ch27 giancoli7e manual

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27-1

Responses to Questions

1. The lightbulb will not produce light as white as the Sun, since the peak of the lightbulb’s emitted light

is in the infrared. The lightbulb will appear more yellowish than the Sun. The Sun has a spectrum that

peaks in the visible range.

2. The difficulty with seeing objects in the dark is that although all objects emit radiation, only a small

portion of the electromagnetic spectrum can be detected by our eyes. Usually objects are so cool that

they only give off very long wavelengths of light (infrared), which our eyes are unable to detect.

3. Bluish stars are the hottest, whitish-yellow stars are hot, and reddish stars are the coolest. This follows

from Wien’s law, which says that stars with the shortest wavelength peak in their spectrum have the

highest temperatures. Blue is the shortest visible wavelength and red is the longest visible wavelength,

so blue is the hottest and red is the coolest.

4. The red bulb used in black-and-white film darkrooms is a very “cool” filament. Thus it does not emit

much radiation in the range of visible wavelengths (and the small amount that it does emit in the

visible region is not very intense), which means that it will not develop the black-and-white film but

will still allow a person to see what’s going on. A red bulb will not work very well in a darkroom that

is used for color film. It will expose and develop the film during the process, especially at the red end

of the spectrum, ruining the film.

5. If the threshold wavelength increases for the second metal, then the second metal has a smaller work

function than the first metal. Longer wavelength corresponds to lower energy. It will take less energy

for the electron to escape the surface of the second metal.

6. According to the wave theory, light of any frequency can cause electrons to be ejected as long as thelight is intense enough. A higher intensity corresponds to a greater electric field magnitude and more

energy. Therefore, there should be no frequency below which the photoelectric effect does not occur.

According to the particle theory, however, each photon carries an amount of energy which depends

upon its frequency. Increasing the intensity of the light increases the number of photons but does not

increase the energy of the individual photons. The cutoff frequency is that frequency at which the

energy of the photon equals the work function. If the frequency of the incoming light is below the

cutoff, then the electrons will not be ejected because no individual photon has enough energy to eject

an electron.

EARLY QUANTUM THEORY AND MODELS OF THE ATOM 27


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27-2 Chapter 27



7. Individual photons of ultraviolet light are more energetic than photons of visible light and will deliver

more energy to the skin, causing burns. UV photons also can penetrate farther into the skin and, once

at the deeper level, can deposit a large amount of energy that can cause damage to cells.

8. Cesium will give a higher maximum kinetic energy for the ejected electrons. Since the incident photons bring in a given amount of energy, and in cesium less of this energy goes to releasing the

electron from the material (the work function), it will give off electrons with a higher kinetic energy.

9. (a) A light source, such as a laser (especially an invisible one), could be directed on a photocell in

such a way that as a burglar opened a door or passed through a window, the beam would be

blocked from reaching the photocell. A burglar alarm could then be triggered to sound when the

current in the ammeter went to 0.

(b) A light source could be focused on a photocell in a smoke detector. As the density of smoke

particles in the air becomes thicker and thicker, more and more of the light attempting to reach

the photocell would be scattered away from the photocell. At some set minimum level the alarm

could be triggered to sound.

(c) The amount of current in the circuit with the photocell depends on the intensity of the light, aslong as the frequency of the light is above the threshold frequency of the photocell material.

The ammeter in the light meter’s circuit could be calibrated to reflect the light intensity.

10. (a) No. The energy of a beam of photons depends not only on the energy of each individual photon

but also on the total number of photons in the beam. It is possible that there could be many more

photons in the IR beam than in the UV beam. In this instance, even though each UV photon has

more energy than each IR photon, the IR beam could have more total energy than the UV beam.

(b) Yes. A photon’s energy depends on its frequency: . E hf = Since infrared light has a lower

frequency than ultraviolet light, a single IR photon will always have less energy than a single UV

photon.

11. No, fewer electrons are emitted, but each one is emitted with higher kinetic energy, when the 400-nm

light strikes the metal surface. The intensity (energy per unit time) of both light beams is the same,

but the 400-nm photons each have more energy than the 450-nm photons. Thus there are fewer

photons hitting the surface per unit time. This means that fewer electrons will be ejected per unit time

from the surface with the 400-nm light. The maximum kinetic energy of the electrons leaving the metal

surface will be greater, though, since the incoming photons have shorter wavelengths and more energy

per photon, and it still takes the same amount of energy (the work function) to remove each electron.

This “extra” energy goes into higher kinetic energy of the ejected electrons.

12. Yes, an X-ray photon that scatters from an electron does have its wavelength changed. The photon

gives some of its energy to the electron during the collision and the electron recoils slightly. Thus, the

photon has less energy and its wavelength is longer after the collision, since the energy and wavelength

are inversely proportional to each other ( / ). E hf hc λ = =

13. In the photoelectric effect, the photons (typically visible frequencies) have only a few eV of energy,

whereas in the Compton effect, the energy of the photons (typically X-ray frequencies) is more than

1000 times greater and their wavelength is correspondingly smaller. In the photoelectric effect, the

incident photons eject electrons completely out of the target material, while the photons are completely

absorbed (no scattered photons). In the Compton effect, the photons are not absorbed but are scattered

from the electrons.


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Early Quantum Theory and Models of the Atom 27-3



14. Light demonstrates characteristics of both waves and particles. Diffraction, interference, and

polarization are wave characteristics and are demonstrated, for example, in Young’s double-slit

experiment. The photoelectric effect and Compton scattering are examples of experiments in which

light demonstrates particle characteristics. We can’t say that light IS a wave or a particle, but it has

properties of each.

15. We say that electrons have wave properties since we see them act like waves when they are diffracted

or exhibit two-slit interference. We say that electrons have particle properties since we see them act

like particles when they are bent by magnetic fields or accelerated and fired into materials where they

scatter other electrons.

16. Both a photon and an electron have properties of waves and properties of particles. They can both be

associated with a wavelength and they can both undergo scattering. An electron has a negative charge

and has mass, obeys the Pauli exclusion principle, and travels at less than the speed of light. A photon is

not charged, has no mass, does not obey the Pauli exclusion principle, and travels at the speed of light.

Property Photon Electron

Mass None 319.11 10 kg−×

Charge None 191.60 10 C−− ×

Speed 83 10 m/s× 83 10 m/s< ×

There are other properties, such as spin, that will be discussed in later chapters.

17. The proton will have the shorter wavelength, since it has a larger mass than the electron and therefore a

larger momentum for the same speed ( / ).h pλ =

18. The particles will have the same kinetic energy. But since the proton has a larger mass, it will have a

larger momentum than the electron 2(KE /2 ). p m= Wavelength is inversely proportional to momentum,

so the proton will have the shorter wavelength.

19. In Rutherford’s planetary model of the atom, the Coulomb force (electrostatic force) keeps the

electrons from flying off into space. Since the protons in the center are positively charged, the

negatively charged electrons are attracted to the center by the Coulomb force and orbit the center just

like the planets orbiting a sun in a solar system due to the attractive gravitational force.

20. At room temperature, nearly all the atoms in hydrogen gas will be in the ground state. When light

passes through the gas, photons are absorbed, causing electrons to make transitions to higher states and

creating absorption lines. These lines correspond to the Lyman series since that is the series of

transitions involving the ground state or 1n = level. Since there are virtually no atoms in higher

energy states, photons corresponding to transitions from 2n ≥ to higher states will not be absorbed.

21. To determine whether there is oxygen near the surface of the Sun, you need to collect light coming

from the Sun and spread it out using a diffraction grating or prism so you can see the spectrum of

wavelengths. If there is oxygen near the surface, then there will be dark (absorption) lines at the

wavelengths corresponding to electron transitions in oxygen.


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27-4 Chapter 27



22. (a) The Bohr model successfully explains why atoms emit line spectra; it predicts the wavelengths

of emitted light for hydrogen; it explains absorption spectra; it ensures the stability of atoms (by

decree); and it predicts the ionization energy of hydrogen.

(b) The Bohr model did not give a reason for orbit quantization; it was not successful for multi-

electron atoms; it could not explain why some emission lines are brighter than others; and itcould not explain the “fine structure” of some very closely spaced spectral lines.

23. The two main difficulties of the Rutherford model of the atom were that (1) it predicted that light of a

continuous range of frequencies should be emitted by atoms and (2) it predicted that atoms would be

unstable.

24. It is possible for the de Broglie wavelength ( / )h pλ = of a particle to be bigger than the dimension of

the particle. If the particle has a very small mass and a slow speed (like a low-energy electron or

proton), then the wavelength may be larger than the dimension of the particle. It is also possible for the

de Broglie wavelength of a particle to be smaller than the dimension of the particle if it has a large

momentum and a moderate speed (like a baseball). There is no direct connection between the size of a

particle and the size of the de Broglie wavelength of a particle. For example, you could also make the

wavelength of a proton much smaller than the size of the proton by making it go very fast.

25. Even though hydrogen only has one electron, it still has an infinite number of energy states for that one

electron to occupy, and each line in the spectrum represents a transition between two of those possible

energy levels. So there are many possible spectral lines. And seeing many lines simultaneously would

mean that there would have to be many hydrogen atoms undergoing energy level transitions—a sample

of gas containing many H atoms, for example.

26. The closely spaced energy levels in Fig. 27–29 correspond to the different transitions of electrons from

one energy state to another—specifically to those that start from closely packed high energy levels,

perhaps with 10n = or even higher. When these transitions occur, they emit radiation (photons) that

creates the closely spaced spectral lines shown in Fig. 27–24.

27. On average, the electrons of helium are closer to the nucleus than are the electrons of hydrogen. The nucleus

of helium contains two protons (positive charges) so attracts each electron more strongly than the single

proton in the nucleus of hydrogen. (There is some shielding of the nuclear charge by the “other” electron,

but each electron still feels an average attractive force of more than one proton’s worth of charge.)

28. The Balmer series spectral lines are in the visible light range and could be seen by early experimenters

without special detection equipment. It was only later that the UV (Lyman) and IR (Paschen) regions

were explored thoroughly, using detectors other than human sight.

29. When a photon is emitted by a hydrogen atom as the electron makes a transition from one energy state

to a lower one, not only does the photon carry away energy and momentum, but to conserve

momentum, the atom must also take away some momentum. If the atom carries away some

momentum, then it must also carry away some of the available energy, which means that the photontakes away less energy than Eq. 27–10 predicts.

30. (a) continuous

(b) line, emission

(c) continuous

(d ) line, absorption

(e) continuous with absorption lines (like the Sun)


5/28




31. No. At room temperature, virtually all the atoms in a sample of hydrogen gas will be in the ground

state. Thus, the absorption spectrum will contain primarily the Lyman lines, as photons corresponding

to transitions from the 1n = level to higher levels are absorbed. Hydrogen at very high temperatures

will have atoms in excited states. The electrons in the higher energy levels will fall to all lower energy

levels, not just the 1n = level. Therefore, emission lines corresponding to transitions to levels higher

than 1n = will be present as well as the Lyman lines. In general, you would expect to see only Lyman

lines in the absorption spectrum of room temperature hydrogen, but you would find Lyman, Balmer,

Paschen, and other lines in the emission spectrum of high-temperature hydrogen.

Responses to MisConceptual Questions

1. (b) A common misconception is that the maximum wavelength increases as the temperature

increases. However, the temperature and maximum wavelength are inversely proportional.

As the temperature increases, the intensity increases and the wavelength decreases.

2. (a) A higher work function requires more energy per photon to release the electrons. Blue light has

the shortest wavelength and therefore the largest energy per photon.

3. (b) The blue light has a shorter wavelength and therefore more energy per photon. Since the beams

have the same intensity (energy per unit time per unit area), the red light will have more photons.

4. (d ) A common misconception is that violet light has more energy than red light. However, the

energy is the product of the energy per photon and the number of photons. A single photon of

violet light has more energy than a single photon of red light, but if a beam of red light has more

photons than the beam of violet light, then the red light could have more energy.

5. (d ) The energy of the photon E is equal to the sum of the work function of the metal and the kinetic

energy of the released photons. If the energy of the photon is more than double the work

function, then cutting the photon energy in half will still allow electrons to be emitted. However,

if the work function is more than half of the photon energy, then no electrons would be emitted ifthe photon energy were cut in half.

6. (b) The momentum of a photon is inversely proportional to its wavelength. Therefore, doubling the

momentum would cut the wavelength in half.

7. (d ) A commons misconception is that only light behaves as both a particle and a wave. De Broglie

postulated, and experiments have confirmed, that in addition to light, electrons and protons (and

many other particles) have both wave and particle properties that can be observed or measured.

8. (d ) A thrown baseball has a momentum on the order of 1kg m/s.⋅ The wavelength of the baseball is

Planck’s constant divided by the momentum. Due to the very small size of Planck’s constant, the

wavelength of the baseball would be much smaller than the size of a nucleus.

9. (d ) This Chapter demonstrates the wave-particle duality of light and matter. Both electrons and

photons have momentum that is related to their wavelength by / . p h λ = Young’s double-slit

experiment demonstrated diffraction with light, and later experiments demonstrated electron

diffraction. Therefore, all three statements are correct.

10. (a, c, d ) Alpha particles are positive. To produce large angle scattering, the nucleus of the atom would

have to repel the alpha particle. Thus the nucleus must be positive. If the nuclear charge was

spread out over a large area, then there would be more small-angle scattering but insufficient


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27-6 Chapter 27



force to create large-angle scattering. With a small nucleus, large scattering forces are possible.

Since most of the alpha particles pass through the foil undeflected, most of the atom must be

empty space. The scattering does not require quantized charge, but it would be possible with a

continuous charge distribution. Therefore, answer (b) is incorrect.

11. (d ) A photon is emitted when the electron transitions from a higher state to a lower state. Photons are

not emitted when an electron transitions from 2 5→ or from 5 8→ . The other two transitions

are to states that are three states below the original state. The energy levels change more rapidly

for lower values of n, so the 5 2→ transition will have a higher energy and therefore a shorter

wavelength than the 8 5→ transition.

12. (b) The lowest energy cannot be zero, if zero energy has been defined as when the electron and

proton are infinitely far away. As the electron and proton approach each other, their potential

energy decreases. The energy levels are quantized and therefore cannot be any value. As shown

in the text after Eq. 27–15b, the lowest energy level of the hydrogen atom is –13.6 eV.

13. (d ) The current model of the atom is the quantum mechanical model. The plum-pudding model was

rejected, as it did not explain Rutherford scattering. The Rutherford atom was rejected, as it didnot explain the spectral lines emitted from atoms. The Bohr atom did not explain fine structure.

Each of these phenomena are explained by the quantum mechanical model.

14. (a) Light is a massless particle. Even though it has no mass, it transports energy and therefore has

kinetic energy and momentum. Light is also a wave with frequency and wavelength.

Solutions to Problems

In several problems, the value of hc is needed. We often use the result of Problem 29, 1240 eV nm.hc = ⋅

1. This scenario fits the experiment that results in Eq. 27–1.

4

2 3 2

(640 V/m)6.2 10 C/kg

(14 10 m)(0.86 T )

e E

m rB −= = = ×

×

2. (a) The velocity relationship is given right before Eq. 27–1 in the text.

46

3

(1.88 10 V/m)7.23 10 m/s

(2.60 10 T)

E

Bυ

−

×= = = ×

×

(b) For the radius of the path in the magnetic field, use an expression derived in Example 20–6.

31 62

19 3

(9.11 10 kg)(7.23 10 m/s)1.58 10 m 1.58 cm

(1.60 10 C)(2.90 10 T)

mr

qB

υ − −− −

× ×= = = × =

× ×

3. The force from the electric field must balance the weight, and the electric field is the potential

difference divided by the plate separation.

2 15 2

19

(1.0 10 m)(2.8 10 kg)(9.80 m/s )5.04 5 electrons

(1.60 10 C)(340 V)

V dmg qE ne mg n

d eV

− −

−→

× ×= = = = = ≈

×


7/28




4. Use Wien’s law, Eq. 27–2, to find the temperature.

3 3

9P

(2.90 10 m K) (2.90 10 m K)5577 K 5600 K

(520 10 m)T

λ

− −

−

× ⋅ × ⋅= = = ≈

×

5. Use Wien’s law, Eq. 27–2.

(a)3 3

5P

(2.90 10 m K) (2.90 10 m K)1.06 10 m 10.6 m, far infrared

(273 K)T λ µ

− −−× ⋅ × ⋅= = = × =

(b)3 3

7P

(2.90 10 m K) (2.90 10 m K)9.35410 m 940 nm, near infrared

(3100 K)T λ

− −−× ⋅ × ⋅= = = ≈

(c)3 3

4P

(2.90 10 m K) (2.90 10 m K)7.25 10 m 0.7 mm, microwave

(4K)T λ

− −−× ⋅ × ⋅= = = × ≈

6. Use Wein’s law, Eq. 27–2.

(a)3 3 5

9P

(2.90 10 m K) (2.90 10 m K) 1.61 10 K (18.0 10 m)

T λ

− −

−× ⋅ × ⋅= = = ×

×

(b)3 3

6P

(2.90 10 m.K) (2.90 10 m K)1.318 10 m 1.3 m

(2200 K)T λ µ

− −−× × ⋅= = = × ≈

7. Because the energy is quantized according to Eq. 27–3, the difference in energy between adjacent

levels is simply . E hf ∆ =

34 13 20 20

20

19

(6.63 10 J s)(8.1 10 Hz) 5.37 10 J 5.4 10 J

1 eV5.37 10 J 0.34 eV

1.60 10 J

E hf − − −

−

−

∆ = = × ⋅ × = × ≈ ×

⎛ ⎞× =⎜ ⎟⎜ ⎟

×⎝ ⎠

8. The potential energy is “quantized” in units of mgh.

(a) 21PE (62.0 kg)(9.80 m/s )(0.200 m) 121.52 J 122 Jmgh= = = ≈

(b) 2 1PE PE2 2 2(121.52 J) 243 Jmg h= = = =

(c) 3 1PE PE3 3 3(121.52 J) 365 Jmg h= = = =

(d ) 1PE PE (121.52 J) (122 ) Jn mgnh n n n= = = =

(e) 2 6PE PE (2 6)(121.52 J) 486 J E ∆ = − = − = −

9. Use Eq. 27–2 with a temperature of 98 F 37 C (273 37)K 310 K (3significant figures).° = ° = + =

3 36

P

(2.90 10 m K) (2.90 10 m K)9.35 10 m 9.35 m

(310 K)T λ µ

− −−× ⋅ × ⋅= = = × =

10. We use Eq. 27–4 to find the energy of the photons.

34 6 26(6.626 10 J s)(91.7 10 Hz) 6.08 10 J E hf

− −= = × ⋅ × = ×


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27-8 Chapter 27



11. We use Eq. 27–4 along with the fact that / f c λ = for light. The longest wavelength will have the

lowest energy.

34 819

1 1 9 19

1

(6.63 10 J s)(3.00 10 m/s) 1 eV4.97 10 J 3.11 eV

(400 10 m) 1.60 10 J

hc E hf

λ

−−

− −

⎛ ⎞× ⋅ ×= = = = × =⎜ ⎟⎜ ⎟

× ×⎝ ⎠

34 819

2 2 9 192

(6.63 10 J s)(3.00 10 m/s) 1 eV2.65 10 J 1.66 eV

(750 10 m) 1.60 10 J

hc E hf

λ

−−

− −

⎛ ⎞× ⋅ ×= = = = × =⎜ ⎟⎜ ⎟× ×⎝ ⎠

Thus the range of energies is 19 192.7 10 J 5.0 10 J , E − −× < < × or 1.7 eV 3.1 eV . E < <

12. Use Eq. 27–4 with the fact that / f c λ = for light.

34 812 3

19 3

(6.63 10 J s)(3.00 10 m/s)3.88 10 m 3.9 10 nm

(1.60 10 J/eV)(320 10 eV)

c hc

f E λ

−− −

−

× ⋅ ×= = = = × ≈ ×

× ×

Significant diffraction occurs when the opening is on the order of the wavelength. Thus there would beinsignificant diffraction through the doorway.

13. Use Eq. 27–6.

3427

7

(6.63 10 J s)1.14 10 kg m/s

(5.80 10 m)

h p

λ

−−

−

× ⋅= = = × ⋅

×

14. The momentum of the photon is found from Eq. 27–6.

3423

9

(6.63 10 J s)4.7 10 kg m/s

(0.014 10 m)

h p

λ

−−

−

× ⋅= = = × ⋅

×

15. Particle Theory Wave Theory

(1) If the light intensity is increased, the

number of electrons ejected should increase.

(1) If the light intensity is increased, the number

of electrons ejected should increase.


maximum kinetic energy of the electrons ejected

should not increase.


maximum kinetic energy of the electrons ejected

should increase.

(3) If the frequency of the light is increased, the

maximum kinetic energy of the electrons should

increase.

(3) If the frequency of the light is increased, the

maximum kinetic energy of the electrons should

not be affected.

(4) There is a “cutoff” frequency, below whichno electrons will be ejected, no matter how

intense the light.

(4) There should be no lower limit to thefrequency—electrons will be ejected for all

frequencies.

16. We use Eq. 27–4 with the fact that / f c λ = for light.

1913 13min

min min min 34

(0.1eV)(1.60 10 J/eV)2.41 10 Hz 2 10 Hz

(6.63 10 J s)

E E hf f

h

−

−

×= → = = = × ≈ ×

× ⋅


9/28




85 5

max 13min

(3.00 10 m/s)1.24 10 m 1 10 m

(2.41 10 Hz)

c

f λ − −

×= = = × ≈ ×

×

17. At the minimum frequency, the kinetic energy of the ejected electrons is 0. Use Eq. 27–5a.

19140

min 0 min 34KE

4.8 10 J0 7.2 10 Hz

6.63 10 J s

W hf W f

h

−

−

×= − = → = = = ×

× ⋅

18. We divide the minimum energy by the photon energy at 550 nm to find the number of photons.

18 9min min

min 34 8

(10 J)(550 10 m)2.77 3 photons

(6.63 10 J s)(3.00 10 m/s)

E E E nhf E n

hf hc

λ − −×= = → = = = = ≈

× ⋅ ×2

19. The longest wavelength corresponds to the minimum frequency. That occurs when the kinetic energy

of the ejected electrons is 0. Use Eq. 27–5a.

0min 0 min

max

8 347

max 190

KE 0

(3.00 10 m/s)(6.63 10 J s)4.29 10 m 429 nm

(2.90 eV)(1.60 10 J/eV)

W chf W f h

ch

W

λ

λ −

−

−

= − = → = = →

× × ⋅= = = × =

×

20. The photon of visible light with the maximum energy has the least wavelength. We use 400 nm as the

lowest wavelength of visible light and calculate the energy for that wavelength.

34 8

max 19 9min

(6.63 10 J s)(3.00 10 m/s)3.11 eV

(1.60 10 J/eV)(400 10 m)

hchf

λ

−

− −

× ⋅ ×= = =

× ×

Electrons will not be emitted if this energy is less than the work function.

The metals with work functions greater than 3.11 eV are copper and iron.

21. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so the work

function is equal to the energy of the photon.

0 maxKE1240 eV nm

2.255 eV 2.3 eV550 nm

hcW hf hf

λ

⋅= − = = = = ≈

(b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum

kinetic energy. We use Eq. 27–5b to calculate the maximum kinetic energy.

max 0 0

max0

KE

KE

1240 eV nm2.255 eV 0.845 eV

400 nm

0.845 eV 0.85 V

hchf W W

V e e

λ

⋅= − = − = − =

= = ≈

22. The photon of visible light with the maximum energy has the minimum wavelength. We use Eq. 27–5b

to calculate the maximum kinetic energy.

max 0 0KE1240 eV nm

2 48 eV 0 62 eV400 nm

hchf W W

λ

⋅= − = − = − . = .


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27-10 Chapter 27



23. We use Eq. 27–5b to calculate the maximum kinetic energy. Since the kinetic energy is much less than

the rest energy, we use the classical definition of kinetic energy to calculate the speed.

max 0 0KE1240 eV nm

2 48 eV 0.92 eV365 nm

hchf W W

λ

⋅= − = − = − . =

192 5max1

max 2 31

KEKE

2 2(0 92 eV)(1.60 10 J/eV)5.7 10 m/s

9.11 10 kgm

mυ υ

−

−

. ×= → = = = ×

×

This speed is only about 0.2% of the speed of light, so the classical kinetic energy formula is correct.

24. We use Eq. 27–5b to calculate the work function.

0 max maxKE KE1240 eV nm

1 40 eV 3 46 eV255 nm

hcW hf

λ

⋅= − = − = − . = .

25. Electrons emitted from photons at the threshold wavelength have no kinetic energy. Use Eq. 27–5b

with the threshold wavelength to determine the work function.

0 maxmax

KE1240 eV nm

3 647 eV340 nm

hc hcW

λ λ

⋅= − = = = .

(a) Now use Eq. 27–5b again with the work function determined above to calculate the kinetic

energy of the photoelectrons emitted by 280-nm light.

max 0KE1240 eV nm

3.647 eV 0 78 eV280 nm

hcW

λ

⋅= − = − ≈ .

(b) Because the wavelength is greater than the threshold wavelength, the photon energy is less than

the work function, so there will be no ejected electrons.

26. The energy required for the chemical reaction is provided by the photon. We use Eq. 27–4 for the

energy of the photon, where / . f c λ =

1240 eV nm1.968 eV 2.0 eV

630 nm

hc E hf

λ

⋅= = = = ≈

Each reaction takes place in a molecule, so we use the appropriate conversions to convert eV/molecule

to kcal/mol.

19 231.968 eV 1.60 10 J 6.02 10 molecules kcal45 kcal/mol

molecule eV mol 4186 J E

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

27. The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic

energy of the photoelectrons. We use Eq. 27–5b to calculate the work function, where the maximum

kinetic energy is the product of the stopping voltage and electron charge.

0 max 0KE1240 eV nm

(1.64 V) 3.32 eV250 nm

hc hcW eV e

λ λ

⋅= − = − = − =


11/28




28. We plot the maximum (kinetic) energy

of the emitted electrons vs. the

frequency of the incident radiation.

Eq. 27–5b says max 0KE hf W = − .

The best-fit straight line is determined

by linear regression in Excel. The

slope of the best-fit straight line to the

data should give Planck’s constant,

the x-intercept is the cutoff frequency,

and the y-intercept is the opposite of

the work function.

(a) 14 19 34(0.4157 eV/10 Hz)(1.60 10 J/eV) 6.7 10 J sh − −= × = × ⋅

(b)140

cutoff 0 cutoff 14

2.3042 eV5.5 10 Hz

(0.4157 eV/10 Hz)

W hf W f

h= → = = = ×

(c) 0 2 3 e VW = .

29. Since / , f c λ = the energy of each emitted photon is / . E hc λ = We insert the values for h and c and

convert the resulting units to eV nm⋅ .

34 8 19

9

(6.626 10 J s)(2.998 10 m/s) (1 eV/1 602 10 J) 1240 eV nm

(in nm)(10 m/1 nm)

hc E

λ λ λ

− −

−

× ⋅ × . × ⋅= = =

30. Use Eq. 27–7.

e e

1 e

31 8 131

34

(1 cos ) (1 cos )

cos 1

(9.11 10 kg)(3 00 10 m/s)(1.7 10 m)cos 1 21 58 22

(6 63 10 J s)

h h

m c m c

m c

h

λ λ φ λ φ

λ φ −

− −−

−

′ = + − → ∆ = − →

∆⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎛ ⎞× . × ×= − = . ° ≈ °⎜ ⎟⎜ ⎟. × ⋅⎝ ⎠

31. The Compton wavelength for a particle of mass m is /h mc.

(a)34

12

31 8e

(6.63 10 J s)2.43 10 m

(9.11 10 kg)(3.00 10 m/s)

h

m c

−−

−

× ⋅= = ×

× ×

(b)34

15

27 8 p

(6.63 10 J s)1.32 10 m

(1.67 10 kg)(3.00 10 m/s)

h

m c

−−

−

× ⋅= = ×

× ×

(c) The energy of the photon is given by Eq. 27–4.

2 photon rest energy

( / )

hc hc E hf mc

h mcλ = = = = =

E = 0.4157 f - 2.3042

R 2 = 0.9999

0.0

0.5

1.0

1.5

2.0

2.5

3.0

6.0 7.0 8.0 9.0 10.0 11.0 12.0

Frequency (1014

Hz)

E n e r g y

( e V )


12/28

27-12 Chapter 27



32. We find the Compton wavelength shift for a photon scattered from an electron, using Eq. 27–7.

The Compton wavelength of a free electron is calculated in Problem 31 above.

3C

e

(1 cos ) (1 cos ) (2.43 10 nm)(1 cos )h

m c

λ λ θ λ θ θ −⎛ ⎞

′ − = − = − = × −⎜ ⎟

⎝ ⎠

(a) 3 4(2.43 10 nm)(1 cos 45 ) 7.12 10 nmaλ λ − −′ − = × − ° = ×

(b) 3 3(2.43 10 nm)(1 cos90 ) 2.43 10 nmbλ λ − −′ − = × − ° = ×

(c) 3 3(2.43 10 nm)(1 cos180 ) 4.86 10 nmcλ λ − −′ − = × − ° = ×

33. The photon energy must be equal to the kinetic energy of the products plus the mass energy of the

products. The mass of the positron is equal to the mass of the electron.

2 photon products products

2 2 products photon products photon electron

KE

KE 2 3.64 MeV 2(0.511 MeV) 2.62 MeV

E m c

E m c E m c

= + →

= − = − = − =

34. The photon with the longest wavelength has the minimum energy in order to create the masses with no

additional kinetic energy. Use Eq. 27–6.

3416

max 2 27 8min

(6.63 10 J s)6.62 10 m

22 2(1.67 10 kg)(3.00 10 m/s)

hc hc h

E mcmcλ

−−

−

× ⋅= = = = = ×

× ×

This must take place in the presence of some other object in order for momentum to be conserved.

35. The minimum energy necessary is equal to the rest energy of the two muons.

2min 2 2(207)(0.511 MeV) 212 MeV E mc= = =

The wavelength is given by Eq. 27–6.

34 815

19 6

(6.63 10 J s)(3.00 10 m/s)5.86 10 m

(1.60 10 J/eV)(212 10 eV)

hc

E λ

−−

−

× ⋅ ×= = = ×

× ×

36. Since 0.001 ,cυ = the total energy of the particles is essentially equal to their rest energy. The particles

have the same rest energy of 0.511 MeV. Since the total momentum is 0, each photon must have half

the available energy and equal momenta.

photon2 photon electron photon0.511 MeV ; 0.511 MeV/

E E m c p c

c= = = =

37. The energy of the photon is equal to the total energy of the two particles produced. The particles have

the same kinetic energy and the same mass.

2 photon KE2( ) 2(0 285 MeV 0 511 MeV) 1.592 MeV E mc= + = . + . =

The wavelength is found from Eq. 27–6.

34 813

19 6

(6.63 10 J s)(3.00 10 m/s)7.81 10 m

(1.60 10 J/eV)(1.592 10 eV)

hc

E λ

−−

−

× ⋅ ×= = = ×

× ×


13/28




38. We find the wavelength from Eq. 27–8.

3432(6.63 10 J s) 3.2 10 m

(0 21kg)(0.10 m/s)

h h

p mλ

υ

−−× ⋅= = = = ×

.

39. The neutron is not relativistic, so use . p mυ = Also use Eq. 27–8.

3412

27 4

(6.63 10 J s)4.7 10 m

(1.67 10 kg)(8.5 10 m/s)

h h

p mλ

υ

−−

−

× ⋅= = = = ×

× ×

40. We assume that the electron is nonrelativistic, and check that with the final answer. We use Eq. 27–8.

346

31 9

(6.63 10 J s)2.695 10 m/s 0 0089

(9.11 10 kg)(0.27 10 m)

h h hc

p m mλ υ

υ λ

−

− −

× ⋅= = → = = = × = .

× ×

The use of classical expressions is justified. The kinetic energy is equal to the potential energy change.

31 6 21

2 212 19

KE (9.11 10 kg)(2.695 10 m/s) 20.7 eV(1.60 10 J/eV)

eV mυ

−

−× ×= = = =

×

Thus the required potential difference is 21 V.

41. Since the particles are not relativistic, we may use2

KE /2 . p m= We then form the ratio of the kinetic

energies, using Eq. 27–8.

2 22 2 27 pe e

2 2 2 31 p e p

KEKE

KE

/2 1.67 10 kg; 1840

2 2 /2 9.11 10 kg

mh m p h

m mm h m

λ

λ λ

−

−

×= = = = = =

×

42. (a) We find the momentum from Eq. 27–8.

3424

10

6.63 10 J s1.5 10 kg m/s

4 5 10 m

h p

λ

−−

−

× ⋅= = = × ⋅

. ×

(b) We assume the speed is nonrelativistic.

346

31 10

6.63 10 J s1.6 10 m/s

(9.11 10 kg)(4.5 10 m)

h h hc

p m mλ υ

υ λ

−

− −

× ⋅= = → = = = ×

× ×

(c) We calculate the kinetic energy classically.

2 2 2 3 2 61 1 12 2 2

KE ( )( / ) (0.511 MeV)(5.39 10 ) 7.43 10 MeV 7.43 eVm mc cυ υ − −= = = × = × =

This is the energy gained by an electron if accelerated through a potential difference of 7.4 V.

43. Because all of the energies to be considered are much less than the rest energy of an electron, we can

use nonrelativistic relationships. We use Eq. 27–8 to calculate the wavelength.

2

KE KEKE

2 ( );2 2 ( )

p h h p m

m p mλ = → = = =


14/28

27-14 Chapter 27



(a)34

10 10

31 19KE

6.63 10 J s3.88 10 m 4 10 m

2 ( ) 2(9.11 10 kg)(10 eV)(1.60 10 J/eV)

h

mλ

−− −

− −

× ⋅= = = × ≈ ×

× ×

(b)34

10 10

31 19KE

6.63 10 J s1.23 10 m 1 10 m

2 ( ) 2(9.11 10 kg)(100 eV)(1.60 10 J/eV)

h

m

λ −

− −

− −

× ⋅= = = × ≈ ×

× ×

(c)34

11

31 3 19KE

6.63 10 J s3.9 10 m

2 ( ) 2(9.11 10 kg)(1.0 10 eV)(1.60 10 J/eV)

h

mλ

−−

− −

× ⋅= = = ×

× × ×

44. Since the particles are not relativistic, use 2KE /2 . p m= Form the ratio of the wavelengths, using

Eq. 27–8.

p p e

e p

e

KE

KE

KE

2; 1

2

2

h

m mh h

h p mm

m

λ λ

λ = = = = <

Thus the proton has the shorter wavelength, since e p .m m<

45. The final KE of the electron equals the negative change in its PE as it passes through the potential

difference. Compare this energy to the electron’s rest energy to determine if it is relativistic.

3 3KE (1 e)(35 10 V) 35 10 eVq V = − ∆ = × = ×

Because this is greater than 1% of the electron rest energy, the electron is relativistic. We use Eq. 26–9

to determine the electron momentum and then Eq.27–6 to determine the wavelength.

2 22 2 2 2 2 2 4

3

2 2 3 2 3 3

KE KEKE

KE KE

2( )[ ]

1240 eV nm6.4 10 nm

2( ) (35 10 eV) 2(35 10 eV)(511 10 eV)

mc E mc p c m c p

c

h hc

p mc

λ −

+= + = + ⇒ =

⋅= = = = ×

+ × + × ×

Because 5 cm,λ diffraction effects are negligible.

46. For diffraction, the wavelength must be on the order of the opening. Find the speed from Eq. 27–8.

3438(6.63 10 J s) 3.9 10 m/s

(1400 kg)(12 m)

h h h

p m mλ υ

υ λ

−−× ⋅= = → = = = ×

This is on the order of39

10−

times ordinary highway speeds.

47. We relate the kinetic energy to the momentum with a classical relationship, since the electrons are

nonrelativistic. We also use Eq. 27–8. We then assume that the kinetic energy was acquired by

electrostatic potential energy.

2 2

2

2 34 2

2 31 19 9 2

KE2 2

(6.63 10 J s)22 V

2 2(9.11 10 kg)(1.60 10 C)(0.26 10 m)

p heV

m m

hV

me

λ

λ

−

− − −

= = = →

× ⋅= = =

× × ×


15/28




48. The kinetic energy is 2850 eV. That is small enough compared to the rest energy of the electron for the

electron to be nonrelativistic. We use Eq. 27–8.

34 8

1/2 2 1/2 19 6 1/2

11

KE KE

(6.63 10 J s) (3.00 10 m/s)

(2 ) (2 ) (1.60 10 J/eV)[2(0.511 10 eV)(2850 eV)]

2.30 10 m 23.0 pm

h h hc

p m mcλ

−

−

−

× ⋅ ×= = = =

× ×

= × =

49. The energy of a level is2

(13 6 eV),n E

n

.= − from Eq. 27–15b for 1. Z =

(a) The transition from 1n = to 3n′ = is an absorption, because the final state, 3,n′ = has a higher

energy. The photon energy is the difference between the energies of the two states.

2 2

1 1(13 6 eV) 12.1 eV

3 1n nhf E E ′

⎡ ⎤⎛ ⎞ ⎛ ⎞= − = − . − =⎢ ⎥⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎣ ⎦

(b) The transition from 6n = to 2n′ = is an emission, because the initial state, 2,n′ = has a higherenergy. The photon energy is the difference between the energies of the two states.

2 2

1 1( ) (13.6 eV) 3.0 eV

2 6n nhf E E ′

⎡ ⎤⎛ ⎞⎛ ⎞= − − = − =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

(c) The transition from 4n = to 5n′ = is an absorption, because the final state, 5,n′ = has a higher

energy. The photon energy is the difference between the energies of the two states.

2 2

1 1(13 6 eV) 0.31 eV

5 4n nhf E E ′

⎡ ⎤⎛ ⎞ ⎛ ⎞= − = − . − =⎢ ⎥⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎣ ⎦

The photon for the transition from 1n = to 3n′ = has the largest energy.

50. Ionizing the atom means removing the electron, or raising it to zero energy.

ionization 2 2

( 13.6 eV) (13.6 eV)0 0 1.51 eV

3n E E

n

−= − = − = =

51. From ,hc

E λ

∆ = we see that the second longest wavelength comes from the transition with the second

smallest energy: 5 to 3n n′= = .

52. Doubly ionized lithium is similar to hydrogen, except that there are three positive charges ( 3) Z = in

the nucleus. Use Eq. 27–15b with 3. Z =

2 2

2 2 2

ionization 1 2

(13.6 eV) 3 (13.6 eV) (122 eV)

(122 eV)0 0 122 eV

(1)

n

Z E

n n n

E E

= = = −

⎡ ⎤= − = − − =⎢ ⎥

⎢ ⎥⎣ ⎦

2 2


16/28

27-16 Chapter 27



53. We use Eq. 27–10 and Eq. 27–15b. Note that 1 2, , E E and 3 E are calculated in the textbook. We also

need 4 2(13 6 eV)

0.85 eV4

E − .= = − and 5 2

(13 6 eV)0.54 eV.

5 E

− .= = −

(a) The second Balmer line is the transition from 4 to 2.n n= =

4 2

1240 eV nm486 nm

( ) [ 0 85 eV ( 3.40 eV)]

hc

E E λ

⋅= = =

− − . − −

(b) The second Lyman line is the transition from 3 to 1.n n= =

3 1

1240 eV nm103 nm

( ) [ 1 51 eV ( 13 6 eV)]

hc

E E λ

⋅= = =

− − . − − .

(c) The third Balmer line is the transition from 5 to 2.n n= =

5 2

1240 eV nm434 nm

( ) [ 0 54 eV ( 3.40 eV)]

hc

E E λ

⋅= = =

− − . − −

54. We evaluate the Rydberg constant using Eq. 27–9 and 27–16. We use hydrogen so 1. Z = We also

substitute0

1.

4k

πε =

2 2 4 2

2 2 3 2 2

2 2 4 2 2 2 4

3 2 2 3

0

2 19 4 31

12 2 2 2 34 3 8

1 1 1 2 1 1

( ) ( ) ( ) ( )

2 2

16

(1) (1.602176 10 C) (9.109382 10 kg)

8(8.854188 10 C /N m ) (6.626069 10 J s) (2.997925 10 m/s)

1.09736

Z e mk R

n n h c n n

Z e mk Z e m R

h c h c

π

λ

π π

π ε − −

− −

⎛ ⎞ ⎛ ⎞= − = − →⎜ ⎟ ⎜ ⎟

′ ′⎝ ⎠ ⎝ ⎠

= =

× ×=

× ⋅ × ⋅ ×

=4

7 7 1

43 3

2 4

C kg1 10 1.0974 10 m

CJ s m/s

N m

−⋅× ≈ ×

⋅

55. The longest wavelength corresponds to the minimum energy, which is the ionization energy.

ion

1240 eV nm91 2 nm

13.6 eV

hc

E λ

⋅= = = .

56. The energy of the photon is the sum of the ionization energy of 13.6 eV and the kinetic energy of 11.5 eV.

The wavelength is found from Eq. 27–4.

totaltotal

1240 eV nm 49.4 nm25.1eV

hc hchf E E

λ λ

⋅= = → = = =

57. Singly ionized helium is like hydrogen, except that there are two positive charges ( 2) Z = in the

nucleus. Use Eq. 27–15b.

2 2

2 2 2

(13 6 eV) 2 (13 6 eV) (54 4 eV)n

Z E

n n n

. . .= − = − = −


17/28




We find the energy of the photon from the 5 to 2n n= = transition in singly ionized helium.

5 2 2 2

1 1(54 4 eV) 12.1 eV

6 2 E E E

⎡ ⎤⎛ ⎞ ⎛ ⎞∆ = − = − . − =⎢ ⎥⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎣ ⎦

Because this is the energy difference for the 1 to 3n n= = transition in hydrogen, the photon can beabsorbed by a hydrogen atom which will jump from 1 to 3 .n n= =

58. Singly ionized helium is like hydrogen, except that there are two

positive charges ( 2) Z = in the nucleus. Use Eq. 27–15b.

2 2

2 2 2

1 2 3 4

5 6

(13 6 eV) 2 (13 6 eV) (54.4 eV)

54.5 eV, 13.6 eV, 6.0 eV, 3.4eV,

2.2 eV, 1.5 eV,

n

Z E

n n n

E E E E

E E

. .= = − = −

= − = − = − = −

= − = − …

2

59. Doubly ionized lithium is like hydrogen, except that there are

three positive charges ( 3) Z = in the nucleus. The square of the

product of the positive and negative charges appears in the

energy term for the energy levels. We can use the results for

hydrogen, if we replace 2e with 2: Ze

2 2

2 2 2

1 2 3

4 5 6

(13.6 eV) 3 (13.6 eV) (122.4 eV)

122 eV, 30.6 eV, 13.6 eV,7.65 eV, 4.90 eV, 3.40 eV,

n

Z E

n n n

E E E E E E

= − = − = −

= − = − = −= − = − = − …

60. The potential energy for the ground state is given by the charge

of the electron times the electric potential caused by the proton.

9 2 2 19 2 19

proton 100 1

PE1 (9.00 10 N m /C )(1.60 10 C) (1 eV/1.60 10 J)

( ) ( )4 (0.529 10 m)

27.2 eV

ee V e

r πε

− −

−

× ⋅ × ×= − = − = −

×

= −

The kinetic energy is the total energy minus the potential energy.

1KE PE

13.6 eV ( 27.2 eV) 13.6 eV E = − = − − − = +

61. The angular momentum can be used to find the quantum number for the orbit, and then the energy can

be found from the quantum number. Use Eqs. 27–11 and 27–15b.

34 2

34

2

2

2 2 5.273 10 kg m /s5.000 5

2 (6.626 10 J s)

13.6 eV(13.6 eV) 0.544 eV

25n

h L L n n

h

Z E

n

π π

π

−

−

( × ⋅ )= → = = = ≈

× ⋅

= − = − = −

122−

30.6−

13.6−7.65−


18/28

27-18 Chapter 27



62. The value of n is found from 2 1,nr n r = and then the energy can be found from Eq. 27–15b.

22

1 101

8

2 2

1.00 10 m13,749 13,700

0.529 10 m

(13 6 eV) (13 6 eV)7.19 10 eV

(13,749)

nn

r r n r n

r

E n

−

−

−

×= → = = = ≈

×

. .= − = − = − ×

63. The velocity is found from Eq. 27–11 evaluated for 1.n =

346 3

10 311 e

2

(6.63 10 J s)2.190 10 m/s 7.30 10

2 2 0.529 10 m 9.11 10 kg

n

nhm r

hc

r m

υ π

υ π π

−−

− −

= →

× ⋅= = = × = ×

( × )( × )

Since ,cυ we say yes , nonrelativistic formulas are justified. The relativistic factor is as follows.

262 2 2 2 51 1

2 2 8

2 190 10 m/s1 / 1 ( / ) 1 1 2.66 10 0.99997

3 00 10 m/sc cυ υ −

⎛ ⎞. ×− ≈ − = − = − × ≈⎜ ⎟⎜ ⎟. ×⎝ ⎠

We see that 2 21 /cυ − is essentially 1. Yes, nonrelativistic formulas are justified.

64. The electrostatic potential energy is given by Eq. 17–2a. Note that the charge is negative. The kinetic

energy is given by the total energy, Eq. 27–15a, minus the potential energy. The Bohr radius is given

by Eq. 27–12.

2 2 2 2 4

2 2 2 2 2

0 0 0 0

2 4

2 2 2 2 2 22 4 2 4 2 4 2 4

0 0

2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 4

0 0 0 0

2 2 2

0

PE

PEKE PE

KE

1 1

4 4 4

4 8; 2

8 4 8 4

8

n

Ze Ze mZe Z e meV

r n h n h

Z e m

n h n h Z e m Z e m Z e m Z e m E

h n n h n h Z e m n h Z e m

n h

π

πε πε ε ε

ε ε

ε ε ε ε

ε

= − = − = − = −

⎛ ⎞= − = − − − = = = =⎜ ⎟

⎝ ⎠

65. When we compare the gravitational and electric forces, we see that we can use the same expression for

the Bohr orbits, Eq. 27–12 and 27–15a, if we replace 2kZe with e pGm m .

20

1 2 2e

2 34 2

12 2 2 11 2 2 31 2 27e p

29

4

(6.626 10 J s)

4 4 (6.67 10 N m /kg )(9.11 10 kg) (1.67 10 kg)

1.20 10 m

hr

m kZe

hr

Gm m

ε

π

π π

−

− − −

= →

× ⋅= =

× ⋅ × ×

= ×

2 2 3 22 2 4 2 2e p2 2e e

1 12 2 2

2 11 2 2 2 31 3 27 297

34 2

22 2( )

2 (6.67 10 N m /kg ) (9.11 10 kg) (1.67 10 kg)4.22 10 J

(6.626 10 J s)

G m m Z e m k m E kZe E

h h h

π π π

π − − − −−

= − = − → = −

× ⋅ × ×= − = − ×

× ⋅


19/28




66. Find the peak wavelength from Wien’s law, Eq. 27–2.

3 33

P

(2.90 10 m K) (2.90 10 m K)1.1 10 m 1.1 mm

(2.7 K)T λ

− −−× ⋅ × ⋅= = = × =

67. To produce a photoelectron, the hydrogen atom must be ionized, so the minimum energy of the photon

is 13.6 eV. We find the minimum frequency of the photon from Eq. 27–4.

1915min

min 34

(13.6 eV)(1.60 10 J/eV)3.28 10 Hz

(6.63 10 J s)

E E E hf f f

h h

−

−

×= → = → = = = ×

× ⋅

68. From Section 25–11, the spacing between planes, d , for the first-order peaks is given by Eq. 25–10,

2 sind λ θ = . The wavelength of the electrons can be found from their kinetic energy. The electrons

are not relativistic at the given energy.

2 2

2

3410

31 19

KEKE

KE

2 sin2 22

(6.63 10 J s)1.2 10 m

2sin 2 2(sin 38 ) 2(9.11 10 kg)(72 eV)(1.60 10 J/eV)

p h hd

m mm

hd

m

λ θ λ

θ

−−

− −

= = → = = →

× ⋅= = = ×

° × ×

69. The power rating is the energy produced per second. If this is divided by the energy per photon, then

the result is the number of photons produced per second. Use Eq. 27–4.

226

photon 34 8 photon

(720 W)(12.2 10 m); 4.4 10 photons/s

(6.63 10 J s)(3.00 10 m/s)

hc P P E hf

E hc

λ

λ

−

−

×= = = = = ×

× ⋅ ×

70. The intensity is the amount of energy per second per unit area reaching the Earth. If that intensity is

divided by the energy per photon, then the result will be the photons per second per unit area reaching

the Earth. We use Eq. 27–4.

photon

2 9sunlight sunlight 21 21

photons 34 8 2 photon

(1300 W/m )(550 10 m) photons3.595 10 3.6 10

(6 63 10 J s)(3.00 10 m/s) s m

hc E hf

I I I

E hc

λ

λ −

−

= =

×= = = = × ≈ ×

. × ⋅ × ⋅

71. The impulse on the wall is due to the change in momentum of the photons. Each photon is absorbed, so

its entire momentum is transferred to the wall.

on wall wall photons photon photon

9 9

1834

(0 )

(5.8 10 N)(633 10 m) 5.5 10 photons/s(6 63 10 J s)

nh F t p p np np

n F t h

λ

λ − −

−

∆ = ∆ = −∆ = − − = = →

× ×= = = ×∆ . × ⋅

72. As light leaves the flashlight, it gains momentum. The momentum change is given by Eq. 22–9a.

Dividing the momentum change by the elapsed time gives the force the flashlight must apply to the

light to produce the momentum, which is equal to the force that the light applies to the flashlight.

9

8

2 5 W8 3 10 N

3 00 10 m/s

p U P

t c t c

−∆ ∆ .= = = = . ×∆ ∆ . ×


20/28

27-20 Chapter 27



73. (a) Since / , f c λ = the photon energy is / E hc λ = and the largest wavelength has the smallest

energy. In order to eject electrons for all possible incident visible light, the metal’s work function

must be less than or equal to the energy of a 750-nm photon. Thus the maximum value for the

metal’s work function 0W is found by setting the work function equal to the energy of the 750-nm

photon.34 8

0 9 19

(6.63 10 J s)(3.00 10 m/s) 1 eV1.7 eV

(750 10 m) 1.60 10 J

hcW

λ

−

− −

⎛ ⎞× ⋅ ×= = =⎜ ⎟⎜ ⎟× ×⎝ ⎠

(b) If the photomultiplier is to function only for incident wavelengths less than 410 nm, then we set

the work function equal to the energy of the 410-nm photon.

34 8

0 9 19

(6.63 10 J s)(3.00 10 m/s) 1 eV3.0 eV

(410 10 m) 1 60 10 J

hcW

λ

−

− −

⎛ ⎞× ⋅ ×= = =⎜ ⎟⎜ ⎟× . ×⎝ ⎠

74. (a) Calculate the energy from the lightbulb that enters the eye by calculating the intensity of the light

at a distance of 1.0 m= by dividing the power in the visible spectrum by the surface area of a

sphere of radius 1 0 m.= . Multiply the intensity of the light by the area of the pupil (diameter = D)to determine the energy entering the eye per second. Divide this energy by the energy of a

photon (Eq. 27–4) to calculate the number of photons entering the eye per second.

22

e2

22 9 3e

34 8

12

( /4)164

0 030(100 W)(550 10 m) 4.0 10 m

/ 16 1.0 m16(6.626 10 J s)(3.00 10 m/s)

8.3 10 photons/s

P P D I P I D

P P Dn

hc hc

π π

λ

λ

− −

−

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠

⎛ ⎞. × ×⎛ ⎞= = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ × ⋅ × ⎝ ⎠

= ×

(b) Repeat the above calculation for a distance of 1.0 km= instead of 1.0 m.

22

e2

22 9 3e

34 8 3

6

( /4)164

0 030(100 W)(550 10 m) 4.0 10 m

/ 16 16(6 626 10 J s)(3.00 10 m/s) 1.0 10 m

8.3 10 photons/s

P P D I P I D

P P Dn

hc hc

π π

λ

λ

− −

−

⎛ ⎞= = = ⎜ ⎟⎝ ⎠

⎛ ⎞. × ×⎛ ⎞= = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ . × ⋅ × ×⎝ ⎠

= ×

75. The total energy of the two photons must equal the total energy (kinetic energy plus mass energy) of

the two particles. The total momentum of the photons is 0, so the momentum of the particles must have

been equal and opposite. Since both particles have the same mass and had the same initial momentum,

they each had the same initial kinetic energy.

2 photons particles e

21 photons e2

KE

KE

2( )

0 85 MeV 0 511 MeV 0 34 MeV

E E m c

E m c

= = + →

= − = . − . = .

76. We calculate the required momentum from de Broglie’s relation, Eq. 27–78.

3422

12

(6.63 10 J s)1.658 10 kg m/s

(4.0 10 m)

h p

λ

−−

−

× ⋅= = = × ⋅

×


21/28




(a) For the proton, we use the classical definition of momentum to determine its speed and kinetic

energy. We divide the kinetic energy by the charge of the proton to determine the required

potential difference.

225

27

2 27 5 2

19

1.658 10 kg m/s

9.93 10 m/s 0.011.67 10 kg

(1.67 10 kg)(9.93 10 m/s)51 V

2 2(1.60 10 C)

p

cm

K mV

e e

υ

υ

−

−

−

−

× ⋅

= = = × ≈×

× ×= = = =

×

(b) For the electron, if we divide the momentum by the electron mass we obtain a speed that is about

60% of the speed of light. Therefore, we must use Eq. 26–9 to determine the energy of the

electron. We then subtract the rest energy from the total energy to determine the kinetic energy

of the electron. Finally, we divide the kinetic energy by the electron charge to calculate the

potential difference.12

12

2 2 2

22 2 8 2 31 2 8 4

14

2 14 31 8 2 14

14

19

KE

KE

[( ) ( ) ]

[(1.658 10 kg m/s) (3.00 10 m/s) (9.11 10 kg) (3.00 10 m/s) ]

9.950 10 J

9.950 10 J (9.11 10 kg)(3.00 10 m/s) 1.391 10 J

1.391 10 J86,9

1.60 10 C

E pc mc

E mc

V e

− −

−

− − −

−

−

= +

= × ⋅ × + × ×

= ×

= − = × − × × = ×

×= = =

×25 V 87 kV≈

77. If we ignore the recoil motion, then at the closest approach the kinetic energy of both particles is zero.

The potential energy of the two charges must equal the initial kinetic energy of the α particle.

Ag

0 min

9 2 2 19 2Ag 14

min 130

KE

KE

( )( )1

4

( )( )1 (9.00 10 N m /C )(2)(79)(1.60 10 C)

4.74 10 m4 (4.8 MeV)(1.60 10 J/MeV)

Z e Z eU

r

Z e Z e

r

α α

α

α

πε

πε

−−

−

= = →

× ⋅ ×

= = = ××

The distance to the “surface” of the gold nucleus is then 14 15 144.74 10 m 7.0 10 m 4.0 10 m .− − −× − × = ×

78. The decrease in mass occurs because a photon has been emitted. We calculate the fractional change.

Since we are told to find the amount of decrease, we use the opposite of the change.

2 228

2 6

1 1(13 6 eV)

1 31.29 10

(939 10 eV)

E

m E c

m m mc

−

⎡ ⎤⎛ ⎞⎛ ⎞−∆⎛ ⎞. −⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟

−∆ −∆ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦= = = = ×

×

79. We calculate the ratio of the forces.e p

2 11 2 2 31 27gravitational e p

2 9 2 2 19 22electric

2

40

(6.67 10 N m /kg )(9.11 10 kg)(1.67 10 kg)

(9.00 10 N m /C )(1.60 10 C)

4.40 10

Gm m

F Gm mr

F keke

r

− − −

−

−

⎛ ⎞⎜ ⎟

× ⋅ × ×⎝ ⎠= = =

× ⋅ ×⎛ ⎞⎜ ⎟⎝ ⎠

= ×

Yes, the gravitational force may be safely ignored.


22/28

27-22 Chapter 27



80. The potential difference gives the electrons a kinetic energy of 12.3 eV, so it is possible to provide this

much energy to the hydrogen atom through collisions. From the ground state, the maximum energy of

the atom is 13 6 eV 12 3 eV 1 3 eV.− . + . = − . From the energy level diagram, Fig. 27–29, we see that this

means that the atom could be excited to the 3n = state, so the possible transitions when the atom

returns to the ground state are 3n = to 2,n = 3n = to 1,n = and 2n = to 1.n = We calculate the

wavelengths from the equation above and Eq. 27–16.

3 23 2

1240 eV nm650 nm

( ) [ 1 5 eV ( 3.4 eV)]

hc

E E λ →

⋅= = =

− − . − −

3 13 1

1240 eV nm102 nm

( ) [ 1 5 eV ( 13 6 eV)]

hc

E E λ →

⋅= = =

− − . − − .

2 12 1

1240 eV nm122 nm

( ) [ 3 4 eV ( 13 6 eV)]

hc

E E λ →

⋅= = =

− − . − − .

81. The stopping potential is the voltage that gives a potential energy change equal to the maximum kinetic

energy. We use Eq. 27–5b to first find the work function and then find the stopping potential for thehigher wavelength.

max 0 0 0 0

0

1 0 0 0

1 1 0 1 0

34 8

19 9 9

KE

1 1

(6.63 10 J s)(3.00 10 m/s) 1 12.10 eV 0.32 eV

(1.60 10 J/ev) 440 10 m 270 10 m

hc hceV W W eV

hc hc hceV W eV hc eV

λ λ

λ λ λ λ λ

−

− − −

= = − → = −

⎛ ⎞ ⎛ ⎞= − = − − = − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞× ⋅ ×= − + =⎜ ⎟

× × ×⎝ ⎠

The potential difference needed to oppose an electron kinetic energy of 0.32 eV is 0.32 V.

82. We assume that the neutron is not relativistic. If the resulting velocity is small, then our assumptionwill be valid. We use Eq. 27–8.

34

27 9

(6.63 10 J s)1323 m/s 1000 m/s

(1.67 10 kg)(0.3 10 m)

h h h

p m mλ υ

υ λ

−

− −

× ⋅= = → = = = ≈

× ×

This is not relativistic, so our assumption was valid.

83. The intensity is the amount of energy hitting the surface area per second. That is found from the

number of photons per second hitting the area and the energy per photon, from Eq. 27–4.

12 12 34 87 2

2 2 2 9

(1 0 10 photons) (1.0 10 ) (6.626 10 J s)(3.00 10 m/s)4.0 10 W/m

(1 m )(1 s) (1 m )(1 m ) (497 10 m)

hc I

λ

−−

−

. × × × ⋅ ×= = = ×

×

The magnitude of the electric field is found from Eq. 22–8.

7 22 21

0 0 02 12 2 2 80

2 2(4.0 10 W/m )1.7 10 V/m

(8.85 10 C /N m )(3.00 10 m/s)

I I cE E

cε

ε

−−

−

×= → = = = ×

× ⋅ ×


23/28




84. The average force on the sail is equal to the impulse (change in momentum) on the sail divided by the

time. Since the photons bounce off the mirror, the impulse is equal to twice the incident momentum.

We use Eq. 27–6 to write the momentum of the photon in terms of the photon energy. The total photon

energy is the intensity of the sunlight multiplied by the area of the sail.

2 3 2

82( / ) 2( / ) 2 2(1350 W/m )(1.0 10 m) 9.0 N

3 00 10 m/s

p E c E t IA F t t c c

∆ ∆ ×= = = = = =∆ ∆ . ×

85. The electrons will be nonrelativistic at that low energy. The maximum kinetic energy of the

photoelectrons is given by Eq. 27–5b. The kinetic energy determines the momentum, and the

momentum determines the wavelength of the emitted electrons. The shortest electron wavelength

corresponds to the maximum kinetic energy.

2 2

electron 0 electron2

electron

0

3410

31 19

KE2 2

2

(6.63 10 J s)

8.2 10 m1240 eV nm2(9.11 10 kg) 2.2 eV (1.60 10 J/eV)

280 nm

hc p h hW

m m hcm W

λ λ λ

λ −

−

− −

= − = = → =⎛ ⎞

−⎜ ⎟⎝ ⎠

× ⋅

= = ×⎛ ⎞⋅× − ×⎜ ⎟

⎝ ⎠

86. We first find the work function from the given data. A photon energy of 6.0 eV corresponds to a

stopping potential of 3.8 V.

0 0 0 0 6 0 eV 3 8 eV 2 2 eVeV hf W W hf eV = − → = − = . − . = .

If the photons’ wavelength is doubled, then the energy is halved, from 6.0 eV to 3.0 eV. The maximum

kinetic energy would be 0.8 eV. If the photon’s wavelength is tripled, then the energy is only 2.0 eV.

Since this is less than the work function, no current flows.

87. The theoretical resolution limit is the wavelength of the electron. We find the wavelength from themomentum and find the momentum from the kinetic energy and rest energy. We use the result from

Chapter 26, Problem 45. The kinetic energy of the electron is 110 keV.

2 2

34 812

19 3 2 6 3

KE KE2

(6 63 10 J s)(3.00 10 m/s)3.5 10 m

(1 60 10 J/eV) (110 10 eV) 2(0 511 10 eV)(110 10 eV)

hc hc

p mc

λ

−−

−

= =+

. × ⋅ ×= = ×

. × × + . × ×

88. Hydrogen atoms start in the 1n = orbit (ground state). Using Eq. 27–10 and Eq. 27–15b, we determine

the state to which the atom is excited when it absorbs a photon of 12.75 eV via collision with an

electron. Then, using Eq. 27–16, we calculate all possible wavelengths that can be emitted as the

electron cascades back to the ground state.

u u 2

13.6 eV 13.6 eV 13.6 eV4

13.6 eV 12.75 eV E E E E E E n

E E n

− −∆ = − → = = + ∆ → = = =

+ ∆ − +

2

Starting with the electron in the 4n = orbit, the following transitions are possible: 4n = to 3;n =

4n = to 2;n = 4n = to 1;n = 3n = to 2;n = 3n = to 1;n = 2n = to 1.n =


24/28


25/28




91. Find the energy of the photon from Eq. 27–4 and Eq. 27–8.

28 8

19

1 eV(3.53 10 kg m/s)(3 00 10 m/s) 0 662 eV

1.60 10 J

c E hf h pc

λ −

−

⎛ ⎞= = = = × ⋅ . × = .⎜ ⎟⎜ ⎟×⎝ ⎠

From Fig. 27–29, the Lyman series photons have 13 6 eV 10 2 eV. E . ≥ ≥ . The Balmer series photons

have 3 4 eV 1.9 eV. E . ≥ ≥ The Paschen series photons have 1.5 eV 0.65 eV. E ≥ ≥ It appears that this

photon belongs to the Paschen series, ejected from energy level 4.

92. (a) Use Eq. 27–5b to calculate the maximum kinetic energy of the electron and set this equal to the

product of the stopping voltage and the electron charge.

0 0max 0 0 0

0

KE/

(1240 eV nm)/(464 nm) 2 28 eV0 3924 V 0 39 V

hf W hc W hf W eV V

e e

V e

λ − −= − = → = =

⋅ − .= = . ≈ .

(b) Calculate the speed from the nonrelativistic kinetic energy equation and the maximum kinetic

energy found in part (a).

21max max2

195 5max

max 31

KE

KE2 2(0 3924 eV)(1.60 10 J/eV)3.713 10 m/s 3.7 10 m/s

9.11 10 kg

m

m

υ

υ −

−

= →

. ×= = = × ≈ ×

×

This is only about 0.001c, so we are justified in using the classical definition of kinetic energy.

(c) We use Eq. 27–8 to calculate the de Broglie wavelength.

349

31 5

6 63 10 J s1 96 10 m 2 0 nm

(9.11 10 kg)(3.713 10 m/s)

h h

p mλ

υ

−−

−

. × ⋅= = = = . × ≈ .

× ×

93. The electron acquires a kinetic energy of 96 eV, which is used to find the speed. We use the classical

expression.

192 61

2 31

KEKE

2 2(96 eV)(1 60 10 J/eV)5.807 10 m/s

(9.11 10 kg)m

mυ υ

−

−

. ×= → = = = ×

×

This is about 2% of the speed of light. The charge-to-mass ratio can be calculated using an expression

derived in Example 20–6.

611

4 2

5 807 10 m/s1.7 10 C/kg

(3 67 10 T)(9 0 10 m)

m qr

qB m Br

υ υ − −

. ×= → = = = ×

. × . ×

94. First find the area of a sphere whose radius is the Earth–Sun distance.

2 9 2 23 24 4 150 10 m 2.83 10 m A r π π = = ( × ) = ×

Multiply this by the given intensity to find the Sun’s total power output.

23 2 2 26(2.83 10 m )(1350 W/m ) 3.82 10 W× = ×


26/28

27-26 Chapter 27



Multiplying by the number of seconds in a year gives the annual energy output.

26 34(3.82 10 W)(3600 s/h)(24 h/d)(365 25 d/yr ) 1.20 10 J/yr E = × . = ×

Finally, we divide by the energy of one photon to find the number of photons per year.

34 952

34 8

(1.20 10 J/yr)(550 10 m)3.3 10 photons/yr

(6 63 10 J s)(3.00 10 m/s)

E E

hf hc

λ −

−

× ×= = = ×

. × ⋅ ×

95. Use Bohr’s analysis of the H atom, replacing the proton mass with Earth’s mass, the electron mass

with the Moon’s mass, and the electrostatic force2

e 2

ke F

r = with the gravitational force, g 2

E M Gm m F r

= .

To account for the change in force, replace2

ke with E MGm m and set 1. Z = With these replacements,

write expressions similar to Eq. 27–12 and Eq. 27–15a for the Bohr radius and energy.

2 2

2 2

2 2 34 22

2 2 2 11 2 2 22 2 24M E

2 129

2 4 2

2 2

2 2 2 3 2 11 2 2 2 24 2E M

2 2

4

(6 626 10 J s)

4 4 (6 67 10 N m /kg )(7 35 10 kg) (5 98 10 kg)

(5 16 10 m)

2

2 2 (6 67 10 N m /kg ) (5.98 10 kg) (7 3

n

n

n

n

h nr

mke

h nr n

Gm m

n

e mk E

n h

G m m E

n h

π

π π

π

π π

−

−

−

−

= →

. × ⋅= =

. × ⋅ . × . ×

= . ×

= →

. × ⋅ × .= =

2

2 2

22 3

2 34 2

165

2

5 10 kg)

(6 626 10 J s)

2 84 10 J

n

n

−

×

. × ⋅

. ×= −

Insert the known masses and Earth–Moon distance into the Bohr radius equation to determine the

Bohr state.

2 2M E

2

2 11 2 2 22 2 24 8

34 2

68

4

4 (6 67 10 N m /kg )(7.35 10 kg) (5 98 10 kg)(3 84 10 m)

(6 626 10 J s)

2 73 10

nGm m r nh

π

π −

−

=

. × ⋅ × . × . ×=

. × ⋅

= . ×

Since 6810 ,n ≈ a value of 1n∆ = is negligible compared to n. Hence the quantization of energy and

radius is not apparent.

96. The kinetic energy of the hydrogen gas would have to be the difference between the 1n = and 2n =

states of the hydrogen atom, 10.2 eV. Use Eq. 13–8.

1943

2 23

KEKE

2 2(10 2 eV)(1.60 10 J/eV)7 88 10 K

3 3(1.38 10 J/K)kT T

k

−

−

. ×= → = = = . ×

×


27/28




Solutions to Search and Learn Problems

1. (a) J. J. Thomson

(b) Robert A. Millikan

(c) Max Planck(d ) A. H. Compton

(e) Louis de Broglie

( f ) C. J. Davisson, L. H. Germer, G. P. Thomson

( g ) Ernest Rutherford

2. The principle of complementarity states that to understand an experiment, sometimes you must use the

wave theory and sometimes you must use the particle theory. That is, to have a full understanding of

various phenomena (like light), you must accept that they have both wave and particle characteristics.

For example, to understand Young’s double-slit experiment for photons or electrons, you must apply

wave theory with interference of the two waves. To understand the photoelectric effect for light or

Compton scattering for electrons, you must use particle theory.

3. (a) We solve Eq. 24–2a for the slit separation with the wavelength given in terms of the electron

momentum from Eq. 27–6. The electron momentum is written in terms of the kinetic energy.

34

31 19

9

sin

6 626 10 J s

sin sin 2 sin sin10 2(9.11 10 kg)(1 602 10 C)(12 V)

2 0 10 m 2 0 nm

m d

h hd

p meV

λ θ

λ

θ θ θ

−

− −

−

= →

. × ⋅= = = =

° × . ×

. × = .=

(b) No. The slit separation distance is the same size as the atoms.

4. (a) particle

(b) wave

(c) particle

(d ) wave

5. (a) The minimum energy necessary to initiate the chemical process on the retina is the energy of a

single photon of red light. Red light has a wavelength of approximately 750 nm.

834

min 9 19

3 00 10 m/s 1 eV(6 626 10 J s) 1 7 eV

750 10 m 1 60 10 J

c E hf h

λ −

− −

⎛ ⎞⎛ ⎞. ×= = = . × ⋅ = .⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟× . ×⎝ ⎠⎝ ⎠

(b) The maximum energy is the energy of a photon of violet light of wavelength 400 nm

834

max 9 19

3 00 10 m/s 1 eV

(6 626 10 J s) 3.1 eV400 10 m 1 60 10 J E −

− −

⎛ ⎞⎛ ⎞. ×= . × ⋅ =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟× . ×⎝ ⎠⎝ ⎠

6. (a) Apply conservation of momentum before and after the emission of the photon to determine the

recoil speed of the atom, where the momentum of the photon is given by Eq. 27–6. The initial

momentum is 0.

343

27 9

(6 63 10 J s)0 6 0 10 m/s

85(1 66 10 kg)(780 10 m)

h hm

mυ υ

λ λ

−−

− −

. × ⋅= − → = = = . ×

. × ×


28/28

27-28 Chapter 27

(b) We solve Eq. 18–5 for the lowest achievable temperature, where the recoil speed is the rms

speed of the rubidium gas.

2 27 3 27

23

3 85(1 66 10 kg)(6.0 10 m/s)1.2 10 K 0 12 K

3 3(1 38 10 J/K)

kT mT

m k

υ υ µ

− −−

−

. × ×= → = = = × = .

. ×

7. (a) Here are standing wave diagrams for the first three modes of vibration.

(b) From the diagram we see that the wavelengths are given by2

,n L

nλ = 1,n = 2, 3, … . The

momentum is ,2

n

h nh p

Lλ = = so the kinetic energy is

2 2 2

2KE .

2 8

nn

p n h

m mL= =

(c) Because the potential energy is zero inside the box, the total energy is the kinetic energy. The

ground state energy has 1.n =

2 2 2 34 2

1 2 31 10 2 19

(1) (6 63 10 J s) 1 eV150 eV

8 8(9 11 10 kg)(0.50 10 m) 1.60 10 J

n h E

mL

−

− − −

⎛ ⎞. × ⋅= = =⎜ ⎟⎜ ⎟. × × ×⎝ ⎠

(d ) Do the same calculation for the baseball, and then find the speed from the kinetic energy.

2 2 2 34 267 67

1 2 2

(1) (6.63 10 J s)9 289 10 J 9.3 10 J

8 8(0 140 kg)(0 65 m)

n h E

mL

−− −× ⋅= = = . × ≈ ×

. .

672 3311

1 2

KE2 2(9.289 10 J)3.6 10 m/s

(0 140 kg) E m

mυ υ

−−×= → = = = ×

.

(e) Find the width of the box from2 2

1 2.

8

n h E

mL=

2 2

1 2

3410

31 191

8

(1)(6.63 10 J s)1 3 10 m 0 13 nm

8 8(9.11 10 kg)(22 eV)(1 60 10 J/eV)

n h E

mL

nh L

mE

−−

− −

= →

× ⋅= = = . × = .

× . ×

ch27 giancoli7e manual

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