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UNIT 10: PHOTONS AND UNIT 10: PHOTONS AND QUANTIZED ENERGYQUANTIZED ENERGY
SF027 2Fig. 10.1aFig. 10.1a
10.1 Planck’s Quantum TheoryThe foundation of the Planck’s quantum theory is a theory of black theory of black body radiationbody 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 radiation emitted by electromagnetic 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, the Rayleigh-Jeans and Wien’stheories failed to fit the experimental curve because this two theories based on classical 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.
Experimental Experimental resultresult
RayleighRayleigh --Jeans theoryJeans theory
WienWien’’ss theorytheory
Classical Classical physicsphysics
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In 1900, Max Planck proposed his theory that is fit with the experimental curve in fig. 10.1a at all wavelengths known as Planck’s quantum theory.The assumptions made by Planck in his theory are :
The e.m. radiation emitted by the black body is a discrete discrete (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.
hfE =
where constantPlanck :hJ s10636 34 . −×=
(10.1a)(10.1a) PlanckPlanck’’s s quantum theoryquantum theory
λfc =
λhcE = (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 the the kinetic energy gained by an electron in being accelerated by a kinetic energy gained by an electron in being accelerated by a potential difference (voltage) of 1 voltpotential 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.
Photon is defined as a particle with zero mass consisting of a a particle with zero mass consisting of a quantum of electromagnetic radiation where its energy is quantum 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 light to have particle property.
J10601eV1 19 . −×=
10.2 Photons and Electromagnetic Waves Energy
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Table 10.2a shows the differences between the photon and electromagnetic wave.
PhotonE.M. Wave
Energy of the e.m. wave depends on the intensity of the wave. Intensity of the wave is proportional to the squared of its amplitude where
Energy of a photon is proportional to the frequency of the e.m.w. where
Its energy is continuouslyand spread out through the medium as shown in figure 10.2a.
Its energy is discrete as shown in figure 10.2b.
2AI ∝
fE ∝hfE =
Fig. 10.2aFig. 10.2aFig. 10.2bFig. 10.2bPhotonPhoton
Table 10.2aTable 10.2a
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Example 1 :A photon of green light has a wavelength of 520 nm. Calculate the photon’s frequency and energy (in joules and electronvolts).(Given the speed of photon in the vacuum, c = 3.00 x 108 m s-1 and Planck constant, h = 6.63 x 10-34 J s)
Solution: λ=520x10-9 mBy applying the equation relates c, f and λ, thus the photon’s frequency is
By using the equation of Planck’s quantum theory, thus the photon’s energy is
In electronvolt :
Example 2 : (exercise)For waves propagating in air, calculate the energy of a photon in electronvolts ofa. 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
fc λ=Hz10775f 14×= .
hfE =J10833E 19−×= .
19
19
1060110833E −
−
××
=.. eV392E .=
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10.3 The Photoelectric EffectDefinition – is defined as the emission of electron from the surface emission of electron from the surface
of a metal when the of 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
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10.3.1 Experiment of Photoelectric EffectFigure 10.3b shows a schematic diagram of an experimental arrangement 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 an evacuated tube (vacuum).The monochromatic light of known frequency and intensity are incident on the cathode.
anodeanodecathode cathode
photoelectronphotoelectronglassglass
-- -- --
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, more photoelectrons reach the anode , hence the photoelectric currentalso increase.As positive voltage becomes sufficiently large, the photoelectric current reaches a maximum constant value Im, called saturation saturation currentcurrent.
Saturation current is defined as the maximum constant value the maximum constant value of photocurrent when all the photoelectrons have reached of photocurrent when all the photoelectrons have reached the 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 I0.
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Finally, when the voltage is made negative by reversing the power supply terminal as shown in figure 10.3c, the photoelectric current decreases even further to very low values since most photoelectronsphotoelectrons are repelledrepelled by anodeanode which is now negativenegativeelectric potential.
As the potential of the anode becomes more negative, less photoelectrons 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) 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 anode.reaching the anode.
anodeanodecathode cathode
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 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 10.3d.
maxKU =
(10.3a)(10.3a)2s mv
21eV =
mI
0I
sV−
IcurrentricPhotoelect ,
VVoltage,0
Fig. 10.3dFig. 10.3d
Before reversing Before reversing the terminalthe terminal
AfterAfter
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10.3.2 Einstein’s theory of Photoelectric EffectA photon is a ‘‘packetpacket’’ of electromagnetic radiationelectromagnetic radiation with particleparticle--like like 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 W0 of a metalIs defined as the minimum energy of minimum energy of e.me.m. radiation required to . radiation required to emit an electron from the surface of the metalemit an electron from the surface of the metal.It depends on the metal used.Equation :
where f0 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 an electron from the surface of the metalelectron from the surface of the metal.Since then
hfE =
hWf 0
0 =
minEW0 =
λ= fc
and 0hfE =min
(10.3b)(10.3b)
00 f
cλ = (10.3c)(10.3c)
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where λ0 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 energy energy EE imparted by a photonimparted 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 kinetic energymaximum kinetic energyof 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
0WKE += max2mv
21K =maxwhere hfE = and
02 Wmv
21hf += (10.3d)(10.3d) EinsteinEinstein’’s s
photoelectric photoelectric eqeq..
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Since then eq. (10.3d) can be written as
Note :
First case : hf >W0 or f >f0
0s WeVhf += (10.3e)(10.3e)
s2 eVmv
21
=
where voltagestopping:sVchargeelectron theof magnitude :e
--hfhfvvmaxmax
--MetalMetal WW00
Second case : hf=W0 or f =f0
--hfhf
v=0v=0
--MetalMetal WW00
Third case : hf<W0 or f <f0
hfhf
--MetalMetal WW00
Electron is emitted with Electron is emitted with maximum kinetic energy.maximum kinetic energy.
KKmaxmaxKKmaxmax=0=0
Electron is emitted but maximum Electron is emitted but maximum kinetic energy is zero.kinetic energy is zero.
No electron is emitted.No electron is emitted.
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Example 3 :Sodium has a work function of 2.30 eV. Calculatea. 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: W0=2.30 x (1.60x10-19)= 3.68 x10-19 Ja. The threshold frequency, f0 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
Hz10555f 140 . ×=
00 hfW =
1-5 m s10562v . ×=
02 Wmv
21hf += and
λcf =
02 Wmv
21hc
+=λ
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c. The stopping voltage Vs 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. Calculatea. the work function, andb. 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: Vs=1.70 V, f=1.70x1015 Hza. By using the Einstein’s photoelectric equation, hence the work
function is
2s mv
21eV =
V1870Vs .=
0s WeVhf +=J10558W 19
0 . −×=
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b. The threshold wavelength is
Example 5 : (exercise)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, 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 70 . −×=λ
00 hfW = and0
0cf
λ=
00
hcWλ
=
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Example 6 : (exercise)In an experiment on the photoelectric effect, the following data were collected.
a. Calculate the maximum velocity of the photoelectrons when thewavelength 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 sExample 7 : (exercise)In a photoelectric effect experiment it is observed that no current flows unless the wavelength is less than 570 nm. Calculatea. 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, Vs (V)
Wavelength of e.m. radiation,λ (nm)
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10.3.3 Graphs in Photoelectric EffectVariation 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 is directly proportionaldirectly proportional to the intensityintensity of the radiation as shown in figure 10.3f.
Intensity 2xIntensity 2x
mI
sV−
IcurrentricPhotoelect ,
VVoltage,0
Fig. 10.3e : graph Fig. 10.3e : graph of of II against against VV Intensity 1xIntensity 1x
mI2
IcurrentricPhotoelect ,
intensityLight 0 ×1
mI2
×2
mI
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For the radiation of different frequenciesdifferent frequencies but its intensity is fixedintensity is fixed.
Explanation: From the Einstein’s photoelectric equation,
ff22
mI
1sV−
IcurrentricPhotoelect ,
VVoltage,0
Fig. 10.3f : graph of Fig. 10.3f : graph of II against against VV
ff11
ff22 > > ff11
2sV−
0s WeVhf +=e
WfehV 0
s −
=
=Y XM C+
2fffrequency,
sVvoltageStopping ,
eW 0−
01f
2sV
1sV
0fWhen When VVss=0,=0, 0W0ehf += )(
hfW0 = 0f
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For the different metals of cathodedifferent metals of cathode but the intensity and intensity and frequencyfrequency of the radiation are fixedfixed.
Explanation: From the Einstein’s photoelectric equation,
WW0101
1sV−
mIIcurrentricPhotoelect ,
VVoltage,0
Fig. 10.3g : graph Fig. 10.3g : graph of of II against against VV
2sV−WW0202
WW0202 > > WW0101
0s WeVhf +=
+
−=
ehfW
e1V 0s
XM C+=Yehf
0W
sVvoltageStopping ,
0 hfE =01W
1sV
02W
2sV
Energy of a photon Energy of a photon in in e.me.m. radiation . radiation
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Variation of stopping voltage Vs with frequency f of the radiation for different metals of cathodedifferent metals of cathode but the intensity intensity is fixedfixed.
Explanation: Since W0=hf0 then
WW0303 >>WW0202 >> WW0101
01f
WW0101
02f
WW0202
03f
WW0303
ffrequency,
sVvoltageStopping ,
0
Fig. 10.3h : graph Fig. 10.3h : graph of of VVss against against ff
00 fW ∝ Threshold (cutThreshold (cut--off) off) frequency frequency
0s WeVhf +=e
WfehV 0
s −
=
=Y XM C+
When When VVss=0,=0, 0W0ehf += )(hfW0 = 0f
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10.4 Quantization of lightTable 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 radiated number of photons radiated per unit time on a unit per unit time on a unit surface areasurface area.Based on Einstein’s photoelectric equation:
The maximum kinetic energykinetic energyof photoelectron depends only on the light frequencyfrequency and the work functionwork function. If the light intensity is doubled, the number of electrons emitted also doubled but the maximum kinetic energy remains unchanged.
0WhfK −=max
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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 depends depends 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=hfEnergy of light depends on its depends on its frequency.frequency.
Classical predictions Experimental observation
Modern theory
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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 there is time interval 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 almostinstantaneouslyinstantaneously 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 on Maxwell’s e.m. theory. 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.
Classical predictions Experimental observation
Modern theory
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Example 8 :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 below.
Based on the graph, for the light frequency of 6.00 x 1014 Hz, calculatea. 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)
Hz10f 14×
02.− )(max eVK
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Solution: f=6.00x1014 Hza. By rearranging Einstein’s photoelectric equation,
From the graph, W0=(2.0)(1.60x10-19)=3.20x10-19 JThe threshold frequency is
b. By applying the Einstein’s photoelectric equation, thus
c. The maximum velocity of the photoelectron is
0WKhf += max
maxKW0 −=
0WhfK −=max
Hz10834f 140 . ×=
=Y XM C+When When f=0,f=0,
00 hfW =
0W0hK −= )(max
Hz10f 14×
02.− )(max eVK
J10787K 20 .max−×=0WKhf += max
1-5 sm10134v . ×=2mv21K =max
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Example 9 : (exercise)A photocell with cathode and anode made of the same metal connected in a circuit as shown in the figure below. 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 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−
)(nAI
)(VV0
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THE END…Next Unit…UNIT 11 :
Wave Particle Duality
