Homework 14
Ch27: P 3, 5, 7, 21, 23
3. (II) An oil drop whose mass is determined to be 2.8 10 15 kg is held at rest between
two large plates separated by 1.0 cm when the potential difference between them is
340 V. How many excess electrons does this drop have?
Solution
The force from the electric field must balance the weight:
mgd
neV
n
qE 
 mg ; which gives
d
ev
n
2.8 10

15


kg 9.80m / s 2 1.0  10 2 m
,
1.6  10 19 C 340V 


which gives n  5.
5. (I) Estimate the peak wavelength for radiation from (a) ice at 0°C, (b) a floodlamp at
3500 K, (c) helium at 4 K, (d) for the universe at T  2.725 K, assuming blackbody
emission. In what region of the EM spectrum is each?
Solution
a) We find the peak wavelength from
P
 2.90  10

3
m K
T
 2.90  10

3
m K
 273K 
 1.06  105 m  10.6  m.
This wavelength is in the near infrared.
(b)
P 
 2.90  10
3
m K
T

 2.90  10
3
m K
 3500K 
 8.29  107 m  829nm.
This wavelength is in the infrared.
(c)
P 
 2.90  10
3
m K
T

 2.90  10
3
 4K 
m K
 7.25  104 m  0.73mm.
This wavelength is in the microwave region.
(d)
P
 2.90  10

T
3
m K
 2.90  10

3
m K
 2.725K 
 1.06  102 m  1.06 cm.
This wavelength is in the microwave region.
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?
Solution
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.
21. (II) What is the maximum kinetic energy of electrons ejected from barium
(W0  2.48 eV) when illuminated by white light,   400 to 750 nm?
Solution
The photon of visible light with the maximum energy has the minimum wavelength:
hf max 
hc
min
 6.63  10

160  10
34
19
J s  3.00  108 m / s 
J/eV  400  109 m 
 3.11 eV.
The maximum kinetic energy of the photoelectrons is
KE max
 hf  W0  3.11 eV  2.48 eV  0.63 eV.
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?
Solution
The energy of the photon is
hf 
hc


 6.63  10
1.60  10
34
19
J s  3.00  108 m / s 
J/eV  285  109 m 
 4.36 eV.
We find the work function from
KE max  hf  W0 ;
W0  hf  KEmax  4.36eV  1.40eV
W0  2.96 eV.