Lecture Notes2

# kt 2 e 1 ln 1 v 3 3 nkt 1 2 2 5 2 iv

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Unformatted text preview: a 0 = 0, a1 = Λ 2 3 2 3/ 2 2 1 1 1 λ = ρΛ3 + ρΛ3 + − 4 3/ 2 3/ 2 2 3 () 1 P 2 Λ3 =ρ+ ρ2 + − 8 kT 25/ 2 35 / 2 3 a1a 2 a1 + = 0,... 21/ 2 2 3 / 2 33 ρΛ + ... () 63 Λ ρ + ... P = ρ + B2 (T ) ρ 2 + B3 (T ) ρ 3 + ... kT l +1 3 1 ∞ (− 1) λl 3 Λ3 = VkT 1 + 5 / 2 ρ + ... E = VkT 3 ∑ 5/ 2 2 2 Λ l =1 l 2 nk = λe − βε k 1 = − βε k − β (ε k − µ ) 1 + λe 1+ e (II) Strongly Degenerate Ideal F - D Gas λe − βε k 1 = − βε k − β (ε k − µ ) 1 + λe 1+ e 1 f (ε ) = − β (ε k − µ ) 1+ e 3/ 2 2m w(ε )dε = 4π 2 Vε 1 / 2 dε h nk = h2 3 µ0 = 2m 8π 2/3 N V 2/3 8 π2 2 µ = µ 0 1 − η + ... 12 2 5π E = E 0 1 + η 2 + ... 12 2 T π ∂E CV = Nk = T 2 ∂ T V F For η = (βµ 0 ) &lt; 1 , −1 (III) Weakly Degenerate Ideal B - E Gas λe − βε k λe − βε k 2m N=∑ ⇒N = + 2π 2 − βε k − βε k 1 − λe h k 1 − λe ( PV = − kT ∑ ln 1 − λe − βε k ∞ 3/ 2 ) V λε 1 / 2 e − βε k dε ∫ − βε k ε &gt; ε 0 1 − λe k N 2m ρ = = 2π 2 V h P 2m = −2π 2 kT h ∞ 3/ 2 3/ 2 ∫ ε &gt;ε 0 λε 1 / 2 e − βε k dε λ + − βε k V (1 − λ ) 1 − λe ∞ ∫ε 1/ 2 ( ) ln 1 − λe − βε k dε − ε &gt;ε 0 1 P 1 g 3 / 2 (λ ) and = 3 g 5 / 2 (λ ), 3 kT Λ Λ 3 P Λ = 1 − 5/ 2 ρ + ... ρkT 2 ρ= E= 1 ln (1 − λ ) V 3 Λ3 NkT 1 − ρ + ... 2 2 5/ 2 (IV) Strongly Degener...
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## This document was uploaded on 03/26/2014 for the course PH 641 at NJIT.

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