1000 Solved Problems in Modern Physics

288 4 Thermodynamics and Statistical Physics

uν=

8 πν^2

c^3

kT (Rayleigh-Jeans law)

(b) Ifhν/kT1 i.ehc/λkT1 then we can ignore 1 in the denominator

in comparison with the exponential term in Planck’s formula

uλdλ=c 1 e−c^2 /λkTdλ (Wien’s distribution law)

where the constants,c 1 = 8 πhcandc 2 =hc

4.71uλdλ=

8 πhc

λ^5

.

1

ehc/λkT− 1

dλ (Planck’s formula)

The wavelengthλmcorresponding to the maximum of the distribution curve

is obtained from the condition

(

duλ

dλ

)

λ=λm

= 0

Differentiating and writinghc/kTλm=β,gives

e−β+

β

5

− 1 = 0

This is a transcental equation and has the solution

β= 4. 9651 ,so that

λmT=

hc

4. 9651 k

=b=constant.

Thus, the constant

b=

6. 626068 × 10 −^34 × 2. 99792 × 108

4. 9651 × 1. 38065 × 10 −^23

= 2. 8978 × 10 −^3 m-K

a value which is in excellent agreement with the experiment.

4.72 By definition

u=

∫

uνdν=aT^4 (1)

Inserting Planck’s formula in (1)

u=aT^4 =

8 πh

c^3

∫∞

0

ν^3 dν

ehν/kT− 1

=

8 πk^4 T^4

h^3 c^3

∫∞

0

x^3 dx

ex− 1

wherex=hν/kT

a=

8 πk^4

h^3 c^3

∫∞

0

x^3 (e−x+e−^2 x+...e−rx+...)

Now,

∫∞

0 x

(^3) e−rxdx= 6
r^4 , andΣ
∞
r= 1
1
r^4 =
π^2
90
a=
48 πk^4
h^3 c^3

.

π^4

90

=

8

15

π^5 k^4

h^3 c^3

∴σ=

ac

4

=

2

15

π^5 k^4

h^3 c^2