Modern Physics Photoelectric Effect

MODERN PHYSICS INTRODUCTION The photoelectric effect or photoemission is the production of electrons or other free corners when light is shone onto a material. Electrons emitted in this manner can be called photoelectrons. It was discovered by manner can be called photoelectrons. It was discovered by Hertz in 1887. He found that when negative plate of on electronic discharge tube was illuminated with ultraviolet light. The electric discharge took place more readily. Further experiments carried out by Hallwachs confirmed that certain negatively charge particles are emitted, whenb a Zn plate is illuminated with ultraviolet light. These particles were identified as electrons. G 1. 2. Photoelectron : The electron emitted in photoelectric effect is called photoelectron. Photoelectric current : If current passes through the circuit in photoelectric effect then the current is called photoelectric current. 3. Work function : The minimum energy required to make an electron free from the metal is called work function. It is constant for a metal and denoted by  or W. It is the minimum for Cesium. It is relatively less for alkali metals. WORK FUNCTIONS OF SOME PHOTOSENSITIVE METALS Metal W ork function (ev) Metal W ork function (eV) Ces ium Po tas s iu m So d iu m Lith ium 1.9 2.2 2.3 2.5 Calciu m Cop p er Silv er Platin u m 3.2 4.5 4.7 5.6 To produce photo electric effect only metal and light is necessary but for observing it, the circuit is completed. Figure shows an arrangement used to study the photoelectric effect. intensity frequency A 1 2 C A V Rheostat cell, few volts Here the plate (1) is called emitter or cathode and other plate (2) is called collector or anode. 1. 2. Saturation current : When all the photo electrons emitted by cathode reach the anode then current flowing in the circuit at that instant is known as saturated current, this is the maximum value of photoelectric current. Stopping potential : Minimum magnitude of negative potential of anode with respect to cathode for which current is zero is called stopping potential. This is also known as cutoff voltage. This voltage is independent of intensity. 3. Retarding potential : Negative potential of anode with respect to cathode which is less than stopping potential is called retarding potential. EXPLANATION OF PHOTOELECTRIC EFFECT 1. On the basis of wave theory : According to wave theory, light is an electromagnetic radiation consisting of oscillating electric field vectors and magnetic field vectors. When electromagnetic radiations are incident on a metal surface, the free electrons [free electrons means the electrons which are loosely bound and free to move inside the metal] absorb energy from the radiation. This occurs by the oscillations of electron under the action of electric field vector of electromagnetic radiation. When an electron acquires sufficiently high energy so that it can overcome its binding energy, it comes out from the metal. 2. On the basis of photon theory: According to photon theory of light, light consists of particles (called photons). Each particle carries a certain amount of energy with it. This energy is given by E=h, where h is the Plank's constant and  is the frequency. When the photons are incident on a metal surface, they collide with electrons. In some of the collisions, a photon is absorbed by an electron. Thus an electron gets energy h. If this energy is greater than the binding energy of the electron, it comes out of the metal surface. The extra energy given to the electron becomes its kinetic energy. EXPERIMENTS 1. Hertz Experiment : Hertz observed that when ultraviolet rays are incident on negative plate of electric discharge tube then conduction takes place easily in the tube. ultraviolet rays cathode anode Evacuated quartz tube A 2. Hallwach Experiment : Hallwach observed that if negatively charged Zn plate is illuminated by U.V. light, its negative charge decreases and it becomes neutral and after some time it gains positive charge. It means in the effect of light, some negative charged particles are emitted from the metal. 3. Lenard Explanation : He told that when ultraviolet rays are incident on cathode, electrons are ejected. These electrons are attracted by anode and circuit is completed due to flow of electrons and current flows. When U.V. rays incident on anode, electrons are ejected but current does not flow. For the photo electric effect the light of short wavelength (or high frequency) is more effective than the light of long wavelength (or low frequency) 4. Experimental study of photoelectric Effect : When light of frequency  and intensity I falls on the cathode, electrons are emitted from it. The electrons are collected by the anode and a current flows in the circuit. This current is called photoelectric current. This experiment is used to study the variation of photoelectric current with different factors like intensity, frequency and the potential difference between the anode and cathode. MODERN PHYSICS Light of intensity I and frequency Photoelectrons Anode A i (Photoelectric current) Cathode (Photosensitive metal) V Potential Divider (i) Variation of photoelectric current with potential difference : With the help of the above experimental setup, a graph is obtained between current and potential difference. The potential difference is varied with the help of a potential divider. The graph obtained is shown below. i -V0 O Anode Potential V The main points of observation are : (a) At zero anode potential, a current exists. It means that electrons are emitted from cathode with some kinetic energy. (b) As anode potential is increased, current increases. This implies that different electrons are emitted with different kinetic energies. (c) After a certain anode potential, current acquires a constant value called saturation current. Current acquires a saturation value because the number of electrons emitted 7 per second from the cathode are fixed. (d) At a certain negative potential, the photoelectric current becomes zero. This is called stopping potential (V0). Stopping potential is a measure of maximum kinetic energy of the emitted electrons. Let KEmax be the maximum kinetic energy of an emitted electron, then KEmax = eV0. i (ii) Variation of current with intensity The photoelectric current is found to be directly proportional to intensity of incident Intensity radiation. i (iii) Effect of intensity on saturation current and stopping potential I2 I1 (a) Saturation current increases with increase in intensity. (b) Stopping potential (and therefore maximum kinetic energy) is independent of intensity. V0 (I2>I1) Anode Potential V (iv) Effect of frequency (a) Stopping potential is found to vary with frequency of incident light linearly. Greater the frequency of incident light, greater the stopping potential. (b) There exists a certain minimum frequency 0 below which no stopping potential is required as no emission of electrons takes place. This frequency is called threshold frequency. For photoelectric emission to take place, > 0. (i) Photo electric effect is an instantaneous process, as soon as light is incident on the metal, photo electrons are emitted. (ii) Stopping potential does not depend on the distance between cathode and anode. (iii) The work function represented the energy needed to remove the least tightly bounded electrons from the surface. It depends only on nature of the metal and independent of any other factors. 1. Failure of wave theory of light (i) According to wave theory when light incident on a surface, energy is distributed continuously over the surface. So that electron has to wait to gain sufficient energy to come out. But in experiment there is no time lag. Emission of electrons takes place in less than 10–9 s. This means, electron does not absorb energy. They get all the energy once. (ii) When intensity is increased, more energetic electrons should be emitted. So that stopping potential should be intensity dependent. But it is not observed. (iii) According to wave theory, if intensity is sufficient then, at each frequency, electron emission is possible. It means there should not be existence of threshold frequency. 2. Einstein's Explanation of Photoelectric Effect Einstein explained photoelectric effect on the basis of photon–electron interaction. The energy transfer takes place due to collisions between an electrons and a photon. The electrons within the target material are held there by electric force. The electron needs a certain minimum energy to escape from this pull. This minimum energy is the property of target material and it is called the work function. When a photon of energy E=h collides with and transfers its energy to an electron, and this energy is greater than the work function, the electron can escape through the surface. 3. Einstein's Photoelectric Equation h     KE max Here h is the energy transferred to the electron. Out of this,  is the energy needed to escape. The remaining energy appears as kinetic energy of the electron. Now KEmax = eV0 (where V0 is stopping potential)   h  h =  + eV0  V0 =      e e Thus, the stopping potential varies linearly with the frequency of incident radiation. h Slope of the graph obtained is . This graph helps in determination of Planck's constant. e MODERN PHYSICS (i) Einstein's Photo Electric equation is based on conservation of energy. (ii) Einstein explained P.E.E. on the basis of quantum theory, for which he was awarded noble prize. (iii) According to Einstein one photon can eject one e– only. But here the energy of incident photon should greater or equal to work function. (iv) In photoelectric effect all photoelectrons do not have same kinetic energy. Their KE range from zero to Emax which depends on frequency of incident radiation and nature of cathode. (v) The photo electric effect takes place only when photons strike bound electrons because for free electrons energy and momentum conservations do not hold together. Ex. In an experiment on photo electric emission, following observations were made; (i) Wavelength of the incident light = 1.98 × 10–7 m; (ii) Stopping potential = 2.5 volt. Find : (a) Kinetic energy of photoelectrons with maximum speed. (b) Work function and (c) Threshold frequency; Sol. (a) Since (b) Energy of incident photon E= (c) vs = 2.5 V, Kmax = eVs 12400 eV = 6.26 eV 1980 So, Kmax = 2.5 eV W = E – Kmax = 3.76 eV hth = W = 3.76 × 1.6 × 10–19 J  th = 3.76  1.6  10 19 6.6  10  34  9.1 1014 Hz Ex. A beam of light consists of four wavelength 4000 Å, 4800 Å, 6000 Å and 7000 Å, each of intensity 1.5 × 10–3 Wm–2. The beam falls normally on an area 10–4 m2 of a clean metallic surface of work function 1.9 eV. Assuming no loss of light energy (i.e. each capable photon emits one electron) calculate the number of photoelectrons liberated per second. Sol. E1 = 12400 = 3.1 eV,, 4000 and E2 = 12400 = 2.58 eV 4800 E4 = 12400 = 1.77 eV 7000 E3 = 12400 = 2.06 eV 6000 Therefore, light of wavelengths 4000 Å, 4800 Å and 6000 Å can only emit photoelectrons.  Number of photoelectrons emitted per second = No. of photons incident per second) = =  1 1 1  3A3 1A1 2A2 + + = A  E  E  E  E1 E2 E3  1 2 3  (1.5  10 3 )(10 4 ) 1.6  10  19 1 1   1      3.1 2.58 2.06  Ex. Sol. Calculate the possible velocity of a photoelectron if the work function of the target material is 1.24 eV and wavelength of light is 4.36 × 10–7 m. What retarding potential is necessary to stop the emission of electrons? 1 hc 2  As KEmax = h –  mvmax  h     2  vmax =  6.6 3  10 34  3  10 8   hc  2  1.24  1.6  10 19  2      = 4.3 6  10 7   = 7.523 × 105 m/s 31 m 9 .11  10  The speed of a photoelectron can be any value between 0 and 7.43 × 105 m/s If V0 is the stopping potential, then eV0 =  V0 = 1 m v2max 2  hc  10 Q e  12 40 0  1 0 V  m    1 mv 2max hc  12400   = – 1.24 = 1.60 V 2 e e  e 4360 Ex. The surface of a metal of work function  is illuminated by light whose electric field component varies with time as E = E0 [ 1 + cos t] sin0t. Find the maximum kinetic energy of photoelectrons emitted from the surface. Sol. The given electric field component is E=E0 sin t + E0 sin 0t cos t = E0 sin 0t + E0 [ sin (0 + ) t + sin(0–)t] 2  The given light comprises three different frequencies viz. , 0 + 0 – The maximum kinetic energy will be due to most energetic photon.  KEmax = h– = Ex. h   0  – 2   Q   2  or  = 2   When light of wavelength  is incident on a metal surface, stopping potential is found to be x. When light of wavelength n is incident on the same metal surface, stopping potential is found to be x . Find the threshold n 1 wavelength of the metal. Sol. Let 0 is the threshold wavelength. The work function is   Now, by photoelectric equation From (i) and (ii) hc hc ex =    0 ...(i) hc . 0 ex hc hc = – n 1 n 0 ...(ii) hc hc hc hc nhc hc   = n2   n  1   n  1   0 n 0 0 n PHOTON THEORY OF LIGHT According to Plank’s Quantum Theory light consist of photon as energy packets having following properties : (i) Each photon i.e. of energy E = h = hc / where h is plank’s constant (ii) A photon is a particle of light moving with speed 299792458 m/s in vacuum. (iii) The speed of a photon is independent of frame of reference. This is the basic postulate of theory of relativity. (iv) The rest mass of a photon is zero. i.e. photons do not exist at rest. MODERN PHYSICS (v) Effective mass of photon m = E hc h 1 = 2 = i.e. m   c2 c  c So mass of violet light photon is greater than the mass of red light photon. (R > V) (vi) According to Planck the energy of a photon is directly proportional to the frequency of the radiation. E  E or hc joule  E = h ( c = ) or E= hc 1 24 00 = eV – Å e   Q  o hc   1 2400(A  eV) e  Here E = energy of photon, c = speed of light, h = Planck's constant, e = charge of electron h = 6.62 × 10–34 J–s,  = frequency of photon, = wavelength of photon E h h   c c  (viii) A photon can collide with material particles like electron. During these collisions, the total energy and total momentum remain constant. (vii) Linear momentum of photon p = (ix) Energy of light passing through per unit area per unit time is known as intensity of light. Intensity of light I  E P  At A ...(i) Here P = power of source, A = Area, t = time taken E = energy incident in t time = Nh N = number of photon incident in t time Intensity I = N(h  ) n(h  ) = At A From equation (i) and (ii), ...(ii)  Q  n N   no. of photon per sec.  t  P P P n(h  ) =  n = = = 5 × 1024 J–1 m–1 × P ×  h hc A A (x) When photons fall on a surface, they exert a force and pressure on the surface. This pressure is called radiation pressure. (xi) Force exerted on perfectly reflecting surface Let 'N' photons are there in time t, Momentum before striking the surface (p1)= Nh  Momentum after striking the surface (p2)=  Nh  Change in momentum of photons = p2 – p1 = But change in momentum of surface = p = F= 2h P  2P    hc c and 2Nh  2Nh 2 Nh 2 h  P  n   but n = ; So that force on surface F = t  hc   Pressure = F 2P 2 I   A cA c P  Q I  A    (xii) Force exerted on perfectly absorbing surface Nh 0 h P  P  p1  p 2 Nh   F= = =n ; F= Q n    c hc  t t t and Pressure = F P I   A Ac c (xiii) When a beam of light is incident at angle  on perfectly reflector surface  F= Ex. 2P F 2 I cos  2IA cos  2 h   cos  = n   cos  Pressure = c A c c   o Violet light (  4000 A) of entensity 4 watt/m2 falls normally on a surface of area 10 cm × 20 cm. Find (a) the energy received by the surface per second. (b) the number of photons hitting the surface per second. (c) If surface is tiled such that plane of the surface makes an angle 30° with light beam, find the number of photons hitting the surface per second. Sol. (a) Energy received per second per unit area E = I A cos  = 4 × 0.02 J × cos 0° = 0.08 J (b) n h (c/) = E  n= (c) n = Ex. 0.08× 4000×10 –10 32×1017 = = 1.699×1017 –34 8 6.63×10 ×3×10 19.89 IAcos60°× λ 1 32×1017 = × = 0.805×1017 2 19.89 hc The intensity of sunlight on the surface of earth is 1400 W/m2. Assuming the mean wavelength of sunlight to be 6000 Å, calculate:– (a) The photon flux arriving at 1 m2 area on earth perpendicular to light radiations and (b) The number of photons emitted from the sun per second (Assuming the average radius of Earth's orbit to be 1.49 × 1011 m) Sol. (a) Energy of a photon E = Photon flux = hc 12 40 0  =2.06 eV = 3.3 × 10–19 J  6 00 0 IA 14 00  1 = = 4.22 × 1021 photons/sec. E 3.3  10 19 (b) Number of photons emitted per second n = P IA 1 4 0 0  4   (1 .4 9  10 11 )2  = = 1.18 × 1045 E E 3 .3  1 0 19 MODERN PHYSICS Ex. In a photoelectric setup, a point source of light of power 3.2 × 10–3 W emits monochromatic photons of energy 5.0 eV. The source is located at a distance 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and radius 8 × 10–3 m. The efficiency of photoelectron emission is one for every 106 incident photons. Assuming that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission, Find (i) the number of photoelectrons emitted per second. (ii) the time t after light source is switched on, at which photoelectron emission stops. Sol. Energy of a single photon E=5.0 eV = 8 × 10–19 J Power of source P = 3.2 × 10–3 W  number of photons emitted per second n = P 3 .2  1 0 3  = 4× 1015/s E 8  1 0 1 9 The number of photons incident per second on metal surface is n0 = n0 = 4  1 0 15 4   0.8  2    8  10 3 Number of electrons emitted =  2 n  r 2 4 R 2  1.0  10 1 1 photon/s 1 .0  1 0 11 = 105 /s 106 Source R=0.8m r=0.8×10–3m KEmax = h– = 5.0 –3.0 = 2.0 eV The photoelectron emission stops, when the metallic sphere acquires stopping potential. q As KEmax = 2.0 eV  Stopping potential V0 = 2V  2 = 4  r q = 1.78 × 10–12 C 0 Now charge q = (number of electrons/second) × t× e t = 1 .7 8  1 0 1 2 = 111s 1 0 5  1 .6  1 0 1 9 PHOTO CELL A photo cell is a practical application of the phenomenon of photo electric effect, with the help of photo cell light energy is converted into electrical energy. 1. Construction : A photo cell consists of an evacuated sealed glass tube containing anode and a concave cathode of suitable emitting material such as Cesium (Cs). 2. Working: When light of frequency greater than the threshold frequency of cathode material falls on the cathode, photoelectrons emitted are collected by the anode and an electric current starts flowing in the external circuit. The current increase with the increase in the intensity of light. The current would stop, if the light does not fall on the cathode. 3. Application (i) In television camera. (ii) In automatic door (iii) Burglar’s alarm (iv) Automatic switching of street light and traffic signals. 1, 2, 3..... represents the I, II & III line of Lymen, Balmer, Paschen series. The hydrogen spectrum (some selected lines) (c) E31 = – 3.08 – (– 15.6) = 12.52 eV Therefore, excitation potential for state n = 3 is 12.52 volt. (d) 1  E 31  31 = 12400 Å–1 = 12.52 –1 Å 12400  1.01 × 107 m–1 (e) (i) E2 – E1 = 10.3 eV > 6 eV. Hence electron cannot excite the atoms. (ii) So, Kmin = 6 eV. E2 – E1 = 10.3 eV < 11 eV. Hence electron can excite the atoms. So, Kmin = (11 – 10.3) = 0.7 eV.