PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY CHAPTER 42 1.
1 = 400 nm to 2 = 780 nm E = h = E1 =
2.
hc
6.63 10 34 3 108 9
4.
8
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 1034
–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/ =
3.
–34
h = 6.63 10
Energy used to take out 1 photon = hc/ =
5.
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 Pdr 1 p 2 2 2r2dr hc 4r ch 4 hc 2r 2 11 –9 In the case = 1.5 10 m, = 500 nm, = 500 10 m
(r + r + dr) =
P 4r 2
3
1.4 103 , No.of photons/m = 3
P
4r 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 ) 4r 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.
7.
–9
19
= 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)
E=
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.
=
8.
m = 20 g The weight of the mirror is balanced. Thus force exerted by the photons is equal to weight P=
h
E=
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. 23 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 =
9.
Pressure =
force 2 107 1 105 2 area 4 3.14 (0.2) 8 3.14 = 0.039 10
–5
= 3.9 10
–7
–7
= 4 10
2
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 =
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
m
hC 6.63 10 34 3 108 1.9 350 109 1.6 10 19 = 1.65 ev = 1.6 ev. –19 15. W 0 = 2.5 10 J a) We know W 0 = h0 Max KE of electrons =
W0 2.5 1019 14 14 = 3.77 10 Hz = 3.8 10 Hz h 6.63 1034 b) eV0 = h – W 0 0 =
h W0 6.63 10 34 6 1014 2.5 10 19 = 0.91 V e 1.6 1019 –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 3.1 107 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 1034 3 108 1.6 10 19
6.63 3 1026 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 1019 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 1034 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 =
=
15
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
14
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 =
1
hc hc 6.63 10 34 3 108 or = = w0 0.6 1.6 10 19 –7
= 20.71 10
2
3
4
v(in 1014 Hz)
m = 2071 nm 42.5
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 =
6
No.of electrons = 1 per 10 photon.
5 400
No.of photoelectrons emitted =
1.6 1242 1019 106 5 400 –6 Photo electric current = 1.6 1019 = 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 2 107
1 10 7 9.945 10
11
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
X X
X X
A
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
S
A
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 – =
V=
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
12
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
6
= 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 80 d 2d 40 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
