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
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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
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.
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
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.
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
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.
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.
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
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.
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.
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
62
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?