Physics 112
Homework 13 (Ch27)
Due July 31
1. 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.
2. 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.
Physics 112
Homework 13 (Ch27)
Due July 31
3. 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.
4. 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:
hc
hf max 
hf max 
min

1243eV  nm
 3.11eV
400nm
 6.63  10

160  10
hc
min
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
K max
 hf  W0  3.11 eV  2.48 eV  0.63 eV.
5. 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
hf 
hc



1243eV  nm
 4.36eV
285nm
 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
K max  hf  W0 ;
W0  hf  K max  4.36eV  1.40eV
W0  2.96 eV.
Physics 112
Homework 13 (Ch27)
Due July 31
6. How much energy is needed to ionize a hydrogen atom in the n  2 state?
Solution
To ionize the atom means removing the electron, or raising it to zero energy:
Eion  0  En 
13.6 eV   13.6 eV  
n2
22
3.4 eV.
7. 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?
Solution
The energy of the photon is
hf  Eion 
K
 13.6 eV  10.0 eV  23.6eV.
We find the wavelength from

hc 1243eV  nm

 52.7nm
hf
23.6eV