Lecture16

# C 3 n e populaeon density of states per state at

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Unformatted text preview: es for photons: D ( E ) = 2 ! =2 dE " (!c )3 4 Total number of photons # in a small energy range dE dN dn = N ( E ) dE dE V E2 D( E ) = 2 ! (!c )3 N (E ) = populaEon Density of states per state at E 1 e E /kT ! 1 note already included degeneracy of spin Total number of photons at temperature T: # V 2E 2 n( E ) = 2 3 3 \$ E /kT dE 2! ! c 0 e "1 5 Energy density ! < E >= " EN ( E ) D( E ) dE 0 # 8! L3 E3 = dE 3 \$ E / kT (hc ) 0 e "1 ! <E> = " u ( f , T ) df 3 L 0 8! hf 3 1 “spectral energy density funcEon” = u ( f , T ) = c 3 ehv/kT " 1 Also called Blackbody spectrum, units are energy per unit frequency per unit volume If we had expressed the density of states in terms of wavelength, we would have derived: 8" hc 1 u (! , T ) = 5 hc/kT ! !e #1 6 Blackbody spectrum If we plot the Planck’s formula for the blackbody spectrum u￿f ,T￿ Blackbody distribuEon peaks at approximately fmax = 2.8 6 ￿ 1017 kT = 5.9 ! 1010 [ K "1s "1 ] T [K] h Or Wien’s empirical displacement law 4000 K 4 ￿ 1017 3000 K 2 ￿ 1017 2000 K 200 400 600 800 1000 f ￿THz￿ 7 Total energy density # # 8! hf 3 1 U (T ) = \$ u ( f , T ) df = \$ df 3 hf / kT ce "1 0 0 x = hf / kT # 8! h ( kT )...
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