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 the light 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.
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, as long 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
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 9.11 10× −31kg
Charge None −1.60 10× −19 C
Speed 3 10 m/s× 8 < ×3 10 m/s8
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 (λ=h p/ ).
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 (KE= p2/2 ).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 n=1 level. Since there are virtually no atoms in higher energy states, photons corresponding to transitions from n≥2 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
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 it could 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 (λ=h p/ ) 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 n=10 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 photon takes away less energy than Eq. 27–10 predicts.
30. (a) continuous (b) line, emission (c) continuous (d) line, absorption
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 n=1 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 n=1 level. Therefore, emission lines corresponding to transitions to levels higher than n=1 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 if the 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 1 kg 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 byp h= / .λ Young’s double-slit experiment demonstrated diffraction with light, and later experiments demonstrated electron diffraction. Therefore, all three statements are correct.
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 did not 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, hc=1240 eV nm.⋅ 1. This scenario fits the experiment that results in Eq. 27–1.
2 (640 V/m)3 2 6.2 10 C/kg4
(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.
4
6 3
(1.88 10 V/m)
7.23 10 m/s (2.60 10 T)
E B
υ= = × − = ×
×
(b) For the radius of the path in the magnetic field, use an expression derived in Example 20–6.
31 6
2
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)
m r
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
− −
−
→ × ×
= = = = = ≈
4. Use Wien’s law, Eq. 27–2, to find the temperature.
3 3
9 P
(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
5
P (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
7
P (2.90 10 m K) (2.90 10 m K) 9.35410 m 940 nm, near infrared
(3100 K)
T
λ = × − ⋅ = × − ⋅ = − ≈
(c)
3 3
4
P (2.90 10T m K) (2.90 10(4 K)m K) 7.25 10 m 0.7 mm, microwave
λ = × − ⋅ = × − ⋅ = × − ≈
6. Use Wein’s law, Eq. 27–2.
(a)
3 3
5 9
P
(2.90 10 m K) (2.90 10 m K) 1.61 10 K (18.0 10 m)
T
λ
− −
−
× ⋅ × ⋅
= = = ×
×
(b)
3 3
6
P (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 Δ =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 eV
5.37 10 J 0.34 eV
1.60 10 J
E hf − − −
−
−
Δ = = × ⋅ × = × ≈ ×
⎛ ⎞
× ⎜⎜ ⎟⎟=
×
⎝ ⎠
8. The potential energy is “quantized” in units of mgh.
(a) PE1=mgh=(62.0 kg)(9.80 m/s )(0.200 m) 121.52 J2 = ≈122 J (b) PE2=mg h2 =2PE1=2(121.52 J)= 243 J
(c) PE3=mg h3 =3PE1=3(121.52 J)= 365 J (d) PEn=mgnh n= PE1=n(121.52 J)= (122 ) Jn (e) Δ =E PE2−PE6= −(2 6)(121.52 J)= −486 J
9. Use Eq. 27–2 with a temperature of 98 F 37 C (273 37)K 310 K (3 significant figures).° = ° = + =
3 3
6
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.
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 8
19
1 1 9 19
1
(6.63 10 J s)(3.00 10 m/s) 4.97 10 J 1 eV 3.11 eV
(400 10 m) 1.60 10 J
hc E hf
λ
−
−
− −
⎛ ⎞
× ⋅ ×
= = = = × ⎜⎜ ⎟⎟=
× ⎝ × ⎠
34 8
19
2 2 9 19
2
(6.63 10 J s)(3.00 10 m/s) 1 eV
2.65 10 J 1.66 eV
(750 10 m) 1.60 10 J
hc E hf
λ
−
−
− −
⎛ ⎞
× ⋅ ×
= = = = × ⎜⎜ ⎟⎟=
× ⎝ × ⎠
Thus the range of energies is 2.7 10× −19 J< <E 5.0 10× −19J , or 1.7 eV< <E 3.1 eV . 12. Use Eq. 27–4 with the fact that f =c/λ for light.
34 8
12 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 be insignificant diffraction through the doorway.
13. Use Eq. 27–6.
34
27 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.
34
23 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.
(2) If the light intensity is increased, the
maximum kinetic energy of the electrons ejected should not increase.
(2) If the light intensity is increased, the
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 which no electrons will be ejected, no matter how intense the light.
(4) There should be no lower limit to the frequency—electrons will be ejected for all frequencies.
16. We use Eq. 27–4 with the fact that f =c/λ for light.
19
13 13
min
min min min (0.1eV)(1.60 1034 J/eV) 2.41 10 Hz 2 10 Hz
(6.63 10 J s)
E
E hf f
h
− − ×
= → = = = × ≈ ×
8
5 5
max 13
min
(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.
19
14 0
min 0 min 34
KE 0 4.8 10 J 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 9
min min
min (10 34J)(550 10 m)8 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.
0
min 0 min
max
8 34
7
max 19
0
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 c
hf 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 9
min
(6.63 10 J s)(3.00 10 m/s) 3.11 eV (1.60 10 J/eV)(400 10 m)
hc hf
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 KEmax 1240 eV nm 2.255 eV 2.3 eV
550 nm
hc
W 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
max 0
KE
KE
1240 eV nm
2.255 eV 0.845 eV 400 nm
0.845 eV 0.85 V
hc
hf 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.
KEmax 0 0 1240 eV nm 2 48 eV 0 62 eV
400 nm
hc
hf W W
λ
⋅
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.
KEmax 0 0 1240 eV nm 2 48 eV 0.92 eV
365 nm
hc
hf W W
λ
⋅
= − = − = − . =
19
2 max 5
1
max 2 31
KE
KE 2 2(0 92eV)(1.60 10 J/eV) 5.7 10 m/s
9.11 10 kg
m
m
υ 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 KEmax KEmax 1240 eV nm 1 40 eV 3 46 eV
255 nm
hc W 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 max
max
KE 1240 eV nm 3 647 eV
340 nm
hc hc
W
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.
KEmax 0 1240 eV nm 3.647 eV 0 78 eV
280 nm
hc W
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 nm 1.968 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 23
1.968 eV 1.60 10 J 6.02 10 molecules kcal
45 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.
28. We plot the maximum (kinetic) energy of the emitted electrons vs. the frequency of the incident radiation. Eq. 27–5b says KEmax =hf W− 0.
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) h=(0.4157 eV/10 Hz)(1.60 1014 × −19 J/eV)= 6.7 10× −34J s⋅
(b) cutoff 0 cutoff 0 2.3042 eV14 5.5 10 Hz14
(0.4157 eV/10 Hz)
W
hf W f
h
= → = = = ×
(c) W0= .2 3 eV
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 13
1
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 −
− −
−
−
′ = + − → Δ = − →
Δ
⎛ ⎞
= ⎜ − ⎟
⎝ ⎠
⎛ × . × × ⎞
= ⎜⎜ − ⎟⎟= . ° ≈ °
. × ⋅
⎝ ⎠
31. The Compton wavelength for a particle of mass m is /h mc.
(a)
34
12
31 8
e
(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.
photon 2 rest energy
( / )
hc hc
E hf mc
h mc
λ
= = = = =
E = 0.4157 f - 2.3042 R2 = 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
ne
rgy (
eV
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.
C 3
e
(1 cos ) (1 cos ) (2.43 10 nm)(1 cos )
h m c
λ λ′ − =⎛⎜ ⎞⎟ − θ =λ − θ = × − − θ
⎝ ⎠
(a) λ′ − =a λ (2.43 10× −3nm)(1 cos 45 )− ° = 7.12 10× −4nm (b) λ′ − =b λ (2.43 10× −3nm)(1 cos 90 )− ° = 2.43 10× −3nm (c) λ′ − =c λ (2.43 10× −3nm)(1 cos180 )− ° = 4.86 10× −3nm
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.
34
16
max 2 27 8
min
(6.63 10 J s) 6.62 10 m
2
2 2(1.67 10 kg)(3.00 10 m/s)
hc hc h
E mc mc
λ − −
− × ⋅
= = = = = ×
× ×
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. Emin =2mc2=2(207)(0.511 MeV)= 212 MeV
The wavelength is given by Eq. 27–6.
34 8
15
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 υ =0.001 ,c 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.
photon electron 2 0.511 MeV ; photon photon 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.
Ephoton =2(KE+mc2) 2(0 285 MeV 0 511 MeV)= . + . = 1.592 MeV
The wavelength is found from Eq. 27–6.
34 8
13
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 − −
−
× ⋅ ×
= = = ×
38. We find the wavelength from Eq. 27–8.
34
32
(6.63 10 J s)
3.2 10 m
(0 21 kg)(0.10 m/s)
h h p m
λ
υ
−
−
× ⋅
= = = = ×
.
39. The neutron is not relativistic, so use p m= υ. Also use Eq. 27–8.
(6.63 1027 34 J s)4 4.7 10 12 m
(1.67 10 kg)(8.5 10 m/s)
h h p m
λ
υ
−
− −
× ⋅
= = = = ×
× ×
40. We assume that the electron is nonrelativistic, and check that with the final answer. We use Eq. 27–8.
34
6
31 9
(6.63 10 J s)
2.695 10 m/s 0 0089 (9.11 10 kg)(0.27 10 m)
h h h
c
p m m
λ υ
υ λ
−
− −
× ⋅
= = → = = = × = .
× ×
The use of classical expressions is justified. The kinetic energy is equal to the potential energy change.
31 6 2
1
2 2
1
2 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 use KE= p2/2 .m We then form the ratio of the kinetic energies, using Eq. 27–8.
2 2
2 2 27
p
e e
2 2 2 31
p p e
KE KE
KE
/2 1.67 10 kg
; 1840
2 2 /2 9.11 10 kg
m h m
p h
m m h m m
λ
λ λ
− − ×
= = = = = =
×
42. (a) We find the momentum from Eq. 27–8.
34
24 10
6.63 10 J s 1.5 10 kg m/s
4 5 10 m
h p
λ
−
− −
× ⋅
= = = × ⋅
. ×
(b) We assume the speed is nonrelativistic.
34
6
31 10
6.63 10 J s
1.6 10 m/s (9.11 10 kg)(4.5 10 m)
h h h
c
p m m
λ υ
υ λ
−
− −
× ⋅
= = → = = = ×
× ×
(c) We calculate the kinetic energy classically.
2 2 2 3 2 6
1 1 1
2 2 2
KE= mυ = (mc )( / )υ c = (0.511 MeV)(5.39 10 )× − =7.43 10× − MeV 7.43 eV=
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 KE
KE
2 ( );
2 2 ( )
p h h
p m
m λ p m
(a)
34
10 10
31 19
KE
6.63 10 J s 3.88 10 m 4 10 m
2 ( ) 2(9.11 10 kg)(10 eV)(1.60 10 J/eV)
h m
λ − − −
− −
× ⋅
= = = × ≈ ×
× ×
(b)
34
10 10
31 19
KE
6.63 10 J s
1.23 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 19
KE
6.63 10 J s 3.9 10 m
2 ( ) 2(9.11 10 kg)(1.0 10 eV)(1.60 10 J/eV)
h m
λ − −
− −
× ⋅
= = = ×
× × ×
44. Since the particles are not relativistic, use KE= p2/2 .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 m
h h
h
p m m
m
λ λ
λ
= = = = <
Thus the proton has the shorter wavelength, since me<mp.
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. KE= − Δ =q V (1 e)(35 10 V) 35 10 eV× 3 = × 3
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 2
2 2 2 2 2 2 4
3
2 2 3 2 3 3
KE KE
KE
KE KE
2( )
[ ]
1240 eV nm
6.4 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 λ 5 cm, diffraction effects are negligible.
46. For diffraction, the wavelength must be on the order of the opening. Find the speed from Eq. 27–8.
34
38
(6.63 10 J s)
3.9 10 m/s (1400 kg)(12 m)
h h h
p m m
λ υ
υ λ
−
−
× ⋅
= = → = = = ×
This is on the order of 10−39 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
KE
2 2
(6.63 10 J s) 22 V
2 2(9.11 10 kg)(1.60 10 C)(0.26 10 m)
p h
eV
m m
h V
me
λ
λ
−
− − −
= = = →
× ⋅
= = =
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 mc
λ − −
−
× ⋅ ×
= = = =
× ×
= × =
49. The energy of a level is En (13 6 eV)2 ,
n .
= − from Eq. 27–15b for Z=1.
(a) The transition from n=1 to n′ =3 is an absorption, because the final state, n′ =3, has a higher energy. The photon energy is the difference between the energies of the two states.
(13 6 eV) 12 12 12.1 eV
3 1
n n
hf =E′−E = − . ⎢⎡⎝⎛⎜ ⎟⎠⎞ ⎛ ⎞−⎜⎝ ⎟⎠⎥⎤=
⎣ ⎦
(b) The transition from n=6 to n′ =2 is an emission, because the initial state,n′ =2, has a higher energy. The photon energy is the difference between the energies of the two states.
( ) (13.6 eV) 12 12 3.0 eV
2 6
n n
hf = − E ′−E = ⎡⎢⎛⎜ ⎞⎟ ⎜−⎛ ⎞⎟⎤⎥=
⎝ ⎠ ⎝ ⎠
⎣ ⎦
(c) The transition from n=4 to n′ =5 is an absorption, because the final state, n′ =5, has a higher energy. The photon energy is the difference between the energies of the two states.
(13 6 eV) 12 12 0.31 eV
5 4
n n
hf =E′−E = − . ⎢⎡⎝⎛⎜ ⎟⎠⎞ ⎛−⎜⎝ ⎟⎞⎠⎥⎤=
⎣ ⎦
The photon for the transition from n=1 to n′ =3 has the largest energy.
50. Ionizing the atom means removing the electron, or raising it to zero energy.
ionization 0 0 ( 13.6 eV)2 (13.6 eV)2 1.51 eV 3
n
E E
n −
= − = − = =
51. From E hc,
λ
Δ = we see that the second longest wavelength comes from the transition with the second
smallest energy: n=5 ton′=3.
52. Doubly ionized lithium is similar to hydrogen, except that there are three positive charges (Z =3) in the nucleus. Use Eq. 27–15b with Z=3.
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
= = = −
⎡ ⎤
= − = − −⎢ ⎥=
⎢ ⎥
⎣ ⎦
53. We use Eq. 27–10 and Eq. 27–15b. Note that E E1, ,2 and E3 are calculated in the textbook. We also
need 4 (13 6 eV)2 0.85 eV
4
E = − . = − and 5 (13 6 eV)2 0.54 eV.
5
E =− . = − (a) The second Balmer line is the transition from n=4 to n=2.
4 2
1240 eV nm 486 nm
( ) [ 0 85 eV ( 3.40 eV)]
hc E E
λ= = ⋅ =
− − . − −
(b) The second Lyman line is the transition from n=3 to n=1.
3 1
1240 eV nm
103 nm ( ) [ 1 51 eV ( 13 6 eV)]
hc E E
λ= = ⋅ =
− − . − − .
(c) The third Balmer line is the transition from n=5 to n=2.
5 2
1240 eV nm 434 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 Z=1. We also substitute
0
1 . 4
k
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
− −
− −
⎛ ⎞ ⎛ ⎞
= ⎜ ′ − ⎟= ⎜ ′ − ⎟ →
⎝ ⎠ ⎝ ⎠
= =
× ×
=
× ⋅ × ⋅ ×
= 7 4 7 1
4 3 3
2 4
C kg
1 10 1.0974 10 m C
J s m/s N m
−
⋅
× ≈ ×
⋅
55. The longest wavelength corresponds to the minimum energy, which is the ionization energy.
ion
1240 eV nm
91 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.
total
total
1240 eV nm 49.4 nm 25.1 eV
hc hc
hf E
E
λ λ
⋅
= = → = = =
57. Singly ionized helium is like hydrogen, except that there are two positive charges (Z=2) 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
. . .
We find the energy of the photon from the n=5 to n=2 transition in singly ionized helium.
5 2 (54 4 eV) 12 12 12.1 eV
6 2
E E E ⎡⎛ ⎞ ⎛ ⎞⎤
Δ = − = − . ⎢⎜ ⎟−⎜ ⎟⎥=
⎝ ⎠
⎝ ⎠
⎣ ⎦
Because this is the energy difference for the n=1 to n=3 transition in hydrogen, the photon can be absorbed by a hydrogen atom which will jump from n=1 to n=3 .
58. Singly ionized helium is like hydrogen, except that there are two positive charges (Z =2) 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 (Z=3) 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 e2 with Ze2:
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 10
0 1
PE ( ) ( ) 1 (9.00 10 N m /C )(1.60 10 C) (1 eV/1.60 10 J)
4 (0.529 10 m)
27.2 eV
e
e V e
r
πε
− −
−
× ⋅ × ×
= − = − = −
× = −
The kinetic energy is the total energy minus the potential energy. KE=E1−PE= −13.6 eV ( 27.2 eV)− − = +13.6 eV
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 /s 5.000 5
2 (6.626 10 J s)
13.6 eV
(13.6 eV) 0.544 eV
25
n
h L
L n n
h Z E
n
π π
π
− −
( × ⋅ )
= → = = = ≈
× ⋅
= − = − = −
122
−
30.6
−
13.6
62. The value of n is found from rn=n r21, and then the energy can be found from Eq. 27–15b. 2 2 1 10 1 8 2 2
1.00 10 m
13,749 13,700
0.529 10 m
(13 6 eV) (13 6 eV)
7.19 10 eV
(13,749)
n
n r
r n r n
r E n − − − × = → = = = ≈ × . . = − = − = − ×
63. The velocity is found from Eq. 27–11 evaluated for n=1.
34
6 3
10 31
1 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 nh m r h c r m υ π υ π π − − − − = → × ⋅ = = = × = × ( × )( × )
Since υ c, we say yes, nonrelativistic formulas are justified. The relativistic factor is as follows.
2 6
2 2 1 2 2 1 5
2 2 8
2 190 10 m/s
1 / 1 ( / ) 1 1 2.66 10 0.99997
3 00 10 m/s
c c
υ υ ⎛ . × ⎞ −
− ≈ − = − ⎜⎜ ⎟⎟ = − × ≈
. ×
⎝ ⎠
We see that 1−υ2/c2 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 2
2 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 PE KE PE KE 1 1
4 4 4
4 8
; 2
8 4 8 4
8
n
Ze Ze mZe Z e m
eV
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 kZe2 with Gm me p.
2 0
1 2 2
e
2 34 2
1 2 2 2 11 2 2 31 2 27
e 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
h r m kZe h r Gm m ε π π π − − − − = → × ⋅ = = × ⋅ × × = ×
2 2 3 2
2 2 4 2 2
e p 2 2
e e
1 2 2 1 2
2 11 2 2 2 31 3 27 2
97
34 2
2
2 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 − − −
66. Find the peak wavelength from Wien’s law, Eq. 27–2.
3 3
3
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.
19
15 min
min (13.6 eV)(1.60 1034 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
34
10
31 19
KE
KE
KE
2 sin
2 2 2
(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 h
d
m m m
h d
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.
photon 34 2 8 26
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 9
sunlight sunlight 21 21
photons 34 8 2
photon
(1300 W/m )(550 10 m) 3.595 10 3.6 10 photons
(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
18 34
(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 − −
−
Δ = Δ = −Δ = − − = = →
× ×
= = = ×
Δ . × ⋅
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.
2 5 W8 8 3 10 9N
3 00 10 m/s
p U P
t c t c
−
Δ =Δ = = . = . ×
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 W0 is found by setting the work function equal to the energy of the 750-nm
photon.
34 8
0 (6.63 10 J s)(3.00 10 m/s)9 1 eV19 1.7 eV
(750 10 m) 1.60 10 J
hc W
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.
0 (6.63 10 34J s)(3.00 10 m/s)9 8 1 eV19 3.0 eV
(410 10 m) 1 60 10 J
hc W
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.
2 2
e 2
2
2 9 3
e
34 8
12
( /4)
16 4
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 m
8.3 10 photons/s
P P D
I P I D
P P D
n
hc hc
π π
λ λ
− −
−
⎛ ⎞
= = = ⎜ ⎟⎝ ⎠
⎛ ⎞
. × ×
⎛ ⎞
= = ⎜ ⎟ = ⎜⎜ ⎟⎟
⎝ ⎠ × ⋅ × ⎝ ⎠
= ×
(b) Repeat the above calculation for a distance of =1.0 kminstead of 1.0 m.
2 2
e 2
2
2 9 3
e
34 8 3
6
( /4)
16 4
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 D
n
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
2 1
photons e 2
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.
34
22 12
(6.63 10 J s) 1.658 10 kg m/s (4.0 10 m)
h p
λ
−
− −
× ⋅
= = = × ⋅
(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.
22
5 27
2 27 5 2
19
1.658 10 kg m/s
9.93 10 m/s 0.01 1.67 10 kg
(1.67 10 kg)(9.93 10 m/s) 51 V
2 2(1.60 10 C)
p c m K m V 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. 1 2 1 2
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 J 86,9
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 2
Ag 14 min 13 0 KE KE ( )( ) 1 4 ( )( )
1 (9.00 10 N m /C )(2)(79)(1.60 10 C)
4.74 10 m
4 (4.8 MeV)(1.60 10 J/MeV)
Z e Z e U
r Z e Z e r α α α α πε πε − − − = = → × ⋅ × = = = × ×
The distance to the “surface” of the gold nucleus is then 4.74 10× −14m 7.0 10− × −15m= 4.0 10× −14m . 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 2 26 2 8
1 1
(13 6 eV)
1 3 1.29 10
(939 10 eV)
E
m c E
m m mc
− ⎡⎛ ⎞ ⎛ ⎞⎤ −Δ ⎛ ⎞ . − ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎜ ⎟
−Δ =⎝ ⎠=−Δ = ⎣⎝ ⎠ ⎝ ⎠⎦= ×
×
79. We calculate the ratio of the forces.
e p
2 11 2 2 31 27
gravitational e p
2 9 2 2 19 2
2 electric
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 r Gm m
F ke ke
r − − − − − ⎛ ⎞ ⎜ ⎟ × ⋅ × × ⎝ ⎠ = = = × ⋅ × ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = ×
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 n=3 state, so the possible transitions when the atom returns to the ground state are n=3 to n=2, n=3 to n=1, and n=2 to n=1. We calculate the wavelengths from the equation above and Eq. 27–16.
3 2
3 2
1240 eV nm 650 nm
( ) [ 1 5 eV ( 3.4 eV)]
hc E E
λ → = = ⋅ =
− − . − −
3 1
3 1
1240 eV nm 102 nm
( ) [ 1 5 eV ( 13 6 eV)]
hc E E
λ → = = ⋅ =
− − . − − .
2 1
2 1
1240 eV nm 122 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 the higher 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 1
2.10 eV 0.32 eV
(1.60 10 J/ev) 440 10 m 270 10 m
hc hc
eV W W eV
hc hc hc
eV 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 assumption will 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 m
λ 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 8
7 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 2
2 2
1
0 0 0
2 12 2 2 8
0
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
c
ε
ε
−
−
− ×
= → = = = ×
