Kapittel 38
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38.23. Model: Assume the blackbodies obey Wein’s law in Equation 38.9: peak  (290 106 nm  K)/T .
Visualize: We want to solve for T in the equation above. We are given peak  300 nm 3000 nm.
Solve:
T
290  106 nm  K
peak
(a)
T
290 106 nm  K
 9667K  9394C
300nm
T
290 106 nm  K
 9667K  694C
3000nm
(b)
38.24. Model: Assume the metal sphere is a blackbody (so the emissivity e 1 ).
Visualize: First use Wein’s law (Equation 38.9) to find the temperature, then use Stefan’s law (Equation 38.8) to
determine the power radiated. We are given R  10 cm and peak  2000 nm.
Solve:
T
290  106 nm  K
 1450 K
2000 nm
Q
 e AT 4  (1)(567 108 W/m2  K 4 )[4 (10 cm)2 ](1450 K) 4  315 W
t
Assess: The sphere radiates more than 3100W light bulbs, but it has a larger surface area than the filaments, so the
answer is reasonable.
38.25. Model: Assume the ceramic cube is a blackbody (so the emissivity e 1 ).
Visualize: First use Stefan’s law (Equation 38.8),
Q
t
 e AT 4 , to find the temperature, then use Wein’s law
(Equation 38.9) to get the peak wavelength. We are given A  6(30 cm  30 cm)  00054 m 2 and Qt  630 W.
Solve: Solve Stefan’s law for T.
T4
Qt
630 W
4
 1198 K
8
e A
(1)(567  10 W/m 2  K 4 )(00054 m 2 )
Now plug this temperature into Wein’s law.
peak 
290  106 nm  K 290  106 nm  K

 2420 nm  242  m
T
1198 K
Kapittel 39
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39.2. Model: Light of frequency f consists of discrete quanta, each of energy E  hf.
Solve: (a) The energy of the light quantum is
E  hf  h
c


 6.63 10
34
J s  3.0  108 m/s 
400  109 m

1 eV
 3.11 eV
1.6  1019 J
From Table 39.1, the work functions for sodium and potassium are smaller than 3.11 eV. That is, light of wavelength
400 nm has enough energy to eject photoelectrons from sodium and potassium.
(b) The energy of the light quantum is
E  hf  h
c


 6.63 10
34
J s  3.0  108 m/s 
250  109 m

1 eV
 4.97 eV
1.6  1019 J
Light of wavelength 250 nm has enough energy to eject photoelectrons from all of the metals on the table except gold.
39.4. Solve: From Equation 39.7, the maximum kinetic energy is
K max  hf  E0  h
 
c

 E0
 6.63 1034 J s 3.0 108 m/s   1 eV  209 nm
hc

E0  K max
4.65 eV  1.30 eV
1.6  10 19 J
Assess:   209 nm is the wavelength of light in the ultraviolet region of the spectrum.
39.5. Model: The threshold frequency for the ejection of photoelectrons is f0  E0 /h where E0 is the work function.
Solve:
The visible region of light extends from 400 nm to 700 nm. For 0  400 nm, the work function is
E0  f 0h 
For 0  700 nm,
hc
0
 6.63 10

34
J s  3.0 108 m/s 
400 109 m
 6.63 10

34

1 eV
 3.11 eV
1.6 1019 J
J s  3.0 108 m/s 
1 eV

 1.78 eV
700  109 m
1.6  1019 J
The cathode that will work in the entire visible range must have a work function of 1.78 eV or less.
E0
39.8. Solve: (a) The frequency of the photon is
f 
c


3.00  108 m/s
 4.29  1014 Hz
700  109 m
From Equation 39.4, the energy is
E  hf   4.14  1015 eV s  4.29  1014 Hz   1.77 eV
(b) The frequency of the photon is
f 
E
5000 eV

 1.208 1018 Hz
h 4.14 1015 eV s
Thus, the wavelength is

c 3.00 108 m/s

 2.5 1010 m  0.25 nm
f 1.208 1018 Hz
Assess: Because x-ray photons are very energetic, their wavelength is small.
39.13. Model: All the light emitted by the light bulb is assumed to have a wavelength of 600 nm.
Solve: (a) The frequency of the emitted light is
f 
c


3.0  108 m/s
 5.0  1014 Hz
600  109 m
(b) The rate of photon emission is
Rphoton 
P 5W
5W
=

 1.5  1019 photons/s  1  1019 photons/s
34
hf
hf
 6.63 10 J s 5.0 1014 Hz 
39.20. Model: To conserve energy, the emission and the absorption photons must have exactly the energy lost or
gained by the atom in the appropriate quantum jumps.
Visualize: The energy of a light quantum is E  hf  hc/.
Solve: (a) The wavelength of the emission photon from the n  2 to n  1 transition is

 4.14 10
hc

E2  E1
15
eV s  3.0  108 m/s 
1.5 eV
 828 nm
Likewise,   497 nm for the 3 2 transition with E  2.5 eV, and   311 nm for the 3  transition with E  4.0
eV.
(b) Because the atom in the ground state is in the n  1 state, the absorption lines correspond to the 1  and
1  3 transitions. The absorption wavelengths are 828 nm and 311 nm. The 2   transition is not seen in absorption.
39.21. Model: To conserve energy, the absorption spectrum must have exactly the energy gained by the atom in
the quantum jumps.
Visualize: Please refer to Figure EX39.20.
Solve: (a) An electron with a kinetic energy of 2.00 eV can collide with an atom in the n  1 state and raise its
energy to the n  2 state. This is possible because E2 – E1  1.50 eV is less than 2.00 eV. On the other hand, the atom
cannot be excited to the n  3 state.
(b) The atom will absorb 1.50 eV of energy from the incoming electron, leaving the electron with 0.50 eV of kinetic
energy.
39.23. Model: The electron must have k  Eatom to cause collisional excitation. The atom is initially in the n  1
ground state.
Visualize:
Solve:
The kinetic energy of the incoming electron is
E  12 mv 2  12  9.111031 kg 1.30 106 m/s   7.698 1019 J  4.81 eV
2
The electron has enough energy to excite the atom to the n  2 stationary state (E2 – E1  4.00 eV). However, it does
not have enough energy to excite the atom into the n  3 state which requires a total energy of 6.00 eV.
39.34. Solve: The laser light delivers 2.50  1017 photons per second and 100  103 J of energy per second. Thus,
the energy of each photon is
100  103 J/s
 4.00  1019 J
2.50  1017 s 1
From Equation 39.4, the wavelength of the photons is

34
8
c hc  6.63 10 J s  3.00 10 m/s 


 4.97 107 m  497 nm
f
E
4.00 1019 J
Assess: The wavelength is in the visible region.
39.51. Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a
lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a higher
energy level. Because most of the atoms are in the n  1 ground state, the only quantum jumps in the absorption
spectrum start from the n  1 state.
Solve: (a) The ionization energy is E1  6.5 eV.
(b) The absorption spectrum consists of the transitions 1  2 and 1  3 from the ground state to excited states.
According to the Bohr model, the required photon frequency and wavelength are
f 
c hc
E
 
f E
h
where E  Ef – Ei is the energy change of the atom. Using the energies given in the figure, we calculated the values in
the table below.
Transition
Ef (eV)
Ei (eV)
E (eV)
(nm)
12
13
3.0
–2.0
6.5
6.5
3.5
4.5
355
276
(c) Both wavelengths are ultraviolet ( < 400 nm).
(d) A photon with wavelength   1240 nm has an energy Ephoton  hf  hc/  1.0 eV. Because Ephoton must exactly
match E of the atom, a 1240 nm photon can be emitted only in a 3  2 transition. So, after the collision the atom
was
in
the
n  3 state. Before the collision, the atom was in its ground state (n  1). Thus, an electron with vi  1.4  106 m/s
collided with the atom in the n  1 state. The atom gained 4.5 eV in the collision as it is was excited from the n  1 to n
 3, so the electron lost 4.5 eV  7.20   J of kinetic energy. Initially, the kinetic energy of the electron was
Ki  12 melecvi2  12  9.111031 kg 1.40 106 m/s   8.93 1019 J
2
After losing 7.20  1019 J in the collision, the kinetic energy is
Kf  Ki  7.20  1019 J  1.73 1019 J  12 melecvf2  vf 
2 1.73 1019 J 
2Kf

 6.16 105 m/s
melec
9.111031 kg
Kapittel 42
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42.21. Solve: Since 1 mW  0.001 W, the laser emits Elight  0.001 J of light energy per second. This energy
consists of N photons. The energy of each photon is
Ephoton  hf 
hc

 3.14 1019 J
Because Elight  NEphoton, the number of photons is
N
Elight
Ephoton

0.001 J
 3.2  1015
3.13  1019 J
So, photons are emitted at the rate 3.2  1015 s1.
42.22. Solve: The energy of each photon is
Ephoton  hf 
hc

 1.876 1020 J
Because Elight  NEphoton,
Elight  (5.0 1022 )(1.876 1020 J)  940 J
So, the energy output is 940 J every second, or 940 W.
42.23. Solve: (a) The wavelength is

hc
1240 eV nm

 1060 nm  1.06  m
E 1.17 eV  0 eV
(b) The energy per photon is Eph  1.17 eV  1.87  1019 J. The power output of the laser is the number of photons per
second times the energy per photon:
P  (1.0  1019 s1)(1.87  1019 J)  1.9 J/s  1.9 W