5.3 Solutions
309
N = n(E)ΔEa
3
= 1.356 × 10
28
× 0.01 × (10
−2 )
3
= 1.356 × 10
20
5.34 E F =
h
2
8m
3n
π
2/3
=
(6.63 × 10
−34 )
2
(8)(9.11 × 10 −31 )
3 × 5.86 × 10
28
π
2/3
= 8.827 × 10
−19 J = 5.517 eV
5.35 P(E) =
1
e ΔE/kT + 1
= 0.9
Substituting kT = 5.52 × 10
−5
× 800 = 0.04416 eV
Solving for ΔE, we get ΔE = E − E F = −2.2 × 0.04416 = −0.097
Therefore, E = 5.52 − 0.10 = 5.42 eV
5.3.4 Semiconductors
5.36 λ =
1241
1.55
= 800 nm
5.37 The number of electrons and holes per unit volume are given by
n e = 2
2π m
kT
h 2
3/2
e
(E F −E g )/kT
(1)
and n h = 2
2πm
kT
h 2
3/2
e
−E F /kT
(2)
Multiplying (1) and (2), one can write
n e n h = 4
mc
2 k
2π 2 c 2
3
T
3 e
−Eg/kT
(3)
= 2.34 × 10
31 T
3 e
−Eg/kT cm
−6
where we have substituted the values of the constants.
5.38 p = k
(1)
E = p
2
/2m = k
2
2
/2m
(2)
1
m ∗ =
1
2
d
2 E
dk 2
(3)
Using (2) in Eq. (3)
1
m ∗ =
1
2
d
2
dk 2
k
2
2
2m
=
2
2
2m 2 =
1
m
∴ m
∗
= m