sol4 - Phys. 3341 Homework Solution Set #3 Yoav Kallus 1...

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Unformatted text preview: Phys. 3341 Homework Solution Set #3 Yoav Kallus 1 Reif 3.1 (a) Each molecule has an independent one-in-four chance of being in the smaller partition. Therefore, the mean number of He and Ne molecules respectively in that partition are 25 and 250 respectively. In the larger partition, consequently, they are 75 and 750. (b) The probability of finding all He molecules in the smaller partition and all Ne molecules in the larger partition is (1 / 4) 100 (3 / 4) 1000 < 10 − 185 . Note that we cannot use the Gaussian approximation here, since we are very far from the maximum of the binomial distribution. 2 Reif 3.2 (a) The number of states of energy E = nμH is given by the number of ways of assigning ( N − n ) / 2 spins parallel to the external field and ( N + n ) / 2 spins anti-parallel to it. Namely, Ω( E ) = N ! N − n 2 ! N + n 2 ! (1) ≃ exp parenleftbigg N log N − N − N − n 2 log N − n 2 + N − n 2 (2) − N + n 2 log N + n 2 + N + n 2 parenrightbigg , (3) where n = E/μH . The inverse-temperature is then β = 1 /k B T = ∂ log Ω( E ) /∂E (4) = (1 /μH ) ∂ log Ω( E ) /∂n (5) = (1 /μH...
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sol4 - Phys. 3341 Homework Solution Set #3 Yoav Kallus 1...

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