1000 Solved Problems in Modern Physics

5.3 Solutions 311

5.44 I=I 0 [exp(eV/kT)−1] (1)

1

re

=

dI

dV

=

eI 0

kT

exp(eV/kT)(2)

But exp (eV/kT)1. Therefore

1

re

=

eI

kT

orre=

kT

eI

=

1. 38 × 10 −^23 × 300

1. 6 × 10 −^19 I

=

25. 875 × 10 −^3

I

If I is in milliamp.

re≈

26

I

(forward bias) (3)

For the reversed bias we note from (2)

1

re

=

dI

dV

=

e

kT

I 0 exp (eV/kT)= 0

For V≤− 4 kT/e,re→∞.(reverse bias) (4)

5.45 For a semiconductor in equilibrium the product ofn(=Nd) andp(=Na)is
equal ton^2 i, the square of the intrinsic concentration.

n×p=n^2 i

p=

n^2 i

n

=

(1. 6 × 1016 )^2

8 × 1021

= 3. 2 × 1010 m−^3

5.3.5 Superconductor ....................................

5.46λ=

1241

E(eV)

nm=

1 , 241

2. 73 × 10 −^3

= 4. 546 × 105 nm

5.47 Eg=

1241

λ(nm)

=

1 , 241

1. 08 × 106

= 1. 15 × 10 −^3 eV

5.48 f=

2eV

h

=

2(1. 602 × 10 −^19 )(1. 5 × 10 −^6 )

6. 626 × 10 −^34

= 7. 253 × 108 Hz= 0 .7253 GHz

5.49 Eg= 3. 53 kTc=(3.53)

(1. 38 × 10 −^23 )

1. 6 × 10 −^19

(3.4)= 1. 035 × 10 −^3 eV

5.50 Tc(B)=Tc

(

1 −

B

Bc

) 1 / 2

2. 0 = 7. 19

(

1 −

0. 074

Bc

) 1 / 2

Solving forBc, we getBc= 0. 079 T.