KVPY PEE.pdf

1. PHOTO ELECTRIC EFFECT

(a) In 1888 Hallwach discovered photo electric effect and experimental verification of this event was done by Hertz.

(b) The phenomenon of emission of electrons from metal surface by the incidence of light photon is called photo electric effect. (c) Emitted e

-

is called photo electron (d) Current produced due

to emitted electron is called photo electric current

(e) Photo electric effect

varifies quantum nature of light.

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

(g) Normaly photo electrons are those electrons which are present freely at the inter molecular places in metal.

(h) Explanation for photo electric effect was given by Einstein. For this excellent work Einstein was honoured by nobel prize in 1921. (i) Photo electric effect is based on law of

conservation of energy Important Definitions : Threshold Frequency (₀) :

Minimum frequency of incident photon below which no ejection of photoelectrons from a metal surface can take place is known as threshold frequency for that metal. Its value is constant for a particular metal but may be different for different 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 kinetic energy of electron is zero.

(c) if > ₀ then photo electrons will come out of the surface along with kinetic energy Threshold wavelength (₀) :

Maximum wavelength of incident photon above which there will be no photoelectric emission from a 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 possess kinetic energy.

(b) if  = ₀ then Photo electric effect will just take place and kinetic energy of ejected photo electron 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 below which no ejection of photo electron from a metal surface will take place is known as work function of threshold energy for that metal .

 = h₀ =

0

hc 

(ii) Work function is the characteristic of given metal

(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 is zero.

(c) if E >  Photo electric effect will take place along with possession of KE by ejected electron

3. LAWS OF PHOTO ELECTRIC EFFECT

On the basis of experiments Lenard gave following 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 not depend upon energy of incident light. (b) Rate of photo electron's emission depends

upon intensity of light which incidents on metal surface

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

(d) Kinetic energy of Emitted electron depends upon frequency or wavelength of incident light. With increasing frequency of incident light, kinetic energy of photo electrons increases but with increasing wavelength it decreases. So

 

K.E. of Emitted electrons



 

K.E. of Emitted electrons



(e) Kinetic energy of emitted photo electrons

(f) Emission of electron from a metal surface is possible only upto a certain minimum frequency (corresponding maximum wavelength) of incident photon. This minimum frequency is called threshold frequency and corresponding wavelength is called threshold wavelength



_



0

c 

(g) Value of threshold frequency or threshold wavelength depends upon photo sensitive nature of metal.

(h) There is no time lag between emission of electron and incidence of photon i.e. the electrons are emitted out as soon as the light 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 8 34 10 26 . 5 10 3 10 6 . 6       =3.764 × 10–7 m

    

   = 3764 Å

Ex.2 The work function of Na is 2.3 eV. What is the maximum wavelength of light that will cause photo electrons to be emitted from sodium?

(A) 539 mm (B) 0.539 mm (C) 539 nm (D) 0.539 nm Sol. The threshold wavelength ₀ = hc_

(

 

= h



₀ = hc/₀) & hc = 1.24 × 10–6 (eV) m ₀ = 3 . 2 10 24 . 1  6 m ; ₀ = 0.539 × 10–6 m = 539 nm

4. EINSTEINS EQUATION OF PHOTO ELECTRIC EFFECT

1. Einstein (1905) explained photo electric effect on the basis of quantum theory . 2. A photon striking the metal surface transfer

whole of its energy h



to any one of the electron present in the metal and it own existence vanished.

The energy supplied to the electrons is used in two ways

:-(a) In form of work function (



) :- To emit electron from the surface of metal (b) To give kinetic energy to emitted electron. 3. If v_max is the maximum velocity of emitted electrons then by law of conservation of energy :-h



=



+ 2 1 mv2 if



₀ : threshold frequency

 

₀ = h



₀ so



h



= h



₀ + 2 1 mv2_max.

This is called Einstein's equation of photo electric effect.

4.1 Einstein’s Equation Explain Following Concepts

-(a) With increasing frequency of light, kinetic energy of electrons increases similarly with decreasing wavelength , Kinetic energy of electrons increases.

If



₀ is threshold frequency then maximum kinetic energy E_max = h



- h



₀



2 1 m

v

_max2 = h(



–



₀)

so maximum velocity of photo electrons

:-

v_max = m ) ( h 2 ₀ m- mass of electron.



- frequency of incident light



₀- threshold frequency ₀- threshold wavelength  - incident wavelength



E _max = hc          ₀ 1 1



2 1 m

v

max 2 = hc ___ _ 0 1 1 (b) If



=



₀ or  = ₀ then v = 0 (c)



<



₀ or  > ₀



There will be no emission of photo electrons.

(d) When Intensity of light is increased it means number of photons have been increased. It does not affect energy of photons. Hence rate of emission increases but there will be no 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 a}

metal whose work function is 2eV. Calculate the maximum possible kinetic energy of the photo 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 10 6 . 1 4 10 8 . 19      = 3.09 eV

Kinetic energy of the emitted electron E_k = 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, then the m axim um kinetic energy of photoelectrons would

be-(A) 0.62 eV (B) 26 eV (C) 62 eV (D) 0.26 eV Sol

E

k_max=      0 0 ] [ hc = 6.62×10–34×3×108 ₂₀ 10 10 10 4500 5800 ] 10 4500 10 5800 [         = 9.9 × 10–20 J

E

_k max = 19 20 10 6 . 1 10 9 . 9     = 0.62 eV



v_max = 0 0 m ) ( hc 2    

5. PHOTO ELECTRIC CURRENT

(1) When light incidents on cathode, electrons are emitted & these are attracted by anode thus current flows in the circuit. It is called photo electric current.

(2) Value of photo electric current depends upon following 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 falling normally on a unit surface area in unit time. or

I

= E

A t.

where

I

= Intensity of light in W m2 E = total energy incidnet = nh = n hc



n = no. of photons A = C/s area

t = time of exposure

(b) Intensity of light is proportional to saturation current

(c) For point source of light

I

 ₂

r 1

(d) For line source of light

I



r 1

where r is the distance of the point from the light source.

6. STOPPING POTENTIAL

(1) When in photo electric cell (+) ve voltage on cathode and negative voltage on anode is applied then with increasing potential difference magnitude of photo electric current decreases.

(2) The negative potential (V₀) applied to the anode at which the current is just reduced to zero is called the stopping potential. (3) Potential on anode equal to greater than

stopping potential give zero current in circuit. (4) If emitted electrons do not reach from cathode to anode then stopping potential is given by

eV₀ =

2 1

m

v

max 2

or E_max = eV₀ eV₀= h (



₀) V₀ = e ) ( h ₀

(5) Value of stopping potential depends upon frequency of incident light.

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

(7) Stopping potential does not depend upon intensity of light

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

2 1

m

v

_max2 = 3 eV

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

2 1

m

v

max

2



5 eV

Because stopping potential is not equal to 5V which cannot be used in einstein equation. Graphs :

(1) Kinetic energy V/s frequency :

At



₀ E_max = 0

(2) V_max V/s

   

:

At



₀ V_max = 0

(3) Saturated Current V/s Intensity :

(4) Stopping potential V/s frequency :



eV₀ = h



-h



₀ tan



= slope =

e h

(constant for all type of metals )

Intercept on x-axis =



₀ Intercept on y-axis = –

e h₀

(5) Potential V/s current : (



: constant)



Stopping potential does not depend upon intensity of light .

(6) Photo electric current V/s Retarding potential :

Ex. 5 If one photon has 25 eV energy and work function of material is 7 eV then value of stopping potential will

be-(A) 32 V (B) 18 V (C) 3.3 V (D) zero Sol V₀ = e E₀ = e 7 25 eV V₀ = 18 V _ _ _ ___ _ _ _ Current Retarding potential E_max _{0 } V_max ₀  Current Intensity I3 I2 I₁ I3>I2>I1 Current Intensity voltage -v₀   v₀

P OINTS TO RE M E MBE R

(1) Photo electric effect is based on the principle of conservation of energy.

(2) Quantum nature of radiations is verified by photo 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 of incident light & the nature of cathode material. (6) The velocity of electrons emitted from the

surface layer is maximum.

(7)







max



₂ 1 max E E =

h

h









1 0 2 0





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

(9) Frequency of untraviolet light is more than that of red light. Hence the kinetic energy of photo electrons emitted by ultraviolet light will be more than that of the electrons emitted by red light.

(10)



_max= m ) ( h 2 ₀

(11) Photoelectric effect was discovered by Hallwachs, experimentally verified by Hertz, and successfully studied and explained by Einstein. (12) V =          ₀ 1 1 e hc

(13) Less energy is required to release the electrons situated in the surface layer of material where as more energy is required to release electrons from the inner layers. (14) One electron is emitted by one photon. (15) Stopping potential does not depend on the

intensity of incident light.

(16) Intensity of light is related to amount of photons and in more simplified terms it is related to no. of photons.

But frequency of light directly indicates toward energy 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 = h



Conclusions

:-(A) If intensity in case 1 is higher so no. of emitted 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 so

kinetic energy of emitted electron will be more in case of II though their intensity is less than case I

(17) One photon can emit only one electron. The whole photo electric effect is based on this simple concept.

(18)



_red <



_violet It means incident violet ray will give more energy to emitted electrons than incident red ray it means



V_{0 (red)} < V_{0 (violet)}



V_{max (red)} < V_{max (violet)}



E_{max (red)} < E_{max (vilolet)}

(19) Photo electric effect is based on energy conservation law.

(20) If light of same frequency incidences on surfaces of various work function then velocity or kinetic energy of that emitted electron will be maximum for which surface that has least work function.

Ex.1 Two different photons of energies, 1 eV and 2.5 eV, fall on two identical metal plates having work function 0.5 eV, Then the ratio of maximum KE of the electrons emitted from the two surface

is-(A) 1 : 2 (B) 1 : 4 (C) 2 : 1 (D) 4 : 1 Sol

K

1max = h



1 –



= 1 – 0.5 = 0.5 eV

K

₂

max = 2.5 – 0.5 = 2.0 eV

Thus

K

1_max :

K

2_max= 0.5 : 2 = 1 : 4

Ex.2 Ultraviolet light of wavelength 280 nm is used in an experiment on photo electric effect with lithium (



= 2.5 eV) cathode. Stopping potential will

be-(A) 1 .9 eV (B) 1.9 V (C) 4.4 eV (D) 4.4 V

Sol The maximum kinetic energy is K_{max= }hc–



= 280 1242 nm nm eV – 2.5 eV = 4.4 eV – 2.5 eV = 1.9 eV

Stopping potential V is given by eV = K_max V = e K_max = e 1.9 eV = 1.9 V

Ex.3 A monochromatic source of light operating at 200 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 = ₂₀

10 4 200  = 5 × 10–19 J Wavelength =



= E hc = ₁₉ 8 34 10 5 ) 10 3 ( ) 10 63 . 6 (      



 = 4.0 × 10-7 = 400 nm

Ex.4 Which metal will be suitable for a photo electric cell using light of wavelength 4000A0. The work functions of sodium and copper are respectively 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 8 34 10 6 . 1 2 10 3 10 6 . 6        = 6188 Å



₀



_1



copper 0 sodium 0 ) ( ) (   = sodium copper ) ( ) (  



(₀)_copper = 4 2 × 6188 = 3094 Å To eject photo-electrons from sodium the longest wavelength is 6188 Å and that for copper is 3094 Å. Hence for light of wavelength 4000 Å, sodium is suitable. Ex.5 The work function for the surface of aluminium

is 4.2 eV. What will be the wavelength of that incident light for which the stopping potential 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 8 34 10 6 . 1 2 . 4 10 3 10 6 . 6        , ₀ = 2.946 × 10–7 m = 2946 Å Ex.6 Slope of V₀ –



curve

is-(A) e (B) e h (C)



₀ (D) h So Relation between V₀ –



., V₀ = e h – e h₀

Put it in the form of y = mx – c, here V₀ = y,



= x, e h₀ = c



y =       e h x – c



m = e h

Ex.7 A radio station is transmitting waves of wavelength 300 m, If diffracting power of transmitter is 10 kw, then numbers of photons diffracted per second

is-(A) 1.5 × 1035 (B) 1.5 × 1031 (C) 1.5 × 1029 (D) 1.5 × 1033 Sol P = 10 × 103 watt n = ?  = 300 m P = t nhc  104 = 1 300 n 10 3 10 62 . 6 34 8       n = _{34 8} 4 10 10 62 . 6 10 300     = 1.5 × 1031

Ex.8 Light of wavelength 332 Å incidents on metal surface (work function = 1.07 eV). To stop emission of photo electron, retarding potential 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 = V_{0 = } e hc – e 0  V₀ = e 10 332 10 3 10 62 . 6 10 8 34       -e 10 1.6 1.07  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 Joule of energy, then what is the number of photons falling 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 8 34 10 5000 10 3 10 6 . 6 n        n = ₂₆ 7 10 10 8 . 19 10 10 5000       = 0.25 × 1012 n = 2.5 × 1011

Ex.10 An electromagnetic radiation of frequency 3 × 1015 cycles per second falls on a photo electric surface whose work function is 4.0 eV. Find out the maximum velocity of the photo 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



= h



₀ + E_k 6.6 × 10-34 × 3 × 108 = 4 × 1.6 × 10–19 + E_k 19.8 × 10-19 – 6.4 × 10-19 = E_k E_k = 13.4 × 10–19 J



2 1 mv2_max = 13.4 × 10–19 v_max = m 10 4 . 13 2  19 = ₃₁ 19 10 9 10 4 . 13 2      = 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 8 34 10 4000 10 3 10 6 . 6       = 4.95 × 10-19 J

Ex.12 W hen ultraviolet light of energy 6.2 eV incidents on a aluminimum surface, it emits photo electrons. If work function for aluminium surface is 4.2 eV, then kinetic energy of emitted 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 E_k = E –



₀ = 6.2 – 4.2 = 2.0 eV, E_k = 2 × 1.6 × 10–19 = 3.2 × 10–19 J Ex.13 Using light of wavelength 6000 Å stopping

potential is obtained 2.4 volt for photo electric cell. If light of wavelength 4000 Å is used then stopping potential would

be-(A) 2.9 V (B) 1.9 V (C) 3.43 V (D) 9.4 V

Sol V₀ = e e hc_0  , 2.4 = e e 10 6000 hc 0 10     ....(1) V₀ = e e 10 4000 hc ₀ 10     ....(2) Eq. (1) - Eq.(2) 2.4 – V₀ = e hc          10 _{4000 10}10 1 10 6000 1 2.4 – V₀ = _  _        24 6 4 10 6 . 1 10 10 3 10 62 . 6 19 7 8 34 V₀ = 2.4 + _{7 19} 8 34 10 6 . 1 10 12 10 3 10 62 . 6          V₀ = 2.4 + 1.03 = 3.43 V

Ex.14 When light source is placed at 1 m distant from photo electric cell, then value of stopping potential is obtained 4 volt. If it is placed at 4 m distant, then value of stopping potential

becomes-(A) 2 volt (B) 1 volt

(C) 4 volt (D) 16 volt

Sol Stopping potential does not depend upon distance from light source.

Ex.15 When monochromatic light of wavelength  illuminates a metal surface then stopping potential for photo electric current is 3V₀. If wavelength changes to 2 then stopping potential becomes V₀ . Threshold wavelength for photo electric emission

is-(A) 4



(B) 8



(C) 4/3



(D) 6



Sol hc_ –



₀ = 3V₀ ...(1)  2 hc –



₀ = V₀ ...(2) eq. (1) - eq.(2) : hc_    2 1 1 _{= 2V} 0  2 hc = 2 V₀

 

= 0 V 4 hc



4 = 0 V hc