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PHYSICS CHAPTER 24
1
CHAPTER 24:
Quantization of light(3 Hours)
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PHYSICS CHAPTER 24
Learning Outcome:
At the end of this chapter, students should be able to:
a. Distinguish between Planks quantum theory and
classical theory of energy
b. Use Einsteins formulae for photon energy,
2
24.1 Plancks quantum theory (1/2 hour)
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PHYSICS CHAPTER 24
24.1 Plancks quantum theory
24.1.1 Classical theory of black body radiation
Black body is defined as an ideal system that absorbs all the
radiation incident on it. The electromagnetic (EM) radiation
emitted by the black body is called black body radiation.
From the black body experiment, the distribution of energy inblack body, Edepends only on the temperature, T.
If the temperature increases thus the energy of the black body
increases and vice versa.
3
(24.1)
constantsBoltzmann':Bkwhere
kelvininetemperatur:T
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PHYSICS CHAPTER 24
The spectrum of EM radiation emitted by the black body
(experimental result) is shown in Figure 24.1.
From the curve, Wiens theory was accurate at short
wavelengths but deviated at longer wavelengths whereas the
reverse was true for the Rayleigh-Jeans theory. 4
Figure 24.1
Experimental
result
Rayleigh -Jeans
theoryWiens theory
Classical
physics
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PHYSICS CHAPTER 24
The Rayleigh-Jeans and Wiens theories failed to fit the
experimental curve because this two theories based on classical
ideas which are Energy of the EM radiation is not depend on its frequency
or wavelength.
Energy of the EM radiation is continuously.
24.1.2 Plancks quantum theory In 1900, Max Planck proposed his theory that is fit with the
experimental curve in Figure 24.1 at all wavelengths known as
Plancks quantum theory.
The assumptions made by Planck in his theory are :
The EM radiation emitted by the black body is in discrete
(separate) packets of energy. Each packet is called a
quantum of energy. This means the energy of EM radiation
is quantised.
The energy size of the radiation depends on its frequency.
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PHYSICS CHAPTER 24
According to this assumptions, the quantum of the energy E
for radiation of frequency fis given by
Since the speed of EM radiation in a vacuum is
then eq. (24.2) can be written as
From eq. (24.3), the quantum of the energy Efor radiation is
inversely proportional to its wavelength.
6
where
(24.2)
(24.3)
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PHYSICS CHAPTER 24
It is convenient to express many quantum energies in electron-
volts.
The electron-volt (eV) is a unit of energy that can be definedas the kinetic energy gained by an electron in being
accelerated by a potential difference (voltage) of 1 volt.
Unit conversion:
In 1905, Albert Einstein extended Plancks idea by proposing
that electromagnetic radiation is also quantised. It consists of
particle like packets (bundles) of energy called photons of
electromagnetic radiation.
7
J101.60eV119
Note:
For EM radiation of n packets, the energyEn is given by
(24.4)
1,2,3,...numberreal: nwhere
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PHYSICS CHAPTER 24
24.1.3 Photon
Photon is defined as a particle with zero mass consisting of aquantum of electromagnetic radiation where its energy is
concentrated.
A photon may also be regarded as a unit of energy equal to
hf.
Photons travel at the speed of light in a vacuum. They arerequired to explain the photoelectric effect and other
phenomena that require light to have particle property.
Table 9.1 shows the differences between the photon andelectromagnetic wave.
8
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PHYSICS CHAPTER 24
EM Wave Photon
9
Energy of the EM wavedepends on the intensityof the wave. Intensity of
the waveIis proportionalto the squared of its
amplitudeA2 where
Energy of a photon isproportional to thefrequency of the EMwave where
Its energy is continuouslyand spread out throughthe medium as shown inFigure 24.2a.
Its energy is discrete asshown in Figure 24.2b.
Table 24.1
2AI
fE
Photon
Figure 24.2a Figure 24.2b
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PHYSICS CHAPTER 24
10
A photon of the green light has a wavelength of 740 nm. Calculate
a. the photons frequency,b. the photons energy in joule and electron-volt.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Plancks constant, h =6.631034 J s)
Solution :a. The frequency of the photon is given by
b. By applying the Plancks quantum theory, thus the photons
energy in joule is
and its energy in electron-volt is
Example 24.1 :
m107409
fc f98 107401000.3
hfE 1434 1005.41063.6 E
101.60
1069.2
19
19
E
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PHYSICS CHAPTER 24
11
For a gamma radiation of wavelength 4.621012 m propagates in
the air, calculate the energy of a photon for gamma radiation inelectron-volt.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Plancks constant, h =6.631034 J s)
Solution :
By applying the Plancks quantum theory, thus the energy of a
photon in electron-volt is
Example 24.2 :
m1062.4
12
hcE
12
834
1062.4
1000.31063.6
E
J1031.4 14E
101.60
1031.419
14
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PHYSICS CHAPTER 24
Learning Outcome:
At the end of this chapter, students should be able to:
a) Explain the phenomenon of photoelectric effect.
b) Define threshold frequency, work function and stopping
potential.
c) Describe and sketch diagram of the photoelectric effect
experimental set-up.
d) Explain the failure of wave theory to justify the
photoelectric effect.
12
24.2 The photoelectric effect (3 hours)
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PHYSICS CHAPTER 24
Learning Outcome ( Cont..):
At the end of this chapter, students should be able to:
c) Explain by using graph and equations the
observation of photoelectric effect experiment in
terms of the dependence of :
i. Kinetic energy of photoelectron on the frequency of
light
ii. Photoelectric current on the intensity of incident
light
iii. Work function and threshold frequency on the types
of metal surface
13
24.2 The photoelectric effect (3 hours)
0s
2
max2
1hfhfeVmv
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PHYSICS CHAPTER 24
Learning Outcome ( Cont..):
At the end of this chapter, students should be able to:
f. Use Einsteins photoelectric equation
14
24.2 The photoelectric effect (3 hours)
WhfeVK s max
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PHYSICS CHAPTER 24
24.2 The photoelectric effect
is defined as the emission of electron from the surfaceof a metal when the EM radiation (light) of higher frequency
strikes its surface.
Figure 24.3 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 itssurface. 15
Figure 24.3
EM
radiation- photoelectron
- - - - - - - - - -
Metal
Free electrons
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PHYSICS CHAPTER 24
24.2.1 Photoelectric experiment
The photoelectric effect can be studied through the experiment
made by Franck Hertz in 1887.
Figure 24.4a shows a schematic diagram of an experimental
arrangement for studying the photoelectric effect.
16
--
-
EM radiation (light)
anodecathode
glass
rheostatpower supply
vacuumphotoelectron
Figure 24.4a
G
V
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PHYSICS CHAPTER 24
The set-up apparatus as follows:
Two conducting electrodes, the anode (positive electric
potential) and the cathode (negative electric potential) areencased 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 photocurrentIwhich is measured bythe galvanometer.
When the positive voltage (potential difference) across the
cathode and anode is increased, more photoelectrons reach the
anode , thus the photoelectric current increases.
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PHYSICS CHAPTER 24
As positive voltage becomes sufficiently large, the photoelectric
current reaches a maximum constant valueIm
, called
saturation current. Saturation current is defined as the maximum constant
value of photocurrent when all the photoelectrons havereached the anode.
If the positive voltage is gradually decreased, the photoelectric
currentIalso decreases slowly. Even at zero voltage there arestill some photoelectrons with sufficient energy reach the anode
and the photoelectric current flows isI0.
Finally, when the voltage is made negative by reversing thepower supply terminal as shown in Figure 24.4b, the
photoelectric current decreases even further to very low valuessince most photoelectrons are repelled by anode which isnow negative electric potential.
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PHYSICS CHAPTER 24
As the potential of the anode becomes more negative, less
photoelectrons reach the anode thus the photoelectric currentdrops 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. 19
Figure 24.4b: reversing power supply terminal
--
-
EM radiation (light)
anodecathode
glass
rheostatpower supply
vacuumphotoelectron
G
V
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PHYSICS CHAPTER 24
The potential energy Udue to this retarding voltage Vs
now
equals the maximum kinetic energyKmax
of the photoelectron.
The variation of photoelectric currentIas a function of the
voltage Vcan be shown through the graph in Figure 9.4c.
20
maxKU(24.5)
electrontheofmass:mwhere
mI
0I
sV
I,currentricPhotoelect
V,Voltage0
Before reversing the terminalAfterFigure 24.4c
Simulation 9.1
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PHYSICS CHAPTER 24
24.2.2 Einsteins theory of photoelectric effect
A photon is a packet of electromagnetic radiation with
particle-like characteristic and carries the energyEgiven 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 requiredto emit an electron from the surface of the metal.
It depends on the metal used.
Its formulae is
wheref0
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. 21
hfE
min0 EW and 0min hfE
(24.6)
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PHYSICS CHAPTER 24
Since c=f then the eq. (24.6) can be written as
where 0
is called threshold wavelength and is defined as the
maximum wavelength of EM radiation required to emit anelectron from the surface of the metal.
Table 24.2 shows the work functions of several elements.
22
(24.7)
Element Work function (eV)
Aluminum 4.3
Sodium 2.3Copper 4.7
Gold 5.1
Silver 4.3
Table 24.2
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PHYSICS CHAPTER 24Einsteins photoelectric equation
In the photoelectric effect, Einstein summarizes that some of the
energy Eimparted by a photon is actually used to release anelectron 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. (24.8) is known as Einsteins photoelectric equation.
SinceKmax
=eVsthen the eq. (24.8) can be written as
23
where and0max WKE hfE
(24.8)
(24.9)
voltagestopping:sVwhere
electronofchargeformagnitude:e
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PHYSICS CHAPTER 24
1st case:
24
Note:
OR0Whf 0ff
Electron is emitted with maximum
kinetic energy.-Metal
hf
0W
-maxv maxK
2nd case: OR0Whf 0ff
Electron is emitted but maximum
kinetic energy is zero.
- 0v 0max K
3rd case: OR0Whf 0ff
No electron is emitted.
-Metal
hf
0W
-Metal 0
W
hf
Figure 24.5a
Figure 24.5b
Figure 24.5c
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PHYSICS CHAPTER 24
25
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 s1
, h =6.63
1034
J s, me=9.11
1031
kg ande=1.601019 C)
Solution :
a. By using the equation of the work function, thus
Example 24.3 :
J1075.61060.122.4 19190 W
00 hfW
03419 1063.61075.6 f
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PHYSICS CHAPTER 24
26
Solution :
b. Given
By applying the Einsteins photoelectric equation, thus
c. The stopping potential is given by
m10275 9
0
2
max2
1Wmv
hc
0max WKE
J1075.61060.122.4 19190 W
192max319834
1075.61011.92
1
10275
1000.31063.6
v
2
maxs2
1mveV
2
maxmax2
1mvK
2531s19 1026.31011.92
11060.1 V
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PHYSICS CHAPTER 24
27
A beam of white light containing frequencies between 4.00
10
14
Hzand 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 24.4 :
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PHYSICS CHAPTER 24
28
Solution :
a. The threshold frequency is
b. The range of the frequencies that eject electrons is
c. For the highestKmax
, take
By applying the Einsteins photoelectric equation, thus
03419 1063.61065.3 f 00 hfW
J1065.31060.128.2 19190 W
Hz1090.7 14f
0
2
max2
1 Wmvhf
0max WKE
19max1434 1065.31090.71063.6 K
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PHYSICS CHAPTER 24
29
Exercise 24.1 :Given c =3.00108 m s1, h =6.631034 J s, m
e=9.111031 kg and
e=1.601019 C1. 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 nocurrent flows when the wavelength of EM radiation is greaterthan 570 nm. Calculate
a. the work function of this material in electron-volts.
b. the stopping voltage required if light of wavelength 400 nmis used.
(Physics for scientists & engineers, 3rd edition, Giancoli, Q15,p.974)
ANS. : 2.18 eV; 0.92 V
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30
Exercise 24.1 :
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
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PHYSICS CHAPTER 24
24.2.3 Graph of photoelectric experiment
Variation of photoelectric currentI
with voltageV
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 24.6b. 31
Intensity 2x
mI
I
V0sV
Intensity 1x
m2I
Figure 24.6a
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PHYSICS CHAPTER 24
for the radiation of different frequencies but its intensity is
fixed.
32
Figure 24.6b
I
intensityLight0 1
mIm2I
2
mI
Figure 24.7a
I
V
0s1V
f1
f2
s2V
f2> f
1
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PHYSICS CHAPTER 24
Reason:
From the Einsteins photoelectric equation,
33
Figure 24.7b
0s WeVhf e
Wf
e
hV 0s
y xm c
e
W0
f,frequency
s,voltageStopping V
02
f
s2V
1f
s1V
IfVs=0, 0)0( Wehf
hfW 0 0f
0f
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PHYSICS CHAPTER 24
For the different metals of cathode but the intensity and
frequency of the radiation are fixed.
Reason: From the Einsteins photoelectric equation,
34
Figure 24.8a
mI
s1V
01W
s2V
02W
W02> W
01
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 24.8b
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PHYSICS CHAPTER 24
Variation of stopping voltage Vs
with frequency fof the radiation
for different metals of cathode but the intensity is fixed.
Reason: Since W0=hf0 then
35
Figure 24.9W03 >W02 > W01
01f
W01
02f
W02
03f
W03
f
sV
0
00 fW
0s WeVhf e
Wf
e
hV 0s
y xm c
IfVs=0, 0)0( Wehf
hfW 0 0f
Threshold (cut-off)frequency
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PHYSICS CHAPTER 24
24.2.4 Failure of wave theory of light
Table 24.3 shows the classical predictions (wave theory),
photoelectric experimental observation and modern theoryexplanation about photoelectric experiment.
36
Classical predictions Experimental
observation
Modern theory
Emission of
photoelectrons occurfor all frequencies of
light. Energy of light is
independent of
frequency.
Emission of
photoelectrons occuronly when frequency
of the light exceeds
the certain frequency
which value is
characteristic of thematerial being
illuminated.
When the light frequency is
greater than thresholdfrequency, a higher rate of
photons striking the metal
surface results in a higher
rate of photoelectrons
emitted. If it is less thanthreshold frequency no
photoelectrons are emitted.
Hence the emission of
photoelectrons depend on
the light frequency
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37
Classical predictions Experimental
observation
Modern theory
The higher theintensity, the greater
the energy imparted to
the metal surface for
emission of
photoelectrons. Whenthe intensity is low, the
energy of the radiation
is too small for
emission of electrons.
Very low intensity buthigh frequency
radiation could emit
photoelectrons. The
maximum kinetic
energy ofphotoelectrons is
independent of light
intensity.
The intensity of light is thenumber of photonsradiated per unit time on aunit surface area.
Based on the Einsteins
photoelectric equation:
The maximum kineticenergy of photoelectrondepends only on the lightfrequency and the workfunction. If the lightintensity is doubled, thenumber of electrons emittedalso doubled but themaximum kinetic energy
remains unchanged.
0WhfK max
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38
Classical predictions Experimental
observation
Modern theory
Light energy is spreadover the wavefront, the
amount of energy
incident on any one
electron is small. An
electron must gathersufficient 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 areemitted from the
surface of the metal
almost
instantaneously
after the surface isilluminated, even at
very low light
intensities.
The transfer of photonsenergy to an electron is
instantaneous as its energy
is absorbed in its entirely,
much like a particle to
particle collision. Theemission of photoelectron
is immediate and no time
interval between
absorption of light energy
and emission.
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PHYSICS CHAPTER 24
39
Classical predictions Experimental
observation
Modern theory
Energy of lightdepends only on
amplitude ( or
intensity) and not on
frequency.
Energy of lightdepends on
frequency.
According to Plancksquantum theory which is
E=hf
Energy of light depends on
its frequency.
Table 24.3Note:
Experimental observations deviate from classical predictions based onwave theory of light. Hence the classical physics cannot explain thephenomenon of photoelectric effect.
The modern theory based on Einsteins photon theory of light canexplain the phenomenon of photoelectric effect.
It is because Einstein postulated that light is quantized and light isemitted, transmitted and reabsorbed as photons.
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40
a. Why does the existence of a threshold frequency in the
photoelectric effect favor a particle theory for light over a wavetheory?
b. In the photoelectric effect, explains why the stopping potential
depends on the frequency of light but not on the intensity.
Solution :
a. Wave theory predicts that the photoelectric effect should occur atany 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 Plancks 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 24.5 :
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41
In a photoelectric experiments, a graph of the light frequencyfisplotted against the maximum kinetic energyK
maxof the
photoelectron as shown in Figure 24.10.
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 =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C)
Example 24.6 :
Hz1014f
83.4
)eV(maxK0Figure 24.10
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42
Solution :
a. By rearranging Einsteins photoelectric equation,
From the graph,
Therefore the threshold wavelength is given by
Hz1014.7 14f
Hz1014f
83.4
)eV(maxK0
0max WKhf hWK
hf 0max
1
y xm c
0max
1fK
hf
Hz1083.4 140 f
0
0f
c
14
8
1083.4
1000.3
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43
Solution :
b. By using the Einsteins photoelectric equation, thus
Hz1014.7 14f
02
max21 Wmvhf
0
2
max2
1hfmvhf
02
max2
1ffhmv
1414342max31 1083.41014.71063.61011.92
1 v
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44
Exercise 24.2 :
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 metalconnected in a circuit as shown in the Figure 24.11a.Monochromatic light of wavelength 365 nm shines on the
cathode and the photocurrentIis measured for various values
of voltage Vacross the cathode and anode. The result isshown in Figure 24.11b
365 nm
V
G 5
1
)nA(I
)V(V0
Figure 24.11a Figure 24.11b
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45
Exercise 24.2 :
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 V2. When EM radiation falls on a metal surface, electrons may be
emitted. This is photoelectric effect.
a. Write Einsteins photoelectric equation, explaining the
meaning of each term.
b. Explain why for a particular metal, electrons are emittedonly 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)
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Next ChapterCHAPTER 25 :
Wave properties of particle
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