1000 Solved Problems in Modern Physics

284 4 Thermodynamics and Statistical Physics

=

e−nω/kT

e−ω/kT

1 −e−ω/kT

=

e−nω/kT

1

eω/kT− 1

=e−nω/kT

(

eω/kT− 1

)

Substituten= 10 ,

ω

k

=

8. 625 × 10 −^5

(1. 38 × 10 −^23 / 1. 6 × 10 −^19 )

= 1. 0

P(10,300)= 3. 2 × 10 −^3

In the limitT→0, the staten=0 alone is populated so thatn=10 state is

unpopulated.

In the limitT →∞, probability forn=10 again goes to zero, as higher

states which are numerous, are likely to be populated.

4.60 Consider a collection ofNmolecules of a large number of energy states,
E 1 ,E 2 ,E 3 etc such that there areN 1 molecules in stateE 1 ,N 2 inE 2 and
so on. The nature of energy is immaterial. The number of ways in whichN
molecules can be accommodated in various states is given by

W=

N!

N 1 !N 2 !...

(1)

The underlying idea is that the state of the system would be state ifWis a

maximum.

Taking logs on both sides and applying Stirling’s approximation lnW =

NlnN−N−ΣNilnNi+ΣNi

=NlnN−ΣNilnNi (2)

becauseΣNi=N (3)

ΣNiEi=E (4)

If the system is in a state of maximum thermodynamic probability, the varia-

tion ofWwith respect to change inNiis zero, that is

ΣδNi=0(5)

ΣEiδNi=0(6)

Σ(1+lnNi)δNi=0(7)

We now use the Lagrange method of undetermined multipliers. Multiplying

(5) byαand (6) byβand adding to (7), we get

Σ{(1+lnNi)+α+βEi}δNi=0(8)

Therefore

lnNi+ 1 +α+βEi=0(9)

orNi=Ce−βEi (10)