1. PHOTO ELECTRIC EFFECT

(a) In 1888 Hallwach discovered photo electriceffect and experimental verification of thisevent was done by Hertz.

(b) The phenomenon of emission of electronsfrom metal surface by the incidence of lightphoton is called photo electric effect.

(c) Emitted e- is called photo electron(d) Current produced due

to emitted electron iscalled photo electriccurrent

(e) Photo electric effectvarifies quantum nature of light.

(f) Photo electric effect can not be explained bywave theory of light

(g) Normaly photo electrons are those electronswhich are present freely at the inter molecularplaces in metal.

(h) Explanation for photo electric effect was givenby Einstein. For this excellent work Einsteinwas honoured by nobel prize in 1921.

(i) Photo electric effect is based on law ofconservation of energy

Important Definitions :Threshold Frequency (0) :Minimum frequency of incident photon belowwhich no ejection of photoelectrons from a metalsurface can take place is known as thresholdfrequency for that metal. Its value is constantfor a particular metal but may be different fordifferent metals.

If = Frequency of incident phtoton& 0 = Threshold Frequency

then(a) if < 0 No ejection of photo electron and

therefore no Photo electric effect.(b) if = 0 Photo electrons are just ejected

from metal surface and in this case kineticenergy of electron is zero.

(c) if > 0 then photo electrons will come outof the surface along with kinetic energy

Threshold wavelength (0) :Maximum wavelength of incident photon abovewhich there will be no photoelectric emission froma metal surface is known as threshold wavelength

0 =0

c

Photon

Metal

e¯

If = wavelength of incident photon then(a) if < 0 then photo electric effect will take

place and ejected electron will possesskinetic energy.

(b) if = 0 then Photo electric effect will justtake place and kinetic energy of ejected photoelectron will be zero

(c) if > 0 there will be no Photo electric effect.

2. WORK FUNCTION OR THRESHOLD ENERGY()

(i) The minimum energy of incident photon belowwhich no ejection of photo electron from a metalsurface will take place is known as work functionof threshold energy for that metal .

= h0 =0

hc

(ii) Work function is the characteristic of givenmetal

(iii) If E = Energy of incident photon, then(a) if E < No photo electric effect will take

place(b) if E = photo electric effect will just take

place but KE of ejected photo electron iszero.

(c) if E > Photo electric effect will takeplace along with possession of KE byejected electron

3. LAWS OF PHOTO ELECTRIC EFFECT

On the basis of experiments Lenard gavefollowing laws regarding photo emission.(a) Rate of photo electron's emission does not

depend upon frequency or wavelength of light(or photon) and in other words it does notdepend upon energy of incident light.

(b) Rate of photo electron's emission dependsupon intensity of light which incidents onmetal surface

(c) Photo electric current depends upon intensityof light but does not depend upon frequency/wavelength or energy.

(d) Kinetic energy of Emitted electron dependsupon frequency or wavelength of incident light.With increasing frequency of incident light,kinetic energy of photo electrons increasesbut with increasing wavelength it decreases.So K.E. of Emitted electrons K.E. of Emitted electrons

(e) Kinetic energy of emitted photo electronsdoes not depend upon intensity of light.

(f) Emission of electron from a metal surface ispossible only upto a certain minimumfrequency (corresponding maximumwavelength) of incident photon. This minimumfrequency is called threshold frequency andcorresponding wavelength is called thresholdwavelength

0

c

(g) Value of threshold frequency or thresholdwavelength depends upon photo sensitivenature of metal.

(h) There is no time lag between emission ofelectron and incidence of photon i.e. theelectrons are emitted out as soon as thelight falls on metal surface.

Ex.1 The work function of silver is 5.26 × 10–19 J.Calculate its threshold wavelength-(A) 3674 Å (B) 3467 Å(C) 3647 Å (D) 3764 Å

Sol. Threshold wavelength = 0 = hc

= 19

834

1026.5103106.6

=3.764 × 10–7 m

= 3764 ÅEx.2 The work function of Na is 2.3 eV. What is

the maximum wavelength of light that willcause photo electrons to be emitted fromsodium?(A) 539 mm (B) 0.539 mm(C) 539 nm (D) 0.539 nm

Sol. The threshold wavelength 0 = hc

( = h0 = hc/0)& hc = 1.24 × 10–6 (eV) m

0 =3.21024.1 6

m ;

0 = 0.539 × 10–6 m = 539 nm

4. EINSTEINS EQUATION OF PHOTO ELECTRICEFFECT

1. Einstein (1905) explained photo electriceffect on the basis of quantum theory .

2. A photon striking the metal surface transferwhole of its energy h to any one of theelectron present in the metal and it ownexistence vanished.

The energy supplied to the electrons is used intwo ways :-

(a) In form of work function () :- To emitelectron from the surface of metal

(b) To give kinetic energy to emitted electron.3. If vmax is the maximum velocity of emitted

electrons then by law of conservation ofenergy :-

h = +21

mv2

if 0 : threshold frequency

0 = h0

so h = h0 +21

mv2max.

This is called Einstein's equation of photoelectric effect.

4.1 Einstein’s Equation Explain FollowingConcepts -(a) With increasing frequency of light, kinetic

energy of electrons increases similarly withdecreasing wavelength , Kinetic energy ofelectrons increases.If 0 is threshold frequency then maximumkinetic energyEmax = h - h0

21

m vmax2 = h(–0)

so maximum velocity of photo electrons :-

vmax = m)(h2 0

m- mass of electron.- frequency of incident light

0- threshold frequency

0- threshold wavelength

- incident wavelength

E max = hc

0

11

21

m vmax2 = hc

0

11

(b) If = 0 or = 0 then v = 0

(c) < 0 or > 0 There will be noemission of photo electrons.

(d) When Intensity of light is increased it meansnumber of photons have been increased. Itdoes not affect energy of photons. Hencerate of emission increases but there will beno change in kinetic energy of electrons.With increasing number of emitted electrons,value of photo electric current increases.

Ex.3 Light of wavelength 4000 A0 is incident on ametal whose work function is 2eV. Calculatethe maximum possible kinetic energy of thephoto electrons.(A) 3.09 eV (B) 1. 9 eV(C) 1.09 eV (D) None

Sol Energy of the incident photon = hc / Energy of the incident photon in

eV = 19

19

106.14108.19

= 3.09 eV

Kinetic energy of the emitted electron

Ek = h – = 3.09 - 2.00 = 1.09 eV

Ex.4 The threshold wavelength of a metal is 5800 Å.If wavelength of incident light is 4500Å, thenthe maximum kinetic energy ofphotoelectrons would be-(A) 0.62 eV (B) 26 eV(C) 62 eV (D) 0.26 eV

Sol Ekmax=

0

0 ][hc

= 6.62×10–34×3×10820

1010

1045005800]104500105800[

= 9.9 × 10–20 J

Ekmax = 19

20

106.1109.9

= 0.62 eV

vmax = 0

0m

)(hc2

5. PHOTO ELECTRIC CURRENT

(1) When light incidents on cathode, electronsare emitted & these are attracted by anodethus current flows in the circuit. It is calledphoto electric current.

(2) Value of photo electric current depends uponfollowing parameters :-

(a) Potential difference between electrodes.

(b) Intensity of incident light.

5.1 Intensity of light (I)

(a) It is the qunantity of light energy fallingnormally on a unit surface area in unit time.

or I =E

A t.

where I = Intensity of light inWm2

E = total energy incidnet = nh = nhc

n = no. of photons

A = C/s area

t = time of exposure

(b) Intensity of light is proportional to saturationcurrent

(c) For point source of light I 2r1

(d) For line source of light I r1

where r is the distance of the point from the lightsource.

6. STOPPING POTENTIAL

(1) When in photo electric cell (+) ve voltage oncathode and negative voltage on anode isapplied then with increasing potentialdifference magnitude of photo electric currentdecreases.

(2) The negative potential (V0 ) applied to theanode at which the current is just reducedto zero is called the stopping potential.

(3) Potential on anode equal to greater thanstopping potential give zero current in circuit.

(4) If emitted electrons do not reach from cathodeto anode then stopping potential is given by

eV0 =21

m vmax2

or Emax = eV0eV0 = h (0)

V0 = e)(h 0

(5) Value of stopping potential depends uponfrequency of incident light.

(6) Stopping potential also depends upon natureof metal (or work function)

(7) Stopping potential does not depend uponintensity of light

(8) Example :-suppose stopping potential= –3 Volt, then

21

m vmax2 = 3 eV

If we apply -5 volt then also there will be zerocurrent in the circuit but

21

m vmax2 5 eV

Because stopping potential is not equal to5V which cannot be used in einstein equation.

Graphs :(1) Kinetic energy V/s frequency :

At 0 Emax = 0

(2) Vmax V/s :At 0 Vmax = 0

(3) Saturated Current V/s Intensity :

(4) Stopping potential V/s frequency :

eV0 = h -h0

tan = slope

= eh

(constant for all type of metals )

Intercept on x-axis = 0

Intercept on y-axis = – eh 0

(5) Potential V/s current : ( : constant)

Stopping potential does not depend uponintensity of light .

(6) Photo electric current V/s Retardingpotential :

Ex. 5 If one photon has 25 eV energy and workfunction of material is 7 eV then value ofstopping potential will be-(A) 32 V (B) 18 V(C) 3.3 V (D) zero

Sol V0 = eE 0 = e

725 eV

V0 = 18 V

Current

Retarding potential

Emax

0

Vmax

0

Current

Intensity

I3I2

I1

I3>I2>I1Current

Intensity

voltage-v0

v0

POINTS TO REMEMBER(1) Photo electric effect is based on the principle

of conservation of energy.

(2) Quantum nature of radiations is verified byphoto electric effect.

(3) Work function depends on nature of metal &impurities present on metal surface.

(4) Work function is the characteristic of matter.

(5) Stopping potential depends on frequency ofincident light & the nature of cathode material.

(6) The velocity of electrons emitted from thesurface layer is maximum.

(7) 2max

1maxEE

=h

h

1 0

2 0

(8) The intensity of light depends on the numberof photons.

(9) Frequency of untraviolet light is more thanthat of red light. Hence the kinetic energy ofphoto electrons emitted by ultraviolet lightwill be more than that of the electronsemitted by red light.

(10) max= m)(h2 0

(11) Photoelectric effect was discovered byHallwachs, experimentally verified by Hertz,and successfully studied and explained byEinstein.

(12) V =

0

11e

hc

(13) Less energy is required to release theelectrons situated in the surface layer ofmaterial where as more energy is required torelease electrons from the inner layers.

(14) One electron is emitted by one photon.

(15) Stopping potential does not depend on theintensity of incident light.

(16) Intensity of light is related to amount ofphotons and in more simplified terms it isrelated to no. of photons.But frequency of light directly indicates towardenergy of one photon.

For eg :Consider the following problem :Case I : 2 photons both have /2 frequency

So energy of one photon =2

h

Case II : One photon having frequency so Energy of this photon = hConclusions :-

(A) If intensity in case 1 is higher so no. ofemitted electrons in case I will be greater.current in case I > current in case II

(B) If energy of one photon is higher in case II sokinetic energy of emitted electron will be morein case of II though their intensity is lessthan case I

(17) One photon can emit only one electron. Thewhole photo electric effect is based on thissimple concept.

(18) red < violet It means incident violet ray willgive more energy to emitted electrons thanincident red ray it means V0 (red) < V0 (violet)

Vmax (red) < Vmax (violet)

Emax (red) < Emax (vilolet)

(19) Photo electric effect is based on energyconservation law.

(20) If light of same frequency incidences onsurfaces of various work function then velocityor kinetic energy of that emitted electron willbe maximum for which surface that has leastwork function.

Ex.1 Two different photons of energies, 1 eV and2.5 eV, fall on two identical metal plateshaving work function 0.5 eV, Then the ratio ofmaximum KE of the electrons emitted fromthe two surface is-(A) 1 : 2 (B) 1 : 4(C) 2 : 1 (D) 4 : 1

Sol K1max = h1 – = 1 – 0.5 = 0.5 eV

K2max = 2.5 – 0.5 = 2.0 eV

Thus K1max : K2max

= 0.5 : 2 = 1 : 4

Ex.2 Ultraviolet light of wavelength 280 nm is usedin an experiment on photo electric effect withlithium ( = 2.5 eV) cathode. Stoppingpotential will be-(A) 1 .9 eV (B) 1.9 V(C) 4.4 eV (D) 4.4 V

Sol The maximum kinetic energy is Kmax= hc

–

= 2801242

nmnmeV

– 2.5 eV

= 4.4 eV – 2.5 eV = 1.9 eVStopping potential V is given by eV = Kmax

V = eKmax = e

1.9 eV = 1.9 V

Ex.3 A monochromatic source of light operating at200 W emits 4 x 1020 photons per second.Find the wavelength of light.(A) 400 mm (B) 200 n(C) 4 × 10–10 Å (D) None

Sol The energy of each photon = 20104200

= 5 × 10–19 J

Wavelength = =Ehc

= 19

834

105)103()1063.6(

= 4.0 × 10-7

= 400 nmEx.4 Which metal will be suitable for a photo

electric cell using light of wavelength 4000A0.The work functions of sodium and copper arerespectively 2.0 eV and 4.0 eV.(A) sodium (B) copper(C) Both (D) None of both

S O LV E D E X A M P L E S

Sol 0 = hc

(0)sodium= 19

834

106.12103106.6

= 6188 Å

0 1

copper0

sodium0)()(

=sodium

copper

)()(

(0)copper =42

× 6188 = 3094 Å

To eject photo-electrons from sodium thelongest wavelength is 6188 Å and that forcopper is 3094 Å. Hence for light ofwavelength 4000 Å, sodium is suitable.

Ex.5 The work function for the surface of aluminiumis 4.2 eV. What will be the wavelength ofthat incident light for which the stoppingpotential will be zero.(A) 2496 Å (B) 2946 × 10-7 m(C) 2649 Å (D) 2946 Å

Sol If the incident light be of threshold wavelength(0), then the stopping potential shall be zero.Thus

0 = hc

, 0 = 19

834

106.12.4103106.6

,

0 = 2.946 × 10–7 m = 2946 Å

Ex.6 Slope of V0 – curve is-

(A) e (B) eh

(C) 0 (D) h

So Relation between V0 – ., V0 = eh

– eh 0

Put it in the form of y = mx – c,

here V0 = y, = x, eh 0 = c

y =

eh

x – c

m = eh

Ex.7 A radio station is transmitting waves ofwavelength 300 m, If diffracting power oftransmitter is 10 kw, then numbers ofphotons diffracted per second is-(A) 1.5 × 1035 (B) 1.5 × 1031

(C) 1.5 × 1029 (D) 1.5 × 1033

Sol P = 10 × 103 wattn = ? = 300 m

P = tnhc

104 =1300

n1031062.6 834

n = 834

4

101062.610300

= 1.5 × 1031

Ex.8 Light of wavelength 332 Å incidents on metalsurface (work function = 1.07 eV). To stopemission of photo electron, retardingpotential required to be-(A) 3.74 V (B) 2.67 V(C) 1.07 V (D) 4.81 V

Sol 0 = 1.07 eV = 1.07 × 1.6 × 10–19 J = 332 × 10–10 m

eV0 =

hc – 0 = V0 =

ehc

– e0

V0 =e103321031062.6

10

834

-

e101.61.07 19

V0 = 3.74 – 1.07 = 2.67 volt.Ex.9 Light of wavelength 5000 Å falls on a sensitive

surface. If the surface has received 10–7 Jouleof energy, then what is the number of photonsfalling on the surface ?(A) 25 × 1011 (B) 25 × 1012

(C) 0.25 × 1011 (D) 2.5 × 1011

Sol Let the energy of one photon = hc/, Energy of n photons E = nhc/

10-7 = 10

834

105000103106.6n

n = 26

710

108.1910105000

= 0.25 × 1012

n = 2.5 × 1011

Ex.10 An electromagnetic radiation of frequency3 × 1015 cycles per second falls on a photoelectric surface whose work function is4.0 eV. Find out the maximum velocity of thephoto electrons emitted by the surface-(A) 13.4 × 10–19 m/s (B) 19.8 × 10-19m/s(C) 1.73 × 106 m/s (D) None

Sol h = h0 + Ek6.6 × 10-34 × 3 × 108

= 4 × 1.6 × 10–19 + Ek19.8 × 10-19 – 6.4 × 10-19 = EkEk = 13.4 × 10–19 J

21

mv2max = 13.4 × 10–19

vmax = m104.132 19

= 31

19

109104.132

= 1.73 × 106 m/s

Ex.11 The wavelength of a photon is 4000 Å.Calculate its energy.(A) 49.5 × 10–19 J(B) 495 × 10–19 J(C) 4.95 × 10–19 K(D) 4.95 × 10–19 J

Sol E =

hc = 10

834

104000103106.6

= 4.95 × 10-19 JEx.12 When ultraviolet light of energy 6.2 eV

incidents on a aluminimum surface, it emitsphoto electrons. If work function for aluminiumsurface is 4.2 eV, then kinetic energy ofemitted electrons is-(A) 3.2 × 10–19 J (B) 3.2 × 10–17 J(C) 3.2 × 10–16 J (D) 3.2 × 10–11 J

Sol Ek = E – 0 = 6.2 – 4.2 = 2.0 eV,Ek = 2 × 1.6 × 10–19 = 3.2 × 10–19 J

Ex.13 Using light of wavelength 6000 Å stoppingpotential is obtained 2.4 volt for photo electriccell. If light of wavelength 4000 Å is usedthen stopping potential would be-(A) 2.9 V (B) 1.9 V(C) 3.43 V (D) 9.4 V

Sol V0 = eehc 0

,

2.4 = ee106000hc 0

10

....(1)

V0 = ee104000hc 0

10

....(2)

Eq. (1) - Eq.(2)2.4 – V0

= ehc

1010 1040001

1060001

2.4 – V0 =

2464

106.1101031062.6

197

834

V0 = 2.4 + 197

834

106.110121031062.6

V0 = 2.4 + 1.03 = 3.43 VEx.14 When light source is placed at 1 m distant

from photo electric cell, then value of stoppingpotential is obtained 4 volt. If it is placed at4 m distant, then value of stopping potentialbecomes-(A) 2 volt (B) 1 volt(C) 4 volt (D) 16 volt

Sol Stopping potential does not depend upondistance from light source.

Ex.15 When monochromatic light of wavelength illuminates a metal surface then stoppingpotential for photo electric current is 3V0. Ifwavelength changes to 2 then stoppingpotential becomes V0 . Threshold wavelengthfor photo electric emission is-(A) 4 (B) 8(C) 4/3 (D) 6

Sol

hc – 0 = 3V0 .....(1)

2hc

– 0 = V0 .....(2)

eq. (1) - eq.(2) :

hc

211 = 2V0

2hc

= 2 V0 =0V4

hc 4 =

0Vhc