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PHYSICS CHAPTER 9
CHAPTER 9: CHAPTER 9: Quantization of lightQuantization of light
(4 Hours)(4 Hours)
PHYSICS CHAPTER 9
2
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Explain brieflyExplain briefly Planck’s quantum theory and classical Planck’s quantum theory and classical
theory of energy.theory of energy. Write and useWrite and use Einstein’s formulae for photon energy, Einstein’s formulae for photon energy,
Learning Outcome:
9.1 Planck’s quantum theory (1 hour)
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PHYSICS CHAPTER 9
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9.1.1 Classical theory of black body radiation Black body is defined as an ideal system that absorbs all the an ideal system that absorbs all the
radiation incident on itradiation incident on it. The electromagnetic (EM) radiation electromagnetic (EM) radiation emitted by the black bodyemitted by the black body is called black body radiationblack body radiation.
From the black body experiment, the distribution of energy in energy in
black body, black body, EE depends only on the temperature, depends only on the temperature, TT.
If the temperature increases thus the energy of the black body increases and vice versa.
9.1 Planck’s quantum theory
TkE B (9.1)(9.1)
constant sBoltzmann': Bkwhere
kelvinin etemperatur: T
PHYSICS CHAPTER 9
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The spectrum of EM radiation emitted by the black body (experimental result) is shown in Figure 9.1.
From the curve, Wien’s theory was accurate at short wavelengths but deviated at longer wavelengths whereas the reverse was true for the Rayleigh-Jeans theory.
Figure 9.1Figure 9.1
Experimental result
Rayleigh -Jeans theory
Wien’s theory
Classical Classical physicsphysics
PHYSICS CHAPTER 9
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The Rayleigh-Jeans and Wien’s theories failed to fit the experimental curve because this two theories based on classical ideas which are EnergyEnergy of the EM radiation is not dependnot depend on its frequencyfrequency
or wavelengthwavelength. EnergyEnergy of the EM radiation is continuouslycontinuously.
9.1.2 Planck’s quantum theory In 1900, Max Planck proposed his theory that is fit with the
experimental curve in Figure 9.1 at all wavelengths known as Planck’s quantum theory.
The assumptions made by Planck in his theory are : The EM radiation emitted by the black body is in discrete discrete
(separate) packets of energy(separate) packets of energy. Each packet is called a quantum of energyquantum of energy. This means the energy of EM radiation is quantisedquantised.
The energy size of the radiation dependsdepends on its frequencyfrequency.
PHYSICS CHAPTER 9
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According to this assumptions, the quantum of the energy quantum of the energy EE
for radiation of frequency for radiation of frequency ff is given by
Since the speed of EM radiation in a vacuum is
then eq. (9.2) can be written as
From eq. (9.3), the quantum quantum of the energy EE for radiation is inversely proportional to its wavelengthinversely proportional to its wavelength.
hfE
s J 1063.6constant sPlanck': 34hwhere
(9.2)(9.2)
fc
hc
E (9.3)(9.3)
PHYSICS CHAPTER 9
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It is convenient to express many quantum energies in electron-volts.
The electron-volt (eV)electron-volt (eV) is a unit of energyunit of energy that can be defined as the kinetic energy gained by an electron in being the kinetic energy gained by an electron in being accelerated by a potential difference (voltage) of 1 voltaccelerated by a potential difference (voltage) of 1 volt.
Unit conversion:
In 1905, Albert Einstein extended Planck’s idea by proposing that electromagnetic radiation is also quantised. It consists of particle like packets (bundles) of energy called photonsphotons of electromagnetic radiation.
J 101.60eV 1 19
Note:Note:
For EM radiation of n packets, the energy En is given by
nhfEn (9.4)(9.4)
1,2,3,...number real: nwhere
PHYSICS CHAPTER 9
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Photon is defined as a particle with zero mass consisting of a particle with zero mass consisting of a quantum of electromagnetic radiation where its energy is a quantum of electromagnetic radiation where its energy is concentratedconcentrated.
A photon may also be regarded as a unit of energy equal to unit of energy equal to
hfhf. Photons travel at the speed of lightspeed of light in a vacuum. They are
required to explain the photoelectric effectexplain the photoelectric effect and other phenomena that require light to have particle propertylight to have particle property.
Table 9.1 shows the differences between the photon and electromagnetic wave.
9.1.3 Photon
PHYSICS CHAPTER 9
9
EM Wave Photon
Energy of the EM wave depends on the intensity of the wave. Intensity of the wave I is proportional to the squared of its amplitude A2 where
Energy of a photon is proportional to the frequency of the EM wave where
Its energy is continuously and spread out through the medium as shown in Figure 9.2a.
Its energy is discrete as shown in Figure 9.2b.
Table 9.1Table 9.1
2AI
fE
PhotonFigure 9.2aFigure 9.2a Figure 9.2bFigure 9.2b
PHYSICS CHAPTER 9
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A photon of the green light has a wavelength of 740 nm. Calculate a. the photon’s frequency,
b. the photon’s energy in joule and electron-volt.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Solution :Solution :
a. The frequency of the photon is given by
b. By applying the Planck’s quantum theory, thus the photon’s
energy in joule is
and its energy in electron-volt is
Example 1 :
m 10740 9
fc f98 107401000.3 Hz 1005.4 14f
hfE 1434 1005.41063.6 EJ 1069.2 19E
101.60
1069.219
19
E eV 66.1E
PHYSICS CHAPTER 9
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For a gamma radiation of wavelength 4.621012 m propagates in the air, calculate the energy of a photon for gamma radiation in electron-volt.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck’s constant, h =6.631034 J s)
Solution :Solution :
By applying the Planck’s quantum theory, thus the energy of a photon in electron-volt is
Example 2 :
m 1062.4 12
hc
E 12
834
1062.4
1000.31063.6
E
J 1031.4 14E
101.60
1031.419
14
eV 10 69.2 5E
PHYSICS CHAPTER 9
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain the phenomenon of photoelectric effect. the phenomenon of photoelectric effect. Define Define threshold frequency, work function and stopping threshold frequency, work function and stopping
potential.potential. Describe and sketchDescribe and sketch diagram of the photoelectric effect diagram of the photoelectric effect
experimental set-up.experimental set-up. Explain by using graph and equationsExplain by using graph and equations the observations the observations
of photoelectric effect experiment in terms of the of photoelectric effect experiment in terms of the dependence of :dependence of : kinetic energy of photoelectron on the frequency of kinetic energy of photoelectron on the frequency of
light; light;
Learning Outcome:
9.2 The photoelectric effect (3 hours)
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0s2
max2
1hfhfeVmv
PHYSICS CHAPTER 9
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: photoelectric current on intensity of incident light;photoelectric current on intensity of incident light; work function and threshold frequency on the types work function and threshold frequency on the types
of metal surface.of metal surface.
ExplainExplain the failure of wave theory to justify the the failure of wave theory to justify the photoelectric effect. photoelectric effect.
Learning Outcome:
9.2 The photoelectric effect (3 hours)
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00 hfW
PHYSICS CHAPTER 9
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is defined as the emission of electron from the surface emission of electron from the surface of a metal when the EM radiation (light) of of a metal when the EM radiation (light) of
higher frequency strikes its surfacehigher frequency strikes its surface. Figure 9.3 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 an electron emitted from the surface of the metal when the EM radiation (light) strikes its surface of the metal when the EM radiation (light) strikes its surfacesurface.
9.2 The photoelectric effect
Figure 9.3Figure 9.3
EM radiation
-- photoelectronphotoelectron
-- -- -- -- -- -- -- -- -- --MetalMetal
Free electronsFree electrons
PHYSICS CHAPTER 9
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The photoelectric effect can be studied through the experiment made by Franck Hertz in 1887.
Figure 9.4a shows a schematic diagram of an experimental arrangement for studying the photoelectric effect.
9.2.1 Photoelectric experiment
---- --
EM radiation (light)
anodecathode
glass
rheostatpower supply
vacuumphotoelectron
Figure 9.4aFigure 9.4a
GG
VV
PHYSICS CHAPTER 9
16
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 experimentExplanation 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.
PHYSICS CHAPTER 9
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As positive voltage becomes sufficiently large, the photoelectric
current reaches a maximum constant value Im, called saturation currentsaturation current. Saturation current is defined as the maximum constant the maximum constant
value of photocurrent when all the photoelectrons have value of photocurrent when all the photoelectrons have reached the anodereached 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 9.4b, the photoelectric current decreases even further to very low values since most photoelectronsphotoelectrons are repelledrepelled by anodeanode which is now negativenegative electric potential.
PHYSICS CHAPTER 9
18
As the potential of the anode becomes more negative, less photoelectrons reach the anode thus the photoelectric currentphotoelectric current drops until its value equals zerozero which the electric potential at
this moment is called stopping potential (voltage)stopping potential (voltage) Vs.
Stopping potential is defined as the minimum value of the minimum value of negative voltage when there are no photoelectrons negative voltage when there are no photoelectrons reaching the anodereaching the anode.
Figure 9.4b: reversing power supply terminalFigure 9.4b: reversing power supply terminal
---- --
EM radiation (light)
anodecathode
glass
rheostatpower supply
vacuumphotoelectron
GG
VV
PHYSICS CHAPTER 9
19
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
maxs 2
1mveV (9.5)(9.5)
electron theof mass: mwhere
mI
0I
sV
I,current ricPhotoelect
V,Voltage0
Before reversing the terminalBefore reversing the terminalAfterAfterFigure 9.4cFigure 9.4c
Stimulation 9.1
PHYSICS CHAPTER 9
20
A photon is a ‘packet’‘packet’ of electromagnetic radiationelectromagnetic radiation with particle-like characteristicparticle-like characteristic and carries the energy E given by
and this energy is not spread out through the mediumnot spread out through the medium.
Work function Work function WW00 of a metalof a metal
Is defined as the minimum energy of EM radiation required minimum energy of EM radiation required to emit an electron from the surface of the metalto emit an electron from the surface of the metal.
It depends on the metal usedmetal used. Its formulae is
where f0 is called threshold frequencythreshold frequency and is defined as the
minimum frequency of EM radiation required to emit an minimum frequency of EM radiation required to emit an electron from the surface of the metalelectron from the surface of the metal.
9.2.2 Einstein’s theory of photoelectric effect
hfE
min0 EW
00 hfW
and 0min hfE
(9.6)(9.6)
PHYSICS CHAPTER 9
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Since c=f then the eq. (9.6) can be written as
where 0 is called threshold wavelengththreshold wavelength and is defined as the maximum wavelengthmaximum wavelength of EM radiation required to emit an of EM radiation required to emit an electron from the surface of the metalelectron from the surface of the metal.
Table 9.2 shows the work functions of several elements.
00
hcW (9.7)(9.7)
ElementElement Work function (eV)Work function (eV)
Aluminum 4.3
Sodium 2.3
Copper 4.7
Gold 5.1
Silver 4.3
Table 9.2Table 9.2
PHYSICS CHAPTER 9
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Einstein’s photoelectric equationEinstein’s photoelectric equation In the photoelectric effect, Einstein summarizes that some of the
energy energy EE imparted by a photon imparted by a photon is actually used to release an release an electronelectron from the surface of a metal (i.e. to overcome the binding force) and that the rest appears as the maximum maximum kinetic energykinetic energy of the emitted electron (photoelectronphotoelectron). It is given by
where eq. (9.8) is known as Einstein’s photoelectric equation.
Since Kmax=eVs then the eq. (9.8) can be written as
where and0max WKE hfE
02
max2
1Wmvhf
2maxmax 2
1mvK
(9.8)(9.8)
0s WeVhf (9.9)(9.9)
voltagestopping: sVwhereelectron of chargefor magnitude: e
PHYSICS CHAPTER 9
23
Note:Note:
1st case: OR0Whf 0ff
Electron is emitted with maximum Electron is emitted with maximum kinetic energykinetic energy.--MetalMetal
hf
0W
--maxv maxK
2nd case: OR0Whf 0ff
Electron is emitted but maximum Electron is emitted but maximum kinetic energy is zerokinetic energy is zero.
-- 0v 0max K
3rd case: OR0Whf 0ff
No electron is emitted.No electron is emitted.
--MetalMetal
hf
0W
--MetalMetal 0W
hf
Figure 9.5aFigure 9.5a
Figure 9.5bFigure 9.5b
Figure 9.5cFigure 9.5c
PHYSICS CHAPTER 9
24
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)
Solution :Solution :
a. By using the equation of the work function, thus
Example 3 :
J 1075.61060.122.4 19190
W
00 hfW
03419 1063.61075.6 f Hz 10 02.1 15
0 f
PHYSICS CHAPTER 9
25
Solution :Solution :
b. Given
By applying the Einstein’s photoelectric equation, thus
c. The stopping potential is given by
m 10275 9
02
max2
1Wmv
hc
0max WKE
15max s m 1026.3 v
J 1075.61060.122.4 19190
W
192max
319
834
1075.61011.92
1
10275
1000.31063.6
v
2maxs 2
1mveV
2maxmax 2
1mvK
2531s
19 1026.31011.92
11060.1 V
V 303.0sV
PHYSICS CHAPTER 9
26
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 :
PHYSICS CHAPTER 9
27
Solution :Solution :
a. The threshold frequency is
b. The range of the frequencies that eject electrons is
5.51 5.51 10101414 Hz and 7.90 Hz and 7.90 10101414 Hz 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
02
max2
1Wmvhf
0max WKE
J 1059.1 19max
K
19max
1434 1065.31090.71063.6 K
PHYSICS CHAPTER 9
28
Exercise 9.1 :Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C1. The energy of a photon from an electromagnetic wave is 2.25
eVa. 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. :ANS. : 553 nm; 1.84553 nm; 1.8410101919 J J2. In a photoelectric effect experiment it is observed that no
current flows when the wavelength of EM radiation is greater than 570 nm. Calculatea. 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, 3(Physics for scientists & engineers, 3rdrd edition, Giancoli, Q15, edition, Giancoli, Q15, p.974)p.974)
ANS. :ANS. : 2.18 eV; 0.92 V2.18 eV; 0.92 V
PHYSICS CHAPTER 9
29
Exercise 9.1 :
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. :ANS. : 7.737.73101055 m s m s11; 6.72; 6.7210103434 J s J s
Wavelength of EM
radiation, (nm)
Stopping potential,
Vs (V)
350 1.70
450 0.900
PHYSICS CHAPTER 9
30
Variation of photoelectric current Variation of photoelectric current II with voltage with voltage VV for the radiation of different intensitiesdifferent intensities but its frequency is frequency is
fixedfixed.
Reason:
From the experiment, the photoelectric currentphotoelectric current is directly directly proportionalproportional to the intensityintensity of the radiation as shown in Figure 9.6b.
9.2.3 Graph of photoelectric experiment
Intensity 2xIntensity 2x
mI
I
V0sV
Intensity 1xIntensity 1x
m2I
Figure 9.6aFigure 9.6a
PHYSICS CHAPTER 9
31
for the radiation of different frequenciesdifferent frequencies but its intensity is intensity is fixedfixed.
Figure 9.6bFigure 9.6b
I
intensityLight 0 1
mIm2I
2
mI
Figure 9.7aFigure 9.7a
I
V0s1V
ff11ff22
s2V
ff22 > > ff11
PHYSICS CHAPTER 9
32
Reason:
From the Einstein’s photoelectric equation,
Figure 9.7bFigure 9.7b
0s WeVhf e
Wf
e
hV 0
s
y xm c
e
W0
f,frequency
s, voltageStopping V
02f
s2V
1f
s1V
If VVss=0=0,, 0)0( Wehf hfW 0 0f
0f
PHYSICS CHAPTER 9
33
For the different metals of cathodedifferent metals of cathode but the intensity and intensity and frequencyfrequency of the radiation are fixedfixed.
Reason: From the Einstein’s photoelectric equation,
Figure 9.8aFigure 9.8a
mI
s1V
01W
s2V02W
WW0202 > > WW0101
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 9.8bFigure 9.8b
PHYSICS CHAPTER 9
34
Variation of stopping voltage Variation of stopping voltage VVss with frequency with frequency ff of the radiationof the radiation for different metals of cathodedifferent metals of cathode but the intensity intensity is fixedfixed.
Reason: Since W0=hf0 then
Figure 9.9Figure 9.9
WW0303 >>WW0202 >> WW0101
01f
WW0101
02f
WW0202
03f
WW0303
f
sV
0
00 fW
0s WeVhf e
Wf
e
hV 0
s
y xm c
If VVss=0=0,, 0)0( Wehf hfW 0 0f
Threshold (cut-off) Threshold (cut-off) frequency frequency
PHYSICS CHAPTER 9
35
Table 9.3 shows the classical predictions (wave theory), photoelectric experimental observation and modern theory explanation about photoelectric experiment.
9.2.4 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 independent of frequency.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 emission of photoelectronsphotoelectrons dependdepend on the light frequencylight frequency
PHYSICS CHAPTER 9
36
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 lightintensity of light is the number of photons number of photons radiated per unit time on a radiated per unit time on a unit surface areaunit surface area.Based on the Einstein’s photoelectric equation:
The maximum kinetic kinetic energyenergy of photoelectron depends only on the light frequencyfrequency and the work work functionfunction. If the light intensity is doubled, the number of electrons emitted also doubled but the maximum kinetic energy remains unchanged.
0WhfK max
PHYSICS CHAPTER 9
37
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 there is time interval 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 instantaneouslyinstantaneously 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 no time intervalinterval between absorption of light energy and emission.
PHYSICS CHAPTER 9
38
Classical predictions Experimental observation
Modern theory
Energy of light depends only on depends only on amplitudeamplitude ( or intensityintensity) and not on frequency.
Energy of light depends on frequency.
According to Planck’s quantum theory which is
E=hfEnergy of light depends on depends on its frequency.its frequency.
Table 9.3Table 9.3Note:Note:
Experimental observations deviate from classical predictions based on wave theory of lightwave theory of light. Hence the classical physics cannot explain the phenomenon of photoelectric effect.
The modern theory based on Einstein’s photon theory of lightmodern theory based on Einstein’s photon theory of light can explain the phenomenon of photoelectric effect.
It is because Einstein postulated that light is quantizedlight is quantized and light is emitted, transmitted and reabsorbed as photonsphotons.
PHYSICS CHAPTER 9
39
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.Solution :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 :
PHYSICS CHAPTER 9
40
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, calculatea. 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 9.10Figure 9.10
PHYSICS CHAPTER 9
41
Solution :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
00 f
c
14
8
1083.4
1000.3
m 1021.6 70
PHYSICS CHAPTER 9
42
Solution :Solution :
b. By using the Einstein’s photoelectric equation, thus
Hz 1014.7 14f
02
max2
1Wmvhf
02
max2
1hfmvhf
02
max2
1ffhmv
1414342max
31 1083.41014.71063.61011.92
1 v
15max s m 1080.5 v
PHYSICS CHAPTER 9
43
Exercise 9.2 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and
e=1.601019 C1. A photocell with cathode and anode made of the same metal
connected in a circuit as shown in the Figure 9.11a. 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 9.11b.
365 nm365 nm
VV
GG5
1
)nA(I
)V(V0Figure 9.11aFigure 9.11a Figure 9.11bFigure 9.11b
PHYSICS CHAPTER 9
44
Exercise 9.2 :
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. :ANS. : 1.601.6010101919 J, 3.85 J, 3.8510101919 J; J; 1.57 V1.57 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, 7(Advanced Level Physics, 7thth edition, Nelkon&Parker, Q6, p.835) edition, Nelkon&Parker, Q6, p.835)
45
PHYSICS CHAPTER 9
Next Chapter…CHAPTER 10 :
Wave properties of particle
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