1000 Solved Problems in Modern Physics

5.3 Solutions 309

N=n(E)ΔEa^3

= 1. 356 × 1028 × 0. 01 ×(10−^2 )^3

= 1. 356 × 1020

5.34 EF=

h^2

8 m

(

3 n

π

) 2 / 3

=

(6. 63 × 10 −^34 )^2

(8)(9. 11 × 10 −^31 )

(

3 × 5. 86 × 1028

π

) 2 / 3

= 8. 827 × 10 −^19 J= 5 .517 eV

5.35 P(E)=

1

eΔE/kT+ 1

= 0. 9

SubstitutingkT= 5. 52 × 10 −^5 × 800 = 0 .04416 eV

Solving forΔE, we getΔE=E−EF=− 2. 2 × 0. 04416 =− 0. 097

Therefore,E= 5. 52 − 0. 10 = 5 .42 eV

5.3.4 Semiconductors...................................

5.36λ=

1241

1. 55

=800 nm

5.37 The number of electrons and holes per unit volume are given by

ne= 2

(

2 πm

kT

h^2

) 3 / 2

e(EF−Eg)/kT (1)

andnh= 2

(

2 πm

kT

h^2

) 3 / 2

e−EF/kT (2)

Multiplying (1) and (2), one can write

nenh= 4

(

mc^2 k

2 π^2 c^2

) 3

T^3 e−Eg/kT (3)

= 2. 34 × 1031 T^3 e−Eg/kTcm−^6

where we have substituted the values of the constants.

5.38 p=k (1)

E=p^2 / 2 m=k^2 ^2 / 2 m (2)

1

m∗

=

1

^2

d^2 E

dk^2

(3)

Using (2) in Eq. (3)

1

m∗

=

1

^2

d^2

dk^2

(

k^2 ^2

2 m

)

=

2 ^2

2 m^2

=

1

m

∴m∗=m