304 5 Solid State Physics
(c)λ=vFτ=(1. 39 × 106 )(3. 7 × 10 −^14 )
= 5. 14 × 10 −^8 m5.15 〈vD〉=
e
mετ=(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.[
〈νT〉=(3kT/me)^1 /^2]
5.16 Current,
i=V
R
(1)
R=
ρl
A(2)
where the resistivity,
ρ=me
ne^2 τ(3)
n=Nod
A× 3 × 104 (4) (4)
wherenis the number of electrons per m^3 , Nobeing Avagardro’s number,A
the atomic weight anddthe density, the factor 3 is for the trivalency.n= 6. 02 × 1023 ×2. 7
27
× 3 × 104 = 1. 806 × 1027
ρ=9. 11 × 10 −^31
1. 806 × 1027 ×(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=Noρ
A=
6. 02 × 1023 × 0. 97
23
cm−^3 = 2. 539 × 1028 m−^3τ=9. 11 × 10 −^31 × 2. 17 × 107
2. 539 × 1028 ×(1. 6 × 10 −^19 )^2
= 3. 04 × 10 −^14 s(b)〈vD〉=e
mετ=(1. 6 × 10 −^19 )(200)(3. 04 × 10 −^14 )
9. 11 × 10 −^31
= 1 .07 m/s