m sphere:

( )( )

2eye 20 12

eye 2sphere

0.0025 mphotons1.5 10 1.0 10 photons/ss 4 15 m

An n

π⎛ ⎞= = × = ×⎜ ⎟⎝ ⎠

Insight: The minimum photon flux that your eye can detect is about 500 photons/s. 24. Picture the Problem: Two radio stations broadcast at the same power level, but station A broadcasts at a lower

frequency than station B.

Strategy: The rate of photon emission is the power level divided by the energy of a single photon. Note that the energy of each photon is proportional to its frequency (equation 30-4).

Solution: 1. (a) The rate of emitted photons is determined by the ratio of the emitted power to the energy of a photon. This ratio is inversely proportional to the frequency of the emitted photons. Therefore, the station with the lower frequency has the highest rate of photon emission, and station A emits more photons per second than station B.

2. (b) Because the energy per photon is directly proportional to the frequency of the photon, the station that broadcasts at the higher frequency emits the higher energy photons. So, station B emits higher energy photons than station A.

Insight: In this example, station A emits 319.7 10 photons/s× that each have an energy of 285.9 10 J.−× Station B emits 316.2 10 photons/s× that each have an energy of 289.3 10 J.−×

Page 8: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 8

25. Picture the Problem: The two atoms of a hydrogen molecule are bound together. When a photon with sufficient energy

is absorbed by the molecule it can disassociate the molecule into two hydrogen atoms.

Strategy: Calculate the energy necessary to disassociate a single hydrogen molecule by dividing the given energy by the number of molecules in a mole. Use equation 30-4 to calculate the frequency of the light and equation 14-1 to calculate the wavelength. Refer to Figure 25-8 to determine the region of the electromagnetic spectrum of this photon.

Solution: 1. (a) Calculate the energy to disassociate one molecule:

( ) ( ) 1923

104.2 kcal mol4186 J/kcal 7.245 10 J

6.02 10 molecules/moleE −= = ×

×

2. Calculate the photon frequency:

1915

34

7.245 10 J 1.09 10 Hz6.63 10 J s

Efh

×= = = ×

× ⋅

3. Calculate the wavelength:

8

15

3.00 10 m/s 275 nm1.09 10 Hz

cf

λ ×= = =

×

4. (b) The photon lies in the ultraviolet region of the electromagnetic spectrum because UV 400 nm.λ <

Insight: Photons with wavelengths greater than 275 nm (such as visible light) do not contain sufficient energy to disassociate the hydrogen molecule.

26. Picture the Problem: The light from the laser can be considered as the emission of many photons of the same wavelength. The power output is the product of the photon emission rate and the energy of each photon.

Strategy: Divide the emitted power by the energy of each photon (equation 30-4) to calculate the rate of photon emission, and then find the frequency using equation 14-1.

Solution: 1. (a) Calculate the rate of photon emission: /

P P P PnE hf hc hc

λλ

= = = =

2. Insert the given wavelength and power:

( )( )( )( )

9 315

34 8

632.8 10 m 1.0 10 W 3.2 10 photons/s

6.63 10 J s 3.00 10 m/sn

− −

× ×= = ×

× ⋅ ×

3. Calculate the frequency from equation 14-1:

8 m14s

9

3.00 104.74 10 Hz

632.8 10 mcfλ −

×= = = ×

×

Insight: The large number of photons in a laser beam can cause permanent damage to the retina before you can blink, which is why you should never look into a laser!

27. Picture the Problem: A 150-watt red light bulb emits photons with wavelength of 650 nm, and a 25-watt blue light bulb emits photons with a wavelength of 460 nm.

Strategy: Combine equations 30-4 and 14-1 to find the energy of the photons for each bulb. Divide the emitted power by the energy of each photon to calculate the rate at which photons are emitted.

Solution: 1. (a) The energy emitted per photon is inversely related to the wavelength, so the red light bulb must emit more photons to produce the same power output. The red bulb also produces more power than the blue bulb, so the red bulb emits more photons per second than does the blue bulb.

2. (b) The energy per photon is inversely proportional to the wavelength. Because the blue bulb has the shorter wavelength, the blue bulb emits photons of higher energy than the red bulb.

3. (c) Calculate the energy of the red photon: ( )

834 19

9

3.00 10 m/s6.63 10 J s 3.06 10 J650 10 m

cE hf hλ

− −−

×= = = × ⋅ = ×

×

4. Divide the power by the energy: 20

red 19

150 W 4.9 10 photons/s3.06 10 J/photon

PnE −= = = ×

×

Page 9: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 9

5. Calculate the energy of the blue photon: ( )

834 19

9

3.00 10 m/s6.63 10 J s 4.32 10 J460 10 m

E − −−

×= × ⋅ = ×

×

6. Divide the power by the energy: 19

blue 19

25 W 5.8 10 photons/s4.32 10 J/photon

PnE −= = = ×

×

Insight: If the blue bulb were to produce the same number of photons per second as the red bulb, it would emit a power of 210 W.

28. Picture the Problem: A photon of wavelength 264 nm will eject an electron from a copper surface. The resulting

electron, however, has no kinetic energy.

Strategy: Set the work function of copper equal to the energy of the photon, given by equations 30-4 and 14-1.

Solution: Calculate the work function of copper:

( )( )34 819

0 9

6.63 10 J s 3.00 10 m/s7.53 10 J

264 10 mhcW hfλ

−−

× ⋅ ×= = = = ×

×

Insight: Electrons ejected from copper by photons with wavelengths less than 264 nm will have kinetic energy. Photons larger than 264 nm cannot eject electrons from copper.

29. Picture the Problem: Photons with energies greater than the work function of a metal can eject electrons from that

metal.

Strategy: Use equation 30-6 to calculate the cut-off frequencies for the two metals.

Solution: 1. (a) Because higher-frequency photons have higher energies, and since Al Ca ,W W> aluminum requires higher-frequency light to produce photoelectrons.

2. (b) Calculate the cutoff frequency for aluminum:

( )( )1915Al

Al 34

4.28 eV 1.60 10 J/eV1.03 10 Hz

6.63 10 J sWf

h

×= = = ×

× ⋅

3. Calculate the cutoff frequency for calcium:

( )( )1914Ca

Ca 34

2.87 eV 1.60 10 J/eV6.93 10 Hz

6.63 10 J sW

fh

×= = = ×

× ⋅

Insight: A photon with frequency 1.03×1015 Hz will eject electrons from either surface. Such a photon has a wavelength of 291 nm and is in the ultraviolet region of the electromagnetic spectrum (see Chapter 25).

30. Picture the Problem: When a photon is absorbed by a metal, its energy is split into the energy needed to eject the

electron and the kinetic energy of the electron. The energy of the photon is inversely proportional to its wavelength.

Strategy: Use equations 30-7 and 14-1 to write an expression of the kinetic energy of the electron in terms of the wavelength of the photon.

Solution: 1. (a) Because max 0 0 ,hcK hf W W

λ= − = − a shorter wavelength implies a larger kinetic energy. So, the beam

with wavelength Bλ produces photoelectrons with greater kinetic energy than the beam with wavelength A .λ

2. (b) Calculate the kinetic energy of the photoelectron when struck by photon A:

( )( )( )( )

34 8

max,A 9 19

6.63 10 J s 3.00 10 m/s1.9 eV 0.1 eV

620 10 m 1.60 10 J/eVK

− −

× ⋅ ×= − =

× ×

3. Calculate the kinetic energy of the photoelectron when struck by photon B:

( )( )( )( )

34 8

max,B 9 19

6.63 10 J s 3.00 10 m/s1.9 eV 1.1 eV

410 10 m 1.60 10 J/eVK

− −

× ⋅ ×= − =

× ×

Insight: The cutoff wavelength that will eject an electron from cesium is 654 nm. Because photon A has a wavelength slightly smaller than the cutoff, the resulting photoelectron has a small kinetic energy.

Page 10: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 10

31. Picture the Problem: When a photon is absorbed by a metal, its energy is split into the energy needed to eject the

electron (the work function) and the kinetic energy of the electron. The energy of the photon is inversely proportional to its wavelength.

Strategy: Calculate the energy of the photon using equations 30-4 and 14-1. Then use equation 30-5 to write an expression for the maximum kinetic energy of photoelectrons ejected from each metal.

Solution: 1. (a) The difference in energy between the incident photons and the work function of the metal is the photoelectron’s kinetic energy. Therefore, if two different metals are illuminated by photons with the same energy, the metal with the smaller work function will emit photoelectrons with a greater maximum kinetic energy. In this case, cadmium has a smaller work function and emits photoelectrons with the greater maximum kinetic energy.

2. (b) Calculate the energy of the photon:

( )( )( )( )

34 8

9 19

6.63 10 J s 3.00 10 m/s4.52 eV

275 10 m 1.60 10 J/eVhcE hfλ

− −

× ⋅ ×= = = =

× ×

3. Insert the work function for zinc into equation 30-5: max, Zn 0 4.52 eV 4.33 eV 0.19 eVK E W= − = − =

4. Repeat for cadmium: max, Cd 4.52 eV 4.22 eV 0.30 eVK = − =

Insight: Because the work function of zinc is closer to the energy of the photon, the electron ejected from zinc has less kinetic energy than the electron ejected from the cadmium.

32. Picture the Problem: When white light is incident upon the potassium, the photons with energies greater than the work

function of potassium will eject electrons. The greater the photon energy, the greater the kinetic energy of the ejected electron.

Strategy: Because the photon energy is proportional to the frequency, the photons with the greatest frequency will eject electrons with the maximum kinetic energy. Insert the maximum frequency into equation 30-7 to calculate the maximum kinetic energy of the photoelectrons. Then use equation 30-6 to calculate the cutoff frequency. All photons with smaller frequencies will not eject electrons.

Solution: 1. (a) Use equation 30-7 to find max :K

( )( )34 14

max 0 19

6.63 10 J s 7.90 10 Hz2.24 eV

1.60 10 J/eV1.03 eV

K hf W−

× ⋅ ×= − = −

×=

2. (b) Calculate the cutoff frequency:

( )( )19140

34

2.24 eV 1.60 10 J/eV5.41 10 Hz

6.63 10 J sW

fh

×= = = ×

× ⋅

3. Write the frequency range for which no electrons are emitted:

14 144.00 10 Hz 5.41 10 Hzf× ≤ < ×

Insight: Photons that are incident upon a potassium surface and that have frequencies greater than 145.41 10 Hz× will emit photoelectrons with kinetic energies ranging from zero to 1.03 eV.

33. Picture the Problem: When the electromagnetic waves are incident upon the aluminum surface, the photons with

energies greater than the work function of aluminum eject electrons. The greater the photon energy, the greater the kinetic energy of the ejected electron.

Strategy: Because the photon energy is proportional to the frequency, the photons with the greatest frequency will eject electrons with the maximum kinetic energy. Insert the maximum frequency into equation 30-7 to calculate the maximum kinetic energy of the photoelectrons. Then use equation 30-6 to calculate the cutoff frequency. All photons with smaller frequencies will not eject electrons.

Page 11: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 11

Solution: 1. (a) Use equation 30-7 to find max :K

( )( )34 16

max 0 19

6.63 10 J s 9.00 10 Hz4.28 eV

1.60 10 J/eV369 eV

K hf W−

× ⋅ ×= − = −

×=

2. (b) Calculate the cutoff frequency:

( )( )19150

34

4.28 eV 1.60 10 J/eV1.03 10 Hz

6.63 10 J sW

fh

×= = = ×

× ⋅

3. Write the frequency range for which no electrons are emitted:

14 154.00 10 Hz 1.03 10 Hzf× ≤ < ×

Insight: Photons with frequencies greater than 151.03 10 Hz× emit electrons with kinetic energies ranging from zero to 369 eV. It is a bit odd to call these photons “electromagnetic waves” as in the problem statement, because the classical description of light as a wave is inadequate to explain the photoelectric effect.

34. Picture the Problem: When photons of the given frequency are incident on the two metals, some of the energy of the

photon will eject the electron from the surface (work function) and the remainder of the energy will become kinetic energy of the electron.

Strategy: Use equation 30-4 to calculate the energy of the photon. Then insert that energy together with the work function of each metal into equation 30-5 to calculate the maximum kinetic energy.

Solution: 1. (a) Because max 0 ,K hf W= − the photoelectrons emitted by the metal with the smaller work function will have the greater kinetic energy. The electrons ejected from the iron surface will have the greater maximum kinetic energy because iron has a smaller work function than platinum.

2. (b) Calculate the photon energy: ( )( )34 15

19

6.63 10 J s 1.88 10 Hz7.79 eV

1.60 10 J/eVE hf

× ⋅ ×= = =

×

3. Calculate the maximum kinetic energy for electrons photoejected from platinum: max, Pt 0 7.79 eV 6.35 eV 1.44 eVK E W= − = − =

4. Calculate the maximum kinetic energy for photoelectrons ejected from iron: max, Fe 7.79 eV 4.50 eV 3.29 eVK = − =

Insight: The cutoff frequency for platinum is 151.53 10 Hz× and for iron is 151.09 10 Hz× . Photons with frequencies between these two values will eject photoelectrons from the iron, but not from the platinum.

35. Picture the Problem: As photons of two different frequencies are incident upon a metal, they eject photoelectrons with

two different maximum kinetic energies. The maximum kinetic energy of photoelectrons is the difference between the energy of the photon and the work function of the metal.

Strategy: Solve equation 30-7 for the work function. Because the work function for the metal is constant, set the work functions for both experiments equal and solve for Planck’s constant.

Solution: 1. Solve equation 30-7 for the work function:

max 0

0 max

K hf WW hf K

= −= −

2. Set the work functions equal and solve for Planck’s constant:

1 max,1 2 max,2

19 19max,1 max,2

12 121 2

34

1.260 10 J 2.480 10 J547.5 10 Hz 738.8 10 Hz

6.377 10 J s

hf K hf KK K

hf f

− −

− = −

− × − ×= =

− × − ×

= × ⋅

Insight: These experiments determine Planck’s constant to within 4% of its accepted value of 346.6260693 10 J s.−× ⋅

Page 12: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 12

36. Picture the Problem: The large pupil in an owl’s eye allows for more photons per second to enter the eye than the

number of photons that would enter the human eye. In addition, the retina of the owl is more sensitive to fewer photons than the human eye.

Strategy: Multiply the minimum intensity of light by the area of the owl’s pupil to determine the power absorbed by the owl’s eye. Divide this power by the energy per photon to determine the minimum photon flux that the owl can detect.

Solution: 1. Calculate the power absorbed in the owl’s eye at minimum intensity: ( ) ( )22 13 2 175.0 10 W/m 0.00425 m 2.837 10 WP IA I rπ π− −= = = × = ×

2. Calculate the energy per photon: ( )( )34 34 196.63 10 J s 7.0 10 Hz 4.64 10 JE hf − − −= = × ⋅ × = ×

3. Divide the power by the energy:

( )17

19

2.837 10 W61 photons/s

4.64 10 JPnE

×= = =

×

Insight: If the human eye were sensitive to 61 photons/s, the smaller pupil of the human eye would require a minimum light intensity of 13 27.4 10 W/m .−×

37. Picture the Problem: The momentum of a photon is doubled, causing its energy to increase.

Strategy: Use the expression p E c= for a photon to answer the conceptual question.

Solution: The momentum of a photon is linearly proportional to its energy. Therefore, doubling the momentum of a photon increases its energy by a multiplicative factor of 2.

Insight: The relationship between p and E is a consequence of the fact that both the energy and the momentum of a photon are linearly proportional to its frequency (equations 30-4 and 30-11).

38. Picture the Problem: The photons in a microwave have a given momentum and wavelength.

Strategy: Use equation 30-11 to calculate the wavelength of the microwaves.

Solution: 1. (a) Solve equation 30-11 for the wavelength:

34

33

6.63 10 J s 13 cm5.1 10 kg m/s

hp

λ−

× ⋅= = =

× ⋅

2. (b) The wavelength is much larger than the size of the holes in the metal screen.

Insight: Because the holes are smaller than the microwaves, the microwaves reflect off the screen and remain inside the microwave oven. However, visible light, which has a much smaller wavelength, can exit through the holes and allow you to see the food cooking.

39. Picture the Problem: A photon and an electron have the same momentum. However, because the electron has mass, its speed will be slower than that of the photon.

Strategy: Use equation 30-11 to calculate the momentum of the photon. Set the momentum of the photon equal to the momentum of the electron. Assume that the speed of the electron is not relativistic ( 0.10v c< ) and use equation 9-1.

Solution: 1. Calculate the momentum of the photon:

3424

9

6.63 10 J s 2.65 10 kg m/s0.25 10 m

hpλ

−−

× ⋅= = = × ⋅

×

2. Set the momentum equal to the momentum of an electron and solve for the velocity:

246

31

2.65 10 kg m/s 2.9 10 m/s 0.00979.11 10 kg

p mvpv cm

=

× ⋅= = = × =

×

Insight: Since the speed of the electron is much less than the speed of light, our assumption of classical momentum is valid. If the speed had been greater than 0.10c, it would have been better to use a relativistic formula (equation 29-5).

Page 13: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 13

40. Picture the Problem: A photon and an electron have the same momentum. The electron’s momentum is determined by

its speed and mass. The photon’s momentum is determined by its wavelength.

Strategy: Calculate the momentum of the electron using equation 9-1. The relativistic formula (equation 29-5) is not necessary because the speed is much lower than the speed of light. Set the electron and photon momenta equal and use equation 30-11 to calculate the photon wavelength.

Solution: 1. Find the momentum of the electron: ( )( )31 279.11 10 kg 1200 m/s 1.09 10 kg m/sp mv − −= = × = × ⋅

2. Calculate the wavelength of the photon:

34

27

6.63 10 J s 610 nm 0.61 m1.09 10 kg m/s

hp

λ μ−

× ⋅= = = =

× ⋅

Insight: This wavelength corresponds to visible, green light.

41. Picture the Problem: A photon and a neutron have the same momentum. The neutron’s momentum is determined by its speed and mass. The photon’s momentum is determined by its wavelength.

Strategy: Calculate the momentum of the neutron using equation 9-1. The relativistic formula (equation 29-5) is not necessary because the speed is much lower than the speed of light. Set this momentum equal to the momentum of the photon and use equation 30-11 to calculate the photon frequency.

Solution: 1. Find p for the neutron: ( )( )27 241.675 10 kg 1500 m/s 2.513 10 kg m/sp mv − −= = × = × ⋅

2. Calculate the frequency of the photon:

( )( )24 818

34

2.513 10 kg m/s 3.00 10 m/s1.1 10 Hz

6.63 10 J spcfh

× ⋅ ×= = = ×

× ⋅

Insight: The photon with this frequency is in the X-ray part of the spectrum.

42. Picture the Problem: A hydrogen atom that is initially at rest emits a 122-nm photon. The atom recoils in the direction opposite to the propagation of the photon.

Strategy: Before the emission the hydrogen atom was at rest. The photon and the hydrogen atom then have equal but opposite momenta. Use equation 30-11 to calculate the momentum of the photon. Set it equal to the momentum of the hydrogen atom and use equation 9-1 to calculate the recoil speed of the hydrogen atom.

Solution 1: Calculate the momentum of the photon:

3427

UV 9

6.63 10 J s 5.43 10 kg m/s122 10 m

hpλ

−−

× ⋅= = = × ⋅

×

2. Solve for the recoil speed: UV H27

UV27

5.43 10 kg m/s 3.25 m/s1.674 10 kg

p p mvp

vm

= =

× ⋅= = =

×

Insight: Note that the speed of the hydrogen atom is inversely proportional to the wavelength of the photon. If the atom had emitted a photon with wavelength 61 nm,λ = the recoil speed would double to 6.5 m/s.

43. Picture the Problem: A blue-green photon collides with a stationary hydrogen atom. The hydrogen atom absorbs the photon and moves forward with the same momentum as the initial photon.

Strategy: Calculate the initial momentum of the photon using equation 30-11. Then set that momentum equal to the momentum of the hydrogen atom after the absorption. Use equation 9-1 to calculate the speed of the hydrogen.

Solution: 1. Calculate the momentum of the photon:

3427

ph 9

6.63 10 J s 1.36 10 kg m/s486 10 m

hpλ

−−

× ⋅= = = × ⋅

×

2. Solve for the speed of the hydrogen atom: ph H

27ph

27

1.364 10 kg m/s 0.815 m/s1.674 10 kg

p p mv

pv

m

= =

× ⋅= = =

×

Insight: The speed of the hydrogen atom is inversely proportional to the wavelength of the incident photon. Decreasing the wavelength of the photon will increase its momentum, and thus increase the speed of the hydrogen atom.

Page 14: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 14

44. Picture the Problem: A red photon and a blue photon each carry momentum in proportion to their frequencies.

Strategy: Calculate the momenta of the two photons using equation 30-11.

Solution: 1. (a) Because blue red ,f f> and since ,hfpc

= blue red .p p> A photon of blue light has the greater momentum.

2. (b) Calculate pred:

( )34 1428

red 8

6.63 10 J s 4.0 10 Hz8.8 10 kg m/s

3.00 10 m/shfpc

−−

× ⋅ ×= = = × ⋅

×

3. Calculate pblue:

( )34 1427

blue 8

6.63 10 J s 7.9 10 Hz1.7 10 kg m/s

3.00 10 m/sp

−−

× ⋅ ×= = × ⋅

×

Insight: Because the frequency of the blue photon is about double the frequency of the red photon, the momentum of the blue photon is about double the momentum of the red photon.

45. Picture the Problem: Photons A and B carry momentum in inverse proportion to their wavelengths.

Strategy: Set the momentum of photon A equal to twice the momentum of photon B. Write the momenta in terms of the wavelengths using equation 30-11. Solve the resulting equation for the wavelength of photon B.

Solution: 1. (a) Because a photon’s momentum is inversely proportional to its wavelength, the photon with the smaller momentum has the longer wavelength. Therefore, photon B has the longer wavelength.

2. (b) Use equation 30-11 to calculate the wavelength of photon B: ( )B A B A

B A

2 2 2 2 333 nm 666 nmh hp p λ λλ λ

= ⇒ = ⇒ = = =

Insight: Because the wavelength is inversely proportional to the momentum, a photon with twice the momentum will have half the wavelength.

46. Picture the Problem: As a beam of photons is absorbed by a black surface, the momentum of the photons is absorbed

by the surface, and a corresponding force is exerted on the surface.

Strategy: Calculate the rate of photon production by dividing the power of the laser beam by the energy in each photon (from equations 30-4 and 14-1). Calculate the change in momentum by using equation 30-11 for the photon momentum.Multiply the photon rate by the momentum change to obtain the net force on the black surface.

Solution: 1. (a) Calculate the energy in each photon:

( )( )34 819

9

6.63 10 J s 3.00 10 m/s3.143 10 J

632.8 10 mhcEλ

−−

× ⋅ ×= = = ×

×

2. Divide the power by the energy per photon:

316

-19photon

5.00 10 W 1.59 10 photons/s3.143 10 J

PnE

−×= = = ×

×

3. (b) Calculate the momentum change when each photon is stopped:

f i i34

279

0

6.63 10 J s 1.05 10 kg m/s632.8 10 m

p p p p

−−

Δ = − = −

× ⋅= − = − = − × ⋅

×

4. (c) Multiply the momentum change by the photon rate:

( )( )16 27

11

1.59 10 photons/s 1.05 10 kg m/s

1.67 10 N

F n p −

= Δ = × × ⋅

= ×

Insight: This force, even though it is small, has applications in fusion research, where it is used to confine hydrogen pellets. It is also employed by an “optical tweezers,” a laser device that allows a researcher to manipulate very tiny objects while looking at them through a microscope.

Page 15: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 15

47. Picture the Problem: As each photon in a laser beam is reflected by a mirror, its momentum is reversed, resulting in an

impulse on the mirror. The impulses from many photons produce an applied force on the mirror.

Strategy: Calculate the rate of photon production by dividing the power of the laser beam by the energy in each photon (from equations 30-4 and 14-1). When the photon is reflected, the magnitude of its change in momentum is equal to twice its momentum. Calculate this change in momentum by using equation 30-11. Multiply the photon rate by the momentum change to obtain the net force on the mirror.

Solution: 1. (a) Calculate the energy in each photon:

( )( )34 819

9

6.63 10 J s 3.00 10 m/s3.143 10 J

632.8 10 mhcEλ

−−

× ⋅ ×= = = ×

×

2. Divide the power by the energy per photon:

316

-19photon

7.50 10 W 2.39 10 photons/s3.143 10 J

PnE

−×= = = ×

×

3. (b) Calculate the momentum change when each photon is reflected: ( )

f i i

3427

9

2

2 6.63 10 J s2 2.10 10 kg m/s632.8 10 m

p p p p

−−

Δ = − =

× ⋅= = = × ⋅

×

4. (c) Multiply the momentum change by the photon rate:

( )( )16 27

11

2.39 10 photons/s 2.10 10 kg m/s

5.00 10 N

F n p −

= Δ = × × ⋅

= ×

Insight: The force on a reflecting mirror is twice the force on an absorbing surface (see problem 46) because the change in momentum is double for the reflecting surface.

48. Picture the Problem: In a Compton scattering experiment, the scattered electron is observed to move in the same

direction as the incident X-ray photon.

Strategy: Use the principles that govern the Compton effect to answer the conceptual question.

Solution: Suppose the initial photon moves in the positive x direction. This means that the initial y component of momentum is zero. After the collision, we are told that the electron moves in the positive x direction—it has no y component of momentum. Therefore, the scattered photon must also have zero y component of momentum, which means it will propagate in either the positive x direction or the negative x direction. If it were to propagate in the positive x direction, the scattering angle would be 0° and the Compton formula (equation 30-15) indicates the photon will transfer no momentum to the electron. We reject that solution because it implies the electron does not scatter at all, and we conclude that the scattering angle of the photon is 180°.

Insight: Inserting θ = 180° into equation 30-15 shows that the wavelength of the scattered photon will increase by the amount ( )2 2 2.426 pm 4.85 pm.eh m cλΔ = = =

49. Picture the Problem: An X-ray photon scatters off of an electron and transfers part of its energy to the electron.

Strategy: By conservation of energy, the kinetic energy of the electron will equal the change in energy of the X-ray.

Solution: Calculate the change in energy of the X-ray: electron photon i f 38.0 keV 33.5 keV 4.5 keVK K K K= −Δ = − = − =

Insight: Inserting the maximum change in wavelength (θ = 180°) into equation 30-15 implies that the energy of the final X-ray must be at least 33.1 keV, for momentum to be conserved.

Page 16: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 16

50. Picture the Problem: The wavelength of a photon increases after it

scatters off an electron and travels at an angle θ from its initial direction.

Strategy: Use the Compton scattering equation (equation 30-15) to find λΔ for each of the scattering angles. A recurring value in similar

Compton scattering problems is 12e 2.426 10 m 2.426 pm.h m c −= × =

Solution: 1. Insert values for the constants in equation 30-15:

( )

( )( )e

1 cos

2.426 pm 1 cos

hm c

λ λ λ θ

θ

′Δ = − = −

= −

2. (a) Insert 30 :θ = ° ( )( )2.426 pm 1 cos30.0 0.325 pmλΔ = − ° =

3. (b) Insert 90 :θ = ° ( )( )2.426 pm 1 cos90.0 2.43 pmλΔ = − ° =

4. (c) Insert 180 :θ = ° ( )( )2.426 pm 1 cos180.0 4.85 pmλΔ = − ° =

Insight: The maximum change in wavelength is 4.85 pm because the maximum scattering angle is 180°. 51. Picture the Problem: The wavelength of a photon increases after it

scatters off an electron and travels at an angle θ from its initial direction.

Strategy: Solve the Compton scattering equation (equation 30-15) for the scattering angle as a function of the change in wavelength. A recurring value in similar Compton scattering problems is

12e 2.426 10 m 2.426 pm.h m c −= × =

Solution: 1. Solve equation 30-15 for θ: ( )

e

1

e

1 cos

cos 1

hm c

h m c

λ θ

λθ −

Δ = −

⎛ ⎞Δ= −⎜ ⎟

⎝ ⎠

2. Insert the given wavelength change: 1 3.13 pmcos 1 107

2.426 pmθ − ⎡ ⎤= − = °⎢ ⎥

⎣ ⎦

Insight: If the change in wavelength had only one-half of this value (1.565 pm), the scattering angle would be 69°, which is greater than one-half of the scattering angle from this problem. The change in wavelength and scattering angles are not linearly proportional to each other.

52. Picture the Problem: The smallest change in wavelength in a Compton scattering is zero, when the photon continues to

travel straight forward. The greatest change in wavelength is when the photon is scattered directly backward. A change in wavelength of one-fourth of the maximum will occur at an angle θ between the minimum and maximum.

Strategy: Set the scattering angle θ in equation 30-15 to 180° to find the maximum change in wavelength. Then solve equation 30-15 for the scattering angle, when the change in wavelength is equal to one-quarter of the maximum. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 2.426 pm.h m c −= × =

Solution: 1. Calculate the maximum change in wavelength: ( ) ( )max

e e e

21 cos 1 cos 180h h hm c m c m c

λ θΔ = − = − ° =⎡ ⎤⎣ ⎦

2. Set the change in wavelength equal to a quarter of the maximum and solve for the scattering angle:

( ) ( )

( )

setmax max

e

1 12

1 cos 1 cos2 4

cos 1 60

hm c

λ λλ θ θ

θ −

Δ ΔΔ = − = − =

= − = °

Insight: This angle is one-third of the maximum angle. If the angle were two-thirds of the maximum, or 120°, the change in wavelength would be 3

max4 .λ

Page 17: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 17

53. Picture the Problem: Two photons of different wavelengths are scattered backward from collisions with free-electrons.

Strategy: Use equation 30-15 to calculate the change in wavelength for a scattering of 180°. Calculate the percent difference by dividing the change in wavelength by the wavelength of each photon. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 2.426 pm.h m c −= × =

Solution: 1. (a) Because the change in wavelength does not depend on the wavelength of the photon, the change in wavelength is the same for both photons.

2. (b) Because the change in wavelength is the same, the photon with the shorter wavelength, 0.030 nm,λ = will experience the greater percent change in wavelength. The X-ray photon thus experiences the greater percent change.

3. (c) Use equation 30-15 to find :λΔ ( )

( ) ( )e

1 cos

2.426 pm 1 cos 180 4.85 pm 0.00485 nm

hm c

λ θΔ = −

= − ° = =⎡ ⎤⎣ ⎦

4. Calculate the percent change in wavelength of the visible photon:

40.00485 nm100% 9.3 10 %520 nm

λλ

−Δ= = ×

5. Calculate the percent change in wavelength of the X-ray:

0.00485 nm100% 16%0.030 nm

λλΔ

= =

Insight: Even though the change in wavelength is constant, the X-ray transfers more energy to the electron in order to conserve momentum in the collision. As the wavelength of the photon decreases, the transferred energy becomes a larger portion of the total energy of the incident photon.

54. Picture the Problem: A photon of wavelength 0.240 nm scatters off a free

electron at rest at an angle of 105°.

Strategy: Calculate the initial momentum of the photon using equation 30-11. Then use equation 30-15 to calculate the final wavelength of the photon. Insert the final wavelength into equation 30-11 to calculate the final momentum. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 0.002426 nm.h m c −= × =

Solution: 1. (a) Calculate the initial momentum:

34

i 9i

24

6.63 10 J s0.240 10 m

2.76 10 kg m/s

hpλ

× ⋅= =

×

= × ⋅

2. (b) Solve equation 30-15 for the final wavelength:

( )

( )

( ) ( )

f ie

f ie

1 cos

1 cos

0.240 nm 0.002426 nm 1 cos 105 0.243 nm

hm c

hm c

λ λ λ θ

λ λ θ

Δ = − = −

= + −

= + − ° =⎡ ⎤⎣ ⎦

3. (c) Calculate the final momentum:

3424

f 9f

6.63 10 J s 2.73 10 kg m/s0.243 10 m

hpλ

−−

× ⋅= = = × ⋅

×

Insight: The final momentum of the electron can be calculated from the vector change in the momentum of the photon. The final electron momentum is 244.36 10 kg m/s−× ⋅ at an angle of φ = −37.3°.

Page 18: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 18

55. Picture the Problem: A photon that has scattered off of an electron at an

angle of 175° has a wavelength of 0.320 nm.

Strategy: The incident photon will have a wavelength that is shorter than 0.320 nm. Solve equation 30-15 for the wavelength of the initial photon. For parts (b) and (c), calculate the initial and final energies of the photons using equations 30-4 and 14-1. The kinetic energy of the electron is equal to the difference in energies of the photons. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 0.002426 nm.h m c −= × =

Solution: 1. (a) Solve equation 30-15 for i :λ

( )

( )

( )( )

f ie

i fe

1 cos

1 cos

0.320 nm 0.002426 nm 1 cos175 0.315 nm

hm c

hm c

λ λ λ θ

λ λ θ

Δ = − = −

= − −

= − − ° =

2. (b) Calculate the energy of the incident photon:

( )( )( )( )

34 8

i 9 16i

6.63 10 J s 3.00 10 m/s3.94 keV

0.315 10 m 1.60 10 J/keVhcE hfλ

− −

× ⋅ ×= = = =

× ×

3. Calculate the energy of the scattered photon:

( )( )( )( )

34 8

f 9 16f

6.63 10 J s 3.00 10 m/s3.88 keV

0.320 10 m 1.60 10 J/keVhcEλ

− −

× ⋅ ×= = =

× ×

4. (c) The kinetic energy of the electron equals the change in photon energy: i f 3940 eV 3880 eV 60 eVK E E= − = − =

Insight: In this situation only 1.4% of the photon energy is transferred to the electron. 56. Picture the Problem: An X-ray scatters 180° of an electron or off a helium atom. As a result of the scattering, some of

the energy of the X-ray is transferred to the particle, and the wavelength of the X-ray is increased.

Strategy: Use equation 30-15 to calculate the change in wavelength, where θ = 180°. For the scattering off of the helium atom, replace the mass of the electron with the mass of helium.

Solution: 1. (a) Because 1

mλΔ ∝ and e He ,m m< the change in wavelength of the X-ray is greater for the electron.

2. (b) Calculate the change in wavelength for scattering off of the electron: ( )

( )( )( )

34

electron 31 8e

6.63 10 J s 1 cos1801 cos 4.85 pm

9.11 10 kg 3.00 10 m/sh

m cλ θ

× ⋅ − °Δ = − = =

× ×

3. Calculate the change in wavelength for scattering off of the helium atom: ( )

( )( )( )

34

He 27 8He

6.63 10 J s 1 cos1801 cos 0.666 fm

6.64 10 kg 3.00 10 m/sh

m cλ θ

× ⋅ − °Δ = − = =

× ×

Insight: Because the helium atom has much more mass than the electron, the helium atom requires less kinetic energy to carry away the same momentum, as such the photon does not need to transfer as much energy to the helium atom.

57. Picture the Problem: When a certain photon scatters off of an electron, its wavelength increases by 10% when its

scattering angle is θ = 135°.

Strategy: Set the change in wavelength equal to 0.10λ i in equation 30-15 and solve for the initial wavelength. Then use equation 30-4 and 14-1 to calculate the initial energy of the photon. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 2.426 pm.h m c −= × =

Solution: 1. Solve eq. 30-15 for the initial wavelength:

( )

( ) ( )( )

ie

ie

0.100 1 cos

1 cos 2.426 pm 1 cos13541.4 pm

0.100 0.100

hm c

hm c

λ λ θ

θλ

Δ = = −

− − °= = =

Page 19: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 19

2. Calculate the initial energy of the photon:

( )( )34 8

i 12i

6.63 10 J s 3.00 10 m/s4.80 fJ

41.4 10 mhcE hfλ

× ⋅ ×= = = =

×

Insight: The final energy of the photon is 4.36 fJ, because 0.44 fJ of energy is transferred to the electron as a result of the collision.

58. Picture the Problem: The image shows a photon with an initial

wavelength of 0.525 nm scattering off an electron. The wavelength of the scattered photon is 3.33 pm longer than the initial wavelength. As a result of the collision, the electron scatters at an angle φ from the positive x-axis.

Strategy: Use equation 30-11 to calculate the magnitude of the initial and final momentum of the photon. Calculate the scattering angle of the photon using equation 30-15. Use the scattering angle to calculate the horizontal and vertical components of the momentum of the scattered X-ray photon. Use conservation of momentum to calculate the components of the electron’s momentum. Finally, use these components to calculate the angle φ that the electron scatters. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 2.426 pm.h m c −= × =

Solution: 1. Calculate the initial momentum of the photon:

3424

ii

6.63 10 J s 1.263 10 kg m/s0.525 nm

hpλ

−−× ⋅

= = = × ⋅

2. Calculate the final momentum of the photon:

3424

fi

6.63 10 J s 1.255 10 kg m/s0.525 nm 0.00333 nm

hpλ λ

−−× ⋅

= = = × ⋅+ Δ +

3. Calculate the scattering angle of the photon:

( )e

1 1e

1 cos

3.33 pm cos 1 cos 1 111.92.426 pm

hm c

m ch

λ θ

λθ − −

Δ = −

Δ ⎛ ⎞⎛ ⎞= − = − = °⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

4. Calculate the horizontal and vertical components of the final photon momentum:

( ) ( )

( ) ( )

24x, f

25

24y, f

24

cos 1.255 10 kg m/s cos 111.9

4.677 10 kg m/s

sin 1.255 10 kg m/s sin 111.9

1.164 10 kg m/s

f

f

p p

p p

θ

θ

= = × ⋅ °

= − × ⋅

= = × ⋅ °

= × ⋅

5. Calculate the horizontal component of the electron momentum: ( )

i x, f e,x

24 25e,x i x, f

24

1.263 10 kg m/s 4.677 10 kg m/s

1.73 10 kg m/s

p p p

p p p − −

= +

= − = × ⋅ − − × ⋅

= × ⋅

6. Calculate the vertical component of the electron momentum: ( )

y, f e, y

24 24e, y y, f

0

1.164 10 kg m/s 1.164 10 kg m/s

p p

p p − −

= +

= − = − × ⋅ = − × ⋅

7. Calculate the angle at which the electron recoils:

24e,y1 1

24e,x

1.164 10 kg m/stan tan 341.73 10 kg m/s

pp

φ−

− −−

⎛ ⎞ ⎛ ⎞− × ⋅= = = − °⎜ ⎟ ⎜ ⎟⎜ ⎟ × ⋅⎝ ⎠⎝ ⎠

Insight: The momentum of the recoiling electron is 2 2 24e, x e, y 2.09 10 kg m/sp p −+ = × ⋅ and its kinetic energy is 14.9 eV.

Page 20: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 20

59. Picture the Problem: Your car gains momentum as it accelerates away from a stoplight.

Strategy: Use the de Broglie wavelength equation (equation 30-16) to answer the conceptual question.

Solution: 1. (a) The de Broglie wavelength h pλ = is inversely proportional to the momentum. Therefore, as the momentum of your car increases, the de Broglie wavelength will decrease.

2. The best explanation is II. The momentum of the car increases. It follows that the de Broglie wavelength will decrease, because it is inversely proportional to the wavelength. Statement I erroneously assumes the de Broglie wavelength is proportional to the momentum, and statement III is false.

Insight: Why doesn’t the de Broglie wavelength of your car increase to infinity when you come to a stop? While this would be the case for a single particle at rest, you car is composed of many billions of particles that are all moving due to their thermal energy. The de Broglie wavelength of each atom is too small to measure unless the atom is nearly at rest (which requires the temperature to be nearly absolute zero).

60. Picture the Problem: The de Broglie wavelength of a particle changes when either its momentum or its kinetic energy

is doubled.

Strategy: Use the de Broglie wavelength equation (equation 30-16) to answer the conceptual question.

Solution: 1. (a) Because the de Broglie wavelength is ,h pλ = it follows that doubling the momentum halves the

wavelength. Therefore, the wavelength changes by a multiplicative factor of 1 2 .

2. Doubling the kinetic energy means that the momentum increases by a factor of 2 . As a result, the de Broglie

wavelength changes by a multiplicative factor of 1 2 .

Insight: In each case the de Broglie wavelength of the particle decreases as the speed increases. 61. Picture the Problem: The de Broglie wavelength of a particle is inversely proportional to its momentum.

Strategy: Replace the momentum in the de Broglie wavelength equation (equation 30-16) with equation 9-1. Solve the resulting expression for the particle’s speed.

Solution: 1. Combine equations 30-16 and 9-1: h h

p mvλ = =

2. Solve for the particle velocity: ( )( )

34

12 27

6.63 10 J s 13.7 km/s7.22 10 m 6.69 10 kg

hvmλ

− −

× ⋅= = =

× ×

Insight: It is appropriate to use the classical equation for the momentum because the particle speed is much less than the speed of light.

62. Picture the Problem: The de Broglie wavelength of a neutron is inversely proportional to its momentum.

Strategy: Replace the momentum in the de Broglie wavelength equation (equation 30-16) with equation 9-1. Solve the resulting equation for the neutron velocity.

Solution: 1. Combine equations 30-16 and 9-1: h h

p mvλ = =

2. Solve for the neutron velocity:

( )( )34

9 27

6.63 10 J s 1.40 km/s0.282 10 m 1.675 10 kg

hvmλ

− −

× ⋅= = =

× ×

Insight: When the wavelength of the neutron is equal to the interatomic spacing, the neutron beam will diffract and produce a maximum at an angle of 60° from the normal.

Page 21: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 21

63. Picture the Problem: If we consider a 79-kg jogger to be a particle we can determine her de Broglie wavelength.

Strategy: Use equation 30-16 to calculate the wavelength of the jogger, where the jogger’s momentum is given by the classical momentum equation (equation 9-1).

Solution: Calculate the de Broglie wavelength:

( )( )34

366.63 10 J s 2.0 10 m79 kg 4.2 m/s

h hp mv

λ−

−× ⋅= = = = ×

Insight: This wavelength is a billion trillion times smaller than the nucleus of a single atom, much too small to measure! Furthermore, the jogger is really a collection of particles (atoms), each of which is moving rapidly in thermal motion and has a de Broglie wavelength that is smaller than an atom.

64. Picture the Problem: The de Broglie wavelength of an electron is related to both the momentum and the kinetic energy of an electron.

Strategy: Write the kinetic energy (equation 7-6) of the electron in terms of the momentum (equation 9-1). Then use equation 30-16 to write the momentum in terms of the de Broglie wavelength.

Solution: 1. Write the kinetic energy in terms of the de Broglie wavelength:

2 2 2 22

2

12 2 2 2

m v p hK mvm m mλ

= = = =

2. Insert the wavelength:

( )( )( )

23417

231 10

6.63 10 J s1.1 10 J

2 9.11 10 kg 1.5 10 mK

−−

− −

× ⋅= = ×

× ×

Insight: In this problem we used the classical equations for the kinetic energy and the momentum. This assumption can be checked solving equation 30-16 for the speed. The resulting 6

e 4.9 10 m/s 0.016v c= × = is small enough that relativistic effects can be neglected.

65. Picture the Problem: A beam of neutrons diffracts off of a crystal due to the de Broglie wavelength of the neutrons.

Strategy: Set the momentum in equation 30-16 equal to the mass times velocity and solve for the velocity. To calculate the second-order diffraction angle, solve equation 30-17 for the angle θ, where m = 2.

Solution: 1. (a) Solve equation 30-16 for v: ( )( )

34

9 27

6.63 10 J s 1.58 km/s0.250 10 m 1.675 10 kg

h h hvp mv m

λλ

− −

× ⋅= = ⇒ = = =

× ×

2. (b) Calculate the second- order diffraction angle: ( )

( )1 1

2 sin

2 0.250 nmsin sin 62.4

2 2 0.282 nm

d m

md

θ λ

λθ − −

=

⎡ ⎤= = = °⎢ ⎥

⎢ ⎥⎣ ⎦

Insight: The neutron beam will exhibit both a first- and a second-order diffraction maximum. However, there can be no higher-order maxima, because for m = 3, 2 1.33m dλ = is outside the acceptable range of the inverse sine function.

66. Picture the Problem: Because the proton is significantly more massive than an electron, it will have a greater momentum when the proton and electron have the same speed. The de Broglie wavelength is inversely proportional to the particle momentum.

Strategy: Calculate the ratio of the de Broglie wavelengths using equation 30-16, where the momentum is the mass times velocity.

Solution: 1. (a) Because h mvλ = and e p ,m m< for identical speeds, an electron has a longer de Broglie wavelength than a proton.

2. (b) Calculate the ratio of wavelengths:

27e pe

31p p e

1.673 10 kg 18369.109 10 kg

mh m vh m v m

λλ

×= = = =

×

Insight: In this problem we assumed a classical speed so that we could use the classical momentum equation. However, because the relativistic factor depends only upon the speed, the ratio of the wavelengths would still be 1840 at relativistic speeds.

Page 22: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 22

67. Picture the Problem: An electron and proton have the same de Broglie wavelength, which means that they must also

have the same momentum. However, their kinetic energies will differ because they have different masses.

Strategy: Use equation 30-16 to write the kinetic energy in terms of the de Broglie wavelength. Then divide the kinetic energy of the proton by the kinetic energy of the electron to calculate their ratio.

Solution: 1. (a) The proton and electron have the same momentum because they have the same de Broglie wavelength. The kinetic energy, 2 2K p m= , is inversely proportional to the mass, and e p .m m< For identical momenta, an electron has a greater kinetic energy than a proton.

2. (b) Write the kinetic energy in terms of the wavelength: 2 2 22 2K p m h mλ= =

3. Calculate the ratio of the kinetic energies:

2 2 27pe e

2 2 31p ep

2 1.673 10 kg 18362 9.109 10 kg

mK h mK mh m

λλ

×= = = =

×

Insight: In this problem we used the classical equation for the kinetic energy. As the speeds approach the speed of light, the ratio of kinetic energies decreases due to relativistic effects.

68. Picture the Problem: When a person walks through a doorway, we typically do not see the person diffract. However, if

the person’s de Broglie wavelength were on the order of the door width, the diffraction would be observable.

Strategy: Solve equation 30-16 for the speed of a person whose de Broglie wavelength equals the width of the door. For part (b), divide one millimeter by the velocity to determine the time elapsed.

Solution: 1. (a) Solve eq. 30-16 for the person’s speed:

( )34

356.63 10 J s 1.3 10 m/s0.76 m 65 kg

h hp mvhvm

λ

λ

−−

= =

× ⋅= = = ×

2. (b) Calculate the time to travel 1.0 mm: 31

35

0.0010 m 7.5 10 s1.342 10 m/s

xtv −= = = ×

×

Insight: This time is about 142.0 10× times longer than the age of the universe. The doorway would not exist long enough to observe the diffraction! In reality, the person is really a collection of many particles (atoms), each of which ismoving rapidly in thermal motion and has a de Broglie wavelength that is smaller than an atom. It would be impossible to slow down every atom in the person’s body to a speed as small as that found in part (a).

69. Picture the Problem: A particle of mass m and charge q gains momentum when it is accelerated through a potential

difference V. The particle’s momentum is inversely proportional to its de Broglie wavelength.

Strategy: Write the kinetic energy of the particle ( 2 2K p m= ) in terms of the de Broglie wavelength using equation 30-16. Then set the kinetic energy equal to the change in electrostatic potential energy (U qV= ) and solve for the wavelength.

Solution: 1. Write the kinetic energy in terms of the de Broglie wavelength:

2 2

22 2p hKm mλ

= =

2. Set K U= and solve for the wavelength:

2

22

2

hK qVm

hmqV

λ

λ

= =

=

Insight: The de Broglie wavelength is inversely proportional to the square root of the potential difference. Quadrupling the potential difference will cut the wavelength in half.

Page 23: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 23

70. Picture the Problem: According to the Heisenberg uncertainty principle, the product of the uncertainty in position and

uncertainty in momentum must be greater than Planck’s constant divided by 2π.

Strategy: The 5.0% uncertainty in speed means that the magnitude of the momentum (mass times speed) of the electron and the baseball are also uncertain by 5.0%. The minimum uncertainty in position is given by equation 30-19.

Solution: 1. Solve the uncertainty principle for the uncertainty in position: 2 y

hypπ

Δ ≥Δ

2. Set 0.050 :yp mvΔ =

( )2 0.050 y

hymvπ

Δ ≥

3. Calculate the uncertainty for the baseball:

( )( )( )34

34baseball

6.63 10 J s 3.4 10 m2 0.050 0.15 kg 41 m/s

−−× ⋅

Δ ≥ = ×

4. Calculate the uncertainty for the electron:

( )( )( )

34

electron 31

6.63 10 J s 57 m2 0.050 9.11 10 kg 41 m/s

y μπ

× ⋅Δ ≥ =

×

Insight: The minimum uncertainty in the position of the baseball is smaller than an atom by a factor of 1024. However, the minimum uncertainty in the position of the electron is measurable, roughly the thickness of a human hair.

71. Picture the Problem: Because the proton is confined within the nucleus, the maximum uncertainty in its position is the

diameter of the nucleus. This maximum uncertainty in position corresponds to a minimum uncertainty in the momentum of the proton because of the Heisenberg uncertainty principle.

Strategy: Solve equation 30-19 for the uncertainty in the proton’s momentum.

Solution: Solve the uncertainty principle for the uncertainty in momentum: ( )

34

15

20

6.63 10 J s2 2 7.5 10 m

1.4 10 kg m/s

y

y

hpy

p

π π

× ⋅Δ ≥ =

Δ ×

Δ ≥ × ⋅

Insight: If we set this uncertainty equal to the proton’s minimum momentum, we conclude that the proton must have a kinetic energy of at least 0.37 MeV inside the nucleus.

72. Picture the Problem: When the position of a cart is measured with a specific accuracy, the Heisenberg uncertainty

principle requires that the momentum must also have a minimum uncertainty.

Strategy: Write the momentum in equation 30-19 in terms of the mass and uncertainty in velocity and solve the inequality for the uncertainty in velocity.

Solution: 1. Write equation 30-19 with y yp m vΔ = Δ :

2 2y yh hp y m v yπ π

Δ Δ ≥ ⇒ Δ Δ ≥

2. Solve for the uncertainty in the velocity: ( )( )

34

31

6.63 10 J s2 2 0.26 kg 0.0022 m

1.8 10 m/s

yhvm yπ π

× ⋅Δ ≥ =

Δ

= ×

Insight: At this speed it would take the cart over a billion years to move the diameter of a proton! The uncertainty principle is not significant for large objects.

Page 24: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 24

73. Picture the Problem: The amount of time required to measure the energy is equal to the maximum uncertainty in the

time measurement. According to the uncertainty principle, this corresponds to a minimum uncertainty in the energy.

Strategy: Divide the inequality in equation 30-20 by the measurement time maxtΔ to solve for the minimum uncertainty in the energy measurement.

Solution: Calculate the minimum uncertainty in energy:

( )34

26min 8

26.63 10 J s 1.1 10 J

2 2 1.0 10 s

hE t

hEt

π

π π

−−

Δ Δ ≥

× ⋅Δ = = = ×

Δ ×

Insight: If an electron does not remain in an excited level of an atom for at least 10−8 seconds, the energy cannot be measured (or known) to an accuracy greater than ±10−26 J.

74. Picture the Problem: When the energy is measured to within a given uncertainty, the time cannot be known to within a

greater uncertainty than is stipulated by the Heisenberg uncertainty principle.

Strategy: Solve equation 30-20 for the minimum uncertainty in the time.

Solution: Calculate the minimum uncertainty in time:

( )( )34

13min 19

max

26.63 10 J s 6.6 10 s

2 2 0.0010 eV 1.60 10 J/eV

hE t

htE

π

π π

−−

Δ Δ ≥

× ⋅Δ = = = ×

Δ ×

Insight: As the energy is measured with greater precision, the time at which the particle had that given energy is known with less certainty. It is impossible to simultaneously measure both energy and time with arbitrary certainty.

75. Picture the Problem: The lifetime of an excited state limits the time over which an energy measurement can be taken.

The Heisenberg uncertainty principle correlates this maximum uncertainty in the time measurement with a minimum uncertainty in the energy measurement.

Strategy: Solve equation 30-20 for the minimum uncertainty in the energy.

Solution: Calculate the minimum uncertainty in energy:

( )34

25min 9

max

26.63 10 J s 1.8 10 J

2 2 0.60 10 s

hE t

hEt

π

π π

−−

Δ Δ ≥

× ⋅Δ = = = ×

Δ ×

Insight: Because the minimum uncertainty in energy is inversely proportional to the mean lifetime of the energy level, energy levels with smaller lifetimes will necessarily have greater uncertainties in energy.

76. Picture the Problem: The lifetime of the particle limits the time over which an energy measurement can be taken. The

Heisenberg uncertainty principle correlates this maximum uncertainty in the time measurement with a minimum uncertainty in the energy measurement.

Strategy: Solve equation 30-20 for the minimum uncertainty in the particle’s energy.

Solution: Calculate the minimum uncertainty in energy:

( )34

25min 10

max

26.63 10 J s 4.2 10 J

2 2 2.5 10 s

hE t

hEt

π

π π

−−

Δ Δ ≥

× ⋅Δ = = = ×

Δ ×

Insight: Because the minimum uncertainty in energy is inversely proportional to the mean lifetime of the particle, particles with smaller lifetimes will necessarily have greater uncertainties in energy.

Page 25: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 25

77. Picture the Problem: When an electron’s location is known to within an uncertainty of 0.15 nm, it will have a

minimum uncertainty in momentum as required by the uncertainty principle. That minimum uncertainty requires the electron to have a nonzero kinetic energy.

Strategy: Use equation 30-19 to calculate the minimum uncertainty Δp in the momentum, and then use p = Δp to find the kinetic energy of the electron according to the kinetic energy equation 2( 2 ).K p m=

Solution: 1. (a) Solve equation 30-19 for the minimum uncertainty in momentum:

( )34

25min 9

26.63 10 J s 7.0 10 kg m/s

2 2 0.15 10 m

hp x

hpx

π

π π

−−

Δ Δ ≥

× ⋅Δ = = = × ⋅

Δ ×

2. (b) Calculate the minimum kinetic energy:

( )( )( )

2252

31 19

7.0 10 kg m/s1.7 eV

2 2 9.11 10 kg 1.60 10 J/eVpKm

− −

× ⋅= = =

× ×

Insight: Decreasing the uncertainty in the position will proportionately increase the uncertainty in the momentum and increase the kinetic energy by the square of the momentum. For example, if the uncertainty were decreased by a factor of three (to 0.05 nm), the momentum would triple and the kinetic energy would increase by a factor of nine.

78. Picture the Problem: When a proton’s location is known to within an uncertainty of 0.15 nm, it will have a minimum

uncertainty in momentum as required by the uncertainty principle. That minimum uncertainty requires the proton to have a nonzero kinetic energy.

Strategy: Use equation 30-19 to calculate the minimum uncertainty Δp in the momentum, and then use p = Δp to find the kinetic energy of the proton according to the kinetic energy equation 2( 2 ).K p m=

Solution: 1. (a) Solve equation 30-19 for the minimum uncertainty in momentum:

( )34

25min 9

26.63 10 J s 7.0 10 kg m/s

2 2 0.15 10 m

hp x

hpx

π

π π

−−

Δ Δ ≥

× ⋅Δ = = = × ⋅

Δ ×

2. (b) Calculate the minimum kinetic energy:

( )( )( )

2252

27 19

7.0 10 kg m/s0.92 meV

2 2 1.673 10 kg 1.60 10 J/eVpKm

− −

× ⋅= = =

× ×

Insight: When we compare the results of this problem with problem 77 (which concerned an electron confined with the same Δx), we see that the uncertainty in momentum for the electron and proton are the same. The minimum kinetic energy of the electron, however, is 1840 times greater than the minimum kinetic energy of the proton.

79. Picture the Problem: As the position of an electron becomes more certain (it has less uncertainty), the uncertainty of

the momentum increases.

Strategy: Calculate the maximum uncertainty in the momentum by multiplying the momentum by 1.0%. Then use equation 30-19 to calculate the minimum uncertainty in the position.

Solution: 1. Calculate the maximum Δp: ( )25 270.010 0.010 1.7 10 kg m/s 1.7 10 kg m/sp p − −Δ = = × ⋅ = × ⋅

2. Solve equation 30-19 for Δx:

( )34

min 27max

26.63 10 J s 62 nm

2 2 1.7 10 kg m/s

hp x

hxp

π

π π

Δ Δ ≥

× ⋅Δ = = =

Δ × ⋅

Insight: The minimum uncertainty in the position could be decreased to 12 nm and still keep the relative uncertainty of the momentum less than 5%.

Page 26: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 26

80. Picture the Problem: You perform an experiment on the photoelectric effect using light with a frequency high enough

to eject electrons. The intensity of the light is increased while the frequency is held constant.

Strategy: Recall the principles that govern the photoelectric effect when answering the conceptual questions. Note that increasing the intensity while holding the frequency constant simply means that more photons (each with the same original energy) strike the metal surface per second.

Solution: 1. (a) Because the energies of the photons stay the same when the frequency is held constant, the maximum kinetic energy of an ejected electron will stay the same.

2. (b) Because the energies of the photons stay the same when the frequency is held constant, the maximum kinetic energy of an ejected electron and therefore its minimum de Broglie wavelength will stay the same.

3. (c) Because the number of photons will increase when the intensity of the light is increased, the number of electrons ejected per second will increase.

4. (d) Because the number of photons will increase when the intensity of the light is increased, the number of electrons ejected per second and therefore the current in the phototube will increase.

Insight: The maximum kinetic energy of an ejected electron will increase only if the frequency of the light is increased. 81. Picture the Problem: You perform an experiment on the photoelectric effect using light with a frequency high enough

to eject electrons. The frequency of the light is increased while the intensity is held constant.

Strategy: Recall the principles that govern the photoelectric effect when answering the conceptual questions. Note that increasing the frequency of the light means that the energy carried by each photon will increase.

Solution: 1. (a) Because the energies of the photons increase when the frequency is increased, the maximum kinetic energy of an ejected electron will increase.

2. (b) Because the energies of the photons increase when the frequency is increased, the maximum kinetic energy of an ejected electron will increase. Because the de Broglie wavelength of an electron is inversely proportional to its momentum (equation 30-16), we conclude that the minimum de Broglie wavelength of ejected electrons will decrease.

3. (c) Because the intensity (energy per area per time) is unchanged, there must be fewer of these higher-energy photons reaching the surface each second. As a result, the number of electrons ejected per second will decrease.

4. (d) Fewer electrons ejected per second means that the current in the phototube will decrease.

Insight: If the number of photons per second were held constant, the intensity of the light would increase if the frequency of the light were increased.

82. Picture the Problem: An electron that is accelerated from rest through a potential difference 0V has a de Broglie

wavelength 0λ .

Strategy: Use the expressions h pλ = (equation 30-16), 2 2K p m= (Conceptual Checkpoint 9-3), and K U qV= Δ = (equation 20-2) to answer the conceptual question.

Solution: To double the electron’s wavelength we must halve its momentum. This means, in turn, that we must decrease its kinetic energy by a factor of four (recall that K = p2/2m). The kinetic energy is linearly proportional to the potential difference. Therefore, the electron should be accelerated from rest through the potential difference V0/4.

Insight: We can also come to this conclusion using a ratio: 2 2 222

0 0 02

0 0 0 0 0 00

2 1 2 4 42

VV K q K p m p h VV K q K p hp m

λ λλλ λ λ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞= = = = = = = = ⇒ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Page 27: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 27

83. Picture the Problem: A beam of particles diffracts from a crystal, producing an interference maximum at the angle θ.

Strategy: Use the expressions h pλ = (equation 30-16) and 2 sind mθ λ= (equation 30-17) to answer the conceptual question.

Solution: 1. (a) Increasing the mass of the particles implies that their momentum increases as well. It follows from h pλ = that their de Broglie wavelength decreases. Therefore, from 2 sind mθ λ= we see that the angle θ of the

interference maximum will decrease. 2. (b) Increasing the energy of the particles also implies that their momentum increases. Therefore, we again expect that

the angle of the interference maximum will decrease. Insight: We can also come to this conclusion using a ratio:

new new new new new old

old old old old old new

sin 2sin 2

m d h p pm d h p p

θ θ λ λθ θ λ λ

= = = = = so that if new old ,p p> then new old .θ θ<

84. Picture the Problem: In order to construct a photocell that works on visible light, the energy from a photon in the

visible spectrum must be sufficient to eject an electron from the metal.

Strategy: Use equations 30-6 and 14-1 to calculate the maximum wavelength that a photon can have so that it is able to eject electrons from each metal. Compare the results with the shortest wavelength of visible light (400 nm).

Solution: 1. Determine the maxλ required to eject electrons from each metal surface:

( )( )( )

34 8

max 190 0 00

6.63 10 J s 3.00 10 m/s 1240 eV nm1.60 10 J/eV

c hcf W WW

λ−

× ⋅ × ⋅= = = =

×

2. Calculate the maxλ for aluminum: max

1240 eV nm 290 nm4.28 eV

λ ⋅= =

3. Calculate the maxλ for lead: max

1240 eV nm 292 nm4.25 eV

λ ⋅= =

4. Calculate the maxλ for cesium: max

1240 eV nm 579 nm2.14 eV

λ ⋅= =

5. Of the three materials, cesium is the only material that will eject electrons from visible light.

Insight: Red light will not eject electrons from the surface because the maximum wavelength for cesium is 579 nm, which is in the yellow portion of the spectrum. The photocell will work for visible light of wavelengths between 579 nm (yellow) and 400 nm (violet).

85. Picture the Problem: The eye must absorb at least 100 photons per second in order to detect green light.

Strategy: Use equations 30-4 and 14-1 to calculate the energy per photon for green light with a wavelength of 545 nm. Multiply this by the rate of 100 photons per second in order to determine the power absorbed by the eye.

Solution: 1. Calculate the photon energy:

834 19

9

3.00 10 m/s6.63 10 J s 3.65 10 J545 10 m

cE hf hλ

− −−

×= = = × ⋅ = ×

×

2. Multiply the energy of the photon by the rate of photon absorption: ( ) ( )( )19 17100 photons/s 3.65 10 J 3.65 10 WP N t E − −= = × = ×

Insight: If the eye were to absorb the same power from blue light of wavelength 484 nm, the eye would only need to

receive 89 photons per second from the source. However, the eye is more sensitive to green light than to blue, and so this amount of power of blue light would not be visible to the eye.

Page 28: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 28

86. Picture the Problem: The length of a pendulum determines the frequency at which it oscillates. The energy of a typical

pendulum oscillation can be considered as a multiple of the quantum oscillation.

Strategy: Calculate the frequency of the oscillation using equations 13-1 and 13-20. Multiply the frequency by Planck’s constant and the quantum number n to calculate the energy of the oscillation. Set the energy equal to the kinetic energy and solve for the velocity.

Solution: 1. (a) Calculate the oscillation frequency:

21 1 1 9.81 m/s 0.56 Hz2 2 0.78 m

gfT Lπ π

= = = =

2. (b) Set the oscillation energy equal to the kinetic energy and solve for the speed: ( )( )( )

212

33 342 1.0 10 6.63 10 J s 0.56 Hz2 2.2 m/s0.15 kg

E nhf mv

nh fvm

= =

× × ⋅= = =

Insight: The n = 1 quantum state corresponds to the minimum energy state for this pendulum oscillator. The maximum speed of the pendulum bob in this state would be 7.1×10−17 m/s.

87. Picture the Problem: A radio antenna receives power by absorbing many photons per second, where each photon is a

quantum of energy determined by the frequency of the station. The received photons exert a force on the antenna when they are absorbed.

Strategy: Divide the power by the energy per photon (from equation 30-4) to calculate the number of photons absorbed per second. Set the power equal to the force times the speed of the wave (equation 7-13) to calculate the force on the antenna.

Solution: 1. (a) Calculate the number of photons absorbed per second: ( )( )

1015

34 6photon

1.0 10 W 1.6 10 photons/s6.63 10 J s 96 10 Hz

P PE hf

×= = = ×

× ⋅ ×

2. (b) Divide the power by the speed of the wave to calculate the force on the antenna: 10

198

1.0 10 W 3.3 10 N3.00 10 m/s

P Fv FcPFc

−−

= =

×= = = ×

×

Insight: The force on the antenna is negligibly small when compared with other forces such as wind and gravity. 88. Picture the Problem: As ice absorbs photons of light, the energy is converted into heat and melts the ice.

Strategy: Use equation 30-4 to calculate the energy of one photon. Divide the energy needed to melt 1.0-kg of ice ( Q mL= ) by the energy of one photon to calculate the number of photons needed. To calculate the number of ice molecules that one photon will melt, divide the number of ice molecules in one kilogram by the number of photons to melt the kilogram of ice. The number of water molecules is equal to Avogadro’s number times 1 kilogram divided by the atomic mass of water (18 g/mole).

Solution: 1. (a) Calculate the energy in one photon: ( )( )34 14 19

photon 6.63 10 J s 6.0 10 Hz 3.978 10 JE h f − −= = × ⋅ × = ×

2. Divide the heat by the photon energy: ( )( )( )f

19photon

23

1.0 kg 80.0 kcal/kg 4186 J/kcal3.978 10 J

8.4 10 photons

mLN

E −= =×

= ×

3. (b) Divide the number of molecules by the number of photons:

( ) ( )( ) ( )

23A A

23

6.022 10 molecules/mole 1.0 kg

8.42 10 photons 0.018 kg/mole

40 molecules/photon

N m M N mN N M

×= =

×

=

Insight: When a photon is absorbed by a water molecule, the intermolecular bonds that attach the molecule to the other molecules in the lattice can be broken. The molecule will then have sufficient remaining kinetic energy that it can collide with 39 other molecules and break their bonds as well.

Page 29: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 29

89. Picture the Problem: As water absorbs photons of light, the energy is converted into heat and raises the temperature of

the water.

Strategy: Use equations 30-4 and 14-1 to calculate the energy of one photon. Find the energy required to heat 1.0 gram of water by one Celsius degree (equation 16-13), and divide this energy by the energy of one photon in order to calculate the number of photons needed to heat the water.

Solution: 1. Find the energy in one photon:

( )( )34 819

photon 9

6.63 10 J s 3.0 10 m/s3.6 10 J

550 10 mhcE h fλ

−−

× ⋅ ×= = = = ×

×

2. Divide the heat by the photon energy:

( )( )( ) 19w19

photon

1.0 g 4.186 J/g C 1 C1.2 10 photons

3.6 10 Jmc T

NE −

⋅ ° °Δ= = = ×

×

Insight: Each photon has enough energy to increase the temperature of 3000 molecules by 1 C°. Note the difference in the two c variables; one represents the speed of light in vacuum and the other represents specific heat.

90. Picture the Problem: As water absorbs microwave photons, the energy is converted into heat and raises the

temperature of the water.

Strategy: Use equations 30-4 and 14-1 to calculate the energy of each photon. Use equation 16-13 to find the energy required to heat the water, and divide this energy by the energy of one photon in order to calculate the number of photons needed to heat the water.

Solution: 1. Calculate the photon energy:

( )( )34 824

photon

6.63 10 J s 3.0 10 m/s1.630 10 J

0.122 mhcE hfλ

−−

× ⋅ ×= = = = ×

2. Calculate the heat added to the water: ( )( )( )4

0.205 kg 4186 J/kg C 90.0 C 20.0 C

6.007 10 J

Q mC T= Δ = ⋅ ° ° − °

= ×

3. Divide the heat by the energy per photon:

428

24photon

6.007 10 J 3.68 10 photons1.630 10 J

QNE −

×= = = ×

×

Insight: If a magnifying glass that is 3.00 inches in diameter captures the entire solar spectrum on a day in which the solar intensity is 1000 W/m2, it would take 3.66 hours to change the temperature of the same mass of water by the same amount.

91. Picture the Problem: When light ejects electrons from a lead surface, part of the photon energy is used to overcome the

work function. The remainder becomes kinetic energy of the photoelectron. The kinetic energy is related to the momentum of the electron, and thus is related to the de Broglie wavelength of the electron.

Strategy: Write the kinetic energy of the electron (from equation 30-7) in terms of the momentum 2 2K p m= and solve for the momentum. Then use equation 30-16 to calculate the wavelength from the momentum.

Solution: 1. Use equation 30-7 to calculate the kinetic energy:

( )( ) ( )34 15 19max 0

19

6.63 10 J s 2.11 10 Hz 4.25 eV 1.6 10 J/eV

7.2 10 J

K hf W − −

= − = × ⋅ × − ×

= ×

2. Calculate the electron momentum: ( )( )

2max

31 19 24max

2

2 2 9.11 10 kg 7.2 10 J 1.15 10 kg m/s

K p m

p mK − − −

=

= = × × = × ⋅

3. Calculate the electron wavelength:

( ) ( )34 246.63 10 J s 1.15 10 kg m/s 0.58 nmh pλ − −= = × ⋅ × ⋅ =

Insight: Increasing the photon frequency will decrease the electron wavelength. For example, if the photon frequency were increased to 154.22 10 Hz,× the wavelength drops to 0.34 nm. Note that doubling the photon frequency does not cut the electron’s de Broglie wavelength in half because of the effect of the work function.

Page 30: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 30

92. Picture the Problem: The de Broglie wavelength of the electron and the wavelength of the photon are both inversely

proportional to their momenta.

Strategy: Use equation 30-16 to calculate the de Broglie wavelength of the electron, where the momentum is given by equation 9-1. Calculate the wavelength of the photon using equation 30-11, setting the momentum of the photon equal to the momentum of the electron.

Solution: 1. (a) Find the de Broglie wavelength:

( )( )34

31 6e e

6.63 10 J s 0.27 nm9.11 10 kg 2.7 10 m/s

h hp m v

λ−

× ⋅= = = =

× ×

2. (b) Calculate the wavelength of the photon: e

e

0.27 nmh hp p

λ λ= = = =

Insight: Any object (electron, photon, proton, person, etc.) with the same momentum will have the same wavelength. 93. Picture the Problem: The spectrum of light emitted by a firefly peaks at a frequency of 145.4 10 Hz,× which

corresponds to visible, yellow-green light.

Strategy: Use equation 30-1 to calculate the temperature of a blackbody that has a peak frequency at 145.4 10 Hz.× Compare this temperature with the probable temperature of a firefly’s abdomen.

Solution: 1. Solve equation 30-1 for T:

14peak10 1 1 10 1 1

5.4 10 Hz 9200 K5.88 10 s K 5.88 10 s K

fT − − − −

×= = =

× ⋅ × ⋅

2. Firefly radiation is definitely not well approximated by blackbody radiation. A temperature of 9200 K is hotter than the surface of the Sun, and is sufficient to completely destroy the firefly.

Insight: The light from the firefly is caused by a low-temperature chemical reaction that produces a specific wavelength of light (characteristic of the species of firefly).

94. Picture the Problem: Light of a given frequency and energy will eject electrons from the surface of the metal. Part of

the energy from the photon overcomes the work function of the metal. The remainder of the energy goes into the kinetic energy of the electron.

Strategy: (a) It is not necessary to use relativistic mechanics because v << c. Substitute 21max2 mv for maxK in the

equation max 0K hc Wλ= − and solve for 0.W Then calculate the cutoff frequency using 0 0 .f W h=

Solution: 1. (b) Solve eq.30-7 for the work function:

( )( ) ( )( )

( )( )

max 0

20 max max

34 8231 5

9

19 190

12

6.63 10 J s 3.00 10 m/s 1 9.11 10 kg 3.10 10 m/s2545 10 m

3.212 10 J 1 eV 1.60 10 J 2.01 eV

hcK W

hc hcW K mv

W

λ

λ λ−

−−

− −

= −

= − = −

⎡ ⎤× ⋅ ×⎢ ⎥= − × ×

×⎢ ⎥⎣ ⎦

= × × =

2. (b) Use equation 30-6 to calculate the cutoff frequency:

19140

0 34

3.212 10 J 4.84 10 Hz6.63 10 J s

Wf

h

×= = = ×

× ⋅

Insight: This cutoff frequency falls in the orange part of the visible spectrum. Therefore, essentially the entire visible spectrum (except red light) will eject electrons from the metal.

Page 31: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 31

95. Picture the Problem: The image shows an atom absorbing a photon of

wavelength 486.2 nm and later emitting a photon of wavelength 97.23 nm.

Strategy: Use equations 30-4 and 14-1 to calculate the energy of the absorbed photon and the energy of the emitted photon. Set the change in energy of the atom equal to the absorbed energy minus the emitted energy.

Solution: 1. (a) Photon energy is inversely proportional to wavelength, so the absorbed photon has less energy than the emitted photon’s energy. Therefore, the net energy of the atom has decreased.

2. (b) Calculate the energy of the absorbed photon:

( )( )( ) ( )

34 9abs

19 19

6.63 10 J s 3.00 10m/s 486.2 10 m

4.091 10 J 1.60 10 J/eV 2.56 eV

E hf hc λ − −

− −

= = = × ⋅ × ×

= × × =

3. Calculate the energy of the emitted photon:

( )( )( ) ( )

34 8 9emit

18 19

6.63 10 J s 3.00 10 m/s 97.23 10 m

2.046 10 J 1.60 10 J/eV 12.8 eV

E hc λ − −

− −

= = × ⋅ × ×

= × × =

4. Calculate ΔE for the atom: abs emit 2.56 eV 12.8 eV 10.2 eVE E EΔ = − = − = −

Insight: It is more common for an atom to absorb a photon of short wavelength and then emit two (or more) photons of longer wavelength, such that the energy of the absorbed photon is equal to the sum of the energies of the emitted photons.

96. Picture the Problem: As the atoms emerge from an oven, their most probable speed is proportional to their most

probable momentum, which is inversely proportional to their most probable wavelength.

Strategy: Write the most probable wavelength of the atoms in terms of the momentum using equation 30-16 and equation 9-1. Solve the resulting equation for the most probable speed.

Solution: 1. Write equation 30-16 in terms of the velocity: mp

mp mp 5h h h

p mv mkTλ = = =

2. Solve for the most probable speed: ( ) ( )23

mp 27

5 1.38 10 J/K 450 K5 4.31 km/s1.677 10 kg

kTvm

×= = =

×

Insight: The most probable speed increases in proportion to the square root of the Kelvin temperature. 97. Picture the Problem: The de Broglie wavelength is inversely proportional to the momentum of the electron. In

classical mechanics, the kinetic energy of the electron is proportional to the square of the momentum.

Strategy: Write the kinetic energy of the electron in terms of the momentum, and then replace the momentum with h λ according to equation 30-16. Solve the resulting equation for the wavelength, writing the kinetic energy K in units of eV.

Solution: 1. (a) The de Broglie wavelength is inversely proportional to momentum, and the momentum increases when the kinetic energy increases. Therefore, the de Broglie wavelength will decrease as the kinetic energy increases.

2. (b) Write K in terms of the de Broglie wavelength:

( ) ( )2 22 2

2 2 22 2 2 2h hcp hK

m m m mcλ

λ λ= = = =

3. Solve for the wavelength:

( )( )( ) ( )

34 5

2 5 19

6.63 10 J s 3.00 10 m/s 1.23 nm eV

2 2 5.11 10 eV 1.60 10 J/eV

hcKmc K K

λ−

⎡ ⎤× ⋅ ×⎛ ⎞ ⋅⎢ ⎥= = =⎜ ⎟⎜ ⎟ ⎢ ⎥× ×⎝ ⎠ ⎢ ⎥⎣ ⎦

Insight: This equation indicates that the wavelength is inversely proportional to the square root of the kinetic energy. For example, an electron with a kinetic energy of 16.3 eV will have a de Broglie wavelength of 0.305 nm.

Page 32: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 32

98. Picture the Problem: Helium atoms in a jar at room temperature and atmospheric pressure are often considered to

behave like point particles. Quantum effects between the particles are generally ignored. This problem tests the assumption that quantum effects are small for the interaction between helium atoms.

Strategy: Use equation 17-12 to calculate the kinetic energy of the helium atoms. Write the kinetic energy in terms of the momentum. Then using equation 30-16 write the momentum in terms of the wavelength. Solve the resulting equation for the average de Broglie wavelength of the atoms. Solve the ideal gas law (equation 17-2) for the volume per atom, and then take the cube root of the volume per atom to calculate the average separation distance.

Solution: 1. (a) Write the kinetic energy in terms of λ and set it equal to 3

2 kT : ( )22 2 set

av 2

32 2 22

hp hK kTm m m

λλ

= = = =

2. Solve for the wavelength in terms of the temperature: ( )( )( )

34

av 27 23

6.63 10 J s3 3 6.65 10 kg 1.38 10 J/K 295.15 K

0.0732 nm

hmkT

λ−

− −

× ⋅= =

× ×

=

3. (b) Use the ideal gas law to calculate the volume occupied per atom: ( )( )23

26 35

1.38 10 J/K 295.15 K4.033 10 m

1.01 10 Pa

PV NkT

V kTN P

−−

=

×= = = ×

×

4. Take the cube root of the volume to calculate the average separation:

3 26 33avg 4.033 10 m 3.44 nmVd

N−= = × =

Insight: The de Broglie wavelength of the helium atom is 47 times smaller than the separation distance between atoms. Therefore, quantum effects are not a significant factor in determining the behavior of the atoms in this helium gas.

99. Picture the Problem: The coefficient of the Compton scattering equation is called the Compton wavelength and is inversely proportional to the mass of the particle.

Strategy: Use the equation C h mcλ = to calculate the Compton wavelength of a proton. Then use equations 30-4 and 14-1 to write the energy of a photon with the same wavelength. Insert the equation for the Compton wavelength into the energy equation of the photon to show that the photon energy is equal to the rest energy of the particle.

Solution: 1. (a) Calculate the Compton wavelength of a proton: ( )( )

34

C 27 8p

6.63 10 J s 1.32 fm1.673 10 kg 3.00 10 m/s

hm c

λ−

× ⋅= = =

× ×

2. (b) Calculate the energy of a photon with the same wavelength:

( )( )34 810

15

6.63 10 J s 3.00 10 m/s1.51 10 J

1.32 10 mhcEλ

−−

× ⋅ ×= = = ×

×

3. (c) Insert the Compton wavelength into the equation for the photon energy:

20

C

hc hcE mc Eh mcλ

= = = =

Insight: When a photon Compton scatters after colliding with a free particle, its wavelength will decrease by up to twice the Compton wavelength of the scattered particle. Note that the Compton wavelength of an electron is 2426 fm or 2.426 pm, a value that is often involved in the solutions to the problems from Section 30-4.

100. Picture the Problem: When a photon is absorbed by the surface of a metal it will eject an electron. The kinetic energy of the ejected electron will equal the difference between the photon energy and the work function of the metal. When photons of the same frequency are incident on two metals of different work functions, the ejected electrons will have different maximum kinetic energies.

Strategy: Solve equation 30-7 for the work function. Then subtract the two work functions to show that the difference in kinetic energies of the ejected electrons is equal to the difference in work functions for any incident frequency.

Solution: 1. (a) The difference in maximum kinetic energy observed from the two surfaces is due to the difference in the work functions of the two materials. Because the maximum kinetic energy depends linearly upon the work function, and because the work function does not depend upon the frequency of the light, the difference in maximum kinetic energy will stay the same.

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Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 33

2. (b) Solve equation 30-8 for the work function:

max 0

0 max

K hf WW hf K

= −

= −

3. Subtract the work functions of the two metals:

( ) ( )0B 0A max,B max,A

19

max,A max,B 19

2.00 10 J 1.25 eV1.60 10 J/eV

W W hf K hf K

K K−

− = − − −

×= − = =

×

Insight: As long as both work functions are below 3.41 eV, the light will eject electrons from both surfaces whose kinetic energies differ by1.25 eV. If one of the work functions were greater than 3.41 eV, that surface would not eject photoelectrons.

101. Picture the Problem: When photons of different wavelengths shine on a lithium surface, they eject electrons with

kinetic energies equal to the difference between the energy of the photon and the work function of lithium.

Strategy: Solve equation 30-7 (writing the frequency in terms of the wavelength) for the constants hc. Set the resulting equations for the two experiments equal to each other and solve for the work function.

Solution: 1. Solve equation 30-7 for hc:

( )max 0

0 max

K hc Whc W K

λλ= −

= +

2. Set the two expressions equal and solve for the work function:

( ) ( )

( )( ) ( )( )

1 0 max, 1 2 0 max, 2

2 max, 2 1 max, 10

1 2

0

253.5 nm 2.57 eV 433.9 nm 0.550 eV433.9 nm 253.5 nm

2.29 eV

W K W K

K KW

W

λ λ

λ λλ λ

+ = +

−=

−=

−=

Insight: The above procedure essentially completes a linear regression between the two ( f, Kmax) data points and finds the x-intercept, which corresponds to the minimum frequency (maximum wavelength) at which photoelectrons are produced. The energy of a photon with that frequency equals the work function of the lithium.

102. Picture the Problem: When photons of different wavelengths shine on a lithium surface, they eject electrons with

kinetic energies equal to the difference between the energy of the photon and the work function of lithium.

Strategy: Insert the data from one of the experiments (and the calculated work function from the previous question) into the equation for hc and solve for Planck’s constant.

Solution: Solve for Planck’s constant: ( ) ( )

( )( )

9

0 max 8

15 19 34

433.9 10 m 2.289 eV 0.550 eV3.00 10 m/s

4.106 10 eV s 1.60 10 J/eV 6.57 10 J s

h W Kcλ −

− − −

×= + = +

×

= × ⋅ × = × ⋅

Insight: Millikan was able to calculate Planck’s constant to within 1% of its presently accepted value using these two experiments.

103. Picture the Problem: When photons of different wavelengths shine on a lithium surface, they eject electrons with

kinetic energies equal to the difference between the energy of the photon and the work function of lithium.

Strategy: Use equation 30-7 together with the work function and Planck’s constant (see the previous two questions) to calculate the maximum kinetic energy of electrons that are ejected when 365-nm light shines on the lithium surface.

Solution: Calculate the maximum K:

( )( )15 8

max 9

4.106 10 eV s 3.00 10 m/s2.29 eV 1.08 eV

365.0 10 mK

× ⋅ ×= − =

×

Insight: Millikan was able to calculate Planck’s constant to within 1% of its presently accepted value using these two experiments.

Page 34: Walker4 Ism Ch30

Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 – 34

104. Picture the Problem: The image shows a 0.6500-nm photon

scattering off of an electron. After the collision, the electron has a kinetic energy of 7.750 eV and the photon scatters at an angle θ.

Strategy: Use equations 20-4 and 14-1 to calculate the initial energy of the photon. Subtract the kinetic energy of the electron from the photon’s energy to calculate the final energy of the photon, and then use equation 20-4 to find the final wavelength. Then use equation 30-15 to calculate the scattering angle. A recurring value in similar Compton scattering problems is

12e 2.426 10 m 0.002426 nm.h m c −= × =

Solution: 1. (a) In this problem the photon has transferred less energy to the electron than it did in Example 30-4, so the scattering angle will be less than 152°.

2. (b) Calculate the initial energy of the photon:

( )( )( )

34 8

19

6.63 10 J s 3.0 10 m/s1912.5 eV

0.6500 nm 1.6 10 J/eVE hc λ

× ⋅ ×= = =

×

3. Calculate the final energy of the photon: 1912.5 eV 7.750 eV 1904.75 eVE E K′ = − = − =

4. Calculate the final wavelength of the photon:

( )( )( )

34 8

19

6.63 10 J s 3.0 10 m/s0.6526 nm

1904.75 eV 1.6 10 J/eVhc Eλ

× ⋅ ×′ ′= = =

×

5. Solve equation 30-15 for the scattering angle:

( )

1

e

1

cos 1

0.6526 nm 0.6500 nmcos 1 94

0.002426 nm

h m cλ λθ −

⎛ ⎞′ −= −⎜ ⎟

⎝ ⎠−⎛ ⎞

= − = °⎜ ⎟⎝ ⎠

Insight: The scattering angle of the photon increases as the momentum of the electron increases. Therefore, if the kinetic energy of the electron decreases, the momentum of the electron also decreases, and the scattering angle of the photon also decreases.

105. Picture the Problem: The image shows a 0.6500-nm photon

scattering off of an electron. After the collision, the photon has a wavelength of 0.6510 nm and scatters at an angle θ.

Strategy: Use equation 30-15 to calculate the scattering angle of the photon. A recurring value in similar Compton scattering problems is 12

e 2.426 10 m 0.002426 nm.h m c −= × =

Solution: 1. (a) The photon will scatter at a smaller angle because the change in wavelength in this problem is less than the change in wavelength of Example 30-4, which means that less energy has been transferred to the electron. Therefore, the scattering angle will be less than 152°.

2. (b) Solve equation 30-15 for the scattering angle: 1

e

1

cos 1

0.6510 nm 0.6500 nmcos 1 540.002426 nm

h m cλ λθ −

⎛ ⎞′ −= −⎜ ⎟

⎝ ⎠−⎛ ⎞= − = °⎜ ⎟

⎝ ⎠

Insight: Scattering angles greater than 152° will occur when the final wavelength of the scattered photon is between 0.6546 nm and 0.6549 nm.