1000 Solved Problems in Modern Physics

282 4 Thermodynamics and Statistical Physics

LetP(E) be the probability function which gives the probability of the state

at the energyEto be occupied. AtT=0 all states below a certain energy are

filled (P=1) and all states above that energy are vacant (P=0). The highest

occupied state under the given conditions is called the Fermi energy.

The product of the densityn(E) of available states and the probabilityP(E)

that those states are occupied, gives the density of occupied statesn 0 (E);

that is

n 0 (E)=n(E)P(E)

The total number of occupied states per unit volume is given by

n=

∫EF

0

n 0 (E)dE

=

8

√

2 πm^3 /^2

h^3

∫EF

0

E^1 /^2 d(E)

=

8

√

2 πm^3 /^2

h^3

.

2

3

E^3 F/^2

or EF=

h^2

8 m

(

3 n

π

) 2 / 3

4.54 P+=

1

e(E−EF)/kT+ 1

=

1

eΔ/kT+ 1

≈

1

2 +Δ/kT

=

1

2

(1−Δ/ 2 kT)

P−=

1

2

(1+Δ/ 2 kT)

∴

P++P−

2

= 1 / 2 =PF

4.55 (a) Fornstates, the number of ways isN=n^2. Therefore, forn=6 states

N= 36

(b) Fornstates the number of ways isN=n^2 −(n−1)or n^2 −n+1. Therefore,

forn= 6 ,N= 31

(c) Fornstates,N=n^2 −n+ 1 −norn^2 − 2 n+1. Therefore forn=6,

N= 25

4.56 If the gas is in equilibrium, the number of particles in a vibrational state is

Nν=N 0 exp

(

−

hν

kT

)

=N 0 exp

(

−

θ

T

)

.

The ratios,N 0 /N 1 = 4. 7619 ,N 1 /N 2 = 4. 8837 ,N 2 /N 3 = 4 .7778, are seen

to be constant at 4.8078. Thus the ratioNν/Nν+ 1 is constant equal to 4.81,

showing the gas to be in equilibrium at a temperature

T= 3 , 350 /(ln 4.81)≈ 2 ,130 K

4.57ΔS=kln(ΔW)
ButΔS=ΔQ/T