PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY CHAPTER 42

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Ashley Preston
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1 PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY CHAPTER nm to 78 nm E h E 1 E h j - s, c m/s, 1 4 nm, 78 nm J 19 So, the range is J to J.. h/p P h/ J J-S kg m/s nm m, 7 nm m E 1 E Energy absorbed by the atom in the process. [1/ 1 1/ ] [1/5 1/7] J 4. P 1 W E in 1 sec 1 J % used to convert into photon 6% Energy used 6 J Energy used to take out 1 photon / No. of photons used a) Here intensity I /m Intensity, I power area /m Let no.of photons/sec emitted n No.of photons/m n/ intensity 3 9 Power Energy emitted/sec n/ P int ensity n b) Consider no.of two parts at a distance r and r + dr from the source. The time interval dt in which the photon travel from one point to another dv/e dt. p dr In this time the total no.of photons emitted N n dt C These points will be present between two spherical shells of radii r and r+dr. It is the distance of the 1 st point from the sources. No.of photons per volume in the shell N Pdr 1 p (r + r + dr) rdr 4r ch 4 r In the case m, 5 nm, m P 4r , No.of photons/m 3 P 4r c) No.of photons (No.of photons/sec/m ) Area ( ) 4r (3.14)( )

2 m, 6, n , h/p P p/ 1 7 Force exerted on the wall n(mv cos ( mv cos )) n mv cos ½ N. 7. Power 1 W P Momentum h p E W Pc/t or Force or, P h or, P h t t or, E Power (W) t t or, P/t W/c force. 7/1 (absorbed) + 3/1 (reflected) 4. 7 W 3 W C 1 C / N. 8. m g The weight of the mirror is balanced. Thus force exerted by the photons is equal to weight P h E P C t t E PC Rate of change of momentum Power/C 3% of light passes through the lens. Thus it exerts force. 7% is reflected. Force exerted (rate of change of momentum) Power/C Power 3% mg C Power 1 w 1 MW Power 1 W Radius cm 6% is converted to light 6 w Now, Force power N. 8 velocity force 1 1 Pressure 1 area (.) N/m. 1. We know, If a perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture if intensity is I, Force r l C I.5 W/m, r 1 cm, C m/s (1) Force N. 5 6

3 For a perfectly reflecting solid sphere of radius r kept in the path of a parallel beam of light of large aperture with intensity I, force exerted r I C 1. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision. We get, hc/ + m c mc and applying conservation of momentum h/ mv m Mass of e m 1 v / c from above equation it can be easily shown that V C or V both of these results have no physical meaning hence it is not possible for a photon to be completely absorbed by a free electron. 13. r 1 m Energy Now, kq kq R 1 kq 1 or kq For max, q should be min, For minimum e C Max kq m For next smaller wavelength m (1.6 ) nn m 1.9 ev Max KE of electrons 1.65 ev 1.6 ev. 15. W J a) We know W h W.5 1 h b) ev h W or, V 34 8 hc Hz Hz h W e ev J a) Threshold wavelength /.91 V hc m 31 nm b) Stopping potential is.5 V E + ev / nm

4 17. Energy of photoelectron ½ mv hv P We know KE m P m KE. P P kg m/s nm m V 1.1 V ev ev.65 ev a) When 35, V s 1.45 and when 4, V s m 6 nm. W (1) and 4 W + 1 () Subtracting () from (1) and solving to get the value of h we get h ev-sec b) Now work function w ev. 35 c) w there cathod nm. w ev - s. The electric field becomes times per second. Frequency h + ke h KE KE ev.48 ev. 1. E E sin[( m 1 ) (x ct)] W C Stopping potential 1/

5 f Hz W 1.9 ev Now ev h W ev ev So, V 1.5 V E 1 sin[( s 1 )t] sin [ s 1 )t] 1 ½ [cos[( s 1 )t] cos [ s 1 )t] The w are and for largest K.E. f max wmax E K.E. hf K.E KE KE ev 3.93 ev. 3. W hv ev (Given V 15 V, No. of photons , Power 5 mw) J ev We have to take two cases : Case I v Hz Case II v Hz We know ; a) ev h w 1.656e h w (1) 5h w () 1.656e 4w w ev.414 ev 4 b) Putting value of w in equation () 5w 5h h 1 14 h ev-s 5. w.6 ev For w to be min becomes maximum. w 34 8 or w m 71 nm 4.5 V (in volts) v(in 1 14 Hz)

6 6. 4 nm, P 5 w E of 1 photon 14 4 ev No.of electrons No.of electrons 1 per 1 6 photon Energy of 1 photon No.of photoelectrons emitted Photo electric current nm 1 7 m E of one photon No.of photons no.s A 1.6 A. 11 Hence, No.of photo electrons Net amount of positive charge q developed due to the outgoing electrons C. Now potential developed at the centre as well as at the surface due to these charger 9 1 Kq r V.3 V ev 1 4 nm, 6 nm for B to the minimum energy should be maximum should be minimum. E ev. The presence of magnetic field will bend the beam there will be no current if the electron does not reach the other plates. r r mv qb me qb 1 cm X X X X X X A B B T 9. Given : fringe width, y 1. mm. mm, D.4 mm, W. ev, D 1. m y or, D d 3 3 yd D m B A B A B S A E ev Stopping potential ev V 4.6

7 ev, nm Stopping potential or energy E WC Minimum 1.7 V is necessary to stop the electron The minimum K.E. ev [Since the electric potential of V is reqd. to accelerate the electron to reach the plates] the maximum K.E. (+1, 7)ev 3.7 ev. 31. Given cm, W (C s ) 1.9 ev, d cm. m, 4 nm we know Electric potential due to a charged plate V E d Where E elelctric field due to the charged plate /E d Separation between the plates. V d 1 E V V e h w w ev or, V 1.5 V As V is much less than V Hence the minimum energy required to reach the charged plate must be.6 ev For maximum KE, the V must be an accelerating one. Hence max KE V + V ev 3. Here electric field of metal plate E P/E v/m accl. de qe / m t y 1 a K.E. w ev sec J [because in previous problem i.e. in problem 31 : KE 1. ev] V KE m Horizontal displacement V t t m 9. cm. 33. When 5 nm Energy of photon K.E. w ev ev 3.6 ev. Velocity to be non positive for each photo electron The minimum value of velocity of plate should be velocity of photo electron Velocity of photo electron KE / m y cm Metal plate 4.7

8 m/sec. 34. Work function, distance d The particle will move in a circle When the stopping potential is equal to the potential due to the singly charged ion at that point. ev 1 ke 1 V e d e Ke Ke Ke d d d d d d 8d. 1 Ke d d e 8d 4 e 35. a) When 4 nm 14 Energy of photon 3.1 ev 4 This energy given to electron But for the first collision energy lost 3.1 ev 1%.31 ev for second collision energy lost 3.1 ev 1%.31 ev Total energy lost the two collision ev K.E. of photon electron when it comes out of metal / work function Energy lost due to collision 3.1 ev ev b) For the 3 rd collision the energy lost.31 ev Which just equative the KE lost in the 3 rd collision electron. It just comes out of the metal Hence in the fourth collision electron becomes unable to come out of the metal Hence maximum number of collision 4. ion d 4.8

The Photoelectric Effect

The Photoelectric Effect Lenard s experiment The photon model Light as photons Einstein s explanation of the photoelectric effect Photon energy Electron volts Electron energy 1 Lenard s experiment Philipp