PS3Sol_f10

# PS3Sol_f10 - Physics 112 Fall 2010 Professor William...

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Physics 112 Fall 2010 Professor William Holzapfel Homework 3 Solutions Problem 1: Energy Fluctuations a) Kittel 3.4 The easiest way to approach to this problem is probably to start with the right side of the answer. By defining β = 1 τ , ∂U ∂τ V = ∂U ∂β V ∂β ∂τ V = - 1 τ 2 ∂U ∂β . Using U = < > = n n e - β n /Z = - 1 Z ∂Z ∂β we can write: τ 2 ∂U ∂τ V = - ∂U ∂β V = ∂β 1 Z ∂Z ∂β = 1 Z 2 Z ∂β 2 - 1 Z 2 ∂Z ∂β 2 . (1) Next, simply note that < > = n n e - β n /Z = - 1 Z ∂Z ∂β < 2 > = n 2 n e - β n /Z = 1 Z 2 Z ∂β 2 and so by (1) τ 2 ∂U ∂τ V = < 2 > - < > 2 = ( ) 2 as desired. b) Using our result from a) and the fact that U = 3 2 for an ideal gas, we have < U ) 2 > = τ 2 ∂U ∂τ V = τ 2 · 3 2 N 1

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and so < U ) 2 > U = 3 2 2 3 2 = 2 3 N . Using 2 3 N = 0 . 1 we get N = 200 3 . So to have the energy fluctuations of an ideal gas to be on the order of 10% or more, you need to have fewer than 67 particles. Problem 2, Kittel 3.6: Diatomic Molecules a) As the problem reminds you, it is important to remember that the partition func- tion is defined as a sum over all microstates , not energy levels. To write the partition function as a sum over energy levels, you must include the degeneracy (or multi- plicity) of each level. We know for a molecule with total angular momentum j the possible values of j z are - j, - j + 1 , · · · , j - 1 , j . This gives a multiplicity of 2 j + 1 for each energy level at fixed j , so the partition function is: Z R = j =0 (2 j + 1) e - j ( j +1) 0 . b) For the high temperature limit τ 0 the thermal energy is much larger than the spacing between energy levels, so we can convert the sum into an integral over j : Z R ( τ 0 ) = j =0 (2 j + 1) e - j ( j +1) 0 0 (2 j + 1) e - j ( j +1) 0 dj = - τ 0 0 ∂j e - j ( j +1) 0 dj = - τ 0 e - j ( j +1) 0 0 . = τ 0 2
c) In the low temperature limit, τ 0 , higher j terms in the partition function quickly decay exponentially, so we can approximate the sum by just keeping the first two terms: Z R ( τ ) = j =0 (2 j + 1) e - j ( j +1) 0 = 1 + 3 e - 2 0 + 5 e - 6 0 + · · · 1 + 3 e - 2 0 . d) U and C V can easily be calculated in the two limits: i) τ 0 U = τ 2 ln Z ∂τ = τ 2 0 1 Z = τ 2 / 0 τ/ 0 = τ C V = ∂U ∂τ V = 1 .

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