Page 1
42.1
PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY
CHAPTER 42
1. ?
1
= 400 nm to ?
2
= 780 nm
E = h ? =
hc
?
h = 6.63 ? 10
–34
j - s, c = 3 ? 10
8
m/s, ?
1
= 400 nm, ?
2
= 780 nm
E
1
=
34 8
19
9
6.63 10 3 10 6.63 3
10
4 400 10
?
?
?
? ? ? ?
? ?
?
= 5 ? 10
–19
J
E
2
=
19
6.63 3
10
7.8
?
?
? = 2.55 ? 10
–19
J
So, the range is 5 ? 10
–19
J to 2.55 ? 10
–19
J. ?
2. ? = h/p
? P = h/ ? =
34
9
6.63 10
500 10
?
?
?
?
J-S = 1.326 ? 10
–27
= 1.33 ? 10
–27
kg – m/s. ?
3. ?
1
= 500 nm = 500 ? 10
–9
m, ?
2
= 700 nm = 700 ? 10
–9
m
E
1
– E
2
= Energy absorbed by the atom in the process. = hc [1/ ?
1
– 1/ ?
2
]
? 6.63 ? 3[1/5 – 1/7] ? 10
–19
= 1.136 ? 10
–19
J ?
4. P = 10 W ? E in 1 sec = 10 J % used to convert into photon = 60%
?Energy used = 6 J
Energy used to take out 1 photon = hc/ ? =
34 8
17
9
6.63 10 3 10 6.633
10
590 590 10
?
?
?
? ? ?
? ?
?
No. of photons used =
17 17
17
6 6 590
10 176.9 10
6.63 3
6.63 3
10
590
?
?
? ? ? ?
?
?
?
= 1.77 ? 10
19
?
5. a) Here intensity = I = 1.4 ? 10
3
?/m
2
Intensity, I =
power
area
= 1.4 ? 10
3
?/m
2
Let no.of photons/sec emitted = n ? Power = Energy emitted/sec = nhc/ ? = P
No.of photons/m
2
= nhc/ ? = intensity
n =
3 9
21
34 8
int ensity 1.9 10 5 10
3.5 10
hc 6.63 10 3 10
?
?
? ? ? ? ?
? ? ?
? ? ?
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.
In this time the total no.of photons emitted = N = n dt =
dr p
C hc
? ? ?
? ?
? ?
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
(r + r + dr) =
2 2 2 2
N P dr 1 p
2 r2dr hc 4 r ch 4 hc r
? ?
? ? ?
? ? ?
In the case = 1.5 ? 10
11
m, ? = 500 nm, = 500 ? 10
–9
m
3
2
P
1.4 10
4 r
? ?
?
, ? No.of photons/m
3
=
2 2
P
4 r hc
?
?
= 1.4 ? 10
3
?
9
13
34 8
500 10
1.2 10
6.63 10 3 10
?
?
?
? ?
? ? ?
c) No.of photons = (No.of photons/sec/m
2
) ? Area
= (3.5 ? 10
21
) ? 4 ?r
2
= 3.5 ? 10
21
? 4(3.14)(1.5 ? 10
11
)
2
= 9.9 ? 10
44
. ?
Page 2
42.1
PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY
CHAPTER 42
1. ?
1
= 400 nm to ?
2
= 780 nm
E = h ? =
hc
?
h = 6.63 ? 10
–34
j - s, c = 3 ? 10
8
m/s, ?
1
= 400 nm, ?
2
= 780 nm
E
1
=
34 8
19
9
6.63 10 3 10 6.63 3
10
4 400 10
?
?
?
? ? ? ?
? ?
?
= 5 ? 10
–19
J
E
2
=
19
6.63 3
10
7.8
?
?
? = 2.55 ? 10
–19
J
So, the range is 5 ? 10
–19
J to 2.55 ? 10
–19
J. ?
2. ? = h/p
? P = h/ ? =
34
9
6.63 10
500 10
?
?
?
?
J-S = 1.326 ? 10
–27
= 1.33 ? 10
–27
kg – m/s. ?
3. ?
1
= 500 nm = 500 ? 10
–9
m, ?
2
= 700 nm = 700 ? 10
–9
m
E
1
– E
2
= Energy absorbed by the atom in the process. = hc [1/ ?
1
– 1/ ?
2
]
? 6.63 ? 3[1/5 – 1/7] ? 10
–19
= 1.136 ? 10
–19
J ?
4. P = 10 W ? E in 1 sec = 10 J % used to convert into photon = 60%
?Energy used = 6 J
Energy used to take out 1 photon = hc/ ? =
34 8
17
9
6.63 10 3 10 6.633
10
590 590 10
?
?
?
? ? ?
? ?
?
No. of photons used =
17 17
17
6 6 590
10 176.9 10
6.63 3
6.63 3
10
590
?
?
? ? ? ?
?
?
?
= 1.77 ? 10
19
?
5. a) Here intensity = I = 1.4 ? 10
3
?/m
2
Intensity, I =
power
area
= 1.4 ? 10
3
?/m
2
Let no.of photons/sec emitted = n ? Power = Energy emitted/sec = nhc/ ? = P
No.of photons/m
2
= nhc/ ? = intensity
n =
3 9
21
34 8
int ensity 1.9 10 5 10
3.5 10
hc 6.63 10 3 10
?
?
? ? ? ? ?
? ? ?
? ? ?
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.
In this time the total no.of photons emitted = N = n dt =
dr p
C hc
? ? ?
? ?
? ?
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
(r + r + dr) =
2 2 2 2
N P dr 1 p
2 r2dr hc 4 r ch 4 hc r
? ?
? ? ?
? ? ?
In the case = 1.5 ? 10
11
m, ? = 500 nm, = 500 ? 10
–9
m
3
2
P
1.4 10
4 r
? ?
?
, ? No.of photons/m
3
=
2 2
P
4 r hc
?
?
= 1.4 ? 10
3
?
9
13
34 8
500 10
1.2 10
6.63 10 3 10
?
?
?
? ?
? ? ?
c) No.of photons = (No.of photons/sec/m
2
) ? Area
= (3.5 ? 10
21
) ? 4 ?r
2
= 3.5 ? 10
21
? 4(3.14)(1.5 ? 10
11
)
2
= 9.9 ? 10
44
. ?
Photo Electric Effect and Wave Particle Quality
42.2
6. ? = 663 ? 10
–9
m, ? = 60°, n = 1 ? 10
19
, ? = h/p
? P = p/ ? = 10
–27
Force exerted on the wall = n(mv cos ? –(–mv cos ?)) = 2n mv cos ?.
= 2 ? 1 ? 10
19
? 10
–27
? ½ = 1 ? 10
–8
N. ?
7. Power = 10 W P ? Momentum
? =
h
p
or, P =
h
?
or,
P h
t t
?
?
E =
hc
?
or,
E hc
t t
?
?
= Power (W)
W = Pc/t or, P/t = W/c = force.
or Force = 7/10 (absorbed) + 2 ? 3/10 (reflected)
=
7 W 3 W
2
10 C 10 C
? ? ? ? ?
8 8
7 10 3 10
2
10 10 3 10 3 10
? ? ? ?
? ?
= 13/3 ? 10
–8
= 4.33 ? 10
–8
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
? ? ?
?
? ?
? ?
? Power =
3 8
20 10 10 3 10 10
2 3
?
? ? ? ? ?
?
= 10 w = 100 MW.
9. Power = 100 W
Radius = 20 cm
60% is converted to light = 60 w
Now, Force =
7
8
power 60
2 10 N
velocity 3 10
?
? ? ?
?
.
Pressure =
7
5
2
force 2 10 1
10
area 8 3.14 4 3.14 (0.2)
?
?
?
? ? ?
? ? ?
= 0.039 ? 10
–5
= 3.9 ? 10
–7
= 4 ? 10
–7
N/m
2
.
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,
Force =
2
r l
C
?
I = 0.5 W/m
2
, r = 1 cm, C = 3 ? 10
8
m/s
Force =
2
8 8
(1) 0.5 3.14 0.5
3 10 3 10
? ? ? ?
?
? ?
= 0.523 ? 10
–8
= 5.2 ? 10
–9
N.
60°
Page 3
42.1
PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY
CHAPTER 42
1. ?
1
= 400 nm to ?
2
= 780 nm
E = h ? =
hc
?
h = 6.63 ? 10
–34
j - s, c = 3 ? 10
8
m/s, ?
1
= 400 nm, ?
2
= 780 nm
E
1
=
34 8
19
9
6.63 10 3 10 6.63 3
10
4 400 10
?
?
?
? ? ? ?
? ?
?
= 5 ? 10
–19
J
E
2
=
19
6.63 3
10
7.8
?
?
? = 2.55 ? 10
–19
J
So, the range is 5 ? 10
–19
J to 2.55 ? 10
–19
J. ?
2. ? = h/p
? P = h/ ? =
34
9
6.63 10
500 10
?
?
?
?
J-S = 1.326 ? 10
–27
= 1.33 ? 10
–27
kg – m/s. ?
3. ?
1
= 500 nm = 500 ? 10
–9
m, ?
2
= 700 nm = 700 ? 10
–9
m
E
1
– E
2
= Energy absorbed by the atom in the process. = hc [1/ ?
1
– 1/ ?
2
]
? 6.63 ? 3[1/5 – 1/7] ? 10
–19
= 1.136 ? 10
–19
J ?
4. P = 10 W ? E in 1 sec = 10 J % used to convert into photon = 60%
?Energy used = 6 J
Energy used to take out 1 photon = hc/ ? =
34 8
17
9
6.63 10 3 10 6.633
10
590 590 10
?
?
?
? ? ?
? ?
?
No. of photons used =
17 17
17
6 6 590
10 176.9 10
6.63 3
6.63 3
10
590
?
?
? ? ? ?
?
?
?
= 1.77 ? 10
19
?
5. a) Here intensity = I = 1.4 ? 10
3
?/m
2
Intensity, I =
power
area
= 1.4 ? 10
3
?/m
2
Let no.of photons/sec emitted = n ? Power = Energy emitted/sec = nhc/ ? = P
No.of photons/m
2
= nhc/ ? = intensity
n =
3 9
21
34 8
int ensity 1.9 10 5 10
3.5 10
hc 6.63 10 3 10
?
?
? ? ? ? ?
? ? ?
? ? ?
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.
In this time the total no.of photons emitted = N = n dt =
dr p
C hc
? ? ?
? ?
? ?
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
(r + r + dr) =
2 2 2 2
N P dr 1 p
2 r2dr hc 4 r ch 4 hc r
? ?
? ? ?
? ? ?
In the case = 1.5 ? 10
11
m, ? = 500 nm, = 500 ? 10
–9
m
3
2
P
1.4 10
4 r
? ?
?
, ? No.of photons/m
3
=
2 2
P
4 r hc
?
?
= 1.4 ? 10
3
?
9
13
34 8
500 10
1.2 10
6.63 10 3 10
?
?
?
? ?
? ? ?
c) No.of photons = (No.of photons/sec/m
2
) ? Area
= (3.5 ? 10
21
) ? 4 ?r
2
= 3.5 ? 10
21
? 4(3.14)(1.5 ? 10
11
)
2
= 9.9 ? 10
44
. ?
Photo Electric Effect and Wave Particle Quality
42.2
6. ? = 663 ? 10
–9
m, ? = 60°, n = 1 ? 10
19
, ? = h/p
? P = p/ ? = 10
–27
Force exerted on the wall = n(mv cos ? –(–mv cos ?)) = 2n mv cos ?.
= 2 ? 1 ? 10
19
? 10
–27
? ½ = 1 ? 10
–8
N. ?
7. Power = 10 W P ? Momentum
? =
h
p
or, P =
h
?
or,
P h
t t
?
?
E =
hc
?
or,
E hc
t t
?
?
= Power (W)
W = Pc/t or, P/t = W/c = force.
or Force = 7/10 (absorbed) + 2 ? 3/10 (reflected)
=
7 W 3 W
2
10 C 10 C
? ? ? ? ?
8 8
7 10 3 10
2
10 10 3 10 3 10
? ? ? ?
? ?
= 13/3 ? 10
–8
= 4.33 ? 10
–8
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
? ? ?
?
? ?
? ?
? Power =
3 8
20 10 10 3 10 10
2 3
?
? ? ? ? ?
?
= 10 w = 100 MW.
9. Power = 100 W
Radius = 20 cm
60% is converted to light = 60 w
Now, Force =
7
8
power 60
2 10 N
velocity 3 10
?
? ? ?
?
.
Pressure =
7
5
2
force 2 10 1
10
area 8 3.14 4 3.14 (0.2)
?
?
?
? ? ?
? ? ?
= 0.039 ? 10
–5
= 3.9 ? 10
–7
= 4 ? 10
–7
N/m
2
.
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,
Force =
2
r l
C
?
I = 0.5 W/m
2
, r = 1 cm, C = 3 ? 10
8
m/s
Force =
2
8 8
(1) 0.5 3.14 0.5
3 10 3 10
? ? ? ?
?
? ?
= 0.523 ? 10
–8
= 5.2 ? 10
–9
N.
60°
Photo Electric Effect and Wave Particle Quality
42.3
11. 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 =
2
r I
C
?
12. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision.
We get, hC/ ? + m
0
c
2
= mc
2
and applying conservation of momentum h/ ? = mv
Mass of e = m =
0
2 2
m
1 v / c ?
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 =
2 2
kq kq
R 1
?
Now,
2
kq
1
=
hc
?
or ? =
2
hc
kq
For max ‘ ?’, ‘q’ should be min,
For minimum ‘e’ = 1.6 ? 10
–19
C
Max ? =
2
hc
kq
= 0.863 ? 10
3
= 863 m.
For next smaller wavelength =
34 8
9 2 38
6.63 3 10 10 863
4 9 10 (1.6 2) 10
?
?
? ? ?
?
? ? ? ?
= 215.74 m
14. ? = 350 nn = 350 ? 10
–9
m
? = 1.9 eV
Max KE of electrons =
34 8
9 19
hC 6.63 10 3 10
1.9
350 10 1.6 10
?
? ?
? ? ?
? ? ? ?
? ? ? ?
= 1.65 ev = 1.6 ev. ?
15. W
0
= 2.5 ? 10
–19
J
a) We know W
0
= h ?
0
?
0
=
19
0
34
W 2.5 10
h 6.63 10
?
?
?
?
?
= 3.77 ? 10
14
Hz = 3.8 ? 10
14
Hz
b) eV
0
= h ? – W
0
or, V
0
=
34 14 19
0
19
h W 6.63 10 6 10 2.5 10
e 1.6 10
? ?
?
? ? ? ? ? ? ?
?
?
= 0.91 V
16. ? = 4 eV = 4 ? 1.6 ? 10
–19
J
a) Threshold wavelength = ?
? = hc/ ?
? ? =
hC
?
=
34 8 27
7
19 9
6.63 10 3 10 6.63 3 10
3.1 10 m
6.4 4 1.6 10 10
? ?
?
? ?
? ? ? ?
? ? ? ?
? ?
= 310 nm.
b) Stopping potential is 2.5 V
? E = ? + eV
? hc/ ? = 4 ? 1.6 ? 10
–19
+ 1.6 ? 10
–19
? 2.5
?? ????
34 8
19
6.63 10 3 10
1.6 10
?
?
? ? ?
? ? ?
= 4 + 2.5
??
26
19
6.63 3 10
1.6 10 6.5
?
?
? ?
? ?
= 1.9125 ? 10
–7
= 190 nm. ?
Page 4
42.1
PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY
CHAPTER 42
1. ?
1
= 400 nm to ?
2
= 780 nm
E = h ? =
hc
?
h = 6.63 ? 10
–34
j - s, c = 3 ? 10
8
m/s, ?
1
= 400 nm, ?
2
= 780 nm
E
1
=
34 8
19
9
6.63 10 3 10 6.63 3
10
4 400 10
?
?
?
? ? ? ?
? ?
?
= 5 ? 10
–19
J
E
2
=
19
6.63 3
10
7.8
?
?
? = 2.55 ? 10
–19
J
So, the range is 5 ? 10
–19
J to 2.55 ? 10
–19
J. ?
2. ? = h/p
? P = h/ ? =
34
9
6.63 10
500 10
?
?
?
?
J-S = 1.326 ? 10
–27
= 1.33 ? 10
–27
kg – m/s. ?
3. ?
1
= 500 nm = 500 ? 10
–9
m, ?
2
= 700 nm = 700 ? 10
–9
m
E
1
– E
2
= Energy absorbed by the atom in the process. = hc [1/ ?
1
– 1/ ?
2
]
? 6.63 ? 3[1/5 – 1/7] ? 10
–19
= 1.136 ? 10
–19
J ?
4. P = 10 W ? E in 1 sec = 10 J % used to convert into photon = 60%
?Energy used = 6 J
Energy used to take out 1 photon = hc/ ? =
34 8
17
9
6.63 10 3 10 6.633
10
590 590 10
?
?
?
? ? ?
? ?
?
No. of photons used =
17 17
17
6 6 590
10 176.9 10
6.63 3
6.63 3
10
590
?
?
? ? ? ?
?
?
?
= 1.77 ? 10
19
?
5. a) Here intensity = I = 1.4 ? 10
3
?/m
2
Intensity, I =
power
area
= 1.4 ? 10
3
?/m
2
Let no.of photons/sec emitted = n ? Power = Energy emitted/sec = nhc/ ? = P
No.of photons/m
2
= nhc/ ? = intensity
n =
3 9
21
34 8
int ensity 1.9 10 5 10
3.5 10
hc 6.63 10 3 10
?
?
? ? ? ? ?
? ? ?
? ? ?
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.
In this time the total no.of photons emitted = N = n dt =
dr p
C hc
? ? ?
? ?
? ?
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
(r + r + dr) =
2 2 2 2
N P dr 1 p
2 r2dr hc 4 r ch 4 hc r
? ?
? ? ?
? ? ?
In the case = 1.5 ? 10
11
m, ? = 500 nm, = 500 ? 10
–9
m
3
2
P
1.4 10
4 r
? ?
?
, ? No.of photons/m
3
=
2 2
P
4 r hc
?
?
= 1.4 ? 10
3
?
9
13
34 8
500 10
1.2 10
6.63 10 3 10
?
?
?
? ?
? ? ?
c) No.of photons = (No.of photons/sec/m
2
) ? Area
= (3.5 ? 10
21
) ? 4 ?r
2
= 3.5 ? 10
21
? 4(3.14)(1.5 ? 10
11
)
2
= 9.9 ? 10
44
. ?
Photo Electric Effect and Wave Particle Quality
42.2
6. ? = 663 ? 10
–9
m, ? = 60°, n = 1 ? 10
19
, ? = h/p
? P = p/ ? = 10
–27
Force exerted on the wall = n(mv cos ? –(–mv cos ?)) = 2n mv cos ?.
= 2 ? 1 ? 10
19
? 10
–27
? ½ = 1 ? 10
–8
N. ?
7. Power = 10 W P ? Momentum
? =
h
p
or, P =
h
?
or,
P h
t t
?
?
E =
hc
?
or,
E hc
t t
?
?
= Power (W)
W = Pc/t or, P/t = W/c = force.
or Force = 7/10 (absorbed) + 2 ? 3/10 (reflected)
=
7 W 3 W
2
10 C 10 C
? ? ? ? ?
8 8
7 10 3 10
2
10 10 3 10 3 10
? ? ? ?
? ?
= 13/3 ? 10
–8
= 4.33 ? 10
–8
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
? ? ?
?
? ?
? ?
? Power =
3 8
20 10 10 3 10 10
2 3
?
? ? ? ? ?
?
= 10 w = 100 MW.
9. Power = 100 W
Radius = 20 cm
60% is converted to light = 60 w
Now, Force =
7
8
power 60
2 10 N
velocity 3 10
?
? ? ?
?
.
Pressure =
7
5
2
force 2 10 1
10
area 8 3.14 4 3.14 (0.2)
?
?
?
? ? ?
? ? ?
= 0.039 ? 10
–5
= 3.9 ? 10
–7
= 4 ? 10
–7
N/m
2
.
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,
Force =
2
r l
C
?
I = 0.5 W/m
2
, r = 1 cm, C = 3 ? 10
8
m/s
Force =
2
8 8
(1) 0.5 3.14 0.5
3 10 3 10
? ? ? ?
?
? ?
= 0.523 ? 10
–8
= 5.2 ? 10
–9
N.
60°
Photo Electric Effect and Wave Particle Quality
42.3
11. 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 =
2
r I
C
?
12. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision.
We get, hC/ ? + m
0
c
2
= mc
2
and applying conservation of momentum h/ ? = mv
Mass of e = m =
0
2 2
m
1 v / c ?
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 =
2 2
kq kq
R 1
?
Now,
2
kq
1
=
hc
?
or ? =
2
hc
kq
For max ‘ ?’, ‘q’ should be min,
For minimum ‘e’ = 1.6 ? 10
–19
C
Max ? =
2
hc
kq
= 0.863 ? 10
3
= 863 m.
For next smaller wavelength =
34 8
9 2 38
6.63 3 10 10 863
4 9 10 (1.6 2) 10
?
?
? ? ?
?
? ? ? ?
= 215.74 m
14. ? = 350 nn = 350 ? 10
–9
m
? = 1.9 eV
Max KE of electrons =
34 8
9 19
hC 6.63 10 3 10
1.9
350 10 1.6 10
?
? ?
? ? ?
? ? ? ?
? ? ? ?
= 1.65 ev = 1.6 ev. ?
15. W
0
= 2.5 ? 10
–19
J
a) We know W
0
= h ?
0
?
0
=
19
0
34
W 2.5 10
h 6.63 10
?
?
?
?
?
= 3.77 ? 10
14
Hz = 3.8 ? 10
14
Hz
b) eV
0
= h ? – W
0
or, V
0
=
34 14 19
0
19
h W 6.63 10 6 10 2.5 10
e 1.6 10
? ?
?
? ? ? ? ? ? ?
?
?
= 0.91 V
16. ? = 4 eV = 4 ? 1.6 ? 10
–19
J
a) Threshold wavelength = ?
? = hc/ ?
? ? =
hC
?
=
34 8 27
7
19 9
6.63 10 3 10 6.63 3 10
3.1 10 m
6.4 4 1.6 10 10
? ?
?
? ?
? ? ? ?
? ? ? ?
? ?
= 310 nm.
b) Stopping potential is 2.5 V
? E = ? + eV
? hc/ ? = 4 ? 1.6 ? 10
–19
+ 1.6 ? 10
–19
? 2.5
?? ????
34 8
19
6.63 10 3 10
1.6 10
?
?
? ? ?
? ? ?
= 4 + 2.5
??
26
19
6.63 3 10
1.6 10 6.5
?
?
? ?
? ?
= 1.9125 ? 10
–7
= 190 nm. ?
Photo Electric Effect and Wave Particle Quality
42.4
17. Energy of photoelectron
? ½ mv
2
=
0
hc
hv ?
?
=
15 8
7
4.14 10 3 10
2.5ev
4 10
?
?
? ? ?
?
?
= 0.605 ev.
We know KE =
2
P
2m
? P
2
= 2m ? KE.
P
2
= 2 ? 9.1 ? 10
–31
? 0.605 ? 1.6 ? 10
–19
P = 4.197 ? 10
–25
kg – m/s
18. ? = 400 nm = 400 ? 10
–9
m
V
0
= 1.1 V
0
0
hc hc
ev ? ?
? ?
?
34 8 34 8
19
9
0
6.63 10 3 10 6.63 10 3 10
1.6 10 1.1
400 10
? ?
?
?
? ? ? ? ? ?
? ? ? ?
? ?
? 4.97 =
26
0
19.89 10
1.76
?
?
?
?
?
26
0
19.89 10
?
?
?
= 4.97 – 17.6 = 3.21
? ?
0
=
26
19.89 10
3.21
?
?
= 6.196 ? 10
–7
m = 620 nm. ?
19. a) When ? = 350, V
s
= 1.45
and when ? = 400, V
s
= 1
?
hc
350
= W + 1.45 …(1)
and
hc
400
= W + 1 …(2)
Subtracting (2) from (1) and solving to get the value of h we get
h = 4.2 ? 10
–15
ev-sec
b) Now work function = w =
hc
?
= ev - s
=
1240
1.45
350
? = 2.15 ev.
c) w =
there cathod
hc hc
w
? ? ?
?
=
1240
2.15
= 576.8 nm. ?
20. The electric field becomes 0 1.2 ? 10
45
times per second.
? Frequency =
15
1.2 10
2
?
= 0.6 ? 10
15
h ? = ?
0
+ kE
? h ? – ?
0
= KE
? KE =
34 15
19
6.63 10 0.6 10
2
1.6 10
?
?
? ? ?
?
?
= 0.482 ev = 0.48 ev.
21. E = E
0
sin[(1.57 ? 10
7
m
–1
) (x – ct)]
W = 1.57 ? 10
7
? C
Stopping
potential
1/ ? ? ?
Page 5
42.1
PHOTO ELECTRIC EFFECT AND WAVE PARTICLE QUALITY
CHAPTER 42
1. ?
1
= 400 nm to ?
2
= 780 nm
E = h ? =
hc
?
h = 6.63 ? 10
–34
j - s, c = 3 ? 10
8
m/s, ?
1
= 400 nm, ?
2
= 780 nm
E
1
=
34 8
19
9
6.63 10 3 10 6.63 3
10
4 400 10
?
?
?
? ? ? ?
? ?
?
= 5 ? 10
–19
J
E
2
=
19
6.63 3
10
7.8
?
?
? = 2.55 ? 10
–19
J
So, the range is 5 ? 10
–19
J to 2.55 ? 10
–19
J. ?
2. ? = h/p
? P = h/ ? =
34
9
6.63 10
500 10
?
?
?
?
J-S = 1.326 ? 10
–27
= 1.33 ? 10
–27
kg – m/s. ?
3. ?
1
= 500 nm = 500 ? 10
–9
m, ?
2
= 700 nm = 700 ? 10
–9
m
E
1
– E
2
= Energy absorbed by the atom in the process. = hc [1/ ?
1
– 1/ ?
2
]
? 6.63 ? 3[1/5 – 1/7] ? 10
–19
= 1.136 ? 10
–19
J ?
4. P = 10 W ? E in 1 sec = 10 J % used to convert into photon = 60%
?Energy used = 6 J
Energy used to take out 1 photon = hc/ ? =
34 8
17
9
6.63 10 3 10 6.633
10
590 590 10
?
?
?
? ? ?
? ?
?
No. of photons used =
17 17
17
6 6 590
10 176.9 10
6.63 3
6.63 3
10
590
?
?
? ? ? ?
?
?
?
= 1.77 ? 10
19
?
5. a) Here intensity = I = 1.4 ? 10
3
?/m
2
Intensity, I =
power
area
= 1.4 ? 10
3
?/m
2
Let no.of photons/sec emitted = n ? Power = Energy emitted/sec = nhc/ ? = P
No.of photons/m
2
= nhc/ ? = intensity
n =
3 9
21
34 8
int ensity 1.9 10 5 10
3.5 10
hc 6.63 10 3 10
?
?
? ? ? ? ?
? ? ?
? ? ?
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.
In this time the total no.of photons emitted = N = n dt =
dr p
C hc
? ? ?
? ?
? ?
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
(r + r + dr) =
2 2 2 2
N P dr 1 p
2 r2dr hc 4 r ch 4 hc r
? ?
? ? ?
? ? ?
In the case = 1.5 ? 10
11
m, ? = 500 nm, = 500 ? 10
–9
m
3
2
P
1.4 10
4 r
? ?
?
, ? No.of photons/m
3
=
2 2
P
4 r hc
?
?
= 1.4 ? 10
3
?
9
13
34 8
500 10
1.2 10
6.63 10 3 10
?
?
?
? ?
? ? ?
c) No.of photons = (No.of photons/sec/m
2
) ? Area
= (3.5 ? 10
21
) ? 4 ?r
2
= 3.5 ? 10
21
? 4(3.14)(1.5 ? 10
11
)
2
= 9.9 ? 10
44
. ?
Photo Electric Effect and Wave Particle Quality
42.2
6. ? = 663 ? 10
–9
m, ? = 60°, n = 1 ? 10
19
, ? = h/p
? P = p/ ? = 10
–27
Force exerted on the wall = n(mv cos ? –(–mv cos ?)) = 2n mv cos ?.
= 2 ? 1 ? 10
19
? 10
–27
? ½ = 1 ? 10
–8
N. ?
7. Power = 10 W P ? Momentum
? =
h
p
or, P =
h
?
or,
P h
t t
?
?
E =
hc
?
or,
E hc
t t
?
?
= Power (W)
W = Pc/t or, P/t = W/c = force.
or Force = 7/10 (absorbed) + 2 ? 3/10 (reflected)
=
7 W 3 W
2
10 C 10 C
? ? ? ? ?
8 8
7 10 3 10
2
10 10 3 10 3 10
? ? ? ?
? ?
= 13/3 ? 10
–8
= 4.33 ? 10
–8
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
? ? ?
?
? ?
? ?
? Power =
3 8
20 10 10 3 10 10
2 3
?
? ? ? ? ?
?
= 10 w = 100 MW.
9. Power = 100 W
Radius = 20 cm
60% is converted to light = 60 w
Now, Force =
7
8
power 60
2 10 N
velocity 3 10
?
? ? ?
?
.
Pressure =
7
5
2
force 2 10 1
10
area 8 3.14 4 3.14 (0.2)
?
?
?
? ? ?
? ? ?
= 0.039 ? 10
–5
= 3.9 ? 10
–7
= 4 ? 10
–7
N/m
2
.
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,
Force =
2
r l
C
?
I = 0.5 W/m
2
, r = 1 cm, C = 3 ? 10
8
m/s
Force =
2
8 8
(1) 0.5 3.14 0.5
3 10 3 10
? ? ? ?
?
? ?
= 0.523 ? 10
–8
= 5.2 ? 10
–9
N.
60°
Photo Electric Effect and Wave Particle Quality
42.3
11. 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 =
2
r I
C
?
12. If the i undergoes an elastic collision with a photon. Then applying energy conservation to this collision.
We get, hC/ ? + m
0
c
2
= mc
2
and applying conservation of momentum h/ ? = mv
Mass of e = m =
0
2 2
m
1 v / c ?
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 =
2 2
kq kq
R 1
?
Now,
2
kq
1
=
hc
?
or ? =
2
hc
kq
For max ‘ ?’, ‘q’ should be min,
For minimum ‘e’ = 1.6 ? 10
–19
C
Max ? =
2
hc
kq
= 0.863 ? 10
3
= 863 m.
For next smaller wavelength =
34 8
9 2 38
6.63 3 10 10 863
4 9 10 (1.6 2) 10
?
?
? ? ?
?
? ? ? ?
= 215.74 m
14. ? = 350 nn = 350 ? 10
–9
m
? = 1.9 eV
Max KE of electrons =
34 8
9 19
hC 6.63 10 3 10
1.9
350 10 1.6 10
?
? ?
? ? ?
? ? ? ?
? ? ? ?
= 1.65 ev = 1.6 ev. ?
15. W
0
= 2.5 ? 10
–19
J
a) We know W
0
= h ?
0
?
0
=
19
0
34
W 2.5 10
h 6.63 10
?
?
?
?
?
= 3.77 ? 10
14
Hz = 3.8 ? 10
14
Hz
b) eV
0
= h ? – W
0
or, V
0
=
34 14 19
0
19
h W 6.63 10 6 10 2.5 10
e 1.6 10
? ?
?
? ? ? ? ? ? ?
?
?
= 0.91 V
16. ? = 4 eV = 4 ? 1.6 ? 10
–19
J
a) Threshold wavelength = ?
? = hc/ ?
? ? =
hC
?
=
34 8 27
7
19 9
6.63 10 3 10 6.63 3 10
3.1 10 m
6.4 4 1.6 10 10
? ?
?
? ?
? ? ? ?
? ? ? ?
? ?
= 310 nm.
b) Stopping potential is 2.5 V
? E = ? + eV
? hc/ ? = 4 ? 1.6 ? 10
–19
+ 1.6 ? 10
–19
? 2.5
?? ????
34 8
19
6.63 10 3 10
1.6 10
?
?
? ? ?
? ? ?
= 4 + 2.5
??
26
19
6.63 3 10
1.6 10 6.5
?
?
? ?
? ?
= 1.9125 ? 10
–7
= 190 nm. ?
Photo Electric Effect and Wave Particle Quality
42.4
17. Energy of photoelectron
? ½ mv
2
=
0
hc
hv ?
?
=
15 8
7
4.14 10 3 10
2.5ev
4 10
?
?
? ? ?
?
?
= 0.605 ev.
We know KE =
2
P
2m
? P
2
= 2m ? KE.
P
2
= 2 ? 9.1 ? 10
–31
? 0.605 ? 1.6 ? 10
–19
P = 4.197 ? 10
–25
kg – m/s
18. ? = 400 nm = 400 ? 10
–9
m
V
0
= 1.1 V
0
0
hc hc
ev ? ?
? ?
?
34 8 34 8
19
9
0
6.63 10 3 10 6.63 10 3 10
1.6 10 1.1
400 10
? ?
?
?
? ? ? ? ? ?
? ? ? ?
? ?
? 4.97 =
26
0
19.89 10
1.76
?
?
?
?
?
26
0
19.89 10
?
?
?
= 4.97 – 17.6 = 3.21
? ?
0
=
26
19.89 10
3.21
?
?
= 6.196 ? 10
–7
m = 620 nm. ?
19. a) When ? = 350, V
s
= 1.45
and when ? = 400, V
s
= 1
?
hc
350
= W + 1.45 …(1)
and
hc
400
= W + 1 …(2)
Subtracting (2) from (1) and solving to get the value of h we get
h = 4.2 ? 10
–15
ev-sec
b) Now work function = w =
hc
?
= ev - s
=
1240
1.45
350
? = 2.15 ev.
c) w =
there cathod
hc hc
w
? ? ?
?
=
1240
2.15
= 576.8 nm. ?
20. The electric field becomes 0 1.2 ? 10
45
times per second.
? Frequency =
15
1.2 10
2
?
= 0.6 ? 10
15
h ? = ?
0
+ kE
? h ? – ?
0
= KE
? KE =
34 15
19
6.63 10 0.6 10
2
1.6 10
?
?
? ? ?
?
?
= 0.482 ev = 0.48 ev.
21. E = E
0
sin[(1.57 ? 10
7
m
–1
) (x – ct)]
W = 1.57 ? 10
7
? C
Stopping
potential
1/ ? ? ?
Photo Electric Effect and Wave Particle Quality
42.5
? f =
7 8
1.57 10 3 10
2
? ? ?
?
Hz W
0
= 1.9 ev
Now eV
0
= h ? – W
0
= 4.14 ? 10
–15
?
15
1.57 3 10
2
? ?
?
– 1.9 ev
= 3.105 – 1.9 = 1.205 ev
So, V
0
=
19
19
1.205 1.6 10
1.6 10
?
?
? ?
?
= 1.205 V. ?
22. E = 100 sin[(3 ? 10
15
s
–1
)t] sin [6 ? 10
15
s
–1
)t]
= 100 ½ [cos[(9 ? 10
15
s
–1
)t] – cos [3 ? 10
15
s
–1
)t]
The w are 9 ? 10
15
and 3 ? 10
15
for largest K.E.
f
max
=
max
w
2 ?
=
15
9 10
2
?
?
E – ?
0
= K.E.
? hf – ?
0
= K.E.
??
34 15
19
6.63 10 9 10
2 KE
2 1.6 10
?
?
? ? ?
? ?
? ? ?
?
? KE = 3.938 ev = 3.93 ev.
23. W
0
= hv – ev
0
=
3
19
15
5 10
1.6 10 2
8 10
?
?
?
? ? ?
?
(Given V
0
= 2V, No. of photons = 8 ? 10
15
, Power = 5 mW)
= 6.25 ? 10
–19
– 3.2 ? 10
–19
= 3.05 ? 10
–19
J
=
19
19
3.05 10
1.6 10
?
?
?
?
= 1.906 eV.
24. We have to take two cases :
Case I … v
0
= 1.656
? = 5 ? 10
14
Hz
Case II… v
0
= 0
? = 1 ? 10
14
Hz
We know ;
a)
0 0
ev h w ? ? ?
1.656e = h ? 5 ? 10
14
– w
0
…(1)
0 = 5h ? 10
14
– 5w
0
…(2)
1.656e = 4w
0
? w
0
=
1.656
4
ev = 0.414 ev
b) Putting value of w
0
in equation (2)
? 5w
0
= 5h ? 10
14
? 5 ? 0.414 = 5 ? h ? 10
14
? h = 4.414 ? 10
–15
ev-s
25. w
0
= 0.6 ev
For w
0
to be min ‘ ?’ becomes maximum.
w
0
=
hc
?
or ? =
0
hc
w
=
34 8
19
6.63 10 3 10
0.6 1.6 10
?
?
? ? ?
? ?
?
= 20.71 ? 10
–7
m = 2071 nm
1 2 3 4 5
1
1.656
2
v(in 10
14
Hz)
V (in volts)
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