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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
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 1
CHAPTER 3:
PHOTOELECTRIC EFFECT
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
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 2
At the end of this chapter, students should be able to:
State and use formulae for wave-particle duality of
de Broglie,
Learning Outcome:
p
h
26.1 de Broglie wavelength (1 hour)
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 3
From the Planck’s quantum theory, the energy of a photon is
given by
From the Einstein’s special theory of relativity, the energy of a
photon is given by
By equating eqs. (10.1) and (10.2), hence
3.1 de Broglie wavelength
hcE (10.1)
2mcE (10.2)
and pmc pcE
pchc
hp particle aspect
wave aspect (10.3)
where momentum: pDR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect
4
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.
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.
Wave Particle
Young’s double slit
experiment
Photoelectric effect
Diffraction experiment Compton effect
Table 1
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 5
He proposed that for any particle of momentum p should
have a wavelength given by
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.
mv
h
p
h
where
(10.4)
h wavelengtBroglie de:
particle a of mass: mparticle a ofvelocity : v
constant sPlanck': h
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 6
7
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 =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Example 1 :
8
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 =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Solution :
By using the de Broglie relation, thus
and the energy of the photon is given by
Example 1 :
m 10550 9
p
h
p
349 1063.6
10550
127 s m kg 1021.1 p
hcE
9
834
10550
1000.31063.6
E
J 1062.3 19E
9
Calculate the de Broglie wavelength for
a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.
b. an electron of mass 9.111031 kg moving at 3.25105 m s1.
(Given the Planck’s constant, h =6.631034 J s)
Example 2 :
10
Calculate the de Broglie wavelength for
a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.
b. an electron of mass 9.111031 kg moving at 3.25105 m s1.
(Given the Planck’s constant, h =6.631034 J s)
Solution :
a. Given
The de Broglie wavelength for the jogger is
b. Given
The de Broglie wavelength for the electron is
Example 2 :
1s m 1.4kg; 77 vm
mv
h
1.477
1063.6 34
m 101.2 361531 s m 1025.3kg; 1011.9 vm
531
34
1025.31011.9
1063.6
m 1024.2 9
11
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C)
Example 3 :
12
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C)
Solution :
a. From de Broglie relation,
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.
Example 3 :
vvv pe
mv
h
13
Solution :
Therefore the ratio of their de Broglie wavelengths is
e
p
m
m
31
27
1011.9
1067.1
1833p
e
vvv pe
vm
h
vm
h
p
e
p
e
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.
Learning Outcome:
26.2 Electron diffraction (1 hour)
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 14
e
+4000 V
Davisson-Germer experiment
Figure 10.1 shows a tube for demonstrating electron diffraction by Davisson and Germer.
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.
3.2 Electron diffraction
screen diffraction
pattern
electron
diffraction
graphite film
anode
cathode
Figure 10.1: electron diffraction tube
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 15
This experiment proves that the de Broglie relation was right and
the wavelength of the electron is given by
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.
mv
h
where electronan of mass: m
(10.5)
electronan ofvelocity : v
KU 2
2
1mveV
m
eVv
2 (10.6)
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 16
By substituting the eq. (10.6) into eq. (10.5), thus
m
eVm
h
2
meV
h
2 (10.7)
Note:
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.
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 17
18
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Example 4 :
19
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Solution :
a. Given
The de Broglie wavelength for the electron is
Example 4 :
V 2000V
meV
h
2
20001060.11011.92
1063.6
1931
34
m 1075.2 11
20
Solution :
b. Given
For an electron,
Its momentum is
and its energy is
m 1021.0 9pe
e
hp
9
34
1021.0
1063.6
p
124 s m kg 1016.3 p2
e2
1vmK
31
224
1011.92
1016.3
19
18
1060.1
1048.5
and
em
pv
e
2
2m
p
eV 3.34K
21
Solution :
b. Given
For a photon,
Its momentum is
and its energy is
m 1021.0 9pe
124 s m kg 1016.3 p
p
hcE
9
834
1021.0
1000.31063.6
19
16
1060.1
1047.9
eV 5919E
22
Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C)
Example 5 :
23
Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C)
Solution :
By using the de Broglie wavelength formulae, thus
Example 5 :
KKK pe
meV
h
2
mK
h
2
and KeV
24
Solution :
Therefore the ratio of their de Broglie wavelengths is
KKK pe
Km
h
Km
h
e
p
p
e
2
2
e
p
m
m
31
27
1011.9
1067.1
8.42p
e
A practical device that relies on the wave properties of electrons
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.
Electron microscope
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 25
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
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.
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 26
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 27
28
Exercise 26.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 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 1.813104, calculate the mass of the particle.
ANS. : 1.621013 J; 1.671027 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)
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;
Learning Outcome:
3.1 The photoelectric effect (3 hours)
0s
2
max2
1hfhfeVmv
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 29
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.
Explain the failure of wave theory to justify the
photoelectric effect.
Learning Outcome:
3.1 The photoelectric effect (3 hours)
00 hfW
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 30
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.
Photoelectron is defined as an electron emitted from the surface of the metal when the EM radiation (light) strikes its surface.
3.1 The photoelectric effect
Figure 1
EM
radiation
- photoelectron
- - - - - - - - - -
Metal
Free electrons
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 31
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.
3.1.1 Photoelectric experiment
- -
-
EM radiation (light)
anode cathode
glass
rheostat power supply
vacuum photoelectron
Figure 2
G
V
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 32
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.
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 33
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.
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 34
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.
Figure 3: reversing power supply terminal
- -
-
EM radiation (light)
anode cathode
glass
rheostat power supply
vacuum photoelectron
G
V
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 35
The potential energy U due to this retarding voltage Vs now
equals the maximum kinetic energy Kmax of the photoelectron.
The variation of photoelectric current I as a function of the
voltage V can be shown through the graph in Figure 9.4c.
maxKU
2
maxs2
1mveV (1)
electron theof mass: mwhere
mI
0I
sV
I,current ricPhotoelect
V,Voltage0
Before reversing the terminal After Figure 4
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 36
A photon is a ‘packet’ of electromagnetic radiation with particle-like characteristic and carries the energy E given by
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
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.
3.1.2 Einstein’s theory of photoelectric effect
hfE
min0 EW
00 hfW
and 0min hfE
(2)
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 37
38
Since c=f then the eq. (9.6) can be written as
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.
0
0
hcW (3)
Element Work function (eV)
Aluminum 4.3
Sodium 2.3
Copper 4.7
Gold 5.1
Silver 4.3
Table 1 DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 38
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
where eq. (4) is known as Einstein’s photoelectric equation.
Since Kmax=eVs then the eq. (4) can be written as
where and 0max WKE hfE
0
2
max2
1Wmvhf
2
maxmax2
1mvK
(4)
0s WeVhf (5)
voltagestopping: sVwhere
electron of chargefor magnitude: e
Note:
1st case: OR 0Whf 0ff
Electron is emitted with maximum
kinetic energy. - Metal
hf
0W
- maxv
maxK
2nd case: OR 0Whf 0ff
Electron is emitted but maximum
kinetic energy is zero.
- 0v 0max K
3rd case: OR 0Whf 0ff
No electron is emitted.
- Metal
hf
0W
- Metal 0W
hf
Figure 5a
Figure 5b
Figure 5c
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 40
41
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Example 3 :
42
Solution :
a. By using the equation of the work function, thus
J 1075.61060.122.4 19190
W
00 hfW
03419 1063.61075.6 f
Hz 10 02.1 150 f
43
Solution :
b. Given
By applying the Einstein’s photoelectric equation, thus
c. The stopping potential is given by
m 10275 9
0
2
max2
1Wmv
hc
0max WKE
15max s m 1026.3 v
J 1075.61060.122.4 19190
W
192
max31
9
834
1075.61011.92
1
10275
1000.31063.6
v
2
maxs2
1mveV
2
maxmax2
1mvK
2531s
19 1026.31011.92
11060.1 V
V 303.0sV
44
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Example 4 :
45
Solution :
a. The threshold frequency is
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
By applying the Einstein’s photoelectric equation, thus
03419 1063.61065.3 f
00 hfW
Hz 1051.5 140 f
J 1065.31060.128.2 19190
W
Hz 1090.7 14f
0
2
max2
1Wmvhf
0max WKE
J 1059.1 19max
K
19max
1434 1065.31090.71063.6 K
46
Exercise 3.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 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; 1.841019 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
47
Exercise 3.1 :
3. In an experiment on the photoelectric effect, the following data
were collected.
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. : 7.73105 m s1; 6.721034 J s
Wavelength of EM
radiation, (nm)
Stopping potential,
Vs (V)
350 1.70
450 0.900
Variation of photoelectric current I with voltage V
for the radiation of different intensities but its frequency is
fixed.
Reason:
From the experiment, the photoelectric current is directly
proportional to the intensity of the radiation as shown in Figure
6.
3.2 Graph of photoelectric experiment
Intensity 2x
mI
I
V0
sV
Intensity 1x
m2I
Figure 6
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 48
49
for the radiation of different frequencies but its intensity is
fixed.
Figure 7
I
intensityLight 0 1
mI
m2I
2
mI
Figure 8
I
V0s1V
f1
f2
s2V
f2 > f1
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 49
Reason:
From the Einstein’s photoelectric equation,
Figure 9
0s WeVhf e
Wf
e
hV 0
s
y xm c
e
W0
f,frequency
s, voltageStopping V
02f
s2V
1f
s1V
If Vs=0, 0)0( Wehf
hfW 0 0f
0f
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 50
For the different metals of cathode but the intensity and
frequency of the radiation are fixed.
Reason: From the Einstein’s photoelectric equation,
Figure 10
mI
s1V
01W
s2V
02W
W02 > W01
0s WeVhf
e
hfW
eV 0s
1
e
hf
0W
sV
0 Ehf 01W
1sV
02W
s2VEnergy of a photon
in EM radiation
I
V0
y xm c
Figure 11
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 51
Variation of stopping voltage Vs with frequency f of the radiation for different metals of cathode but the intensity is fixed.
Reason: Since W0=hf0 then
Figure 12
W03 >W02 > W01
01f
W01
02f
W02
03f
W03
f
sV
0
00 fW
0s WeVhf e
Wf
e
hV 0
s
y xm c
If Vs=0, 0)0( Wehf
hfW 0 0f
Threshold (cut-off)
frequency
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 52
53
Table 2 shows the classical predictions (wave theory),
photoelectric experimental observation and modern theory
explanation about photoelectric experiment.
3.3 Failure of wave theory of light
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
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 53
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:
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.
0WhfK max
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 54
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.
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 55
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:
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.
DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 56
57
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.
Example 5 :
58
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.
Example 5 :
59
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.
Based on the graph, for the light of frequency 7.141014 Hz, calculate
a. the threshold wavelength,
b. the maximum speed of the photoelectron.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Example 6 :
Hz1014f
83.4
)eV(maxK0
Figure 13
60
Solution :
a. By rearranging Einstein’s photoelectric equation,
From the graph,
Therefore the threshold wavelength is given by
Hz 1014.7 14f
Hz1014f
83.4
)eV(maxK0
0max WKhf h
WK
hf 0
max
1
y xm c
0max
1fK
hf
Hz 1083.4 140 f
0
0f
c
14
8
1083.4
1000.3
m 1021.6 70
61
Solution :
b. By using the Einstein’s photoelectric equation, thus
Hz 1014.7 14f
0
2
max2
1Wmvhf
0
2
max2
1hfmvhf
0
2
max2
1ffhmv
1414342
max31 1083.41014.71063.61011.9
2
1 v
15max s m 1080.5 v
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Exercise 25.2 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 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.
365 nm
V
G 5
1
)nA(I
)V(V0Figure 14 Figure 15
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. : 1.601019 J, 3.851019 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)