306
5 Solid State Physics
Now, ρ =
m e V F
e 2 nλ
or λ =
m e V F
ρe 2 n
=
(9.11 × 10
−31 )(1.39 × 10
6 )
(1.5 × 10 −8 )(1.6 × 10 −19 ) 2 (5.85 × 10 28 )
= 9.02 × 10
−8 m
∴
λ
d
=
9.02 × 10
−8
2.58 × 10 −10 = 350
5.24 =
2kT
K
1/2
=
2 × 1.38 × 10
−23
× 500
20
1/2
= 2.63 × 10
−11 m = 0.26 ˚
A
5.25 (a) =
3
5
E F =
3 × 5.54
5
= 3.32 eV
(b) v = c
2E
Mc 2
1/2
= 3 × 10
8
2 × 3.32
0.511 × 10 6
1/2
= 1.08 × 10
6 m/s
(c)
3
2
kT =
3
5
E F = 3.32 eV = 3.32 × 1.6 × 10
−19 J
T =
2
3
×
3.32 × 1.6 × 10
−19
1.38 × 10 −23
= 2.56 × 10
4 K
5.26 E F =
h
2
2m ∗
3N
8π V
2/3
where m
∗ is the effective mass.
Density of Cu atoms =
N 0 ρ
A
=
6.02 × 10
23
× 8.94
63.5
= 8.475 × 10
22 atoms/cm
3
= 8.475 × 10
28 atoms/m
3
= 8.475 × 10
28 e/m
3
(∵ each atom gives one electron)
E F =
(6.625 × 10
−34 )
2
2 × 1.01 × 9.11 × 10 −31
3
8π
× 8.475 × 10
28
2/3
= 11.157 × 10
−19 J = 6.97 eV
5.27 The probability is given by Fermi-Dirac distribution p(E) =
1
e (E−E F )/kT + 1
(a) K = 1.38 × 10
−23 J = 8.625 × 10
−5 eV K
−1
E − E F
kT
=
0.05
(8.625 × 10 −5 )(300)
= 1.932
p(E) =
1
e 1.932 + 1
= 0.126