Chapter27 Solution

1000

Chapter 27

EARLY QUANTUM PHYSICS AND THE PHOTON

Conceptual Questions

1. The temperature of the filament drops as the bulb is dimmed. The frequency of maximum radiation output from a blackbody decreases as its temperature drops. Thus, as the temperature of the bulb decreases, less high-frequency light is radiated and the emitted radiation becomes increasingly red.

2. The photoelectric effect describes the mechanism by which electrons are ejected from materials when certain frequencies of electromagnetic radiation are incident upon them. The first puzzling feature of this effect was that brighter light increases the current of electrons but does not affect their kinetic energy. This is understood with the photon model of light because increasing the intensity of the radiation simply increases the number of photons colliding with the material but does not increase their energy. The second puzzling feature was that the maximum kinetic energy of the electrons depends upon the wavelength of the photons. This too is explained by the photon theory because a shorter wavelength photon has a greater energy, and therefore, will provide a larger “kick” to the photoelectron. The third puzzling feature was that for a given metal there is a threshold frequency below which no electrons will be emitted. This is an understandable phenomenon because some minimum amount of energy is required to set the electrons free from the material to which they are bound, and if the photon energy is too low, an electron will not be ejected after absorption of a photon. The final puzzling feature of the photoelectric effect was that electrons are emitted virtually instantaneously after turning on a photon source. This too makes sense as a result of the photon theory because each individual photon carries enough energy to cause a photoelectron to be emitted—only a single photon is required to start a photocurrent.

3. In the photon model of light, the energy of a photon is determined solely by its frequency. UV photons are thus more energetic than visible photons, and therefore, have the potential to inflict more damage to skin cells. Furthermore, the energy of UV photons is sufficient to ionize atoms or molecules in the skin, causing significant damage to individual skin cells. Even high intensity visible light would not be as much of a threat, because the energy of the visible photons is insufficient to cause ionization.

4. The maximum possible Compton shift is 0.00486 nm. Visible light photons have wavelengths in the range 400-700 nm. The Compton shift is thus at most 0.001% of the incident wavelength—the scattered photons are not noticeably shifted in wavelength.

5. A blackbody at a higher temperature emits radiation at higher frequencies than does a blackbody at a lower temperature. Reddish stars are therefore cooler than bluish stars, with yellow and white stars falling between.

6. According to the Bohr model, the difference in energy between two levels of the hydrogen atom is proportional to the value of the electron charge. If the electron charge varied from particle to particle, the photons emitted from a hydrogen atom would not all have the same energy. Furthermore, when the emission spectrum of hydrogen was observed, photons from the same transition would have slightly different wavelengths and would therefore not form sharp lines.

7. As the frequency of light incident upon the surface increases, the maximum kinetic energy of the photoelectrons also increases. Specifically, the maximum kinetic energy of emitted electrons is equal to the difference of the energy of the incident photon (as determined by its frequency) and the work function of the material.

8. When the energy corresponding to the frequency of the incident photons is below the threshold level, no photoelectrons are emitted. This occurs because the amount of energy needed to overcome the attraction between the material and the electrons has not been reached.

College Physics Chapter 27: Early Quantum Physics and the Photon

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9. A continuous spectrum of x-rays is produced via the process of bremsstrahlung. In this process, electrons are accelerated as they are deflected by positively charged nuclei. As the electrons accelerate, they emit x-ray radiation. The spectrum has no maximum wavelength. The minimum wavelength is equal to the wavelength associated with the value of the electron’s initial kinetic energy—in other words, the energy of the photon that brings the electron to rest.

10. A red lightbulb emits long wavelengths, and thus, low energy photons. Such photons are not energetic enough to ruin the film by triggering the chemical reactions that expose it.

11. The Bohr model contains four assumptions: 1) The electron can exist without radiating only in certain circular orbits; 2) The laws of Newtonian mechanics apply to the motion of the electron in any of these stationary states; 3) The electron can make a transition between stationary states through the emission or absorption of a single photon; 4) The stationary states are those circular orbits in which the electron’s angular momentum is quantized in integral multiples of /(2 ).h π

12. Yellow light has a lower frequency (longer wavelength) than green light, and red light has a lower frequency still. Since the photons of yellow light don’t have sufficient energy to eject electrons from the metal, photons of red light certainly won’t either. More intense yellow light would not eject electrons either, since it is only the photon energy that matters. Violet light on the other hand has a higher frequency than green light, so we would expect electrons to be ejected by violet light regardless of the intensity.

13. In Compton scattering, a photon scatters off a free electron and loses some energy in the process. In the photoelectric effect, a photon is completely absorbed by an electron in a metal, not just scattered. A free electron would be unable to totally absorb a photon and still conserve energy and momentum in the process. When an electron absorbs a photon in the photoelectric effect, the atoms in the metal take part in the collision, allowing energy and momentum to be conserved by the process.

14. Some of the energy of the photon is transferred to the electron in the collision. Since a photon’s energy is proportional to its frequency, a decrease in the photon’s energy translates to an increase in its wavelength. The change in wavelength of the scattered photon is independent of the photon’s initial energy. However, an x-ray photon has a much smaller wavelength than a photon of visible light, and therefore, suffers a much greater fractional change in wavelength.

15. The process of pair production becomes especially important for photons with energies in excess of 1.02 MeV.

16. In the plum pudding model, the charge and mass of the atom are spread evenly throughout its volume. Alpha particles incident on a thin gold foil would not encounter any dense concentrations of charge and would therefore be expected to pass right through the foil, barely deflected at all. Rutherford found, however, that some of the alpha particles were deflected through large angles, indicating that there are regions of dense positive charge in the atoms of the foil—i.e., atomic nuclei.

17. The potential across the metal plate and collecting wire in a photoelectric effect apparatus is adjusted until the current in the collecting wire is zero; this determines the stopping potential. At this point, the potential barrier is just sufficient to stop the most energetic of the ejected electrons, so the stopping potential allows us to calculate the maximum kinetic energy of the electrons.

18. A fluorescent substance absorbs UV light and emits visible light. UV light has a higher frequency than visible light; the emitted radiation therefore has the longer wavelength. Another way to see this is to note that the energy of the emitted photons must be less than or equal to the energy of the absorbed photons. Therefore, the wavelength of the emitted radiation must be greater than or equal to that of the absorbed.

Chapter 27: Early Quantum Physics and the Photon College Physics

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19. In a hydrogen gas discharge tube, the hydrogen atoms are excited to higher energy states in a random process of collisions with energetic electrons. Once in an excited state, the atoms decay back to the ground state, possibly passing through several intermediate states, and emitting one or more photons. Because of the random nature of the excitations in this process, the atoms in the tube are excited, on average, to every possible state, and pass through every possible intermediate state during the decays. The emission spectrum of hydrogen therefore contains a line for every possible atomic transition. In an absorption spectrum, the atoms absorb incident photons in a similarly random process that puts them on average in every excited state. Once excited, however, the atom quickly decays back to the ground state by emitting one or more photons. The excited states do not live long enough to absorb a significant amount of the incident radiation and undergo transitions from intermediate states to higher energy states. For this reason, an absorption spectrum shows lines corresponding to transitions from the ground state only, while an emission spectrum contains lines from transitions between any two states.

20. A solar cell operates on the basis of the photoelectric effect. The higher the intensity of the light, the greater the number of photons incident on the solar cell in any interval of time. We therefore expect the current produced by a solar cell to increase with increasing intensity. Furthermore, we would expect the current to depend very little on the wavelength of the incident light, so long as the wavelength was below some maximum value (the threshold wavelength). If the wavelength were greater than this maximum value, we would expect the current to become essentially zero, independent of the intensity of the light.

21. In the wave model, the energy of light is spread out evenly over a wavefront. A low light intensity would mean each individual molecule in the retina would receive only a very small amount of energy per unit time. It would seem strange that individual molecules could be excited quickly enough for someone to be able to see in low light. In the photon model on the other hand, the minimum energy supplied to a molecule in the retina is the energy of a photon of the light. Furthermore, even with low intensity light, one of the molecules in the retina would absorb a photon all at once, while many other molecules would receive no energy at all. As long as the photon energy is sufficient to excite the molecule (visible light) and trigger the neural response, one should be able to see in low light.

22. The process of electron-positron annihilation must conserve both energy and momentum. Furthermore, these quantities must be conserved in every inertial reference frame. In particular, in the center of mass frame of the electron and positron, the total momentum is zero. If they were to annihilate and emit only one photon there would be no way to have zero momentum in the final state. By creating a pair of photons traveling in opposite directions in the center of mass frame, the momentum is conserved.

23. Since UV light has a higher frequency than blue, and both metals produce photoelectrons for blue light, both metals will also produce photoelectrons for UV light. Metal 1 produces photoelectrons for red light, so it might produce them for infrared as well. Metal 2 will not produce them for infrared light, since it doesn’t produce them for red light. Photoelectrons are produced whenever the photon energy is sufficient to overcome the work function of the metal. Therefore, the metal with the higher threshold frequency has the larger work function, which is metal 2.

24. The sharp peaks are characteristic of the target material being bombarded in the x-ray tube. When an incident electron knocks one of the electrons from an inner shell of the target to a higher energy level, the characteristic x-ray appears as an electron from an upper level drops down into the vacant level, emitting a photon that represents the difference in energy levels in the process.


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Problems

1. (a) Strategy and Solution The energy of a photon is inversely proportional to its wavelength, so the photons with the shorter wavelength have the greater energy. Therefore, a single ultraviolet photon has the greater energy.

(b) Strategy The energy of a photon of EM radiation with frequency f is .E hf= The frequency and wavelength are related by .f cλ = Solution Compute the energy of a single infrared photon.

34 820

infrared 6(6.626 10 J s)(3.00 10 m s) 9.9 10 J

2.0 10 mhcE hfλ

−−

−× ⋅ ×= = = = ×

×

Compute the energy of a single ultraviolet photon. 34 8

18ultraviolet 8

(6.626 10 J s)(3.00 10 m s) 2.8 10 J7.0 10 m

hcE hfλ

−−

−× ⋅ ×= = = = ×

×

(c) Strategy Use the definition of power. Let N be the number of photons. The total energy is NE, where E is the energy of a single photon. Solution Find the number of infrared photons emitted per second.

infrared 2120

200 W, so 2.0 10 photons s .9.9 10 J photon

NNE PPt t E −= = = = ×

∆ ∆ ×

Find the number of ultraviolet photons emitted per second. ultraviolet 19

18200 W 7.0 10 photons s

2.84 10 J photonN P

t E −= = = ×∆ ×

2. Strategy The energy of a photon of EM radiation with frequency f is .E hf= The frequency and wavelength are related by .f cλ = Solution Compute the energy.

21240 eV nm 1.8 eV7.0 10 nm

hcE hfλ

⋅= = = =×

3. Strategy The energy of a photon of EM radiation with frequency f is .E hf= The frequency and wavelength are related by .f cλ =

Solution

(a) Calculate the wavelength of a photon with energy 3.1 eV. 1240 eV nm, so 400 nm .

3.1 eVhc hcE hf

Eλ

λ⋅= = = = =

(b) Calculate the frequency of a photon with energy 3.1 eV. 8

149

3.00 10 m/s 7.5 10 Hz400 10 m

cfλ −

×= = = ××


1004

4. Strategy The minimum energy required to remove an electron from silver is the work function for silver. The work function is related to the threshold frequency 0f by 0.hfφ = Solution Compute the minimum energy.

15 150 (4.136 10 eV s)(1.04 10 Hz) 4.30 eVhfφ −= = × ⋅ × =

5. (a) Strategy Use Einstein’s photoelectric equation. Solution Calculate the maximum kinetic energy.

max1240 eV nm 2.16 eV 0.84 eV

413 nmhcK hf φ φλ

⋅= − = − = − =

(b) Strategy The threshold wavelength is related to the threshold frequency by 0 0 .f cλ = Use Eq. (27-8). Solution Find the threshold wavelength.

0 00

1240 eV nm, so 574 nm .2.16 eV

hc hchf φ λλ φ

⋅= = = = =

6. (a) Strategy The threshold wavelength is 0 288 nm.λ = The threshold wavelength is related to the threshold frequency by 0 0 .f cλ = The work function is given by Eq. (27-8). Solution Find the work function.

00

1240 eV nm 4.31 eV288 nm

hchfφλ

⋅= = = =

(b) Strategy Use Einstein’s photoelectric equation. Solution Calculate the maximum kinetic energy.

max1240 eV nm 4.306 eV 4.6 eV


⋅= − = − = − =

7. Strategy The work function for the metal is 2.60 eV.φ = Use Eq. (27-8) to find the maximum wavelength of the photons that will eject electrons from the metal. The frequency and wavelength are related by .f cλ = Solution Find the longest wavelength.

01240 eV nm, so 477 nm .

2.60 eVc hcf

hφ λ

λ φ⋅= = = = =

8. Strategy Use Eq. (27-7) to find the work function by setting the kinetic energy equal to s ,eV where sV is the stopping potential. Then, use Eq. (27-8) to find the maximum wavelength of the photons that will eject electrons from the metal. Solution Find the work function.

max s s1240 eV nm, so 1.10 eV 2.4 eV.

350 nmhc hcK hf eV eVφ φ φλ λ

⋅= − = − = = − = − =

Find the maximum wavelength of the photons.

0 maxmax

1240 eV nm, so 510 nm .2.44 eV

c hcfhφ λ

λ φ⋅= = = = =


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9. Strategy The stopping potential sV is related to the maximum kinetic energy of the electrons by max s.K eV= Use Einstein’s photoelectric equation and .f cλ = Solution Find the work function.

max s s1240 eV nm, so 1.1 eV 4.5 eV .

220 nmhc hcK eV hf eVφ φ φλ λ

⋅= = − = − = − = − =

10. Strategy Use Einstein’s photoelectric equation and .f cλ =

Solution

(a) Find the maximum kinetic energy of an electron ejected from the metal assuming it is cesium.

max1240 eV nm 1.8 eV 1.3 eV


⋅= − = − = − =

Yes; the metal could be cesium, since the maximum kinetic energy of ejected photons is nonnegative.

(b) Find the maximum kinetic energy of an electron ejected from the metal assuming it is tungsten.



⋅= − = − = − = −

No; the metal could not be tungsten, since the maximum kinetic energy of ejected photons is negative.

(c) Find the maximum kinetic energy of an electron ejected from the metal assuming it is cesium.



⋅= − = − = − =

Find the maximum kinetic energy of an electron ejected from the metal assuming it is tungsten.



⋅= − = − = − =

11. Strategy The frequency and wavelength are related by .f cλ = Use Eq. (27-8).

Solution

(a) The three relevant frequencies are 8 8

14 14yellow violet9 9

140

3.00 10 m/s 3.00 10 m/s5.2 10 Hz, 7.06 10 Hz, and580 10 m 425 10 m

6.20 10 Hz.

c cf f

fλ λ− −

× ×= = = × = = = ×× ×

= ×

No; only the source of violet light can eject electrons from the metal surface. Since yellow 0,f f< the yellow light source cannot eject electrons from the metal. Electrons are ejected from the metal by the violet light, since violet 0.f f>

(b) The amount of energy required to eject an electron from the metal is the work function for the metal. 15 14

0 (4.136 10 eV s)(6.20 10 Hz) 2.56 eVhfφ −= = × ⋅ × =


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12. Strategy Assume the electrons are nonrelativistic. Use Einstein’s photoelectric equation and .f cλ =

Solution

(a) Find the maximum speed of the emitted electrons. 2

max

195

31

1 , so2

2 2(1.602 10 J eV) 1240 eV nm 1.4 eV 9.8 10 m s .300 nm9.109 10 kg

hcK hf mv

hcvm

φ φλ

φλ

−

−

= − = − =

× ⋅⎛ ⎞ ⎛ ⎞= − = − = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠×

(b) Find the maximum speed of the emitted electrons.

max1240 eV nm 1.6 eV 0.05 eV


⋅= − = − = − = −

max 0,K < which is impossible, so no electrons are ejected from the metal.

(c) Increasing the intensity of the light increases the number of electrons ejected from the metal, but does not increase their kinetic energy, since the photon energy remains the same. Therefore, doubling the intensity has no effect on the electrons’ speed.

13. Strategy The frequency intercept gives the threshold frequency, so 130 43.9 10 Hz.f = × The stopping potential is

related to the maximum kinetic energy of the ejected electrons by max s.K eV= Einstein’s photoelectric equation is

max ,K hf φ= − so s .hfVe e

φ= − The work function φ is given by the stopping potential intercept 0 / ,V eφ= −

where s 0.V mf V= + s 0V = when 0.f f= Solution

(a) The slope of the graph is / .m h e= Compute the numerical value. 15

13 131.50 V 0 V 4.155 10 V s

80.0 10 Hz 43.9 10 Hzm −−= = × ⋅

× − ×

Planck’s constant is 19 15 34(1.602 10 C)(4.155 10 V s) 6.66 10 J s .h em − − −= = × × ⋅ = × ⋅

(b) Find the work function. 15 13

0 00 , so (4.155 10 V s)(43.9 10 Hz) 1.82 eV .mf emf eeφ φ −= − = = × ⋅ × =

14. Strategy The laser emits a cylindrical pulse of volume 2length area ( 2) .V c t dπ= × = ∆ × The energy of a photon within this pulse is photon ,E hc E Nλ= = where E is the total energy of the pulse and N is the number of

photons within the pulse. The energy density of the pulse is 20 rms .u E E V⟨ ⟩ = =� Since the pulse is emitted in

one second, the number N of photons in the pulse is the number of photons emitted by the laser per second. Solution Find the number of photons emitted per second by the laser.

2 212 2 20 rms0 rms 0 rms4

photon9 12 2 2 2 2

1434

, so 4

(640 10 m)[8.854 10 C (N m )](120 V m) (0.0015 m) 2.2 10 photons s .4(6.626 10 J s)

E c t dE V E dE NNE hc hc t h

λ π λ πλ

π− −

−

∆= = = =

∆

× × ⋅= = ×× ⋅

��


1007

15. Strategy The maximum kinetic energy K of an electron is equal to max .hf min max/c fλ = and ,K eV= where V is the applied voltage. Solution Find the applied voltage.

maxmin min

1240 eV nm, so 2.7 kV .0.46 nm

hc hchf K eV Ve eλ λ

⋅= = = = = =⋅

16. Strategy The minimum potential difference is found by equating the energy of an x-ray photon and the electron’s kinetic energy, which is equal to the electric potential energy of the electron, min .eV The energy of a photon of EM radiation with frequency f is .E hf= The frequency and wavelength are related by .f cλ = Solution Find the minimum potential difference.

min min1240 eV nm, so 4.96 kV .0.250 nm

hc hchf K eV Ve eλ λ

⋅= = = = = =⋅

17. Strategy The cutoff (maximum) frequency occurs when the photon’s energy and the electron’s kinetic energy are equal. The kinetic energy of the electron is equal to the electric potential energy of the electron, eV. Use Eq. (27-9). Solution Find the cutoff frequency.

319

max max 1546 10 eV, so 1.1 10 Hz .

4.136 10 eV seVhf K eV fh −

×= = = = = ×× ⋅

18. Strategy The minimum wavelength can be found by equating the photon’s energy and the electron’s kinetic energy. The kinetic energy of the electron is equal to the electric potential energy of the electron, eV. Use Eq. (27-9). Solution Find the minimum wavelength.

max min 3min

1240 eV nm, so 31.0 pm .40.0 10 eV

hc hchf K eVeV

λλ

⋅= = = = = =×

19. Strategy The smallest (minimum) wavelength can be found by equating the photon’s energy and the electron’s kinetic energy. The kinetic energy of the electron is equal to the electric potential energy of the electron, eV. Use Eq. (27-9). Solution Find the smallest wavelength.

max min 3min

1240 eV nm, so 62.0 pm .20.0 10 eV

hc hchf K eVeV

λλ

⋅= = = = = =×

20. Strategy Let V be the potential difference in the x-ray tube. The maximum frequency (or cutoff frequency) occurs when the photon’s energy is equal to the electron’s kinetic energy. The kinetic energy of the electron is equal to the electric potential energy of the electron, eV. Use Eq. (27-9). Solution

max max, so .ehf K eV f Vh

= = =

The cutoff frequency is proportional to the potential difference V, since e and h are constant.


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21. (a) Strategy Use the Compton shift, Eq. (27-14).

Solution Find the wavelength of the incident photon.

e e34

1231 8

12

(1 cos ), so (1 cos )

6.626 10 J s2.81 10 m (1 cos 29.5 )(9.109 10 kg)(2.998 10 m s )

2.50 10 m .

h hm c m c

λ λ θ λ λ θ

−−

−

−

′ ′− = − = − −

× ⋅= × − − °× ×

= ×

= 29.5˚e−

λ

λ

′

φ

θ

(b) Strategy The kinetic energy of the electron plus the energy of the scattered photon is equal to the energy of the incident photon. Use Eq. (27-10). Solution Find the kinetic energy of the electron.

e

9e 12 12

, so1 1(1240 10 eV m) 55.6 keV .

2.4954 10 m 2.81 10 m

K E Ehc hcK E Eλ λ

−− −

′+ =⎛ ⎞′= − = − = × ⋅ − =⎜ ⎟′ × ×⎝ ⎠

22. Strategy Use the Compton shift, Eq. (27-14).

Solution

(a) Find the wavelength of the scattered x-rays.

e e(1 cos ), so (1 cos ) 10.0 pm (2.426 pm)(1 cos 45.0 ) 10.7 pm .h h

m c m cλ λ θ λ λ θ′ ′− = − = + − = + − ° =

(b) e

(1 cos ) 10.0 pm (2.426 pm)(1 cos90.0 ) 12.4 pmhm c

λ λ θ′ = + − = + − ° =


Solution

(a) Find the Compton shift in wavelength.

e(1 cos ) (2.426 pm)(1 cos80.0 ) 2.00 pmh

m cλ λ θ′ − = − = − ° =

(b) Find the wavelength of the scattered photon. 2, so 2.00 pm 1.50 10 pm 2.00 pm 152 pm .λ λ λ λ λ′ ′− = ∆ = + = × + =

= 80.0˚e−

λ

λ

′

φ

θ


Solution Find the wavelength of the incident photons.

e

e

(1 cos ), so

(1 cos ) 124.65 pm (2.426 pm)(1 cos100.0 ) 121.80 pm .

hm c

hm c

λ λ θ

λ λ θ

′ − = −

′= − − = − − ° = = 100.0˚

e−λφ

θ

124.65 pm


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25. Strategy Use the Compton shift, Eq. (27-14). The frequency and wavelength are related by .f cλ =

Solution Find the frequency of the scattered photon.

8

e 19

e8

193.00 10 m s 123.0 10 Hz

(1 cos ), so

3.00 10 m s 2.4 10 Hz .(1 cos ) (2.426 10 m)(1 cos90 )

c hf m c

c c hf f m c

cf

λ λ θ

θ × −×

′ − = − = −′

×′ = = = ×+ − + × − °

= 90˚

e− φ

θ

26. (a) Strategy Use the Compton shift, Eq. (27-14).

Solution Find the scattering angle. e

e2

e e1 1

(1 cos ), so 1 ( ) cos or

cos 1 ( ) cos 1 ( )

m chm c h

m c m ch hc

λ λ θ λ λ θ

θ λ λ λ λ− −

′ ′− = − − − =

⎡ ⎤⎡ ⎤′ ′= − − = − −⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

e−

θN

0.14800 nm

0.14900 nm

psouth

6

1 0.511 10 eVcos 1 (0.14900 nm 0.14800 nm) 54.0 .1240 eV nm

− ⎡ ⎤×= − − = °⎢ ⎥⋅⎢ ⎥⎣ ⎦

(b) Strategy The momentum of the electron plus the momentum of the scattered photon is equal to the momentum of the incident photon. The incident photon has no north-south component of momentum, so the south component of the electron’s momentum is equal to the north component of the scattered photon’s momentum.

Solution Find the south component of the electron’s momentum. 34

24e,south north 9

6.626 10 J ssin sin 54.0 3.60 10 kg m s0.14900 10 m

hp p θλ

−−

−× ⋅′= = = ° = × ⋅

′ ×

27. Strategy Use conservation of momentum to find the velocity of the electron. Assume that the velocity of the electron is sufficiently less than the speed of light to use the nonrelativistic relationship between momentum and velocity. From Problem 26, 54.0 .θ = °

Solution Find the speed of the electron.

e,south e s north

346

s 31 9e

sin , so

6.626 10 J ssin sin 54.0 3.95 10 m s.(9.109 10 kg)(0.14900 10 m)

hp m v p

hvm

θλ

θλ

−

− −

′= = =′

× ⋅= = ° = ×′ × ×

e−

θN

0.14800 nm

0.14900 nm

φ

v

e,east e e east

346

e 31 9 9e

cos , so

1 cos 6.626 10 J s 1 cos54.0 2.05 10 m s.9.109 10 kg 0.14800 10 m 0.14900 10 m

h hp m v p p

hvm

θλ λ

θλ λ

−

− − −

′= = − = −′

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

2 2 6 2 6 2 6e s (2.05 10 m s) (3.95 10 m s) 4.45 10 m sv v v= + = × + × = ×

Find the angle. s1 1

e

3.95tan tan 62.62.05

vv

θ − −− −= = = − °

Thus, the velocity of the electron is 64.45 10 m s at 62.6 south of east .× ° Since 1.000110,γ ≈ our initial assumption was reasonable.


1010

28. Strategy The energy of the scattered photon is the initial photon energy minus the kinetic energy of the electron. The energy of the scattered photon is related to its wavelength by .E hf hc λ′′ = =

Solution Find the energy of the scattered photon. 240.0 keV 190.0 keV 50.0 keVE E K′ = − = − =

Find the wavelength.

31240 eV nm 0.0248 nm 24.8 pm50.0 10 eV

hcE

λ ⋅′ = = = =′ ×

e−λ

λ′

29. Strategy The change in kinetic energy of the electron is the difference in the energies of the incident photon and the scattered photon. Solution Find the change in kinetic energy.

41240 eV nm 1240 eV nm 2.4 10 eV0.0100 nm 0.0124 nm

hc hcK E Eλ λ

⋅ ⋅′∆ = − = − = − = ×′

30. Strategy Since a photon’s energy is inversely proportional to its wavelength, the scattered photons with the smallest energy will have the longest wavelength. Use the Compton shift, Eq. (27-14).

Solution

(a) The wavelength of the scattered photons is

e(1 cos )h

m cλ λ θ′ = + −

Since 1 cos 1θ− ≤ ≤ , the maximum value for ′ λ is

e2 h

m cλ λ′ = +

Find the value of θ that maximizes .λ′ 11 cos 2, so cos ( 1) 180 .θ θ −− = = − = °

λ

λ

′

θ

(b) For 180 ,θ = °e

2 10.0 pm 2 2.426 pm 14.9 pm.hm c

λ λ′ = + = + × = The ratio is 14.9 pm 1.49 .10.0 pm

λλ′

= =

31. Strategy The orbital radius in the nth state is given by 20.nr n a=

Solution Compute the orbital radius.

23 3 52.9 pm 476 pm 0.476 nmr = ⋅ = =

32. Strategy The energy for a hydrogen atom in the nth stationary state is given by 2( 13.6 eV) .nE n= − Solution Find the energy.

4 213.6 eV 0.850 eV

4E −= = −


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33. Strategy The orbital radius in the nth state is given by 20.nr n a=

Solution

(a) Find the difference in the radii of the 1 and 2n n= = states for hydrogen. 2 2 2 12 10

0 2 1 0 0, so (2 1 ) 3 3(52.9 10 m) 1.59 10 m .nr n a r r a a − −= − = − = = × = ×

(b) Find the difference in the radii of the 100 and 101n n= = states for hydrogen. 2 2 2 12 8

0 2 1 0 0, so (101 100 ) 201 201(52.9 10 m) 1.06 10 m .nr n a r r a a − −= − = − = = × = ×

The result for (b) is much larger than that for (a), so the orbital separations are much larger for larger n values.

34. Strategy Lithium has three protons in its nucleus. Doubly-ionized lithium has only one electron. Use Eq. (27-25). Solution Find the Bohr radius.

2 2

1 01 (52.9 pm) 17.6 pm3

nr aZ

= = =

35. Strategy Lithium has three protons in its nucleus. Assume the electron is initially in the ground state. Use Eq. (27-26). Solution Find the ground-state energy of the remaining electron in a doubly-ionized lithium atom.

2 2

1 2 23(13.6 eV) (13.6 eV) 122 eV1

ZEn

= − = − = −

It will take 122 eV of energy to remove the electron.

36. Strategy The amount of energy required to cause a transition from the ground state to the n = 4 state is equal to the difference in the energy between the two states. The energy for a hydrogen atom in the nth stationary state is given by 2( 13.6 eV) .nE n= − Solution The energy of a hydrogen atom in the n = 4 state is

14 2

13.6 eV 0.850 eV.164

EE −= = = −

The energy that must be supplied is 4 1 0.850 eV ( 13.6 eV) 12.8 eV .E E− = − − − =

37. Strategy The energy supplied by the photon raises the atom from the ground state to a higher energy level. The energy for a hydrogen atom in the nth stationary state is given by 2( 13.6 eV) .nE n= − Solution Find the energy level to which the atom is excited.

1 11 12

1

13.6 eV12.1 eV, so 3.12.1 eV 12.1 eV 13.6 eVn

E EE E E nEn

−− = − = = = =+ −

The atom is excited to the 3n = energy level.


1012

38. Strategy Use Coulomb’s law and Newton’s law of universal gravitation. Solution Compute the magnitude of the electric force.

2 9 2 2 19 28

E 2 12 21

(8.988 10 N m C )(1.602 10 C) 8.24 10 N.(52.9 10 m)

keFr

−−

−× ⋅ ×= = = ×

×

Compute the magnitude of the gravitational force. 11 2 2 31 27

e p 47G 2 12 2

1

(6.674 10 N m kg )(9.109 10 kg 1.673 10 kg) 3.63 10 N(52.9 10 m)

Gm mF

r

− − −−

−× ⋅ × )( ×= = = ×

×

Compute the ratio of these forces. 47

40G8

E

3.63 10 N 4.41 108.24 10 N

FF

−−

−×= = ××

39. Strategy The minimum energy for an ionized atom is ionized 0.E = Use the ground-state energy. Solution The energy needed to remove a ground-state electron from a hydrogen atom is

ionized 1 0 ( 13.6 eV) 13.6 eV .E E E= − = − − =

40. Strategy The minimum energy for an ionized atom is ionized 0.E = Use Eq. (27-24). Solution The energy needed to ionize a hydrogen atom initially in the 2n = state is

1ionized 2 ionized 2

13.6 eV0 3.40 eV .42

EE E E E −⎛ ⎞= − = − = − =⎜ ⎟⎝ ⎠

41. Strategy The smallest transition that a ground-state electron can make is to the n = 2 state. Use Eq. (27-24). Solution The energy of the photon is

12 1 12 2

13.6 eV ( 13.6 eV) 10.2 eV .2 2EE E E E −= − = − = − − =

42. Strategy The energy of a photon is related to its wavelength by .E hc λ= Use Eq. (27-24) to find the energy of the transition. Solution The amount of energy released when a hydrogen atom makes a transition from the 6n = to the 3n =

state is 6 3 2 213.6 eV 13.6 eV 1.13 eV.

6 3E E E − −= − = − =

The wavelength of the radiation emitted is 1240 eV nm 1.09 µm .1.133 eV

hcE

λ ⋅= = =

43. (a) Strategy Use Coulomb’s law. Solution Find the magnitude of the force on the electron due to the proton.

21 22 2 2

k q q k e e keFr r r

+ −= = =


1013

(b) Strategy The electron is undergoing uniform circular motion, so the acceleration is radial. Use Newton’s second law. Solution Find the speed of the electron.

2 2 2

r r e2e

, so .ke v keF ma m vr m rr

Σ = = = =

(c) Strategy Use the Bohr condition for angular momentum and the result of part (b). Solution Solve for the speed.

ee

, so .nn

nm vr n vm r

= =

Find the radius of the nth Bohr orbit. 2 2 2 2 2

22

e e e e, so or .n

n n n

n ke n nke rm r m r m r m ke

= = =

44. (a) Strategy Use the classical kinetic energy and the expression for the electron’s speed found in Problem 43. Solution Find the kinetic energy.

22 2 2

2e e e

e e

1 1 12 2 2 2

ke ke keK m v m mm r m r r

⎛ ⎞⎜ ⎟= = = =⎜ ⎟⎝ ⎠

(b) Strategy U eV= − is the potential energy of the electron and V ke r= is the electric potential of the proton. Solution Find the electric potential energy.

2ke keU eV er r

⎛ ⎞= − = − = −⎜ ⎟⎝ ⎠

(c) Strategy Use conservation of energy and the results of parts (a) and (b). Solution Find the mechanical energy.

2 2 2 2 222 2 2 2ke ke ke ke keE K U

r r r r r= + = − = − = −

(d) Strategy Use Eq. (27-19) and the result of part (c). Solution Find the energy of the nth Bohr orbit.

2 2

2e

2 42 2e

2 22 22n nn

m ke

m k eke keEr n

−= − = = −

45. Strategy Use Eq. (27-20) and the values of the fundamental constants. Solution Compute the numerical value of the Bohr radius.

2 2 34 211

0 2 2 2 2 31 9 2 2 19 2e e

(6.626 10 J s) 5.29 10 m4 4 (9.109 10 kg)(8.988 10 N m C )(1.602 10 C)

ham ke m keπ π

−−

− −× ⋅= = = = ×

× × ⋅ ×


1014

46. Strategy Use Eq. (27-23) and the values of the fundamental constants. Solution Compute the numerical value of the ground-state energy for hydrogen in the Bohr model.

2 4 2 2 4e e

1 2 2

2 31 9 2 2 2 19 4

34 2 19

22

2 (9.109 10 kg)(8.988 10 N m C ) (1.602 10 C) 1 eV 13.6 eV(6.626 10 J s) 1.602 10 J

m k e m k eEh

π

π − −

− −

= − = −

× × ⋅ × ⎛ ⎞= − = −⎜ ⎟× ⋅ ×⎝ ⎠

47. Strategy Use the Bohr condition for angular momentum, Eq. (27-18). Solution Calculate the speed of the electron.

346

e 31 11e 1 e 0

(1) 6.626 10 J s, so 2.19 10 m s .2 2 (9.109 10 kg)(5.29 10 m)n

hm vr n vm r m aπ π

−

− −× ⋅= = = = = ×

× ×

48. Strategy Use the Bohr condition for angular momentum, Eq. (27-18), and Eq. (27-25). Solution The Bohr equation for angular momentum is e .nm vr n=

Solving for v with n = 1 givese 1 e 1

.2

hvm r m rπ

= =

In this case, 1r is the radius of the ground state of singly-ionized helium. The atomic number for helium is 2.Z =

The orbital radii equation is2

0.nnr aZ

= For the ground state of He ,+ 1n = and 2,Z = so2

01 0

1 .2 2

ar a= =

Substituting this value of r1 into the speed equation gives the result. 34

631 11

e 0

6.626 10 J s 4.38 10 m s(9.109 10 kg)(5.29 10 m)

hvm aπ π

−

− −× ⋅= = = ×

× ×

49. Strategy The energy of a photon with the given wavelength is equal to the energy difference between the two helium levels. The energy of a photon of EM radiation with wavelength λ is .E hc λ= Solution Find the energy difference.

1240 eV nm 2.11 eV587.6 nm

hcEλ

⋅= = =

50. Strategy The 15.0 eV of energy transferred to the atom provides more than enough energy to allow the electron to escape the potential well of the hydrogen nucleus. The left-over energy becomes the kinetic energy of the free electron. Solution Determine the energy required for the electron to escape the potential well (n = 2 to ).n = ∞

f i 2 213.6 eV0 3.40 eV

2E E E E E∞

−∆ = − = − = − =

The kinetic energy of the electron when it is far from the nucleus is 15.0 eV − 3.40 eV = 11.6 eV.


1015

51. Strategy and Solution

(a) The electron can return to the ground state directly, emitting one photon; thus, the minimum number of photons is one.

(b) The electron can return to the ground state by stopping at each intermediate state: n = 5 to n = 4, 4 to 3, 3 to 2, and then finally, from n = 2 to the ground state, n = 1. One photon is emitted for each transition, for a total of four photons.

52. Strategy Use Eq. (27-24) to find the energies of the transitions. The energy of a photon is related to its wavelength by .E hc λ=

Solution

(a) A photon with the longest wavelength will have the least energy. The transition from the n = 4 state to the n = 3 state will release the least energy.

1 14 3 2 2

13.61 eV 13.61 eV 0.6616 eV16 94 3

E EE E E − −= − = − = − =

Calculate the wavelength.

( )19

34 8

1.602 10 JeV

(6.626 10 J s)(2.998 10 m s), so 1874 nm .(0.6616 eV)

hc hcEE

λλ −

−

×

× ⋅ ×= = = =

(b) The energy released by a transition from the n = ∞ state to the n = 3 state is

213.61 eV0 1.5122 eV.

3E −= − =

The wavelength of a photon with this energy is

( )19

34 8

1.602 10 JeV

(6.626 10 J s)(2.998 10 m s) 820.0 nm .(1.5122 eV)

hcE

λ−

−

×

× ⋅ ×= = =

(c) The range of wavelengths from 820.0 nm to 1874 nm is in the IR (infrared) part of the EM spectrum.

53. Strategy The energy of a photon of EM radiation with wavelength λ is .E hc λ= Solution The energy of the UV photon is 1 1 .E hc λ= The solid dissipates 0.500 eV, so the energy available for

the emitted photon is 2 11

0.500 eV 0.500 eV.hcE Eλ

= − = −

Calculate the wavelength of the emitted photon.

1 1

0.500 eV 0.500 eV1 12 320 nm 1240 eV nm

1 1 370 nm0.500 eVhc

hc

hc hcE λ λ

λ⋅

= = = = =− − −


1016

54. (a) Strategy Lithium has three protons in its nucleus. Doubly-ionized lithium has only one electron. Use Eq. (27-26). Solution Find the first four energy levels of doubly-ionized lithium.

2 2 2

1 22 2 22 2

3 42 2

3 3(13.6 eV) (13.6 eV) 122.4 eV 122 eV , (13.6 eV) 30.6 eV ,1 2

3 3(13.6 eV) 13.6 eV , and (13.6 eV) 7.65 eV .3 4

ZE En

E E

= − = − = − = − = − = −

= − = − = − = −

(b) Strategy Compute the energy of the photon emitted due to each transition between the first four energy levels of doubly-ionized lithium. Solution

4 1: 7.65 eV ( 122.4 eV) 115 eV→ − − − = 3 1: 13.6 eV ( 122.4 eV) 109 eV→ − − − =

4 2 : 7.65 eV ( 30.6 eV) 23.0 eV→ − − − = 3 2 : 13.6 eV ( 30.6 eV) 17.0 eV→ − − − =

4 3 : 7.65 eV ( 13.6 eV) 6.0 eV→ − − − = 2 1: 30.6 eV ( 122.4 eV) 92 eV→ − − − =

(c) Strategy Compute the energy of photons with wavelengths corresponding the limits of the visible EM spectrum. Then, compare these with those energies found in part (b). Solution Photons in the visible range have wavelengths from 400 nm to 700 nm.

max minmin max

1240 eV nm 1240 eV nm3.10 eV and 1.77 eV.400 nm 700 nm

hc hcE Eλ λ

⋅ ⋅= = = = = =

The lowest-energy photon created by one of the transitions in part (b) is 6.0 eV. Since 6.0 eV > 3.10 eV, of the photons generated, none are in the visible region.

55. Strategy Compute the energies of electron transitions between consecutive energy levels until an energy matches that of a visible photon. Use Eq. (27-26). Solution Find the energy levels of doubly-ionized lithium in terms of .n

2 2

2 2 23 122.4 eV(13.6 eV) (13.6 eV)n

ZEn n n

= − = − = −

Photons in the visible range have wavelengths from 400 nm to 700 nm. Compute the maximum and minimum visible-photon energies.

max minmin max

1240 eV nm 1240 eV nm3.10 eV and 1.77 eV.400 nm 700 nm

hc hcE Eλ λ

⋅ ⋅= = = = = =

The energy levels get closer together for greater n, so compute the transition energies until one is found that is less than 3.10 eV.E = 2 1: 30.6 eV ( 122.4 eV) 92 eV→ − − − = 3 2 : 13.6 eV ( 30.6 eV) 17.0 eV→ − − − = 4 3 : 7.65 eV ( 13.6 eV) 6.0 eV→ − − − = 5 4 : 4.90 eV ( 7.65 eV) 2.75 eV→ − − − = So, the lowest value of n for which a transition from the to the 1n n + state is caused by a photon in the visible region is 4 .n =


1017

56. Strategy For a photon, the maximum wavelength corresponds to the minimum energy. The minimum energy needed is the rest energy of an electron-positron pair. Use 2

0 and .E mc E hc λ= = Solution Compute the required energy.

2 31 8 2 13e2 2(9.109 10 kg)(3.00 10 m s) 1.64 10 JE m c − −= = × × = ×

Compute the wavelength. 34 8

13(6.626 10 J s)(3.00 10 m s) 1.21 pm

1.64 10 JhcE

λ−

−× ⋅ ×= = =

×

57. Strategy The energy of the photon must equal the sum of the total energy of the electron and the positron. Use 2

0 .E mc= Solution Find the energy of the photon.

2 2photon e p e e p p 0.511 MeV 0.22 MeV 0.511 MeV 0.22 MeV 1.46 MeVE E E m c K m c K= + = + + + = + + + =

58. Strategy The energy of the photon must equal the sum of the total energy of the electron and the positron. The energy of a photon of EM radiation with wavelength λ is .E hc λ= Solution Find the wavelength of the photon.

1514

photon electron positronelectron positron

1240 10 MeV m, so 7.75 10 m .8.0 MeV 8.0 MeV

hc hcE E EE E

λλ

−−× ⋅= + = = = = ×

+ +

59. Strategy The rest energies of the electron and positron are converted into the combined energy of the two photons. The energies of the photons are the same, since the frequencies of the photons are identical. Thus, the energy of each photon will equal the rest energy of an electron, 2

e 511 keV.E m c= = The energy of a photon of EM radiation with wavelength λ is .E hc λ= Solution Compute the wavelength.

31240 eV nm 0.00243 nm 2.43 pm511 10 eV

hcE

λ ⋅= = = =×

60. Strategy The rest energy of the muon and antimuon are converted into the energy of the two photons of equal energy. Since the photons have equal energy, their wavelengths are the same. The energy of a photon of EM radiation with wavelength λ is .E hc λ= Use 2

0 .E mc= Solution Find the wavelength.

2 2 2photon muon antimuon e e e

1514

2e

2 207 207 414 2 , so

1240 10 MeV m 1.17 10 m .207(0.511 MeV)207

hcE E E m c m c m c

hcm c

λ

λ−

−

= + = + = =

× ⋅= = = ×


1018

61. Strategy and Solution

(a) Since the electron and positron are at rest, their combined momentum is zero. Conservation of momentum requires that the total momentum also be zero after the annihilation. A single photon has momentum /p E c= in the direction that the photon is moving, so a single photon can never have a total momentum of zero. Therefore, the annihilation of an electron and positron can never produce a single photon. Two photons of equal energy moving in opposite directions will have zero total momentum, so the reaction e− + e+ → 2γ can occur.

(b) The two photons must have equal energy, so each photon will have the energy of an electron at rest, 2

e 511 keV .E m c= =

62. (a) Strategy The energy E of a pulse is equal to the power of the pulse P times the emission time .t∆

Solution Compute the energy of each pulse. (0.500 W)(0.0200 s) 0.0100 J 10.0 mJE P t= ∆ = = =

(b) Strategy The ratio of the energy per pulse to the energy per photon gives the number of photons per pulse.

Solution Compute the ratio. 9pulse 16

34 8photon

(0.500 W)(0.0200 s)(643 10 m) 3.23 10(6.626 10 J s)(3.00 10 m s)hc

E P t P tE hcλ

λ −

−∆ ∆ ×= = = = ×

× ⋅ ×

63. (a) Strategy Plot the data and draw a best-fit line.

Solution

1.61.41.21.00.80.60.40.2

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0–0.2–0.4–0.6–0.8–1.0–1.2–1.4–1.6–1.8

(µm–1)

V (V)

1λ

(b) Strategy The intercept of the stopping potential times e gives the work function. The 1 -interceptλ gives the threshold wavelength.

Solution The V-intercept is about 1.57 V.− This equates to a work function of 1.57 eV .

The 1 -interceptλ is about 11.35 m .−µ The corresponding value for the threshold wavelength is

11 741 nm .

1.35 mλ −= =

µ


1019

(c) Strategy Use Einstein’s photoelectric equation. Solution The slope is approximately

( ) 111.57 V 1.16 V m 1160 V nm .

1.35 mV

λ−

∆ = = ⋅µ = ⋅µ∆

To calculate the theoretical value of the slope, start with Einstein’s photoelectric equation. maxK hf φ= −

Substitute maxK eV= and .f c λ= 1, so .hc hceV V

e eφφ

λ λ= − = ⋅ −

This equation relates V to 1/λ, as does the graph. The theoretical slope is 1240 eV nm 1240 V nm.hc

e e⋅= = ⋅

This value is reasonably close to the measured value.

64. (a) Strategy The energy of a photon of EM radiation with frequency f is .E hf= Solution Compute the energy of each photon.

34 6 719

1 eV(6.626 10 J s)(89.3 10 Hz) 3.69 10 eV1.602 10 J

E − −−

⎛ ⎞= × ⋅ × = ×⎜ ⎟×⎝ ⎠

(b) Strategy The ratio of the total radiated power to the energy per photon is the rate of photons emitted. Solution Compute the rate of photon emission.

34 6

329

(6.626 10 J s)(89.3 10 Hz)1 photon

50.0 10 W 8.45 10 photons/sPE −× ⋅ ×

×= = ×

65. Strategy Power is equal to intensity times area. The energy of a photon of EM radiation with wavelength λ is .E hc λ= The minimum detected number of photons per second is given by the ratio of the power to the energy per photon. Solution The minimum amount of power that an owl’s eye can detect is

2 13 2 3 2 17(5.0 10 W m ) (4.25 10 m) 2.84 10 W.P IA I rπ π− − −= = = × × = × The energy of a photon of wavelength 510 nm is

34 819

9(6.626 10 J s)(3.00 10 m s) 3.9 10 J.

510 10 mhcEλ

−−

−× ⋅ ×= = = ×

×

Compute the minimum number of photons per second that an owl eye can detect. 17

192.84 10 W 73 photons/s

3.9 10 J

−

−× =×

66. Strategy The shortest wavelength x-ray corresponds to the highest frequency, which occurs when the energy of the photon is equal to the electron’s kinetic energy. Solution Find the wavelength.

3photon electron 6

1240 eV nm, so 6.2 10 nm 6.2 pm .0.20 10 V

hc hcE K eVeV e

λλ

−⋅= = = = = = × =× ×


1020

67. Strategy Use Eq. (27-24).

Solution The energy of the n = 4 state of the hydrogen atom is

4 213.6 eV 0.850 eV.

4E −= = −

To remove an electron from the n = 4 state requires 0.850 eV of energy.

68. Strategy After the collision the electron recoils in the direction of motion of the incident photon. Its motion has no component perpendicular to the motion of the incident photon. Conservation of momentum requires the scattering angle must be 180°. Use Eq. (27-10) and the Compton shift, Eq. (27-14).

Solution Solve the Compton shift formula for λ.

e

e

e

(1 cos ), so

(1 cos180 ) 2(2.426 pm) 4.852 pm.

The kinetic energy of the electron is .

hm c

hm c

hc hcK E E

λ λ θ

λ λ λ λ

λ λ

′ − = −

′ ′ ′= − − ° = − = −

′= − = −′

e−

Before :

After :

e−

v

Substituting 4.852 pmλ λ′= − and e 0.20 keV,K = and solving for λ′ gives the wavelength of the scattered x-ray.

e

3

4 1

2 4 2

2 4 2

1 14.852 pm

0.20 10 eV 1 1(1240 eV nm)(1000 pm/nm) 4.852 pm

(1.613 10 pm )( 4.852 pm) ( 4.852 pm)(4.852 pm) 3.008 10 pm

(4.852 pm) 3.008 10 pm 0

Khc λ λ

λ λλ λ λ λ

λ λλ λ

− −

= −′ ′−

× = −′ ′⋅ −

′ ′ ′ ′× − = − −′ ′− = ×

′ ′− − × =

Solve for λ′ using the quadratic formula. 2 4 24.852 pm ( 4.852 pm) 4 1 ( 3.008 10 pm )

176 pm2 1

λ± − − ⋅ ⋅ − ×

′ = =⋅

or −171 pm, which is extraneous.

The wavelength of the incident x-ray can now be calculated. 4.852 pm 176 pm 4.852 pm 171 pmλ λ′= − = − =

69. Strategy Use Eqs. (27-21), (27-24), (27-25), and (27-26).

Solution Equate the energy levels in He+ and H atoms.

H He2

2 2

2 2 2

13.6 eV 13.6 eV 2

22

n mE E

n mm n

m n

+=

− − ×=

==

The ratio of the orbital radii of H and He + is 20+

2 2H 0

2He 2

2 .nm a

m

r n a nr m

= =

Now substitute m = 2n.

+ +

2 2H H

2 2He He

2 2 12(2 ) 4

n

m

r n n Zr Zn n

= = = =

For levels of equal energy, the ratio of orbital radii appears to equal the ratio of atomic numbers.


1021

70. Strategy Use Eq. (27-25). From the Bohr condition of quantized angular momentum, the electron speeds are e( ).nv n m r= Solution The orbital radius of the n = 1 state of He ( 2)Z+ = is

+

2 20

0 0He1 .2 2

n ar a aZ

= = =

The orbital radius of the n = 2 state of 3Be ( 4)Z+ = is 3

2

0 0Be2 .4

r a a+ = =

Compute the speeds.

( )+ 3+0He Be

e 0 e 0e 2

(1) 2 (2) and .a

v vm a m am

= = =

The expressions for the two speeds are identical.

3+ +6

Be He 4.4 10 m sv v= = ×

71. (a) Strategy The energy of a photon of EM radiation with wavelength λ is .E hc λ= The momentum of a photon is related to its energy by .E pc= Solution Calculate the energy of a single photon.

1240 eV nm 1.9 eV670 nm

hcEλ

⋅= = =

Calculate the momentum. 19

288

(1.85 eV)(1.6 10 J eV) 9.9 10 kg m s3.00 10 m s

Epc

−−×= = = × ⋅

×

(b) Strategy The power of the laser divided by the energy per photon gives the rate of photons emitted. Solution Compute the rate of photon emission.

315

191 10 J s 3 10 photons/s

(1.85 eV)(1.6 10 J eV)PE

−

−×= = ×

×

(c) Strategy Use the impulse-momentum theorem, av .F t p∆ = ∆ Solution Find the average force.

312

av 81 10 W 3 10 N3 10 m s

p E c PFt E P c

−−∆ ×= = = = = ×

∆ ×


1022

72. Strategy Use Einstein’s photoelectric equation.

Solution

(a) Find the work function for sodium.

max max s1240 eV nm, so 0.28 eV 1.9 eV.

570 nmhcK hf hf K eVφ φλ

⋅= − = − = − = − =

Find the stopping potential.

s s1 1240 eV nm, so 1.9 V 1.2 V .

400.0 nmhceV hf Ve e e

φφλ

⋅= − = ⋅ − = − =⋅

(b) The stopping potential does not depend upon the intensity of the light, so it is still 0.28 V .

(c) 1.9 eV , as found in part (a).φ =

73. (a) Strategy The photons that can be absorbed have energies equal to the energy differences between energy levels. Solution Calculate the energies and associated wavelengths needed for a transition from the ground state to the n = 2, 3, 4, and 5 states.

11 1 2n

EE E E E

n= − = −

1 2: 13.6 eV 3.4 eV 10.2 eV1240 eV nm 122 nm

10.2 eV1 3: 13.6 eV 1.51 eV 12.09 eV

1240 eV nm 103 nm12.09 eV

EhcE

E

λ

λ

→ = − =⋅= = =

→ = − =⋅= =

1 4: 13.6 eV 0.85 eV 12.75 eV1240 eV nm 97.3 nm

12.75 eV1 5: 13.6 eV 0.544 eV 13.056 eV

1240 eV nm 95.0 nm13.056 eV

E

E

λ

λ

→ = − =⋅= =

→ = − =⋅= =

The wavelengths that can be absorbed are 97.3 nm and 103 nm.

(b) Strategy By conservation of momentum, the momentum of the atom must equal the momentum of the photon. Solution Compute the recoil speeds.

H H photon

34

27 9H

, so

6.626 10 J s 4.07 m s for the 97.3-nm photon(1.674 10 kg)(97.3 10 m)

E hp m v pc

hvm

λ

λ

−

− −

= = = =

× ⋅= = =× ×

and34

27 96.626 10 J s 3.86 m s for the 103-nm photon .

(1.674 10 kg)(102.6 10 m)v

−

− −× ⋅= =

× ×

(c) Strategy and Solution The transition from the n = 3 state to the ground state (n = 1) can occur two ways: 3 to 2, then 2 to 1 or 3 to 1. The transition from the n = 4 to the ground state can occur four ways: 4 to 3, then 3 to 2, then 2 to 1 or 4 to 3, then 3 to 1 or 4 to 2, then 2 to 1 or 4 to 1. Therefore, there are six ways total for the atom to return to the ground state: two ways when the 103-nm wavelength is absorbed and four ways when the 97.3-nm wavelength is absorbed.


1023

74. Strategy Intensity is equal to the power radiated divided by the area. The power radiated is equal to the energy per photon times the rate of photon emission N.

Solution

(a) Compute the intensity.

( )

2 2 2

34 813 2

2 90.007 m2

energy per photon photons per second

(6.626 10 J s)(3.00 10 m s)(20 photons s) 1.7 10 W m(600 10 m)

hc NP hcNIA r r r

λπ π π λ

π

−−

−

×= = = =

× ⋅ ×= = ××

Find the distance required for a 10-W source to have this intensity.

2 13 210 W, so 2000 km .

44 4 (1.7 10 W m )P P PI rA Ir ππ π −= = = = =

×

(b) We don’t see lightbulbs at this distance because light is scattered by the atmosphere and other light sources overwhelm the light from a single bulb.

75. Strategy The energy of a photon of EM radiation with wavelength λ is .E hc λ= Use the Compton shift, Eq. (27-14).

Solution

(a) The scattered photons have half the energy of the incident photons. Relate the wavelengths. 1 1 , so 2 .2 2

hc hcE E λ λλ λ

′ ′= = = =′

Find the wavelength of the scattered photons.

e2 (1 cos ), so (2.426 pm)(1 cos90.0 ) 2.426 pm .h

m cλ λ λ λ θ λ′ − = − = − = − ° =

Detector

Target

(b) e

(1 cos ) 2.426 pm (2.426 pm)(1 cos180 ) 7.278 pmhm c

λ λ θ′ = + − = + − ° =

(c) There would be no change. Electrons are the same in any material, and the photons are scattered by electrons.

76. Strategy The minimum wavelength corresponds to the maximum frequency, which occurs when the energy of the photon is equal to the electron’s kinetic energy. Solution Find the potential difference.

max e 3min min

1240 eV nm, so 27.6 kV .45.0 10 nm

hc hchf K eV Ve eλ λ −

⋅= = = = = =× ×

77. Strategy and Solution When the potential difference accelerating electrons in the x-ray tube is doubled, the energies of the characteristic x-rays remain the same because they are characteristic of the target material’senergy transitions.


1024

78. Strategy The energy of a photon of EM radiation with wavelength λ is .E hc λ= Use the Compton shift, Eq. (27-14). Solution The wavelength of an incident photon is

31240 eV nm 6.667 pm.186 10 eV

hcE

λ ⋅= = =×

Calculate the Compton shifts for θ = 90.0° and 180.0°.

90 180e

(1 cos90.0 ) 2.426 pm and (2.426 pm)(1 cos180.0 ) 4.852 pm.hm c

λ λ∆ = − ° = ∆ = − ° =

The wavelengths of the scattered x-rays are 90 1806.667 pm 2.426 pm 9.09 pm and 6.667 pm 4.852 pm 11.52 pm.λ λ λ λ′ ′= + ∆ = + = = + =

The energies of the scattered x-rays are

90 1801240 eV nm 1240 eV nm136 keV and 108 keV .0.00909 nm 0.01152 nm

hcE Eλ

⋅ ⋅= = = = =

79. Strategy Use Eq. (27-26).

Solution

(a) The ground state (n = 1) energy for He + (Z = 2) is

+

2 2

He 2 2213.6 eV 13.6 eV 54.4 eV .1

ZEn

= − × = − × = −

(b) The ground state energy for Li 2+ (Z = 3) is

2+

2

Li 23 13.6 eV 122 eV .1

E = − × = −

(c) Since deuterium has an atomic number of Z = 1, the energy levels for deuterium will be the same as for hydrogen.

deuterium 13.6 eVE = −

80. (a) Strategy The energy per second incident is the power incident, which is equal to the intensity times the area. Solution Compute the energy per second falling on the atom.

2 9 2 22 22(0.01 W m )(0.1 10 m) 1 10 W 1 10 J sP IA − − −= = × = × = ×

(b) Strategy Since ,P E t= ∆ ∆ and E φ∆ = in this case, / .t Pφ∆ = Solution Compute the time lag.

193

22(2.0 eV)(1.6 10 J eV) 3.2 10 s

1 10 J st

Pφ −

−×∆ = = = ×

×

(c) Strategy and Solution Whether or not an electron is ejected has nothing to do with the intensity of the light; ejection of the electron depends upon the energy of individual photons. An electron is ejected immediately when a single photon of sufficient energy strikes it.


1025

81. (a) Strategy The momentum of the atom must equal the momentum of the incident photon.

Solution Find the recoil speed. 34

H H photon 27 9H

6.626 10 J s, so 4.1 m s .(1.674 10 kg)(97 10 m)

h hp m v p vmλ λ

−

− −× ⋅= = = = = =

× ×

(b) Strategy The energy of a photon of EM radiation with wavelength λ is .E hc λ= Use Eq. (27-24).

Solution Compute the energy of the incident photon. 1240 eV nm 12.8 eV

97 nmhcEλ

⋅= = =

This energy corresponds to a transition from the n = 1 state to the n = 4 state. 1

1 2 =13.6 eV 0.850 eV 12.8 eV4E

E E= − − =

The transition from the n = 4 state back to the n = 1 state can occur in 4 ways emitting six different photons: 4 1, 4 2 1, 4 3 1, and 4 3 2 1.→ → → → → → → →

(c) Strategy There are six photons possible. The classifications are UV for 400 nm,λ < visible for 400 nm 700 nm,λ≤ ≤ and IR for 700 nm.λ > The wavelengths are given by

2 2 2 2f i f i

7 11 1 1 1

1 1 .(1.097 10 m )

n n n nR

λ−

= =⎛ ⎞ ⎛ ⎞− × −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Solution The results are given in the table.

Transition λ (nm) Class

4 3→ 1875 IR

4 2→ 486 visible

4 1→ 97 UV

3 2→ 656 visible

3 1→ 103 UV

2 1→ 122 UV 82. Strategy Use Eq. (27-8) and Einstein’s photoelectric equation.

Solution

(a) Calculate the threshold frequency. 19

150 34

(4.5 eV)(1.602 10 J eV) 1.1 10 Hz6.626 10 J s

fhφ −

−×= = = ×

× ⋅

(b) The maximum kinetic energy is max s.K eV= Find the stopping potential.

max s s, so 4.8 V 4.5 V 0.3 V .hfK eV hf Ve e

φφ= = − = − = − =

(c)

Classical theory predicts that electrons can absorb whatever EM radiation available, regardless of thefrequency.


1026

83. Strategy The maximum possible kinetic energy is equal to the difference of the initial and final photon energies. Use the Compton shift, Eq. (27-14).

Solution Find the wavelength of the scattered photon.

e(1 cos ) (2.426 pm)(1 cos )h

m cλ λ λ θ θ′∆ = − = − = −

λ∆ is a maximum for 180 ,θ = ° so (2.426 pm)(2) 6.00 pm 4.852 pm 10.85 pm.λ λ′ = + = + = Compute the maximum possible kinetic energy.

max1 1(1240 eV nm) 92.4 keV

0.00600 nm 0.01085 nmhc hcK E Eλ λ

⎛ ⎞′= − = − = ⋅ − =⎜ ⎟′ ⎝ ⎠

84. (a) Strategy Substitute max sK eV= into Einstein’s photoelectric equation and solve for s .V

Solution Calculate the stopping potential.

max s s 31240 eV nm 4.5 eV, so 1.7 V .

1 0.20 10 nmhc hcK eV hf V

e e eeφφ φ

λ λ⋅= = − = − = − = − =

× ×

(b) Strategy Compute the energy of a photon and compare it to the work function.

Solution The energy of a photon with λ = 400 nm is 1240 eV nm 3.1 eV.

400 nmhcE hfλ

⋅= = = =

This energy is less than the work function for tungsten, 4.5 eV,φ = so the current is zero.

85. Strategy The minimum wavelength corresponds to the maximum frequency, which occurs when the photon’s energy equals the kinetic energy of the electron. Use Eq. (27-9).

Solution Find the minimum wavelength.

max e min 3min e

1240 eV nm, so 0.62 nm .2.0 10 eV

hc hchf KK

λλ

⋅= = = = =×

86. Strategy In general, for the hydrogen atom, the wavelengths are given by

2 2 2f i i

7 11 1 1

1 1

(1.097 10 m ) 1n n n

Rλ

−= =

⎛ ⎞ ⎛ ⎞− × −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

for transitions to the ground state f( 1).n =

Solution A larger in results in a smaller ,λ so i 2, 3, and 4n = give the longest wavelengths: 121.5 nm, 102.6 nm, and 97.23 nm, respectively.

87. Strategy The largest possible energy change occurs when the scattered photon has the largest possible wavelength. This happens when the Compton shift has its maximum value, (2.426 pm)(1 cos180 )λ∆ = − ° = 4.852 pm.

Solution The wavelength of the scattered photon is

.hcE

λ λ λ λ′ = + ∆ = + ∆

The energy of the scattered photon is 11 .hc

E E hc

hc hcE λλ λ ∆′ = = =′ + ∆ +

The fractional change is 3 3(4.000 10 eV)(4.852 10 nm)1240 eV nm

1 11 1 1 0.01541 .1 1

Ehc

E E EE E λ −∆ × ×

⋅

′ ′− = − = − = − =+ +


1027

88. Strategy For +He , the energy of the photon emitted when the electron goes from state in to state f 2n = (first

excited state) is 2i 2 1 2 2 2 2

i i i

1 1 1 1 1 14( 13.6 eV) (54.4 eV) .4 42

hcE E E Z En n nλ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − = − = − − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

So, the wavelengths are given by 2

2 ii

124054.4

1 11 1 44

nm1240 eV nm .(54.4 eV) nn

λ ⋅= =⎛ ⎞ −−⎜ ⎟⎝ ⎠

Solution A larger in results in a smaller ,λ so i 3, 4, and 5n = give the longest wavelengths: 164 nm, 122 nm, and 109 nm, respectively.

89. (a) Strategy The Balmer series f( 2)n = gives visible wavelengths. The wavelengths are given by

2 2i

7 1 1 12

1 .(1.097 10 m )

n

λ−

=⎛ ⎞× −⎜ ⎟⎝ ⎠

Solution i 3n = gives 656.3 nm and i 4n = gives 486.2 nm, both of which are visible. So, the incident radiation excites the ground-state atoms into the n = 4 state. The energy difference between these states is equal to the energy of the incident radiation.

4 1 1 21 11 ( 13.6 eV) 1 12.75 eV

164E E E E ⎛ ⎞ ⎛ ⎞∆ = − = − = − − =⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Calculate the wavelength. 1240 eV nm 97.3 nm

12.75 eVhcE

λ ⋅= = =

(b) Strategy Since photon1 ,Eλ ∝ the longest wavelength corresponds to the smallest energy for the incident

radiation. So, the incident radiation must excite the ground-state atom to its n = 3 state for visible light to be emitted. Solution Calculate the wavelength of the incident radiation.

( ) ( )27 1 11

93

1 1 102.6 nm(1.097 10 m ) 11R

λ −= = =× −−

As found in part (a), the wavelength of the emitted radiation for the n = 3 to the n = 2 state is 656.3 nm.

(c) Strategy The incident photon must be energetic enough to excite an electron from any state to f .n = ∞ The case where the electron is in the ground state, i 1,n = represents the minimum energy required, 13.6 eV. Solution Find the range of wavelengths.

13.6 eV

13.6 eV

1240 eV nm13.6 eV

Ehcλ

λ

≥

≥

⋅≤

91.2 nmλ ≤

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Chapter27 Solution

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