42.photo electric effect by Aditya Bhardwaj

PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY CHAPTER 42 1.

1 = 400 nm to 2 = 780 nm E = h = E1 =

hc 

6.63  10 34  3  108 9



j - s, c = 3  10 m/s, 1 = 400 nm, 2 = 780 nm

6.63  3 –19  10 19 = 5  10 J 4

400  10 6.63  3 –19 E2 =  10 19 = 2.55  10 J 7.8 –19 –19 So, the range is 5  10 J to 2.55  10 J.  = h/p 6.63  1034

–27 –27 J-S = 1.326  10 = 1.33  10 kg – m/s. 500  10 9 –9 –9 1 = 500 nm = 500  10 m, 2 = 700 nm = 700  10 m E1 – E2 = Energy absorbed by the atom in the process. = hc [1/1 – 1/2] –19 –19  6.63  3[1/5 – 1/7]  10 = 1.136  10 J P = 10 W  E in 1 sec = 10 J % used to convert into photon = 60% Energy used = 6 J

 P = h/ =

–34

h = 6.63  10

Energy used to take out 1 photon = hc/ =

6.63  10 34  3  108 9



6.633  10 17 590

590  10 6 6  590 19   1017  176.9  1017 = 1.77  10  No. of photons used = 6.63  3  6.63 3  10 17 590 power 3 2 3 2 a) Here intensity = I = 1.4  10 /m Intensity, I = = 1.4  10 /m area Let no.of photons/sec emitted = n  Power = Energy emitted/sec = nhc/ = P 2 No.of photons/m = nhc/ = intensity int ensity   1.9  103  5  10 9   3.5  1021 hc 6.63  10 34  3  108 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. n=

 p  dr In this time the total no.of photons emitted = N = n dt =    hc  C These points will be present between two spherical shells of radii ‘r’ and r+dr. It is the distance of the st 1 point from the sources. No.of photons per volume in the shell

N Pdr 1 p    2 2 2r2dr hc 4r ch 4 hc 2r 2 11 –9 In the case = 1.5  10 m,  = 500 nm, = 500  10 m

(r + r + dr) =

P 4r 2

 1.4  103 ,  No.of photons/m = 3



4r 2 hc 2

500  10 9

 1.2  1013 6.63  10 34  3  108 2 c) No.of photons = (No.of photons/sec/m )  Area 21 2 = (3.5  10 )  4r 21 11 2 44 = 3.5  10  4(3.14)(1.5  10 ) = 9.9  10 . = 1.4  10 

42.1

Photo Electric Effect and Wave Particle Quality 6.

–9

 = 663  10 m,  = 60°, n = 1  10 ,  = h/p –27  P = p/ = 10 Force exerted on the wall = n(mv cos  –(–mv cos )) = 2n mv cos . 19 –27 –8 = 2  1  10  10  ½ = 1  10 N. Power = 10 W P  Momentum h h P h = or, P = or,  p  t t

hc  W = Pc/t or Force

60°

E hc = Power (W)  t t or, P/t = W/c = force. = 7/10 (absorbed) + 2  3/10 (reflected)

or,

7 W 3 W 7 10 3 10    2    2  10 3  108 10 3  108 10 C 10 C –8 –8 = 13/3  10 = 4.33  10 N.

m = 20 g The weight of the mirror is balanced. Thus force exerted by the photons is equal to weight P=

h 

hc  PC 

E P  C t t  Rate of change of momentum = Power/C 30% of light passes through the lens. Thus it exerts force. 70% is reflected.  Force exerted = 2(rate of change of momentum) = 2  Power/C 

 2  Power  30%    mg C  

20  10 3  10  3  108  10 = 10 w = 100 MW. 23 Power = 100 W Radius = 20 cm 60% is converted to light = 60 w power 60 Now, Force =   2  10 7 N . velocity 3  108  Power =

Pressure =

force 2  107 1    105 2 area 4  3.14  (0.2) 8  3.14 = 0.039  10

–5

= 3.9  10

–7

= 4  10

N/m .

10. 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,

r 2l C 2 8 I = 0.5 W/m , r = 1 cm, C = 3  10 m/s

Force =

  (1)2  0.5 8



3.14  0.5

3  10 3  108 –8 –9 = 0.523  10 = 5.2  10 N. 42.2

Photo Electric Effect and Wave Particle Quality 11. For a perfectly reflecting solid sphere of radius ‘r’ kept in the path of a parallel beam of light of large r 2I aperture with intensity ‘I’, force exerted = C 12. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision. 2 2 We get, hC/ + m0c = mc and applying conservation of momentum h/ = mv m0 Mass of e = m = 1 v2 / c2 from above equation it can be easily shown that V=C or V=0 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,

kq2 kq2  R 1

kq2 hc = 1 

or  =

hc kq2

For max ‘’, ‘q’ should be min, –19 For minimum ‘e’ = 1.6  10 C hc 3 = 0.863  10 = 863 m. Max  = kq2 For next smaller wavelength = 14.  = 350 nn = 350  10  = 1.9 eV

–9

6.63  3  10 34  108 9  109  (1.6  2)2  10 38



863 = 215.74 m 4

hC 6.63  10 34  3  108   1.9  350  109  1.6  10 19 = 1.65 ev = 1.6 ev. –19 15. W 0 = 2.5  10 J a) We know W 0 = h0 Max KE of electrons =

W0 2.5  1019 14 14 = 3.77 10 Hz = 3.8  10 Hz  h 6.63  1034 b) eV0 = h – W 0 0 =

h  W0 6.63  10 34  6  1014  2.5  10 19  = 0.91 V e 1.6  1019 –19 16.  = 4 eV = 4  1.6  10 J a) Threshold wavelength =   = hc/ or, V0 =

hC 6.63  10 34  3  108 6.63  3 10 27 =   9  3.1 107 m = 310 nm.  6.4 4  1.6  10 19 10 b) Stopping potential is 2.5 V  E =  + eV –19 –19  hc/ = 4  1.6  10 + 1.6  10  2.5  =

  

6.63  1034  3  108   1.6  10 19

6.63  3  1026 1.6  10

19

 6.5

= 4 + 2.5

= 1.9125  10

–7

= 190 nm. 42.3

Photo Electric Effect and Wave Particle Quality 17. Energy of photoelectron 2

 ½ mv =

hc 4.14  10 15  3  108  2.5ev = 0.605 ev.  hv 0 =  4  10 7

P2 2  P = 2m  KE. 2m 2 –31 –19 P = 2  9.1  10  0.605  1.6  10 –25 P = 4.197  10 kg – m/s –9 18.  = 400 nm = 400  10 m V0 = 1.1 V hc hc   ev 0  0 We know KE =



6.63  10 34  3  108 400  10

 4.97 = 

9



6.63  10 34  3  108  1.6  10 19  1.1 0

19.89  10 26  1.76 0

19.89  10 26 = 4.97 – 17.6 = 3.21 0

19.89  10 26 –7 = 6.196  10 m = 620 nm. 3.21 19. a) When  = 350, Vs = 1.45 and when  = 400, Vs = 1  0 =



hc = W + 1.45 350

…(1)

hc =W+1 …(2) 400 Subtracting (2) from (1) and solving to get the value of h we get –15 h = 4.2  10 ev-sec

Stopping potential

and

b) Now work function = w = =

hc = ev - s 

1240  1.45 = 2.15 ev. 350

c) w =

hc hc   there cathod   w

1240 = 576.8 nm. 2.15 45 20. The electric field becomes 0 1.2  10 times per second. =

 Frequency =

1.2  1015 15 = 0.6  10 2

h = 0 + kE  h – 0 = KE

6.63  10 34  0.6  1015

2 1.6  10 19 = 0.482 ev = 0.48 ev. 7 –1 21. E = E0 sin[(1.57  10 m ) (x – ct)] 7 W = 1.57  10  C  KE =

42.4

1/ 

Photo Electric Effect and Wave Particle Quality

1.57  107  3  108 Hz 2 Now eV0 = h – W 0  f=

W 0 = 1.9 ev

1.57  3  1015 – 1.9 ev 2 = 3.105 – 1.9 = 1.205 ev = 4.14  10

So, V0 =

–15



1.205  1.6  10 19

= 1.205 V. 1.6  1019 15 –1 15 –1 22. E = 100 sin[(3  10 s )t] sin [6  10 s )t] 15 –1 15 –1 = 100 ½ [cos[(9  10 s )t] – cos [3  10 s )t] 15 15 The w are 9  10 and 3  10 for largest K.E. fmax =

w max 9  1015 = 2 2

E – 0 = K.E.  hf – 0 = K.E. 

6.63  1034  9  1015

 2  KE  2  1.6  10 19  KE = 3.938 ev = 3.93 ev. 23. W 0 = hv – ev0 5  10 3

 1.6  10 19  2 (Given V0 = 2V, No. of photons = 8  10 , Power = 5 mW) 8  1015 = 6.25  10–19 – 3.2  10–19 = 3.05  10–19 J =

3.05  10 19

= 1.906 eV. 1.6  10 19 24. We have to take two cases : Case I … v0 = 1.656 14  = 5  10 Hz Case II… v0 = 0 14  = 1  10 Hz We know ; a) ev 0  h  w 0

V (in volts) 2 1.656 1

1.656e = h  5  10 – w0 …(1) 14 …(2) 0 = 5h  10 – 5w0 1.656e = 4w0 1.656  w0 = ev = 0.414 ev 4 b) Putting value of w0 in equation (2) 14  5w0 = 5h  10 14  5  0.414 = 5  h  10 –15  h = 4.414  10 ev-s 25. w0 = 0.6 ev For w0 to be min ‘’ becomes maximum. w0 =

hc hc 6.63  10 34  3  108 or  = =  w0  0.6  1.6  10 19 –7

= 20.71  10

v(in 1014 Hz)

m = 2071 nm 42.5

Photo Electric Effect and Wave Particle Quality 26.  = 400 nm, P = 5 w E of 1 photon =

hc  1242  =   ev   400 

5 5  400  Energy of 1 photon 1.6  10 19  1242

No.of electrons =

No.of electrons = 1 per 10 photon.

5  400

No.of photoelectrons emitted =

1.6  1242  1019  106 5  400 –6 Photo electric current =  1.6  1019 = 1.6  10 A = 1.6 A. 1.6  1242  106  10 19 –7 27.  = 200 nm = 2  10 m E of one photon = No.of photons =

hc 6.63  10 34  3  108 –19 = = 9.945  10  2  107

1 10 7 9.945  10

19

= 1  10

no.s

1 1011

7 = 1  10 10 4 Net amount of positive charge ‘q’ developed due to the outgoing electrons 7 –19 –12 = 1  10  1.6  10 = 1.6  10 C. Now potential developed at the centre as well as at the surface due to these charger

Hence, No.of photo electrons =

Kq 9  109  1.6  10 12 –1 = 3  10 V = 0.3 V.  r 4.8  10 2 28. 0 = 2.39 eV 1 = 400 nm, 2 = 600 nm for B to the minimum energy should be maximum   should be minimum. =

hc  0 = 3.105 – 2.39 = 0.715 eV.  The presence of magnetic field will bend the beam there will be no current if the electron does not reach the other plates. mv r= qB E=

10 cm

X X

2mE qB

 r=  0.1 =

2  9.1 10 31  1.6  10 19  0.715

1.6  10 19  B  B = 2.85  10 T 29. Given : fringe width, y = 1.0 mm  2 = 2.0 mm, D = 0.24 mm, W 0 = 2.2 ev, D = 1.2 m –5

D y= d or,  =

yd 2  10 3  0.24  10 3 –7  = 4  10 m D 1.2

B A B A B

hc 4.14  10 15  3  108 = 3.105 ev   4  10 Stopping potential eV0 = 3.105 – 2.2 = 0.905 V E=

42.6

Photo Electric Effect and Wave Particle Quality 30.  = 4.5 eV,  = 200 nm

WC   Minimum 1.7 V is necessary to stop the electron The minimum K.E. = 2eV [Since the electric potential of 2 V is reqd. to accelerate the electron to reach the plates] the maximum K.E. = (2+1, 7)ev = 3.7 ev. 31. Given –9 –2  = 1  10 cm , W 0 (Cs) = 1.9 eV, d = 20 cm = 0.20 m,  = 400 nm we know  Electric potential due to a charged plate = V = E  d Where E  elelctric field due to the charged plate =/E0 d  Separation between the plates. Stopping potential or energy = E –  =

 1 10 9  20 = 22.598 V = 22.6 d  E0 8.85  10 12  100

hc 4.14  10 15  3  108  w0   1.9   4  10 7 = 3.105 – 1.9 = 1.205 ev or, V0 = 1.205 V As V0 is much less than ‘V’ Hence the minimum energy required to reach the charged plate must be = 22.6 eV For maximum KE, the V must be an accelerating one. Hence max KE = V0 + V = 1.205 + 22.6 = 23.8005 ev 32. Here electric field of metal plate = E = P/E0 V0e = h – w0 =

1 10 19

= 113 v/m 8.85  10 12 accl. de =  = qE / m =

1.6  10 19  113 9.1 10 31

= 19.87  10

Metal plate

y = 20 cm

2y 2  20  10 2 –7 t= = 1.41  10 sec  a 19.87  10 31 hc K.E. =  w = 1.2 eV  –19 = 1.2  1.6  10 J [because in previous problem i.e. in problem 31 : KE = 1.2 ev]  V=

2KE  m

2  1.2  1.6  10 19

–6

= 0.665  10 4.1 10 31  Horizontal displacement = Vt  t –6 –7 = 0.655  10  1.4  10 = 0.092 m = 9.2 cm. 33. When  = 250 nm Energy of photon =

hc 1240 = 4.96 ev   250

hc  w = 4.96 – 1.9 ev = 3.06 ev.  Velocity to be non positive for each photo electron The minimum value of velocity of plate should be = velocity of photo electron

 K.E. =

 Velocity of photo electron =

2KE / m

42.7

Photo Electric Effect and Wave Particle Quality =

3.06 9.1 10 31

3.06  1.6  10 19



9.1 10 31

= 1.04  10 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. hc eV0 =  

ke  hc  hc 1 1  V0 =        2d     e e 

ion d

Ke2 hc hc Ke2 Ke2  2d       2d 2d 2d

 =

hc 2d 2

Ke  2d



8 hcd 2hcd .  2 0 1 e  80 d  2d  40 e2

35. a) When  = 400 nm

hc 1240 = 3.1 eV   400 This energy given to electron But for the first collision energy lost = 3.1 ev  10% = 0.31 ev for second collision energy lost = 3.1 ev  10% = 0.31 ev Total energy lost the two collision = 0.31 + 0.31 = 0.62 ev K.E. of photon electron when it comes out of metal = hc/ – work function – Energy lost due to collision = 3.1 ev – 2.2 – 0.62 = 0.31 ev rd b) For the 3 collision the energy lost = 0.31 ev Which just equative the KE lost in the 3rd 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.  Energy of photon =



42.8