photon and quantized enrgy

8/2/2019 Photon and Quantized Enrgy

http://slidepdf.com/reader/full/photon-and-quantized-enrgy 1/15

SF027 1

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



SF027 3

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)

SF027 4

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



SF027 5

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

SF027 6

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 .=



SF027 7

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



SF027 9

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



SF027 11

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)



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.



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 . −×=



SF027 17

b. The threshold wavelength is


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


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


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.


(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)



SF027 19

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 −


V Voltage,0

Fig. 10.3e : graphFig. 10.3e : graph

of of I I againstagainst V V Intensity 1xIntensity 1x

m I 2


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 −


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



SF027 21

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


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


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,


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



SF027 23

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.


observation

Modern theory



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.


observation

Modern theory

SF027 26

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.


Hz 10 f 14×

02.−

)(max eV K



SF027 27

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


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.


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



SF027 29

THE END…

Next Unit…

UNIT 11 :

Wave Particle Duality

photon and quantized enrgy

Documents