SUGGESTED SOLUTIONS FOR TUTORIAL 10
(PHOTON AND QUANTIZED ENERGY)
1.
Define
i.
Photoelectric Effect
ii.
Work Function
iii.
Stopping Potential
iv.
Threshold Frequency
Solution:
i.
Photoelectric – The phenomenon of electron emission from cold metal
surface when shined by the light of high frequency.
ii.
Work Function – Minimum energy required to emit/remove an electron
from metallic surface.
iii.
Stopping Potential – Potential value to stop the whole electron (maximum
kinetic energy) to reach the collector plate (anode) and the instantaneous current
is zero.
iv.
Threshold Frequency – Minimum frequency required to emit/remove an
electron from metallic surface.
2.
a.
b.
c.
d.
e.
What is energy of photon light of wavelength 0.70 μm?
Ans:(2.8 x 10-19 J or 1.78 eV)
Find the wavelength and frequency of a 3.1 eV photon.
Ans:(7.8 x 1014 Hz)
Determine the vacuum wavelength corresponding to γ-ray energy of 1019
eV.
Ans:(1.24 x 10-25 m)
What is the momentum of a single photon of red light (f = 400 x 1012 Hz)
moving through free space.
Ans:(8.84 x 10-28 kg m s-1)
What is the work function of sodium metal if the photoelectric threshold
wavelength is 680 nm.
Ans:(2.93 x 10-19 J or 1.83 eV)
Solutions:
  0.7  10 6 m
a.
hc 6.63  10 34 3.0  10 8
E

 2.84  10 19 J or 1.78 eV
6

0.7  10

b.


E  3.1 eV  3.11.6  10 19   4.96  10 19 J



hc 6.63  10 34 3.0  10 8
E
 

 4.01  10 7 m
19

E
4.96  10
c
3.0  10 8
 7.8  1014 Hz
From eq. c  f  f  
 4.01  10 7
hc
c.
E  1019 eV  1.6 J
hc
hc 6.63  10 34 3.0  10 8
E
 

 1.24  10 25 m

E
1.6
d.
f  400  1012 Hz

p
e.
h






hf
6.63  10 34 400  1012

 8.84  10 28 kg m s-1
c
3.0  10 8
o  680  10 9 m
Wo  hf o 
3.

hc

o
6.63  10 3.0  10   2.93  10
34
8
680  10
19
9
J or 1.83 eV
A rubidium surface has work function of 2.16 eV.
i.
What is the maximum kinetic energy of ejected electrons if the incident
radiation is of wavelength 413 nm?
Ans: (  4.78  10 17 J or 299 eV)
ii.
What is the threshold wavelength for this surface?
Ans:(5.75x107m)
Solutions:
Wo  2.16 eV  2.16 1.6  10 19  3.46  10 19 J,   4.13  10 9 m


i. T .K max  hf  Wo 
ii. Wo  hf o 
4.
hc
o
hc

 o 
 Wo 
6.63  10 3.0  10   3.46  10
34
8
9
4.13  10
 4.78  10 17 J or 299 eV


19
J

hc
6.63  10 34 3.0  10 8

 5.75  10 7 m
Wo
3.46  10 19
When a Cesium metal is shined by the light of wavelength 500 nm, electron
emitted has a maximum kinetic energy of 0.5 eV. Calculate
a.
work function for Cesium.
Ans: (  3.178  10 19 J)
b.
stopping voltage emitted electron produced by the light of wavelength 600
nm.
Ans:( 0.085 V)
Solutions:
a.
  500 nm, T .K max  hf  Wo  Wo  hf  T .K max

6.63  10 3.0  10   0.571.6 10 
34
500  10
 3.178  10 19 J
8
9
19
b.
5.
  600  10 9 m
E  T .K max  Wo
hc
 eVs  Wo

6.63  10 34 3.0  108   3.178  10 19
hc
 eVs 
 Wo 

600  10 9
Vs  0.085 V where e =1.6 x 10-19 J
Use suitable figure to describe how the way to measure Planck Constant, h, by
using Photoelectric Effect experiment.
Solutions:
Anode
Catode
Vs
e
I
V
A
Wo/e
E  T .K max  Wo  hf  eVs  Wo
eVs  hf  Wo
h
hf Wo
, then plot graph Vs vs f, gradient
then the

e
e
e
Planck Constant can be determined.
Vs 
6.
In an experiment of Photoelectric Effect, a metal surface is shined with
electromagnetic ray of wavelength 300 nm. If work function of the metal is 3.6
eV, calculate the maximum velocity of emitted electron.
Ans:(4.37 x 105 m s-1)
9
  300  10 m
Wo  3.6 eV
hc
E  T .K max  Wo 
 T .K max  Wo

hc
6.63  10 34 3.0  10 8
T .K max 
 Wo 
 3.6  1.6  10 19  8.7  10 20 J

300  10 9
1
T .K max  mv 2 max
2



f
vmax 
2T .K max
, m  9.11  10 31 kg
m
2  8.7  10 20
9.11  10 31
 4.37  10 5 m s-1

7.
Stopping Potential for emitted electron when electromagnetic ray with frequency
6.8 x 1014 Hz is shined to a piece of a metal is 1.8 V. What is the maximum
kinetic energy of emitted electrons. Calculate work function of the metal.
Ans:(1.8 eV, 1.02 eV)
Solutions:
Vs  1.8 eV
f  6.8  1014 Hz
T .K max  eVs  1.8 eV
Wo  E  T .K max  hf  T .K max





 6.63  10 34 6.8  1014  1.8 1.6  10 19  1.02
8.
Figure (a) below show a cell of photoelectric that used to investigate the change
of current and voltage. Catode is shined with monochromatic light with
wavelength 360 nm. The result is shown in figure (b).
λ = 360 nm
I (A)
2
V
A
-2
0
2
4
V(volt)
a.
Calculate the maximum kinetic energy for emitted photoelectron.
b.
Determine the work function for the metal.
c.
If the experiment is repeated by using monochromatic light with
wavelength 320 nm, determine the interception at V-axis.
Ans:a)3.20 x 10-19 J,b)2.33 x 10-19 J, c)2.34 V at negative V-axis
Solutions:
  360  10 9 m
a.
Vs  2 V
T .K max  eVs  2  1.6  10 19  3.20  10 19 J
b.
E  T .K max  Wo
Wo  E  T .K max 

 T .K max 
6.63  10 3.0  10   3.20  10
34
8
19
19
360  10
19
 2.33  10 J
Wo  const  2.33  10 19 J
J
  320  10 9 m
hc
T .K max 
 Wo


hc
6.63  10 34 3.0  10 8 
eVs 
 Wo 
 2.33  10 19
9

320  10
Vs  2.34 V at negative V-axis
c.
9.
hc
Determine the maximum kinetic energy ejected from a potassium surface by
ultraviolet light of wavelength 2000 Å. What retarding potential difference is
required to stop the emission of electrons? The photoelectric threshold
wavelength for potassium is 4400 Å.
Ans:(3.39 eV, 3.39 V)
Solutions:

  2000 A  200 nm  200  10 9 m

o  4400 A  440 nm  440  10 9 m
K max  E  Wo
 hf  hf o

hc


hc
o
1 1
 hc 
  o





1
1
 6.63  10 34 3.0  10 8 

9
9 
440  10 
 200  10
 5.42  10 19 J #

 3.39 eV

#

 

K max  eVs
K max
e
3.39 eV

e
 Vs  3.39 V #
Vs 