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Unformatted text preview: 12 Average energy in each mode for light quanta
In statistical mechanics one learns that at temperature T, H  the probability P for a state to have energy E is: P e kT Since P is normalized to 1, 01/28/2005 , k = 1.381 10  23 J K P=e  H kT 1
e
all H  H kT Average energy in this mode, < H >= H e
all H  H kT 1
e
all H  H kT Energy of the mode is H = n ( ) for n light quanta, each with energy ( ) all H e  H kT = [e
n =0  ( ) kT n = 1 1 e
(  kT ) , He  H
all H d e   H = = e d all H (1  e  ) 2 Average energy in each mode: e  ( ) < H >= = ( ) / kT  1 e e 1
[email protected] 13 The energy of light quanta and Plank's constant dZ ( ) = 8 ( L )3 2 d Number of modes in a frequency interval: c
Average energy density in that interval: Wien's displacement law: 01/28/2005 u ( )d = 8 c 3 2 ( )
e ( ) / kT 1 d u ( ) / T 3 only depends on
c 4 /T ( ) = h and RT ( ) = u = T 3 2 h ( / T )3 c 2 exp( h ) 1 kT There is no longer an ultraviolet catastrophe R
0 T ( )d = T 4 2 k c 2 h3
4 h 3 ) kT h h kT exp( ) 1 0 kT ( d = T 4 2 k c 2 h3
4 0 x3 exp( x ) 1 dx = T 4 2 k 4 4 c 2 h 3 15 RT ( )d = T 4 0 15c 2 h=3 5 4 2 k
[email protected] Plank's constant: h = 6.626 10 34 Js 14 Interpretation 01/28/2005 A electromagnetic wave with frequency Q contains light quanta (photons) with the energy hQ. The energy of the wave determines the number of such photons that make up the wave. For small Q, the wave can have nearly all energies nhQ and one obtains the classical limit RT ( ) = T 3 2 h ( / T )3 c 2 exp( 1 ) 1 kT 2 h c2 3 h / kT + O( h ) = kT 2 c2 kT 2 + O( h ) kT RT ( ) T=6000 K visible max = 0.06 THz T K 0 THz 430 THz 700 nm 1.8 eV 750 THz 400 nm 3.1 eV 1000 THz
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This note was uploaded on 09/28/2008 for the course PHYS 316 taught by Professor Hoffstaetter during the Spring '05 term at Cornell.
 Spring '05
 HOFFSTAETTER
 Physics, Energy, Statistical Mechanics, Light

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