HC Verma Solutions: Chapter 41 - Electric Current through Gases | Physics Class 11 - NEET PDF Download

 Page 1


41.1
ELECTRIC CURRENT THROUGH GASES
CHAPTER 41
1. Let the two particles have charge ‘q’
Mass of electron m
a
= 9.1 ? 10
–31
kg
Mass of proton m
p
= 1.67 ? 10
–27
kg
Electric field be E
Force experienced by Electron = qE
accln. = qE/m
e
For time dt 
S
e
= 
2
e
1 qE
dt
2 m
? ? …(1)
For the positive ion, 
accln. = 
p
qE
4 m ?
S
p
= 
2
p
1 qE
dt
2 4 m
? ?
?
…(2)
p
e
p e
4m
S
S m
? = 7340.6
2. E = 5 Kv/m = 5 ? 10
3
v/m ; t = 1 ?s = 1 ? 10
–6
s
F = qE = 1.6 ? 10
–9
? 5 ? 10
3
a = 
16
31
qE 1.6 5 10
m 9.1 10
?
?
? ?
?
?
a) S = distance travelled
= 
2
1
at
2
= 439.56 m = 440 m
b) d = 1 mm = 1 ? 10
–3
m
1 ? 10
–3
= 
5 2
1 1.6 5
10 t
2 9.1
?
? ?
? t
2
= 
18
9.1
10
0.8 5
?
?
?
? t = 1.508 ? 10
–9
sec ? 1.5 ns.
3. Let the mean free path be ‘L’ and pressure be ‘P’
L ? 1/p for L = half of the tube length, P = 0.02 mm of Hg
As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space.
Hence the required pressure = 0.02/2 = 0.01 m of Hg.
4. V = f(Pd)
v
s
= P
s
d
s
v
L
= P
l
d
l
?
s s s
l l l
V P d
V P d
? ? ?
100 10 1 mm
100 20 x
? ?
? x = 1 mm / 2 = 0.5 mm
5. i = ne or n = i/e
‘e’ is same in all cases.
We know, 
A C 
Page 2


41.1
ELECTRIC CURRENT THROUGH GASES
CHAPTER 41
1. Let the two particles have charge ‘q’
Mass of electron m
a
= 9.1 ? 10
–31
kg
Mass of proton m
p
= 1.67 ? 10
–27
kg
Electric field be E
Force experienced by Electron = qE
accln. = qE/m
e
For time dt 
S
e
= 
2
e
1 qE
dt
2 m
? ? …(1)
For the positive ion, 
accln. = 
p
qE
4 m ?
S
p
= 
2
p
1 qE
dt
2 4 m
? ?
?
…(2)
p
e
p e
4m
S
S m
? = 7340.6
2. E = 5 Kv/m = 5 ? 10
3
v/m ; t = 1 ?s = 1 ? 10
–6
s
F = qE = 1.6 ? 10
–9
? 5 ? 10
3
a = 
16
31
qE 1.6 5 10
m 9.1 10
?
?
? ?
?
?
a) S = distance travelled
= 
2
1
at
2
= 439.56 m = 440 m
b) d = 1 mm = 1 ? 10
–3
m
1 ? 10
–3
= 
5 2
1 1.6 5
10 t
2 9.1
?
? ?
? t
2
= 
18
9.1
10
0.8 5
?
?
?
? t = 1.508 ? 10
–9
sec ? 1.5 ns.
3. Let the mean free path be ‘L’ and pressure be ‘P’
L ? 1/p for L = half of the tube length, P = 0.02 mm of Hg
As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space.
Hence the required pressure = 0.02/2 = 0.01 m of Hg.
4. V = f(Pd)
v
s
= P
s
d
s
v
L
= P
l
d
l
?
s s s
l l l
V P d
V P d
? ? ?
100 10 1 mm
100 20 x
? ?
? x = 1 mm / 2 = 0.5 mm
5. i = ne or n = i/e
‘e’ is same in all cases.
We know, 
A C 
Electric current through gases
41.2
i = AST
2 / RT
e
? ?
? = 4.52 eV, K = 1.38 ? 10
–23
J/k
n(1000) = As ? (1000)
2
?
19 23
4.52 1.6 10 /1.38 10 1000
e
? ?
? ? ? ? ?
? 1.7396 ? 10
–17
a) T = 300 K
19 23
2 4.52 1.6 10 /1.38 10 300
17
n(T) AS (300) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 7.05 ? 10
–55
b) T = 2000 K
?
19 23
2 4.52 1.6 10 /1.38 10 2000
17
n(T) AS (2000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 9.59 ? 10
11
c) T = 3000 K
?
19 23
2 4.52 1.6 10 /1.38 10 3000
17
n(T) AS (3000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 1.340 ? 10
16
?
6. i = AST
2 / KT
e
? ?
i
1
= i i
2
= 100 mA
A
1
= 60 ? 10
4
A
2
= 3 ? 10
4
S
1
= S S
2
= S
T
1
= 2000 T
2
= 2000
?
1
= 4.5 eV ?
2
= 2.6 eV
K = 1.38 ? 10
–23
J/k
i = (60 ? 10
4
) (S) ? (2000)
2
19
23
4.5 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
100 = (3 ? 10
4
) (S) ? (2000)
2
19
23
2.6 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
Dividing the equation 
4.5 1.6 10 2.6 1.6 10
( )
1.38 2 1.38 20
i
e
100
? ? ? ? ? ? ? ?
? ?
? ? ? ?
?
?
11.014
i i
20 e 20 0.000016
100 100
?
? ? ? ? ?
? i = 20 ? 0.0016 = 0.0329 mA = 33 ?A ?
7. Pure tungsten Thoriated tungsten 
? = 4.5 eV ? = 2.6 eV
A = 60 ? 10
4
A/m
2
– k
2
A = 3 ? 10
4
A/m
2
– k
2
i = AST
2 / KT
e
? ?
i
Thoriated Tungsten
= 5000 i
Tungsten
So, 5000 ? S ? 60 ? 10
4
? T
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? S ? 3 ? 10
4
? T
2
?
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
8
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
4
Taking ‘ln’
? 9.21 T = 220.29
? T = 22029 / 9.21 = 2391.856 K
Page 3


41.1
ELECTRIC CURRENT THROUGH GASES
CHAPTER 41
1. Let the two particles have charge ‘q’
Mass of electron m
a
= 9.1 ? 10
–31
kg
Mass of proton m
p
= 1.67 ? 10
–27
kg
Electric field be E
Force experienced by Electron = qE
accln. = qE/m
e
For time dt 
S
e
= 
2
e
1 qE
dt
2 m
? ? …(1)
For the positive ion, 
accln. = 
p
qE
4 m ?
S
p
= 
2
p
1 qE
dt
2 4 m
? ?
?
…(2)
p
e
p e
4m
S
S m
? = 7340.6
2. E = 5 Kv/m = 5 ? 10
3
v/m ; t = 1 ?s = 1 ? 10
–6
s
F = qE = 1.6 ? 10
–9
? 5 ? 10
3
a = 
16
31
qE 1.6 5 10
m 9.1 10
?
?
? ?
?
?
a) S = distance travelled
= 
2
1
at
2
= 439.56 m = 440 m
b) d = 1 mm = 1 ? 10
–3
m
1 ? 10
–3
= 
5 2
1 1.6 5
10 t
2 9.1
?
? ?
? t
2
= 
18
9.1
10
0.8 5
?
?
?
? t = 1.508 ? 10
–9
sec ? 1.5 ns.
3. Let the mean free path be ‘L’ and pressure be ‘P’
L ? 1/p for L = half of the tube length, P = 0.02 mm of Hg
As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space.
Hence the required pressure = 0.02/2 = 0.01 m of Hg.
4. V = f(Pd)
v
s
= P
s
d
s
v
L
= P
l
d
l
?
s s s
l l l
V P d
V P d
? ? ?
100 10 1 mm
100 20 x
? ?
? x = 1 mm / 2 = 0.5 mm
5. i = ne or n = i/e
‘e’ is same in all cases.
We know, 
A C 
Electric current through gases
41.2
i = AST
2 / RT
e
? ?
? = 4.52 eV, K = 1.38 ? 10
–23
J/k
n(1000) = As ? (1000)
2
?
19 23
4.52 1.6 10 /1.38 10 1000
e
? ?
? ? ? ? ?
? 1.7396 ? 10
–17
a) T = 300 K
19 23
2 4.52 1.6 10 /1.38 10 300
17
n(T) AS (300) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 7.05 ? 10
–55
b) T = 2000 K
?
19 23
2 4.52 1.6 10 /1.38 10 2000
17
n(T) AS (2000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 9.59 ? 10
11
c) T = 3000 K
?
19 23
2 4.52 1.6 10 /1.38 10 3000
17
n(T) AS (3000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 1.340 ? 10
16
?
6. i = AST
2 / KT
e
? ?
i
1
= i i
2
= 100 mA
A
1
= 60 ? 10
4
A
2
= 3 ? 10
4
S
1
= S S
2
= S
T
1
= 2000 T
2
= 2000
?
1
= 4.5 eV ?
2
= 2.6 eV
K = 1.38 ? 10
–23
J/k
i = (60 ? 10
4
) (S) ? (2000)
2
19
23
4.5 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
100 = (3 ? 10
4
) (S) ? (2000)
2
19
23
2.6 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
Dividing the equation 
4.5 1.6 10 2.6 1.6 10
( )
1.38 2 1.38 20
i
e
100
? ? ? ? ? ? ? ?
? ?
? ? ? ?
?
?
11.014
i i
20 e 20 0.000016
100 100
?
? ? ? ? ?
? i = 20 ? 0.0016 = 0.0329 mA = 33 ?A ?
7. Pure tungsten Thoriated tungsten 
? = 4.5 eV ? = 2.6 eV
A = 60 ? 10
4
A/m
2
– k
2
A = 3 ? 10
4
A/m
2
– k
2
i = AST
2 / KT
e
? ?
i
Thoriated Tungsten
= 5000 i
Tungsten
So, 5000 ? S ? 60 ? 10
4
? T
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? S ? 3 ? 10
4
? T
2
?
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
8
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
4
Taking ‘ln’
? 9.21 T = 220.29
? T = 22029 / 9.21 = 2391.856 K
Electric current through gases
41.3
8. i = AST
2 / KT
e
? ?
i ? = AST
12 / KT
e
? ? ?
2 / KT
12 /KT
i T e
i T e
? ?
? ? ?
?
?
?
2 2
/ KT KT KT / KT
i T T
e e
i T T
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
= 
19
23
4.5 1.6 10
2
1.38 10
i 2000 1 1
e
i 2010 2010 2000
?
?
? ?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
= 0.8690
?
i 1
1.1495
i 0.8699
? ?
?
= 1.14
9. A = 60 ? 10
4 
A/m
2
– k
2
? = 4.5 eV ? = 6 ? 10
–8
?/m
2
– k
4
S = 2 ? 10
–5
m
2
K = 1.38 ? 10
–23
J/K
H = 24 ??
The Cathode acts as a black body, i.e. emissivity = 1
? E = ? A T
4
(A is area)
? T
4
= 
13 12
8 5
E 24
2 10 K 20 10 K
A 6 10 2 10
? ?
? ? ? ? ?
? ? ? ?
? T = 2.1147 ? 10
3
= 2114.7 K
Now, i = AST
2 / KT
e
? ?
= 6 ? 10
5
? 2 ? 10
–5
? (2114.7)
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 1.03456 ? 10
–3
A = 1 mA ?
10. i
p
= 
3 / 2
p
CV …(1)
? di
p
= C 3/2 
(3 / 2) 1
p p
V dv
?
?
p 1/ 2
p
p
di
3
CV
dv 2
? …(2)
Dividing (2) and (1)
1/ 2
p p
3 / 2
p p
di 3 / 2CV
i
i dv CVp
?
?
p
p p
di
1 3
i dv 2V
?
?
p
p p
dv
2V
di 3i
?
? R = 
3
3
p
2V 2 60
4 10 4k
3i 3 10 10
?
?
? ? ? ? ?
? ?
11. For plate current 20 mA, we find the voltage 50 V or 60 V.
Hence it acts as the saturation current. Therefore for the same temperature, the plate current is 20 mA 
for all other values of voltage.
Hence the required answer is 20 mA.
12. P = 1 W, p = ?
V
p
= 36 V, V
p
= 49 V, P = I
p
V
p
Page 4


41.1
ELECTRIC CURRENT THROUGH GASES
CHAPTER 41
1. Let the two particles have charge ‘q’
Mass of electron m
a
= 9.1 ? 10
–31
kg
Mass of proton m
p
= 1.67 ? 10
–27
kg
Electric field be E
Force experienced by Electron = qE
accln. = qE/m
e
For time dt 
S
e
= 
2
e
1 qE
dt
2 m
? ? …(1)
For the positive ion, 
accln. = 
p
qE
4 m ?
S
p
= 
2
p
1 qE
dt
2 4 m
? ?
?
…(2)
p
e
p e
4m
S
S m
? = 7340.6
2. E = 5 Kv/m = 5 ? 10
3
v/m ; t = 1 ?s = 1 ? 10
–6
s
F = qE = 1.6 ? 10
–9
? 5 ? 10
3
a = 
16
31
qE 1.6 5 10
m 9.1 10
?
?
? ?
?
?
a) S = distance travelled
= 
2
1
at
2
= 439.56 m = 440 m
b) d = 1 mm = 1 ? 10
–3
m
1 ? 10
–3
= 
5 2
1 1.6 5
10 t
2 9.1
?
? ?
? t
2
= 
18
9.1
10
0.8 5
?
?
?
? t = 1.508 ? 10
–9
sec ? 1.5 ns.
3. Let the mean free path be ‘L’ and pressure be ‘P’
L ? 1/p for L = half of the tube length, P = 0.02 mm of Hg
As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space.
Hence the required pressure = 0.02/2 = 0.01 m of Hg.
4. V = f(Pd)
v
s
= P
s
d
s
v
L
= P
l
d
l
?
s s s
l l l
V P d
V P d
? ? ?
100 10 1 mm
100 20 x
? ?
? x = 1 mm / 2 = 0.5 mm
5. i = ne or n = i/e
‘e’ is same in all cases.
We know, 
A C 
Electric current through gases
41.2
i = AST
2 / RT
e
? ?
? = 4.52 eV, K = 1.38 ? 10
–23
J/k
n(1000) = As ? (1000)
2
?
19 23
4.52 1.6 10 /1.38 10 1000
e
? ?
? ? ? ? ?
? 1.7396 ? 10
–17
a) T = 300 K
19 23
2 4.52 1.6 10 /1.38 10 300
17
n(T) AS (300) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 7.05 ? 10
–55
b) T = 2000 K
?
19 23
2 4.52 1.6 10 /1.38 10 2000
17
n(T) AS (2000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 9.59 ? 10
11
c) T = 3000 K
?
19 23
2 4.52 1.6 10 /1.38 10 3000
17
n(T) AS (3000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 1.340 ? 10
16
?
6. i = AST
2 / KT
e
? ?
i
1
= i i
2
= 100 mA
A
1
= 60 ? 10
4
A
2
= 3 ? 10
4
S
1
= S S
2
= S
T
1
= 2000 T
2
= 2000
?
1
= 4.5 eV ?
2
= 2.6 eV
K = 1.38 ? 10
–23
J/k
i = (60 ? 10
4
) (S) ? (2000)
2
19
23
4.5 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
100 = (3 ? 10
4
) (S) ? (2000)
2
19
23
2.6 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
Dividing the equation 
4.5 1.6 10 2.6 1.6 10
( )
1.38 2 1.38 20
i
e
100
? ? ? ? ? ? ? ?
? ?
? ? ? ?
?
?
11.014
i i
20 e 20 0.000016
100 100
?
? ? ? ? ?
? i = 20 ? 0.0016 = 0.0329 mA = 33 ?A ?
7. Pure tungsten Thoriated tungsten 
? = 4.5 eV ? = 2.6 eV
A = 60 ? 10
4
A/m
2
– k
2
A = 3 ? 10
4
A/m
2
– k
2
i = AST
2 / KT
e
? ?
i
Thoriated Tungsten
= 5000 i
Tungsten
So, 5000 ? S ? 60 ? 10
4
? T
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? S ? 3 ? 10
4
? T
2
?
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
8
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
4
Taking ‘ln’
? 9.21 T = 220.29
? T = 22029 / 9.21 = 2391.856 K
Electric current through gases
41.3
8. i = AST
2 / KT
e
? ?
i ? = AST
12 / KT
e
? ? ?
2 / KT
12 /KT
i T e
i T e
? ?
? ? ?
?
?
?
2 2
/ KT KT KT / KT
i T T
e e
i T T
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
= 
19
23
4.5 1.6 10
2
1.38 10
i 2000 1 1
e
i 2010 2010 2000
?
?
? ?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
= 0.8690
?
i 1
1.1495
i 0.8699
? ?
?
= 1.14
9. A = 60 ? 10
4 
A/m
2
– k
2
? = 4.5 eV ? = 6 ? 10
–8
?/m
2
– k
4
S = 2 ? 10
–5
m
2
K = 1.38 ? 10
–23
J/K
H = 24 ??
The Cathode acts as a black body, i.e. emissivity = 1
? E = ? A T
4
(A is area)
? T
4
= 
13 12
8 5
E 24
2 10 K 20 10 K
A 6 10 2 10
? ?
? ? ? ? ?
? ? ? ?
? T = 2.1147 ? 10
3
= 2114.7 K
Now, i = AST
2 / KT
e
? ?
= 6 ? 10
5
? 2 ? 10
–5
? (2114.7)
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 1.03456 ? 10
–3
A = 1 mA ?
10. i
p
= 
3 / 2
p
CV …(1)
? di
p
= C 3/2 
(3 / 2) 1
p p
V dv
?
?
p 1/ 2
p
p
di
3
CV
dv 2
? …(2)
Dividing (2) and (1)
1/ 2
p p
3 / 2
p p
di 3 / 2CV
i
i dv CVp
?
?
p
p p
di
1 3
i dv 2V
?
?
p
p p
dv
2V
di 3i
?
? R = 
3
3
p
2V 2 60
4 10 4k
3i 3 10 10
?
?
? ? ? ? ?
? ?
11. For plate current 20 mA, we find the voltage 50 V or 60 V.
Hence it acts as the saturation current. Therefore for the same temperature, the plate current is 20 mA 
for all other values of voltage.
Hence the required answer is 20 mA.
12. P = 1 W, p = ?
V
p
= 36 V, V
p
= 49 V, P = I
p
V
p
Electric current through gases
41.4
? I
p
= 
p
P 1
V 36
?
I
p
? (V
p
)
3/2
I ?
p
? (V ?
p
)
3/2
?
3 / 2
p p
p p
I (V )
I V
?
? ?
?
3 / 2
p
1/ 36 36
I 49
? ?
?
? ?
?
? ?
?
p
p
1 36 6
I 0.4411
36 I 49 7
? ? ? ? ?
?
P ? = V ?
p
I ?
p
= 49 ? 0.4411 = 2.1613 W = 2.2 W
13. Amplification factor for triode value
= ? = 
p
g
V
Charge in Plate Voltage
Charge in Grid Voltage V
?
?
?
= 
250 225 25
12.5
2.5 0.5 2
?
? ?
?
[ ? ?Vp = 250 – 225, ?Vg = 2.5 – 0.5] ?
14. r
p
= 2 K ? = 2 ? 10
3
?
g
m
= 2 milli mho = 2 ? 10
–3
mho
? = r
p
? g
m
= 2 ? 10
3
? 2 ? 10
–3
= 4 Amplification factor is 4.
15. Dynamic Plate Resistance r
p
= 10 K ? = 10
4
?
?I
p
= ?
?V
p
= 220 – 220 = 20 V
?I
p
= ???V
p
/ r
p
) / V
g
= constant.
= 20/10
4
= 0.002 A = 2 mA
16. r
p
= 
p
p
V
I
? ? ?
? ?
? ?
?
? ?
at constant V
g
Consider the two points on V
g
= –6 line
r
p
= 
3
3
(240 160)V 80
10 8K
10 (13 3) 10 A
?
?
? ? ? ? ?
? ?
g
m
= 
p
p
g
I
v
V
? ? ?
? ?
? ?
?
? ?
= constant
Considering the points on 200 V line,
g
m
= 
3 3
(13 3) 10 10 10
A 2.5
[( 4) ( 8)] 4
? ?
? ? ?
? ?
? ? ?
milli mho
? = r
p
? gm
= 8 ? 10
3
? ? 2.5 ? 10
–3
?
–1
= 8 ? 1.5 = 20
17. a) r
p
= 8 K ? = 8000 ?
?V
p
= 48 V ?I
p
= ?
?I
p
= ??V
p
/ r
p
) / V
g
= constant.
So, ?I
p
= 48 / 8000 = 0.006 A = 6 mA
b) Now, V
p
is constant.
?I
p
= 6 mA = 0.006 A
Page 5


41.1
ELECTRIC CURRENT THROUGH GASES
CHAPTER 41
1. Let the two particles have charge ‘q’
Mass of electron m
a
= 9.1 ? 10
–31
kg
Mass of proton m
p
= 1.67 ? 10
–27
kg
Electric field be E
Force experienced by Electron = qE
accln. = qE/m
e
For time dt 
S
e
= 
2
e
1 qE
dt
2 m
? ? …(1)
For the positive ion, 
accln. = 
p
qE
4 m ?
S
p
= 
2
p
1 qE
dt
2 4 m
? ?
?
…(2)
p
e
p e
4m
S
S m
? = 7340.6
2. E = 5 Kv/m = 5 ? 10
3
v/m ; t = 1 ?s = 1 ? 10
–6
s
F = qE = 1.6 ? 10
–9
? 5 ? 10
3
a = 
16
31
qE 1.6 5 10
m 9.1 10
?
?
? ?
?
?
a) S = distance travelled
= 
2
1
at
2
= 439.56 m = 440 m
b) d = 1 mm = 1 ? 10
–3
m
1 ? 10
–3
= 
5 2
1 1.6 5
10 t
2 9.1
?
? ?
? t
2
= 
18
9.1
10
0.8 5
?
?
?
? t = 1.508 ? 10
–9
sec ? 1.5 ns.
3. Let the mean free path be ‘L’ and pressure be ‘P’
L ? 1/p for L = half of the tube length, P = 0.02 mm of Hg
As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space.
Hence the required pressure = 0.02/2 = 0.01 m of Hg.
4. V = f(Pd)
v
s
= P
s
d
s
v
L
= P
l
d
l
?
s s s
l l l
V P d
V P d
? ? ?
100 10 1 mm
100 20 x
? ?
? x = 1 mm / 2 = 0.5 mm
5. i = ne or n = i/e
‘e’ is same in all cases.
We know, 
A C 
Electric current through gases
41.2
i = AST
2 / RT
e
? ?
? = 4.52 eV, K = 1.38 ? 10
–23
J/k
n(1000) = As ? (1000)
2
?
19 23
4.52 1.6 10 /1.38 10 1000
e
? ?
? ? ? ? ?
? 1.7396 ? 10
–17
a) T = 300 K
19 23
2 4.52 1.6 10 /1.38 10 300
17
n(T) AS (300) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 7.05 ? 10
–55
b) T = 2000 K
?
19 23
2 4.52 1.6 10 /1.38 10 2000
17
n(T) AS (2000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 9.59 ? 10
11
c) T = 3000 K
?
19 23
2 4.52 1.6 10 /1.38 10 3000
17
n(T) AS (3000) e
n(1000K) AS 1.7396 10
? ?
? ? ? ? ?
?
? ?
?
? ?
= 1.340 ? 10
16
?
6. i = AST
2 / KT
e
? ?
i
1
= i i
2
= 100 mA
A
1
= 60 ? 10
4
A
2
= 3 ? 10
4
S
1
= S S
2
= S
T
1
= 2000 T
2
= 2000
?
1
= 4.5 eV ?
2
= 2.6 eV
K = 1.38 ? 10
–23
J/k
i = (60 ? 10
4
) (S) ? (2000)
2
19
23
4.5 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
100 = (3 ? 10
4
) (S) ? (2000)
2
19
23
2.6 1.6 10
1.38 10 2000
e
?
?
? ? ?
? ?
Dividing the equation 
4.5 1.6 10 2.6 1.6 10
( )
1.38 2 1.38 20
i
e
100
? ? ? ? ? ? ? ?
? ?
? ? ? ?
?
?
11.014
i i
20 e 20 0.000016
100 100
?
? ? ? ? ?
? i = 20 ? 0.0016 = 0.0329 mA = 33 ?A ?
7. Pure tungsten Thoriated tungsten 
? = 4.5 eV ? = 2.6 eV
A = 60 ? 10
4
A/m
2
– k
2
A = 3 ? 10
4
A/m
2
– k
2
i = AST
2 / KT
e
? ?
i
Thoriated Tungsten
= 5000 i
Tungsten
So, 5000 ? S ? 60 ? 10
4
? T
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? S ? 3 ? 10
4
? T
2
?
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
8
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 
19
23
2.65 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
? 3 ? 10
4
Taking ‘ln’
? 9.21 T = 220.29
? T = 22029 / 9.21 = 2391.856 K
Electric current through gases
41.3
8. i = AST
2 / KT
e
? ?
i ? = AST
12 / KT
e
? ? ?
2 / KT
12 /KT
i T e
i T e
? ?
? ? ?
?
?
?
2 2
/ KT KT KT / KT
i T T
e e
i T T
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ?
= 
19
23
4.5 1.6 10
2
1.38 10
i 2000 1 1
e
i 2010 2010 2000
?
?
? ?
?
? ? ? ?
? ?
? ? ? ?
?
? ? ? ?
= 0.8690
?
i 1
1.1495
i 0.8699
? ?
?
= 1.14
9. A = 60 ? 10
4 
A/m
2
– k
2
? = 4.5 eV ? = 6 ? 10
–8
?/m
2
– k
4
S = 2 ? 10
–5
m
2
K = 1.38 ? 10
–23
J/K
H = 24 ??
The Cathode acts as a black body, i.e. emissivity = 1
? E = ? A T
4
(A is area)
? T
4
= 
13 12
8 5
E 24
2 10 K 20 10 K
A 6 10 2 10
? ?
? ? ? ? ?
? ? ? ?
? T = 2.1147 ? 10
3
= 2114.7 K
Now, i = AST
2 / KT
e
? ?
= 6 ? 10
5
? 2 ? 10
–5
? (2114.7)
2
?
19
23
4.5 1.6 10
1.38 T 10
e
?
?
? ? ?
? ?
= 1.03456 ? 10
–3
A = 1 mA ?
10. i
p
= 
3 / 2
p
CV …(1)
? di
p
= C 3/2 
(3 / 2) 1
p p
V dv
?
?
p 1/ 2
p
p
di
3
CV
dv 2
? …(2)
Dividing (2) and (1)
1/ 2
p p
3 / 2
p p
di 3 / 2CV
i
i dv CVp
?
?
p
p p
di
1 3
i dv 2V
?
?
p
p p
dv
2V
di 3i
?
? R = 
3
3
p
2V 2 60
4 10 4k
3i 3 10 10
?
?
? ? ? ? ?
? ?
11. For plate current 20 mA, we find the voltage 50 V or 60 V.
Hence it acts as the saturation current. Therefore for the same temperature, the plate current is 20 mA 
for all other values of voltage.
Hence the required answer is 20 mA.
12. P = 1 W, p = ?
V
p
= 36 V, V
p
= 49 V, P = I
p
V
p
Electric current through gases
41.4
? I
p
= 
p
P 1
V 36
?
I
p
? (V
p
)
3/2
I ?
p
? (V ?
p
)
3/2
?
3 / 2
p p
p p
I (V )
I V
?
? ?
?
3 / 2
p
1/ 36 36
I 49
? ?
?
? ?
?
? ?
?
p
p
1 36 6
I 0.4411
36 I 49 7
? ? ? ? ?
?
P ? = V ?
p
I ?
p
= 49 ? 0.4411 = 2.1613 W = 2.2 W
13. Amplification factor for triode value
= ? = 
p
g
V
Charge in Plate Voltage
Charge in Grid Voltage V
?
?
?
= 
250 225 25
12.5
2.5 0.5 2
?
? ?
?
[ ? ?Vp = 250 – 225, ?Vg = 2.5 – 0.5] ?
14. r
p
= 2 K ? = 2 ? 10
3
?
g
m
= 2 milli mho = 2 ? 10
–3
mho
? = r
p
? g
m
= 2 ? 10
3
? 2 ? 10
–3
= 4 Amplification factor is 4.
15. Dynamic Plate Resistance r
p
= 10 K ? = 10
4
?
?I
p
= ?
?V
p
= 220 – 220 = 20 V
?I
p
= ???V
p
/ r
p
) / V
g
= constant.
= 20/10
4
= 0.002 A = 2 mA
16. r
p
= 
p
p
V
I
? ? ?
? ?
? ?
?
? ?
at constant V
g
Consider the two points on V
g
= –6 line
r
p
= 
3
3
(240 160)V 80
10 8K
10 (13 3) 10 A
?
?
? ? ? ? ?
? ?
g
m
= 
p
p
g
I
v
V
? ? ?
? ?
? ?
?
? ?
= constant
Considering the points on 200 V line,
g
m
= 
3 3
(13 3) 10 10 10
A 2.5
[( 4) ( 8)] 4
? ?
? ? ?
? ?
? ? ?
milli mho
? = r
p
? gm
= 8 ? 10
3
? ? 2.5 ? 10
–3
?
–1
= 8 ? 1.5 = 20
17. a) r
p
= 8 K ? = 8000 ?
?V
p
= 48 V ?I
p
= ?
?I
p
= ??V
p
/ r
p
) / V
g
= constant.
So, ?I
p
= 48 / 8000 = 0.006 A = 6 mA
b) Now, V
p
is constant.
?I
p
= 6 mA = 0.006 A
Electric current through gases
41.5
g
m
= 0.0025 mho
?V
g
= ? ?I
p
/ g
m
) / V
p
= constant.
= 
0.006
0.0025
= 2.4 V
18. r
p
= 10 K ? = 10 ? 10
3
?
? = 20 V
p
= 250 V
V
g
= –7.5 V I
p
= 10 mA
a) g
m
= 
p
p
g
I
V
V
? ? ?
? ?
? ?
?
? ?
= constant
? ?V
g
= 
3 3
p
m p
I
15 10 10 10
g /r
? ?
?
? ? ?
?
?
= 
3
3
5 10 5
2.5
2 20 /10 10
?
?
? ?
?
r ?
g
= +2.5 – 7.5 = –5 V
b) r
p
= 
p
p
V
I
? ? ?
? ?
? ?
?
? ?
V
g
= constnant
? 10
4
= 
p
3 3
V
(15 10 10 10 )
? ?
?
? ? ?
? ?V
p
= 10
4
? 5 ? 10
–3
= 50 V
V ?
p
– V
p
= 50 ? V ?
p
= –50 + V
p
= 200 V
19. V
p
= 250 V, V
g
= –20 V
a) i
p
= 41(V
p
+ 7V
g
)
1.41
? 41(250 – 140)
1.41
= 41 ? (110)
1.41
= 30984 ?A = 30 mA
b) i
p
= 41(V
p
+ 7V
g
)
1.41
Differentiating,
di
p
= 41 ? 1.41 ? (V
p
+ 7V
g
)
0.41
? (dV
p
+ 7dV
g
)
Now r
p
= 
p
g
p
dV
V
di
= constant.
or
6
p
0.41
p
dV
1 10
di 41 1.41 110
?
?
? ?
= 10
6
? 2.51 ? 10
–3
? 2.5 ? 10
3
? = 2.5 K ?
c) From above,
dI
p
= 41 ? 1.41 ? 6.87 ? 7 d V
g
g
m
= 
p
g
dI
dV
= 41 ? 1.41 ? 6.87 ? 7 ? mho
= 2780 ? mho = 2.78 milli mho.
d) Amplification factor
? = r
p
? g
m
= 2.5 ? 10
3
? 2.78 ? 10
–3
= 6.95 = 7 ?
20. i
p
= K(V
g
+ V
p
/ ?)
3/2
…(1)
Diff. the equation :
di
p
= K 3/2 (V
g
+ V
p
/ ?)
1/2
d V
g
?
1/ 2
p 0
g
g
di V 3
V K
dV 2
? ?
? ?
? ?
?
? ?