1000 Solved Problems in Modern Physics

5.3 Solutions 305

5.18 σ=

ne^2 τ

m

n=

6. 02 × 1023 × 8. 88 × 106

63. 57

= 8. 38 × 1028 m−^3

σ=

8. 38 × 1028 ×(1. 6 × 10 −^19 )^2 × 2. 3 × 10 −^14

9. 11 × 10 −^31

= 5. 422 × 107 Ω−^1 m−^1

K=σCWFT=(5. 422 × 107 )(2. 31 × 10 −^8 )(300)=376 Wm−^1 K−^2

5.19 (a)σ=

1

ρ

=

1

1. 72 × 10 −^8

= 5. 8 × 107 Ω−^1 m−^1

(b)μ=RHσ=(0. 55 × 10 −^10 )(5. 8 × 107 )= 0 .0032 m^2 V−^1 s−^1

(c)τ=

μm

e

=

(0.0032)(9. 11 × 10 −^31 )

1. 6 × 10 −^19

= 1. 82 × 10 −^14 s

(d)n=

σ

eμ

=

5. 8 × 107

(1. 6 × 10 −^19 )(0.0032)

= 1. 13 × 1029 m−^3

5.20 (a)RH=

VHt

iB

=

(4. 5 × 10 −^6 )(2× 10 −^5 )

(1.5)(2)

= 0. 3 × 10 −^10 m^3 C

− 1

(b)n=

1

RHe

=

1

(0. 3 × 10 −^10 )(1. 6 × 10 −^19 )

= 2. 08 × 1029 m−^3

5.21 Integratingn(E)dEfrom zero toEF:

13. 6 × 1027

∫EF

0

E^1 /^2 dE= 8. 5 × 1028

E^3 F/^2 = 9. 375

OrEF= 4 .445 eV

5.22v=

√

2 E

m

=

√

2 × 4. 445 × 1. 6 × 10 −^19

9. 11 × 10 −^31

= 1. 25 × 106 m/s

5.23vF=

(

2 EF

mc^2

) 1 / 2

c=

(

2 × 5. 5

0. 511 × 106

) 1 / 2

(3× 108 )= 1. 39 × 106 m/s

Since each atom contributes one electron, the density of electrons is equal to

that of atoms.

n=

6. 02 × 1026 × 10. 5 × 103

108

= 5. 85 × 1028 e/m^3

Each atom occupies approximately a volumed^3. Therefore

d=

(

1

5. 85 × 1028

) 1 / 3

= 2. 576 × 10 −^10 m