Chapter 27
7.
(I) An HCl molecule vibrates with a natural frequency of 8.1  1013 Hz. What is the difference
in energy (in joules and electron volts) between possible values of the oscillation energy?
7. Because the energy is quantized, E = nhf, the difference in energy between adjacent levels is
E  hf   6.63  1034 J s 8.1  1013Hz   5.4  1020 J  0.34eV.
8.
(II) The steps of a flight of stairs are 20.0 cm high (vertically). If a 68.0-kg person stands with
both feet on the same step, what is the gravitational potential energy of this person, relative to
the ground, on (a) the first step, (b) the second step, (c) the third step, (d) the nth step? (e) What
is the change in energy as the person descends from step 6 to step 2?
8. (a) The potential energy on the first step is
2
PE1  mgh   68.0kg   9.80m s   0.200m   133J.
(b) The potential energy on the second step is
PE 2  mg 2h  2PE1  2 133J   267J.
(c) The potential energy on the third step is
PE3  mg 3h  3PE1  3133J   400J.
(d) The potential energy on the nth step is
PE n  mgnh  nPE1  n 133J   133n J.
(e) The change in energy is
E  PE 2  PE6   2  6133J    533J.
9.
(II) Estimate the peak wavelength of light issuing from the pupil of the human eye (which
approximates a blackbody) assuming normal body temperature.
9. We use a body temperature of 98 F  37 C  310K. We find the peak wavelength from
P
 2.90 10

T
3
m K
 2.90 10

3
m K
 310K 
 9.4  106 m  9.4  m.
13.
(I) About 0.1 eV is required to break a “hydrogen bond” in a protein molecule. Calculate the
minimum frequency and maximum wavelength of a photon that can accomplish this.
13. We find the minimum frequency from
Emin  hf min ;
 0.1eV  1.60  1019 J
eV    6.63  1034 J s  f min , which gives f min  2.4  1013 Hz.
The maximum wavelength is
 3.00  108 m s   1.2  105 m.
c
max 

f min
 2.4  1013 Hz 
20.
(II) In a photoelectric-effect experiment it is observed that no current flows unless the
wavelength is less than 570 nm. (a) What is the work function of this material? (b) What is the
stopping voltage required if light of wavelength 400 nm is used?
20. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have
KE  hf  W0  0;
W0 
hc
max
 6.63  10

1.60  10
34
19
J s  3.00  108 m / s 
J/eV  570  109 m 
 2.18 eV.
(b) The stopping voltage is the voltage that gives a potential energy change equal to the
maximum
kinetic energy:
KE max  eV0  hf  W0 ;
  6.63  1034 J s  3.00  108 m / s  
  2.18 eV  3.11 eV  2.18 eV  0.93 eV,
1e V0  
19
9
 1.60  10 J / eV  400  10 m  
so the stopping voltage is 0.93 V.
23.
(II) When UV light of wavelength 285 nm falls on a metal surface, the
maximum kinetic energy of emitted electrons is 1.40 eV. What is the work
function of the metal?
23. The energy of the photon is
34
8
hc  6.63  10 J s  3.00  10 m / s 
hf 

 4.36 eV.
 1.60  1019 J/eV  285  109 m 
We find the work function from
KE max  hf  W0 ;
1.40 eV  4.36 eV  W0 , which gives W0  2.96 eV.
33.
(II) What is the longest wavelength photon that could produce a proton–antiproton pair? (Each
has a mass of 1.67 10 27 kg. )
33. The photon with the longest wavelength has the minimum energy to create the masses:
hc
hf min 
 2m0 c 2 ;
max
 6.63  10
34
J s
max
 2 1.67  1027 kg  3.00  108 m s  , which gives
max  6.62  1016 m.
36.
(II) A gamma-ray photon produces an electron–positron pair, each with a kinetic energy of 245
keV. What was the energy and wavelength of the photon?
36. The energy of the photon is
hf  2  KE  m0 c 2   2  0.245MeV  0.511MeV   1.51MeV.
The wavelength is
 6.63  1034 J s  3.00  108 m / s   8.23  1013 m.

1.60  1019 J/eV 1.51  106 eV 
41.
(II) An electron has a de Broglie wavelength   5.0  10 10 m. (a) What is its momentum? (b)
What is its speed? (c) What voltage was needed to accelerate it to this speed?
41. (a) We find the momentum from
h 6.63  1034 J s
p 
 1.3  1024 kg m s.
10
 5.0  10 m
(b) We find the speed from
h
h
 
;
p mv
5.0  1010 m 
6.63  1034 J s
, which gives v  1.5  106 m s.
31
 9.11  10 kg  v
(c) With v  0.005 c, we can calculate
KE
 mv
1
2
KE
classically:
2
 12  9.11  1031 kg 1.5  106 m s   9.7  1019 J, which converted to electron2
 9.7  10 
19
voltsequals
1.06  10
19
J eV 
 6.0 eV. This is the energy gained by an electron as it is
accelerated through a potential difference of 6.0V.
48.
(I) For the three hydrogen transitions indicated below, with n being the initial state and n 
being the final state, is the transition an absorption or an emission? Which is higher, the initial
state energy or the final state energy of the atom? Finally, which of these transitions involves
the largest energy photon? (a) n  1, n   3 (b) n  6, n   2 (c) n  4, n   5.
48. The energy of a level is En  
13.6 eV  .
n2
(a) The transition from n = 1 to n' = 3 is an absorption, because the final state, n' = 3, has
a
higher energy. The energy of the photon is
 1   1  
hf  En  En   13.6 eV   2    2    12.1 eV.
 3   1  
(b) The transition from n = 6 to n' = 2 is an emission, because the initial state, n' = 2, has a
higher energy. The energy of the photon is
 1   1  
hf    En  En   13.6 eV   2    2    3.0 eV.
 2   6  
(c) The transition from n = 4 to n' = 5 is an absorption, because the final state, n' = 5, has
a
higher energy. The energy of the photon is
 1   1  
hf  En  En   13.6 eV   2    2    0.31 eV.
 5   4  
The photon for the transition from n = 1 to n' = 3 has the largest energy.
54.
(II) What is the longest wavelength light capable of ionizing a hydrogen atom in the ground
state?
54. The longest wavelength corresponds to the minimum energy, which is the ionization energy:
34
8
hc  6.63  10 J s  3.00  10 m / s 


 9.14  108 m  91.4 nm.
19
Eion
1.60  10 J/eV  13.6 eV 
55.
(II) What wavelength photon would be required to ionize a hydrogen atom in the ground state
and give the ejected electron a kinetic energy of 10.0 eV?
55. The energy of the photon is
hf  Eion  KE  13.6 eV  10.0 eV  23.6eV.
We find the wavelength from
34
8
hc  6.63  10 J s  3.00  10 m / s 


 5.27  108 m  52.7 nm.
19
hf
1.60  10 J/eV   23.6 eV 