306 5 Solid State Physics
Now,ρ=meVF
e^2 nλorλ=meVF
ρe^2 n=
(9. 11 × 10 −^31 )(1. 39 × 106 )
(1. 5 × 10 −^8 )(1. 6 × 10 −^19 )^2 (5. 85 × 1028 )
= 9. 02 × 10 −^8 m∴
λ
d=
9. 02 × 10 −^8
2. 58 × 10 −^10
= 350
5.24=
(
2 kT
K) 1 / 2
=
(
2 × 1. 38 × 10 −^23 × 500
20
) 1 / 2
= 2. 63 × 10 −^11 m= 0. 26 A ̊5.25 (a)<E>=3
5
EF=
3 × 5. 54
5
= 3 .32 eV(b)v=c(
2 E
Mc^2) 1 / 2
= 3 × 108
(
2 × 3. 32
0. 511 × 106
) 1 / 2
= 1. 08 × 106 m/s(c)3
2
kT=3
5
EF= 3 .32 eV= 3. 32 × 1. 6 × 10 −^19 JT=
2
3
×
3. 32 × 1. 6 × 10 −^19
1. 38 × 10 −^23
= 2. 56 × 104 K
5.26 EF=
h^2
2 m∗(
3 N
8 πV) 2 / 3
wherem∗is the effective mass.Density of Cu atoms=N 0 ρ
A=
6. 02 × 1023 × 8. 94
63. 5
= 8. 475 × 1022 atoms/cm^3
= 8. 475 × 1028 atoms/m^3
= 8. 475 × 1028 e/m^3
(∵each atom gives one electron)EF=(6. 625 × 10 −^34 )^2
2 × 1. 01 × 9. 11 × 10 −^31
(
3
8 π× 8. 475 × 1028
) 2 / 3
= 11. 157 × 10 −^19 J= 6 .97 eV5.27 The probability is given by Fermi-Dirac distributionp(E)=
1
e(E−EF)/kT+ 1
(a) K= 1. 38 × 10 −^23 J= 8. 625 × 10 −^5 eV K−^1
E−EF
kT=
0. 05
(8. 625 × 10 −^5 )(300)
= 1. 932
p(E)=1
e^1.^932 + 1