304
5 Solid State Physics
(c) λ = v F τ = (1.39 × 10
6 )(3.7 × 10
−14 )
= 5.14 × 10
−8 m
5.15 v D =
e
m
ετ
=
(1.6 × 10
−19 )(20)(10
−14 )
9.11 × 10 −31
= 0.0351 m/s
= 3.51 cm/s
Note that the drift velocities are much smaller than the average thermal
velocities which are of the order of 10
5 m/s.
ν T = (3kT /m e )
1/2
5.16 Current,
i =
V
R
(1)
R =
ρl
A
(2)
where the resistivity,
ρ =
m e
ne 2 τ
(3)
n =
N o d
A
× 3 × 10
4 (4)
(4)
where n is the number of electrons per m
3
, N o being Avagardro’s number, A
the atomic weight and d the density, the factor 3 is for the trivalency.
n = 6.02 × 10
23
×
2.7
27
× 3 × 10
4
= 1.806 × 10
27
ρ =
9.11 × 10
−31
1.806 × 10 27 × (1.6 × 10 −19 ) 2 × 4 × 10 −14 = 4.92 × 10
−9
R =
4.92 × 10
−9
× 20
2 × 10 −6
= 0.0492 Ω
i =
3
0.0492
= 61 ˚
A
5.17 (a) τ =
mσ
ne 2
Assuming that one conduction electron will be available for each sodium
atom,
n =
N o ρ
A
=
6.02 × 10
23
× 0.97
23
cm
−3
= 2.539 × 10
28 m
−3
τ =
9.11 × 10
−31
× 2.17 × 10
7
2.539 × 10 28 × (1.6 × 10 −19 ) 2 = 3.04 × 10
−14 s
(b) v D =
e
m
ετ
=
(1.6 × 10
−19 )(200)(3.04 × 10
−14 )
9.11 × 10 −31
= 1.07 m/s