1. CHAPTER 3:
PHOTOELECTRIC EFFECT
Dr Ahmad Taufek Abdul Rahman
School of Physics & Material Studies
Faculty of Applied Sciences
Universiti Teknologi MARA Malaysia
Campus of Negeri Sembilan
72000 Kuala Pilah, NS
1
PHY310 – Photoelectric Effect
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2. is a phenomenon
where under certain
circumstances a particle exhibits wave properties
and under other conditions a wave exhibits
properties of a particle.
Wave properties of particle
2
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3. Learning Outcome:
26.1 de Broglie wavelength (1 hour)
At the end of this chapter, students should be able to:

State and use formulae for wave-particle duality of
de Broglie,
h

p
3
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4. 3.1 de Broglie wavelength
 From the Planck’s quantum theory, the energy of a photon is
given by
E
hc
(10.1)

 From the Einstein’s special theory of relativity, the energy of a
photon is given by
2
E  mcand mc  p
(10.2)
E  pc
 By equating eqs. (10.1) and (10.2), hence
hc
particle aspect
 pc

h
p

where
4
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(10.3)
wave aspect
p : momentum
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5.  From the eq. (10.3), thus light has momentum and exhibits
particle property. This also show light is dualistic in nature,
behaving is some situations like wave and in others like
particle (photon) and this phenomenon is called wave particle
duality of light.
 Table 10.1 shows the experiment evidences to show wave
particle duality of light.
Wave
Particle
Young’s double slit
experiment
Photoelectric effect
Diffraction experiment
Compton effect
Table 1
 Based on the wave particle duality of light, Louis de Broglie
suggested that matter such as electron and proton might also
have a dual nature.
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6.  He proposed that for any particle of momentum
p should
have a wavelength  given by
h
h
(10.4)
 
p mv
where  : de Broglie wavelength
h : Planck's constant
m : mass of a particle
v : velocityof a particle
Eq. (10.4) is known as de Broglie relation (principle).
 This wave properties of matter is called de Broglie waves or
matter waves.
 The de Broglie relation was confirmed in 1927 when Davisson
and Germer succeeded in diffracting electron which shows that
electrons have wave properties.
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7. Example 1 :
In a photoelectric effect experiment, a light source of
wavelength 550 nm is incident on a sodium surface. Determine the
momentum and the energy of a photon used.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
7

8. Example 1 :
In a photoelectric effect experiment, a light source of
wavelength 550 nm is incident on a sodium surface. Determine the
momentum and the energy of a photon used.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Solution :
  550 10 9 m
By using the de Broglie relation, thus
h

p
6.63  10 34
550  10 9 
p
p  1.21 10 27 kg m s 1
and the energy of the photon is given by
E
hc

6.63 10 3.00 10 
E
34
550  10 9
E  3.62  10 19 J
8
8

9. Example 2 :
Calculate the de Broglie wavelength for
a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.
b. an electron of mass 9.111031 kg moving at 3.25105 m s1.
(Given the Planck’s constant, h =6.631034 J s)
9

10. Example 2 :
Calculate the de Broglie wavelength for
a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.
b. an electron of mass 9.111031 kg moving at 3.25105 m s1.
(Given the Planck’s constant, h =6.631034 J s)
Solution :
a. Given m  77 kg; v  4.1 m s 1
The de Broglie wavelength for the jogger is
6.63  10 34

77 4.1
36
  2.110 m
b. Given m  9.11 10 31 kg; v  3.25  10 5 m s 1
h

mv
The de Broglie wavelength for the electron is
6.63  10 34

9.11  10 31 3.25  10 5



  2.24 10 9 m
10

11. Example 3 :
An electron and a proton have the same speed.
a. Which has the longer de Broglie wavelength? Explain.
b. Calculate the ratio of e/ p.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,
mp=1.671027 kg and e=1.601019 C)
11

12. Example 3 :
An electron and a proton have the same speed.
a. Which has the longer de Broglie wavelength? Explain.
b. Calculate the ratio of e/ p.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,
mp=1.671027 kg and e=1.601019 C)
Solution :
a. From de Broglie relation,
ve  vp  v
h

mv
the de Broglie wavelength is inversely proportional to the
mass of the particle. Since the electron lighter than the mass
of the proton therefore the electron has the longer de Broglie
wavelength.
12

13. Solution :
ve  vp  v
Therefore the ratio of their de Broglie wavelengths is
 h 

m v
e  e 


p  h 


 mp v 


mp

me
1.67  10 27

9.11  10 31
e
 1833
p
13

14. Learning Outcome:
26.2 Electron diffraction (1 hour)
At the end of this chapter, students should be able to:

Describe Davisson-Germer experiment by using a
schematic diagram to show electron diffraction.

Explain the wave behaviour of electron in an electron
microscope and its advantages compared to optical
microscope.
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15. 3.2 Electron diffraction
Davisson-Germer experiment
 Figure 10.1 shows a tube for demonstrating electron diffraction
by Davisson and Germer.
graphite film
anode
screen
diffraction
pattern
e
+4000 V
cathode
electron
diffraction
Figure 10.1: electron diffraction tube
 A beam of accelerated electrons strikes on a layer of graphite
which is extremely thin and a diffraction pattern consisting of
rings is seen on the tube face.
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16.  This experiment proves that the de Broglie relation was right and
the wavelength of the electron is given by
h

mv
where
(10.5)
m : mass of an electron
v : velocity of an electron
 If the velocity of electrons is increased, the rings are seen to
become narrower showing that the wavelength of electrons
decreases with increasing velocity as predicted by de broglie
(eq. 10.5).
 The velocity of electrons are controlled by the applied voltage
V
across anode and cathode i.e.
U K
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PHY310 – Photoelectric Effect
1 2
eV  mv
2
2eV
v
m
(10.6)
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17.  By substituting the eq. (10.6) into eq. (10.5), thus

Note:



17
h
 2eV
m
 m

h

2meV




(10.7)
Electrons are not the only particles which behave as waves.
The diffraction effects are less noticeable with more massive particles
because their momenta are generally much higher and so the
wavelength is correspondingly shorter.
Diffraction of the particles are observed when the wavelength is of the
same order as the spacing between plane of the atom.
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18. Example 4 :
a. An electron is accelerated from rest through a potential difference
of 2000 V. Determine its de Broglie wavelength.
b. An electron and a photon has the same wavelength of 0.21 nm.
Calculate the momentum and energy (in eV) of the electron and
the photon.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
18

19. Example 4 :
a. An electron is accelerated from rest through a potential difference
of 2000 V. Determine its de Broglie wavelength.
b. An electron and a photon has the same wavelength of 0.21 nm.
Calculate the momentum and energy (in eV) of the electron and
the photon.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Solution :
a. Given
V  2000 V
The de Broglie wavelength for the electron is

h
2meV


6.63  10 34


2 9.11  10 31 1.60  10 19 2000
  2.75 10
11
m
19

20. Solution :
b. Given e  p  0.21  10 9 m
For an electron,
h
Its momentum is p 
and its energy is
6.63  10 34
p
0.21  10 9
e
p  3.16 10 24 kg m s 1
p
1
2
K  me vand v 
me
2 2
p

2me
 24 2
3.16  10

2 9.11  10 31
5.48  10 18

1.60  10 19
20
K  34.3 eV





21. Solution :
b. Given e  p  0.21  10 9 m
For a photon,
Its momentum is p  3.16  10 24
and its energy is
hc
kg m s 1
E
p
6.63  10 34 3.00  108

0.21  10 9
9.47  10 16

1.60  10 19
E  5919 eV


21


22. Example 5 :
Compare the de Broglie wavelength of an electron and a proton if
they have the same kinetic energy.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,
mp=1.671027 kg and e=1.601019 C)
22

23. Example 5 :
Compare the de Broglie wavelength of an electron and a proton if
they have the same kinetic energy.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,
mp=1.671027 kg and e=1.601019 C)
Solution :
Ke  Kp  K
By using the de Broglie wavelength formulae, thus


h
2meV
h
and
eV  K
2mK
23

24. Solution :
Ke  Kp  K
Therefore the ratio of their de Broglie wavelengths is

h 


e  2me K 


p  h 


 2m K 
p


mp

me
1.67  10 27

9.11  10 31
e
 42.8
p
24

25. Electron microscope
 A practical device that relies on the wave properties of electrons




25
is electron microscope.
It is similar to optical compound microscope in many aspects.
The advantage of the electron microscope over the optical
microscope is the resolving power of the electron microscope
is much higher than that of an optical microscope.
This is because the electrons can be accelerated to a very high
kinetic energy giving them a very short wavelength λ typically
100 times shorter than those of visible light. Therefore the
diffraction effect of electrons as a wave is much less than that
of light.
As a result, electron microscopes are able to distinguish details
about 100 times smaller.
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26.  In operation, a beam of electrons falls on a thin slice of sample.
 The sample (specimen) to be examined must be very thin (a few




26
micrometres) to minimize the effects such as absorption or
scattering of the electrons.
The electron beam is controlled by electrostatic or magnetic
lenses to focus the beam to an image.
The image is formed on a fluorescent screen.
There are two types of electron microscopes:
 Transmission – produces a two-dimensional image.
 Scanning – produces images with a three-dimensional
quality.
Figures 10.2 and 10.3 are diagram of the transmission electron
microscope and the scanning electron microscope.
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27. 27
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28. Exercise 26.1 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C
1.
a. An electron and a photon have the same wavelengths and
the total energy of the electron is 1.0 MeV. Calculate the
energy of the photon.
b. A particle moves with a speed that is three times that of an
electron. If the ratio of the de Broglie wavelength of this
particle and the electron is 1.813104, calculate the mass
of the particle.
ANS. : 1.621013 J; 1.671027 kg
2. a. An electron that is accelerated from rest through a
potential difference V0 has a de Broglie wavelength 0. If
the electron’s wavelength is doubled, determine the
potential difference requires in terms of V0.
b. Why can an electron microscope resolve smaller objects
than a light microscope?
(Physics, 3rd edition, James S. Walker, Q12 & Q11, p.1029)
28

29. Learning Outcome:
3.1
The photoelectric effect (3 hours)
At the end of this chapter, students should be able to:

Explain the phenomenon of photoelectric effect.

Define threshold frequency, work function and stopping
potential.

Describe and sketch diagram of the photoelectric effect
experimental set-up.

Explain by using graph and equations the observations
of photoelectric effect experiment in terms of the
dependence of :

kinetic energy of photoelectron on the frequency of
light;
29
1
2
mv max  eVs  hf  hf0
2
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30. Learning Outcome:
3.1
The photoelectric effect (3 hours)
At the end of this chapter, students should be able to:

photoelectric current on intensity of incident light;

work function and threshold frequency on the types
of metal surface.
W0  hf 0

30
Explain the failure of wave theory to justify the
photoelectric effect.
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31. 3.1 The photoelectric effect
 is defined as the emission of electron from the surface
of a metal when the EM radiation (light) of higher frequency
strikes its surface.
 Figure 1 shows the emission of the electron from the surface of
the metal after shining by the light.
-
EM
radiation
-
-
- - -
-
photoelectron
-
- -
-
Metal
31
Free electrons
Figure 1
 Photoelectron is defined as an electron emitted from the
surface of the metal when the EM radiation (light) strikes its
surface.
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32. 3.1.1 Photoelectric experiment
 The photoelectric effect can be studied through the experiment
made by Franck Hertz in 1887.
 Figure 2 shows a schematic diagram of an experimental
arrangement for studying the photoelectric effect.
EM radiation (light)
cathode
anode
-
photoelectron
-
-
vacuum
glass
G
V
power supply
32
rheostat
Figure 2
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33.  The set-up apparatus as follows:
 Two conducting electrodes, the anode (positive electric
potential) and the cathode (negative electric potential) are
encased in an evacuated tube (vacuum).
 The monochromatic light of known frequency and intensity is
incident on the cathode.
Explanation of the experiment
 When a monochromatic light of suitable frequency (or
wavelength) shines on the cathode, photoelectrons are emitted.
 These photoelectrons are attracted to the anode and give rise to
the photoelectric current or photocurrent I which is measured by
the galvanometer.
 When the positive voltage (potential difference) across the
cathode and anode is increased, more photoelectrons reach the
anode , thus the photoelectric current increases.
33
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34.  As positive voltage becomes sufficiently large, the photoelectric
current reaches a maximum constant value Im, called saturation
current.
 Saturation current is defined as the maximum constant
value of photocurrent when all the photoelectrons have
reached the anode.
 If the positive voltage is gradually decreased, the photoelectric
current I also decreases slowly. Even at zero voltage there are
still some photoelectrons with sufficient energy reach the anode
and the photoelectric current flows is I0.
 Finally, when the voltage is made negative by reversing the
power supply terminal as shown in Figure 2, the photoelectric
current decreases even further to very low values since most
photoelectrons are repelled by anode which is now negative
electric potential.
34
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35. EM radiation (light)
cathode
anode
-
photoelectron
-
-
vacuum
glass
G
V
power supply
rheostat
Figure 3: reversing power supply terminal
 As the potential of the anode becomes more negative, less
photoelectrons reach the anode thus the photoelectric current
drops until its value equals zero which the electric potential at
this moment is called stopping potential (voltage) Vs.
 Stopping potential is defined as the minimum value of
negative voltage when there are no photoelectrons
reaching the anode.
35
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36. U due to this retarding voltage Vs now
equals the maximum kinetic energy Kmax of the photoelectron.
 The potential energy
U  K max
1
2
(1)
eVs  mv max
2
where m : mass of theelectron
 The variation of photoelectric current
I as a function of the
voltage V can be shown through the graph in Figure 9.4c.
Photoelectric current, I
Im
I0
 Vs
Figure 4
36
PHY310 – Photoelectric Effect
After
0
Voltage ,V
Before reversing the terminal
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37. 3.1.2 Einstein’s theory of photoelectric effect
 A photon is a ‘packet’ of electromagnetic radiation with
particle-like characteristic and carries the energy E given by
E  hf
and this energy is not spread out through the medium.
Work function W0 of a metal
 Is defined as the minimum energy of EM radiation required to
emit an electron from the surface of the metal.
 It depends on the metal used.
 Its formulae is
and Emin  hf 0
W0  Emin
W0  hf 0
(2)
where f0 is called threshold frequency and is defined as the
minimum frequency of EM radiation required to emit an
electron from the surface of the metal.
37
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38.  Since
c=f then the eq. (9.6) can be written as
hc
W0 
(3)
0
where 0 is called threshold wavelength and is defined as the
maximum wavelength of EM radiation required to emit an
electron from the surface of the metal.
 Table 1 shows the work functions of several elements.
Element
Aluminum
2.3
Copper
4.7
Gold
5.1
Silver
PHY310 – Photoelectric Effect
4.3
Sodium
38
Work function (eV)
4.3
Table 1
38
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39. Einstein’s photoelectric equation
 In the photoelectric effect, Einstein summarizes that some of the
energy E imparted by a photon is actually used to release an
electron from the surface of a metal (i.e. to overcome the
binding force) and that the rest appears as the maximum kinetic
energy of the emitted electron (photoelectron). It is given by
E  K max  W0 where
1
2
hf  mv max  W0
2
E  hf and K max
1
2
 mv max
2
(4)
where eq. (4) is known as Einstein’s photoelectric equation.
 Since
Kmax=eVs then the eq. (4) can be written as
hf  eVs  W0
where
(5)
Vs : stoppingvoltage
e : magnitude for charge of electron

40. Note:
 1st case:
hf  WOR
0
f  f0
hf
2nd case: hf
-
 WOR
0

3rd case:
Electron is emitted with maximum
kinetic energy.
f  f0
hf
Figure 5b
K max
W0
Metal
Figure 5a

-
vmax
Metal
hf  WOR
0
-
K max  0
v0
W0 Electron is emitted but maximum
kinetic energy is zero.
f  f0
hf
No electron is emitted.
Figure 5c
40
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Metal
-
W0
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41. Example 3 :
Cadmium has a work function of 4.22 eV. Calculate
a. its threshold frequency,
b. the maximum speed of the photoelectrons when the cadmium is
shined by UV radiation of wavelength 275 nm,
c. the stopping potential.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
41

42. Solution :
a. By using the equation of the work function, thus


W0  4.22 1.60 10 19  6.75 10 19 J
W0  hf 0


6.75 10 19  6.63 10 34 f 0
15
f 0  1.02 10 Hz
42

43. 

Solution : W  4.22 1.60  10 19  6.75  10 19 J
0
b. Given   275  10 9 m
By applying the Einstein’s photoelectric equation, thus
E  K max  W0
hc 1
2
 mv max  W0
 2
6.63  10 34 3.00  108 1
2
 9.11 10 31 vmax  6.75  10 19
2
275  10 9
vmax  3.26 105 m s 1


 

c. The stopping potential is given by
K max
1
2
 mv max
2
1
2
eVs  mv max
2
1
19
1.60  10 Vs  9.11 10 31 3.26 10 5
2
43
Vs  0.303 V





2

44. Example 4 :
A beam of white light containing frequencies between 4.00 1014 Hz
and 7.90 1014 Hz is incident on a sodium surface, which has a
work function of 2.28 eV.
a. Calculate the threshold frequency of the sodium surface.
b. What is the range of frequencies in this beam of light for which
electrons are ejected from the sodium surface?
c. Determine the highest maximum kinetic energy of the
photoelectrons that are ejected from this surface.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
44

45. 
Solution : W  2.28 1.60  10 19
0
a. The threshold frequency is
  3.65 10
19
J
W0  hf 0
3.65 10 19  6.63 10 34 f 0
f 0  5.51 1014 Hz


b. The range of the frequencies that eject electrons is
5.51 1014 Hz and 7.90 1014 Hz
c. For the highest Kmax, take f  7.90  1014 Hz
By applying the Einstein’s photoelectric equation, thus
E  K max  W0
1
2
hf  mv max  W0
2
6.63 10 7.90 10   K
34
14
max
 3.65 10 19
K max  1.59 10 19 J
45

46. Exercise 3.1 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C
1.
The energy of a photon from an electromagnetic wave is
2.25 eV
a. Calculate its wavelength.
b. If this electromagnetic wave shines on a metal, electrons
are emitted with a maximum kinetic energy of 1.10 eV.
Calculate the work function of this metal in joules.
ANS. : 553 nm; 1.841019 J
2. In a photoelectric effect experiment it is observed that no
current flows when the wavelength of EM radiation is greater
than 570 nm. Calculate
a. the work function of this material in electron-volts.
b. the stopping voltage required if light of wavelength 400 nm
is used.
(Physics for scientists & engineers, 3rd edition, Giancoli, Q15, p.974)
ANS. : 2.18 eV; 0.92 V
46

47. Exercise 3.1 :
3.
In an experiment on the photoelectric effect, the following data
were collected.
Wavelength of EM
radiation,  (nm)
Stopping potential,
Vs (V)
350
1.70
450
0.900
a. Calculate the maximum velocity of the photoelectrons
when the wavelength of the incident radiation is 350 nm.
b. Determine the value of the Planck constant from the above
data.
ANS. : 7.73105 m s1; 6.721034 J s
47

48. 3.2
Graph of photoelectric experiment
Variation of photoelectric current I with voltage V
 for the radiation of different intensities but its frequency is
fixed.
I
2I m
Im
 Vs
Intensity 2x
Intensity 1x
0
V
Figure 6
Reason:
From the experiment, the photoelectric current is directly
proportional to the intensity of the radiation as shown in Figure
6.
48
PHY310 – Photoelectric Effect
DR.ATAR @ UiTM.NS

49. I
2I m
Im
0
1
2
Light intensity
Figure 7
 for the radiation of different frequencies but its intensity is
fixed.
I
Im
f2 > f1
f2
f1
 Vs2 Vs1 0
49
PHY310 – Photoelectric Effect
V
Figure 8
49
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50. Reason:
From the Einstein’s photoelectric equation,
Stopping voltage ,Vs hf  eV  W
s
0
W0
h
Vs    f 
e
e
y m x c
Vs2
Vs1
0
W0

e
f 0 f1 f 2
frequency, f
If Vs=0,
hf  e(0)  W0
W0  hf f 0
Figure 9
50
PHY310 – Photoelectric Effect
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51.  For the different metals of cathode but the intensity and
frequency of the radiation are fixed.
I
Im
W02 > W01
W01
Figure 10
W02
V
 Vs1  Vs20
Reason: From the Einstein’s photoelectric equation,
Vs
hf
e
hf  eVs  W0
Vs1
Vs2
0
 1
 hf 
Vs    W0   
 e
 e 
y m x  c
W01 W02
hf  E
W0
Energy of a photon
in EM radiation
Figure 11
51
PHY310 – Photoelectric Effect
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52. Variation of stopping voltage Vs with frequency f of the radiation

for different metals of cathode but the intensity is fixed.
Vs
W01
W02
W03
W03 >W02 > W01
Figure 12
0
f 01
f 02 f 03
f
W0  f 0 Threshold (cut-off)
Reason: Since W0=hf0 then
frequency
W0 If Vs=0,
h
hf  e(0)  W0
Vs    f 
hf  eVs  W0
e
e
W0  hf f 0
y m x c
52
PHY310 – Photoelectric Effect
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53. 3.3
Failure of wave theory of light
 Table 2 shows the classical predictions (wave theory),
photoelectric experimental observation and modern theory
explanation about photoelectric experiment.
Classical predictions
Experimental
observation
Modern theory
Emission of
photoelectrons occur
for all frequencies of
light. Energy of light is
independent of
frequency.
Emission of
photoelectrons occur
only when frequency
of the light exceeds
the certain frequency
which value is
characteristic of the
material being
illuminated.
When the light frequency is
greater than threshold
frequency, a higher rate of
photons striking the metal
surface results in a higher
rate of photoelectrons
emitted. If it is less than
threshold frequency no
photoelectrons are emitted.
Hence the emission of
photoelectrons depend on
the light frequency
53
53
PHY310 – Photoelectric Effect
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54. Classical predictions
Experimental
observation
Modern theory
The higher the
intensity, the greater
the energy imparted to
the metal surface for
emission of
photoelectrons. When
the intensity is low, the
energy of the radiation
is too small for
emission of electrons.
Very low intensity but
high frequency
radiation could emit
photoelectrons. The
maximum kinetic
energy of
photoelectrons is
independent of light
intensity.
The intensity of light is the
number of photons radiated
per unit time on a unit
surface area.
Based on the Einstein’s
photoelectric equation:
54
PHY310 – Photoelectric Effect
K max  hf  W0
The maximum kinetic
energy of photoelectron
depends only on the light
frequency and the work
function. If the light intensity
is doubled, the number of
electrons emitted also
doubled but the maximum
kinetic energy remains
unchanged.
DR.ATAR @ UiTM.NS

55. Classical predictions
Experimental
observation
Modern theory
Light energy is spread
over the wavefront, the
amount of energy
incident on any one
electron is small. An
electron must gather
sufficient energy
before emission, hence
there is time interval
between absorption of
light energy and
emission. Time interval
increases if the light
intensity is low.
Photoelectrons are
emitted from the
surface of the metal
almost
instantaneously after
the surface is
illuminated, even at
very low light
intensities.
The transfer of photon’s
energy to an electron is
instantaneous as its energy
is absorbed in its entirely,
much like a particle to
particle collision. The
emission of photoelectron
is immediate and no time
interval between absorption
of light energy and
emission.
55
PHY310 – Photoelectric Effect
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56. Classical predictions
Experimental
observation
Modern theory
Energy of light
depends only on
amplitude ( or
intensity) and not on
frequency.
Energy of light
depends on
frequency.
According to Planck’s
quantum theory which is
E=hf
Energy of light depends on
its frequency.
Table 2
Note:



56
Experimental observations deviate from classical predictions based on
wave theory of light. Hence the classical physics cannot explain the
phenomenon of photoelectric effect.
The modern theory based on Einstein’s photon theory of light can
explain the phenomenon of photoelectric effect.
It is because Einstein postulated that light is quantized and light is
emitted, transmitted and reabsorbed as photons.
PHY310 – Photoelectric Effect
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57. Example 5 :
a. Why does the existence of a threshold frequency in the
photoelectric effect favor a particle theory for light over a wave
theory?
b. In the photoelectric effect, explains why the stopping potential
depends on the frequency of light but not on the intensity.
57

58. Example 5 :
Solution :
a. Wave theory predicts that the photoelectric effect should occur at
any frequency, provided the light intensity is high enough.
However, as seen in the photoelectric experiments, the light must
have a sufficiently high frequency (greater than the threshold
frequency) for the effect to occur.
b. The stopping voltage measures the kinetic energy of the most
energetic photoelectrons. Each of them has gotten its energy
from a single photon. According to Planck’s quantum theory , the
photon energy depends on the frequency of the light. The
intensity controls only the number of photons reaching a unit area
in a unit time.
58

59. Example 6 :
In a photoelectric experiments, a graph of the light frequency f is
plotted against the maximum kinetic energy Kmax of the
photoelectron as shown in Figure 9.10.
f 1014 Hz
4.83
0
K max (eV)
Figure 13
Based on the graph, for the light of frequency 7.141014 Hz,
calculate
a. the threshold wavelength,
b. the maximum speed of the photoelectron.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
59

60. Solution : f  7.14  1014 Hz
a. By rearranging Einstein’s photoelectric equation,
f 1014 Hz
hf  K max  W0
4.83
0
K max (eV)
W0
1
f    K max 
h
h
1
f    K max  f 0
h
y m x  c
From the graph, f 0  4.83  1014 Hz
Therefore the threshold wavelength is given by
c
0 
f0
3.00  108

4.83  1014
0  6.21 10 7 m
60

61. Solution : f  7.14  1014 Hz
b. By using the Einstein’s photoelectric equation, thus

1
2
hf  mv max  W0
2
1
2
hf  mv max  hf0
2
1
2
mv max  h f  f 0 
2


1
2
9.11 10 31 vmax  6.63 10 34 7.14 1014  4.83 1014
2
vmax  5.80 105 m s 1
61


62. Exercise 25.2 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C
1. A photocell with cathode and anode made of the same metal
connected in a circuit as shown in the Figure 14.
Monochromatic light of wavelength 365 nm shines on the
cathode and the photocurrent I is measured for various values
of voltage V across the cathode and anode. The result is
shown in Figure 15.
I (nA)
365 nm
5
G
V
1
Figure 14
0
Figure 15
V ( V)
62

63. Exercise 25.2 :
1.
a. Calculate the maximum kinetic energy of photoelectron.
b. Deduce the work function of the cathode.
c. If the experiment is repeated with monochromatic light of
wavelength 313 nm, determine the new intercept with the
V-axis for the new graph.
ANS. : 1.601019 J, 3.851019 J; 1.57 V
2. When EM radiation falls on a metal surface, electrons may be
emitted. This is photoelectric effect.
a. Write Einstein’s photoelectric equation, explaining the
meaning of each term.
b. Explain why for a particular metal, electrons are emitted
only when the frequency of the incident radiation is greater
than a certain value?
c. Explain why the maximum speed of the emitted electrons
is independent of the intensity of the incident radiation?
(Advanced Level Physics, 7th edition, Nelkon&Parker, Q6, p.835)
63