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UNIT 10: PHOTONS ANDUNIT 10: PHOTONS AND
QUANTIZED ENERGYQUANTIZED ENERGY
SF027 2Fig. 10.1aFig. 10.1a
10.1 Planck’s Quantum Theory The foundation of the Planck’s quantum theory is a theory of blacktheory of black
body radiationbody radiation.
Black body is defined as an ideal system that absorbs all thean ideal system that absorbs all theradiation incident on itradiation incident on it. The electromagnetic radiation emitted byelectromagnetic radiation emitted by
the black bodythe black body is called black body radiationblack body radiation.
The spectrum of electromagnetic radiation emitted by the black body(experimental result) is shown in figure 10.1a.
From the fig. 10.1a, theRayleigh-Jeans and Wien’stheories failed to fit theexperimental curve because
this two theories based onclassical ideas.
The classical ideas are
EnergyEnergy of the e.m.
radiation is not dependnot depend on
its frequencyfrequency or
wavelengthwavelength.
EnergyEnergy of the e.m.
radiation is continuouslycontinuously.
ExperimentalExperimental
resultresult
RayleighRayleigh --
Jeans theoryJeans theory
WienWien’’ss theorytheory
ClassicalClassical
physicsphysics
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In 1900, Max Planck proposed his theory that is fit with theexperimental curve in fig. 10.1a at all wavelengths known as Planck’squantum theory.
The assumptions made by Planck in his theory are :
The e.m. radiation emitted by the black body is a discretediscrete
(separate) packets of energy(separate) packets of energy known as quantaquanta. This means the
energy of e.m. radiation is quantisedquantised. The energy size of the radiation dependeddepended on its frequencyfrequency.
According to this assumptions, the quantum E of the energy for
radiation of frequency f is given by
Since the speed of electromagnetic wave in a vacuum is ,
then eq. (10.1a) can be written as
From the eq. (10.1b), the quantum E of the energy for radiation is
inversely proportional to its wavelength.
hf E =
where constantPlanck :h J s10636 34 . −×=
(10.1a)(10.1a) PlanckPlanck’’ss
quantum theoryquantum theory
λ f c =
λ hc E = (10.1b)(10.1b)
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It is convenient to express many quantum energies in electronvolts.
The electronvoltelectronvolt ((eVeV)) is a unit of energyunit of energy that can be defined as thethe
kinetic energy gained by an electron in being accelerated by akinetic energy gained by an electron in being accelerated by apotential difference (voltage) of 1 voltpotential difference (voltage) of 1 volt.
Unit conversion :
In 1905, Albert Einstein extended Planck’s idea by proposing thatelectromagnetic radiation is also quantised. It consists of particle likepackets (bundles) of energy called photonsphotons of electromagneticradiation.
Photon is defined as a particle with zero mass consisting of aa particle with zero mass consisting of aquantum of electromagnetic radiation where its energy isquantum of electromagnetic radiation where its energy is
concentrated.concentrated.
A photon may also be regarded as a unit of energy equal to hf . Photons travel at the speed of lightspeed of light in a vacuum. They are required to
explain the photoelectric effect and other phenomena that require lightto have particle property.
J 10601eV 1 19 . −×=
10.2 Photons and Electromagnetic Waves
Energy
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Table 10.2a shows the differences between the photon andelectromagnetic wave.
PhotonE.M. Wave
Energy of the e.m. wave
depends on the intensityof the wave. Intensity of the wave is proportionalto the squared of itsamplitude where
Energy of a photon is
proportional to thefrequency of the e.m.w.where
Its energy is continuouslyand spread out throughthe medium as shown infigure 10.2a.
Its energy is discrete asshown in figure 10.2b.
2 A I ∝
f E ∝hf E =
Fig. 10.2aFig. 10.2aFig. 10.2bFig. 10.2b
PhotonPhoton
Table 10.2aTable 10.2a
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Example 1 :
A photon of green light has a wavelength of 520 nm. Calculate thephoton’s frequency and energy (in joules and electronvolts).
(Given the speed of photon in the vacuum, c = 3.00 x 108 m s-1 andPlanck constant, h = 6.63 x 10-34 J s)
Solution: λ =520x10-9 m
By applying the equation relates c, f and λ , thus the photon’sfrequency is
By using the equation of Planck’s quantum theory, thus the photon’senergy is
In electronvolt :
Example 2 : (exercise)
For waves propagating in air, calculate the energy of a photon inelectronvolts of
a. gamma rays of wavelength 4.61 x 10-14 m.
b. visible light of wavelength 5.21 x 10-7 m.
Ans. :2.70 x 107 eV, 2.39 eV
f c λ = Hz 1077 5 f
14×= .
hf E = J 10833 E 19−×= .
19
19
10601
10833 E
−
−
×
×=
.
.eV 392 E .=
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10.3 The Photoelectric Effect Definition – is defined as the emission of electron from the surfaceemission of electron from the surface
of a metal when theof a metal when the e.me.m. radiation (light) of higher . radiation (light) of higher
frequency strikes its surface.frequency strikes its surface.
Figure 10.3a 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 an electron emitted from the surface of
the metal when light strikes its surface.the metal when light strikes its surface. The photoelectric effect can be studied through the experiment made
by Hertz in 1887.
--lightlight photoelectronphotoelectron
-- -- -- -- -- -- -- -- -- --
MetalMetal
Free electronsFree electronsFig. 10.3aFig. 10.3a
SF027 8
10.3.1 Experiment of Photoelectric Effect
Figure 10.3b shows a schematic diagram of an experimentalarrangement for studying the photoelectric effect.
The set-up as follows :
Two conducting electrodes, the anode (positive electric potential)and the cathode (negative electric potential) are encased in anevacuated tube (vacuum).
The monochromatic light of known frequency and intensity areincident on the cathode.
anodeanodecathodecathode
photoelectronphotoelectron
glassglass
----
--
rheostatrheostatpower supplypower supply
e.me.m. radiation (light). radiation (light)
vacuumvacuum GG
VV
Fig. 10.3bFig. 10.3b
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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 a
photoelectric current or photocurrent I which is detected by the
galvanometer.
When the positive voltage (potential difference) is increased, morephotoelectrons reach the anode , hence the photoelectric currentalso increase.
As positive voltage becomes sufficiently large, the photoelectric
current reaches a maximum constant value I m, called saturationsaturation
currentcurrent.
Saturation current is defined as the maximum constant valuethe maximum constant value
of photocurrent when all the photoelectrons have reachedof photocurrent when all the photoelectrons have reachedthe anode.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 I 0.
SF027 10
Finally, when the voltage is made negative by reversing the power
supply terminal as shown in figure 10.3c, the photoelectric currentdecreases even further to very low values since mostphotoelectronsphotoelectrons are repelledrepelled by anodeanode which is now negativenegativeelectric potential.
As the potential of the anode becomes more negative, lessphotoelectrons reach the anode thus the photoelectric currentphotoelectric currentdrops until its value equals zerozero which the electric potential at this
moment is called stopping potential (voltage)stopping potential (voltage) V s. Stopping potential is defined as the minimum value of the minimum value of
negative voltage when there are no photoelectronsnegative voltage when there are no photoelectrons
reaching the anode.reaching the anode.
anodeanodecathodecathode
photoelectronphotoelectronvacuumvacuum
glassglass
----
--
GG
VV
rheostatrheostatpower supplypower supply
e.me.m. radiation (light). radiation (light)
Fig. 10.3c : reversing power supply terminalFig. 10.3c : reversing power supply terminal
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The potential energy U due to this retarding voltage V s now
equals the maximum kinetic energy K max of the photoelectron.
The variation of photoelectric current I as a function of the voltage
V can be shown through the graph in figure 10.3d.
max K U =
(10.3a)(10.3a)2
s mv
2
1eV =
m I
0 I
sV −
I current ric Photoelect ,
V Voltage,0
Fig. 10.3dFig. 10.3d
Before reversingBefore reversing
the terminalthe terminal
After After
SF027 12
10.3.2 Einstein’s theory of Photoelectric Effect
A photon is a ‘‘packetpacket’’ of electromagnetic radiationelectromagnetic radiation with particleparticle--likelike
characteristiccharacteristic and carries energy E given by
and this energy is not spread out through the mediumnot spread out through the medium.
Work functionWork function W 0 of a metal
Is defined as the minimum energy of minimum energy of e.me.m. radiation required to. radiation required toemit an electron from the surface of the metalemit an electron from the surface of the metal.
It depends on the metal used.
Equation :
where f 0 is called threshold frequencythreshold frequency and is defined as the
minimum frequencyminimum frequency of of e.me.m. radiation required to emit an. radiation required to emit anelectron from the surface of the metalelectron from the surface of the metal.
Since then
hf E =
h
W
f 0
0=
min E W 0 =
λ= f c
and 0hf E =min
(10.3b)(10.3b)
0
0 f
c λ = (10.3c)(10.3c)
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SF027 13
where λ 0 is called threshold wavelengththreshold wavelength and is defined as
the maximum wavelengthmaximum wavelength of of e.me.m. radiation required to emit. radiation required to emit
an electron from the surface of the metal.an electron from the surface of the metal.
Table 10.3a shows the work functions of several elements.
Einstein’s photoelectric equation :
In photoelectric effect, Einstein summarizes that some of the
energyenergy E E imparted by a photonimparted by a photon is actually used to release anrelease an
electronelectron from the surface of a metal (i.e. to overcome the bindingforce) and that the rest appears as the maximum kinetic energymaximum kinetic energy
of the emitted electron (photoelectron). It given by
4.3Silver
5.1Gold
4.7Copper
2.7Sodium
4.3Aluminum
Work function (Work function (eVeV))ElementElement
Table 10.3aTable 10.3a
0W K E += max
2mv2
1 K =maxwhere hf E = and
0
2 W mv2
1hf += (10.3d)(10.3d) EinsteinEinstein’’ss
photoelectricphotoelectric eqeq..
SF027 14
Since then eq. (10.3d) can be written as
Note :
First case : hf >W 0 or f >f 0
0 s W eV hf += (10.3e)(10.3e)
s
2
eV mv2
1
=
where voltagestopping: sV chargeelectrontheof magnitude:e
--hf hf vvmax max
--MetalMetalW W 00
Second case : hf=W 0 or f =f 0
--hf hf
v=0v=0
--MetalMetalW W 00
Third case : hf<W 0 or f <f 0
hf hf
--MetalMetalW W 00
Electron is emitted withElectron is emitted with
maximum kinetic energy.maximum kinetic energy.
K K max max
K K max max =0=0
Electron is emitted but maximumElectron is emitted but maximum
kinetic energy is zero.kinetic energy is zero.
No electron is emitted.No electron is emitted.
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SF027 15
Example 3 :
Sodium has a work function of 2.30 eV. Calculate
a. its threshold frequency,
b. the maximum speed of the photoelectrons produced when the
sodium is illuminated by light of wavelength 500 nm,
c. the stopping potential with light of this wavelength.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
Solution: W 0=2.30 x (1.60x10-19 )= 3.68 x10-19 J
a. The threshold frequency, f 0 is given by
b. Given λ =500 x 10-9 mBy using the Einstein’s photoelectric equation, hence the maximum
speed of the photoelectrons is
Hz 10555 f 14
0 . ×=00 hf W =
1-5m s1056 2v . ×=
0
2 W mv21hf += and
λ
c f =
0
2 W mv2
1hc+=
λ
SF027 16
c. The stopping voltage V s is given by
Example 4 :
In an experiment of photoelectric effect, no current flows through the
circuit when the voltage across the anode and cathode is -1.70 V.
Calculate
a. the work function, and
b. the threshold wavelength of the metal (cathode) if it is illuminated by
ultraviolet radiation of frequency 1.70 x 1015 Hz.(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,
mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
Solution: V s=1.70 V, f=1.70x1015 Hz a. By using the Einstein’s photoelectric equation, hence the work
function is
2
s mv2
1eV =
V 187 0V s .=
0 s W eV hf +=
J 10558W 19
0 . −×=
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SF027 17
b. The threshold wavelength is
Example 5 : (exercise)
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, photoelectrons
are emitted with a maximum kinetic energy of 1.10 eV. Calculate
the work function of this metal in joules.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s ,
1 eV=1.60 x 10-19 J, mass of electron m = 9.11 x 10-31 kg,
e = 1.60 x 10-19 C)
Ans. : 553 nm, 1.84 x 10-19 J
m10332
7
0 .
−×=λ
00 hf W = and
0
0
c f
λ=
0
0
hcW
λ=
SF027 18
Example 6 : (exercise)
In an experiment on the photoelectric effect, the following data werecollected.
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.
(Given c = 3.00 x 108 m s-1, 1 eV=1.60 x 10-19 J, mass of electron m =
9.11 x 10-31
kg, e = 1.60 x 10-19
C)Ans. : 7.73 x 105 m s-1, 6.72 x 10-34 J s
Example 7 : (exercise)
In a photoelectric effect experiment it is observed that no current flowsunless the wavelength is less than 570 nm. Calculate
a. the work function of this material in electronvolts.
b. the stopping voltage required if light of wavelength 400 nm is used.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
(Giancoli,pg. 974,no.15)
Ans. : 2.18 eV, 0.92 V
0.900450
1.70350
Stopping
potential, V s (V)
Wavelength of e.m.
radiation,λ (nm)
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10.3.3 Graphs in Photoelectric Effect
Variation of photoelectric current I with voltage V for the radiation of different intensitiesdifferent intensities but its frequency is fixedfrequency is fixed.
Explanation: From the experiment, the photoelectric currentphotoelectric current isdirectly proportionaldirectly proportional to the intensityintensity of the radiation asshown in figure 10.3f.
Intensity 2xIntensity 2x
m I
sV −
I current ric Photoelect ,
V Voltage,0
Fig. 10.3e : graphFig. 10.3e : graph
of of I I againstagainst V V Intensity 1xIntensity 1x
m I 2
I current ric Photoelect ,
intensityLight0 ×1
m I 2
×2
m I
SF027 20
For the radiation of different frequenciesdifferent frequencies but its intensity is fixedintensity is fixed.
Explanation: From the Einstein’s photoelectric equation,
f f 22
m I
1 sV −
I current ric Photoelect ,
V Voltage,0
Fig. 10.3f : graph of Fig. 10.3f : graph of
I I againstagainst V V
f f 11
f f 22 >> f f 11
2 sV −
0 s W eV hf += eW f
ehV 0
s − =
=Y X C +
2 f f frequency,
sV voltageStopping ,
e
W 0−
01 f
2 sV
1 sV
0 f WhenWhen V V s s=0,=0, 0W 0ehf += )(
hf W 0 = 0 f
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For the different metals of cathodedifferent metals of cathode but the intensity andintensity andfrequencyfrequency of the radiation are fixedfixed.
Explanation: From the Einstein’s photoelectric equation,
W W 0101
1 sV −
m I
I current ric Photoelect ,
V Voltage,0
Fig. 10.3g : graphFig. 10.3g : graph
of of I I againstagainst V V
2 sV −
W W 0202
W W 0202 >> W W 0101
0 s W eV hf +=
+
−=
e
hf W
e
1V 0 s
X C +=Y e
hf
0W
sV voltageStopping ,
0 hf E =01W
1 sV
02W
2 sV
Energy of a photonEnergy of a photon
inin e.me.m. radiation. radiation
SF027 22
Variation of stopping voltage V s with frequency f of the radiation for different metals of cathodedifferent metals of cathode but the intensityintensity is fixedfixed.
Explanation: Since W 0=hf 0 then
W W 0303 >>W W 0202 >> W W 0101
01 f
W W 0101
02 f
W W 0202
03 f
W W 0303
f frequency,
sV voltageStopping ,
0
Fig. 10.3h : graphFig. 10.3h : graph
of of V V s s againstagainst f f
00 f W ∝ Threshold (cutThreshold (cut--off)off)
frequencyfrequency
0 s W eV hf +=e
W f
e
hV 0
s −
=
=Y X C +
WhenWhen V V s s=0,=0, 0W 0ehf += )(hf W 0 = 0 f
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10.4 Quantization of light Table below shows the classical predictions , photoelectric
experimental observation and modern theory explanation of experimental observation.
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 radiatednumber of photons radiated
per unit time on a unitper unit time on a unit
surface areasurface area.
Based on Einstein’s
photoelectric equation:
The maximum kinetic energykinetic energy
of photoelectron depends only
on the light frequencyfrequency and the
work functionwork function. If the lightintensity is doubled, the
number of electrons emitted
also doubled but the maximum
kinetic energy remains
unchanged.
0W hf K −=max
SF027 24
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
Energy of light dependsdepends
only on amplitudeonly on amplitude ( or
intensityintensity) 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 itsdepends on its
frequency.frequency.
Classical predictions Experimental
observation
Modern theory
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SF027 25
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 isthere is
time intervaltime 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 intervalno time interval between
absorption of light energy and
emission.
Experimental observations deviate from classical predictions based onMaxwell’s e.m. theory. Hence the classical physics cannot explain thephenomenon of photoelectric effect.
The modern theory based on Einstein’s 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.
Classical predictions Experimental
observation
Modern theory
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Example 8 :
In a photoelectric experiments, a graph of the light frequency f isplotted against the maximum kinetic energy K max of the photoelectronas shown in figure below.
Based on the graph, for the light frequency of 6.00 x 1014 Hz, calculate
a. the threshold frequency.
b. the maximum kinetic energy of the photoelectron.
c. the maximum velocity of the photoelectron.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
Hz 10 f 14×
02.−
)(max eV K
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Solution: f=6.00x1014 Hz a. By rearranging Einstein’s photoelectric equation,
From the graph, W 0=(2.0)(1.60x10-19 )=3.20x10-19 J The threshold frequency is
b. By applying the Einstein’s photoelectric equation, thus
c. The maximum velocity of the photoelectron is
0W K hf += max
max K W 0 −=
0W hf K −=max
Hz 10834 f 14
0 . ×=
=Y C +
WhenWhen f=0, f=0,
00 hf W =
0W 0h K −= )(max
Hz 10 f 14×
02.−)(max eV K
J 10787 K 20 .max
−×=0
W K hf +=max
1-5 sm10134v . ×=2mv2
1 K =max
SF027 28
Example 9 : (exercise)
A photocell with cathode and anode made of the same metalconnected in a circuit as shown in the figure below. Monochromaticlight of wavelength 365 nm shines on the cathode and the photocurrent
I is measured for various values of voltage V across the cathode andanode. The result is shown in the graph.
a. Calculate the maximum kinetic energy of the 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.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
Ans. : 1.60 x 10-19 J, 3.85 x 10-19 J, -1.57 V
365 nm365 nm
VV
GG5
1−
)(nA I
)(V V 0