PHYS 212-071
Solutions to HW-3
(Numbers refer to 2nd Edition)
Pb.16 Consider the metals lithium, beryllium, and mercury, which have work functions of
2.3 eV, 3.9 eV, and 4.5 eV, respectively. If light of wavelength 300 nm is incident on
each of these metals, determine (a) which metal exhibits the photoelectric effect and (b)
the maximum kinetic energy for the photoelectron in each case.
A. 16 The metals which exhibit the photoelectric effect are those for which the maximum
kinetic energy of the ejected electrons is positive. Namely
K max  h    0
We have h  4.136  10 15 eV.s and  
c


3  10 8 m / s
 1015 Hz . Therefore we get the
300  10 9 m
following table:
Metal
 [eV]
Kmax [eV]
Exhibit Photoelectric Effect
Lithium
2.3
1.836
YES
Beryllium
3.9
0.236
YES
Mercury
4.5
 0.364
NO
Pb. 20. Figure P2.20 shows the stopping potential versus incident photon frequency for
the photoelectric effect for sodium. Use these data points to find (a) the work function,
(b) the ratio h / e , and (c) the cutoff wavelength. (d) Find the percent difference between
your answer to (b) and the accepted value.
A 20 (a) The work function should be found using the graph by extrapolating the linear
curve back to zero frequency. At this point we have Vs   / e . Knowing the electronic
charge, we can deduce the work function   eVs [eV]. (b) Similarly the ratio h / e is to
be found from the slope of the linear graph, namely
1
h Vs

[V.s]. Knowing the
e
f
electronic charge we can find Planck’s constant from the graph h graph . (c) The cutoff
wavelength is the wavelength for which
K max  hf 0   
hc
0
   0.  0 [nm] 
hc 1240.8[eV .nm]

.

[eV ]
(d) The % difference can be found by calculating
| hgraph  haccepted |
h
[%]  100 
h
haccepted
Pb. 25. X-rays with a wavelength of 0.040 nm undergo Compton scattering. (a) Find the
wavelength of photons scattered at angles of 30o, 60o, 120o, 150o, 180o, and 210o. (b) Find
the energy of the scattered electrons corresponding to these scattered x-rays. (c) Which
one the given scattering angles provides the electron with the greatest energy?
A 25. (a) The wavelengths   of the scattered photons can be calculated using the
Compton scattering formula    0 
h
1  cos   , where h  0.00243 nm is the
me c
me c
Compton wavelength of the electron. The kinetic energies of the scattered electrons can
be found using the following formula K e  h 0  h  
hc
0

hc
. We therefore get the

following table:
 [degrees]
 [nm]
  [nm]
30
0.000326
0.040326
60
0.001215
0.041215
120
0.003645
0.043645
150
0.004534
0.044534
180
0.00486
0.04486
210
0.004534
0.044534
K e [eV]
250
914
2591
3158
3361
3158
The scattering angle of 180o provides the electron with the greatest kinetic energy of 3.4
keV.
2
Pb. 32. Show that Compton formula
   
h
1  cos  
me c
results when expressions for the electron energy (Equation 2.34) and momentum
(Equation 2.35) are substituted into the relativistic energy expression,
Ee  pe c 2  me c 4
2
2
2
A 32. Equation 2.34 gives the (total) electron energy: Ee  hf  hf   me c 2 . Equation
2 h 2 ff 
 hf    hf 


cos  .
momentum: p e2  
  
c2
 c   c 
2
(2.35)
gives
the
electron’s
2
By
substitution of these expressions into the relativistic energy expression we get:
hf  hf   m c 
2 2
e
 hf   2  hf  2 2h 2 ff 
 2
2 4
 
cos

   
  c  me c
2
c
c
c
  


Expanding the square on the left-hand side of the equation and after eliminating equal
terms we get:
f  f   ff 
Multiplying through by
h
1  cos  
me c 2
c
c
c
, and recalling that   and   
, we get
f
ff 
f
   
h
1  cos  
me c
QED.
3