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