Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture2
Examples on blackbody radiation
Example 1:
Show that (,T)d in both Eqs. (1-1) and (1-2) has units of energy per
unit volume , Jouls per cubic meter.
Solution:
- For The Reyliegh-Jeans Law (1-1),
p ( ,T )d  
8 kT
2


d
3
C
(J  K 1 )K
1 2
1
3

(
s
)
(
s
)

J
.
m
= (m  s 1 )3
- For the plank distribution (equation 1-2)
8 h 3
d
p ( ,T )d 

C3
e h kT  1
(J  S )(S 1 )3 1
 (s )  J .m 3
1 3
=
(m  s )
Where p ( ,T )d  is energy denisty
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture2
Example (2):
Equation (1-2) expresses planck’s radiation Law in terms of frequency.
Express Planck’s constant radiation law in terms of wave length.
Solution:
h=c, d  c  d   2
Substitute d into equation (1-2)
 ( ,T )d  
8 hc
5

d
e hc  kT  1
(1-3)
Where  ( ,T )d  is the energy density between and +d
* we can use eqn. (1-3) to derive an empirical relationship was known
at time as Wien displacement law.
* The Wien displacement law: Says that if max is the wavelength at
 ( ,T ) is a maximum , then:
maxT=2.90  10-3 m.K  (1-4)
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture2
By differentiating  ( ,T ) with respect to 
hc
maxT= 4.965K
(1-5)
Ex. If we have max = 500 nm
T 
2.90 103
max
2.90 103

 5800K
9
500 10
* Steven-Boltzmann Law:
From eq. (1-2) Stefan-Boltzmann showed hat the total energy
radiated per unit area per unit time from a black body is given by
R
C
E   T
4
4
 (1-6)
Where E is the total radiation energy
is the Stefan-Boltzmann Constant
The experimental value of  is 5.6697  10-8 J.m-2K-4. S-1
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture2
Example 3
Planck’s distribution of black body radiation gives the energy
density between n and +d. Integrate the Planck distribution overall
frequencies and compare the result to Eq. (1-6)
Solution:
The integral of q. 1-2 over all frequencies is


8 h  3d
E    ( ,T )d  3  h kT
c 0e
1
0
(1-7)
If we use the fact that :

x 3dx  4
0 e x  1  15
8 5 k 4T 4
 E 
15h 3c 3
(1-8)
c
E    T 4
From eq. (1-6)
4
 by substituting E from (1-8) into (1-6)
8 5 k 4T 4 c
  T
3 3
15h c
4
4
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture2
2 5 k 4
 
15h 3c 2
(1-9)
Using the values of k, h and c , the calculated value of  is 5.670  10-8
J.m-2K-4.S-1, in excellent agreement with the experimental value.
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