Matriculation Physics ( Quantization of Light )

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PHYSICS CHAPTER 9

CHAPTER 9: CHAPTER 9: Quantization of lightQuantization of light

(4 Hours)(4 Hours)

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: 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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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

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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

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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.

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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

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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

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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)


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

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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

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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


EM radiation

-- photoelectronphotoelectron

-- -- -- -- -- -- -- -- -- --MetalMetal

Free electronsFree electrons

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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

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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

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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

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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 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

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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

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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 9.5bFigure 9.5b

Figure 9.5cFigure 9.5c

PHYSICS CHAPTER 9

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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)


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

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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

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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.


e=1.601019 C)

Example 4 :

PHYSICS CHAPTER 9

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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

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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

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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

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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


PHYSICS CHAPTER 9

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for the radiation of different frequenciesdifferent frequencies but its intensity is intensity is fixedfixed.


I

intensityLight 0 1

mIm2I

2

mI


I

V0s1V

ff11ff22

s2V

ff22 > > ff11

PHYSICS CHAPTER 9

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Reason:

From the Einstein’s photoelectric equation,


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

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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.

Reason: From the Einstein’s photoelectric equation,


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


PHYSICS CHAPTER 9

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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


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

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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

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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

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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

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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

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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

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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.


e=1.601019 C)

Example 6 :

Hz1014f

83.4

)eV(maxK0


PHYSICS CHAPTER 9

41


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

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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

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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

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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)

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PHYSICS CHAPTER 9

Next Chapter…CHAPTER 10 :

Wave properties of particle

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Matriculation Physics ( Quantization of Light )