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PS9Sol_f10

# PS9Sol_f10 - Physics 112 Fall 2010 Professor William...

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Physics 112 Fall 2010 Professor William Holzapfel Homework 9 Solutions Problem 1, Kittel 7.13: Chemical Potential versus Concentration (a) We know for a classical ideal gas μ/τ = ln n n Q , so for n n Q 1 we expect a logarithmic behavior of the chemical potential. We’ve also seen that for a degenerate Bose gas ( n n Q 1) the chemical potential acts like μ/τ ≈ - 1 N . In the thermodynamic limit, N 1 so we essentially have μ 0 for n n Q > 1. This behavior is plotted in figure 1. Figure 1: μ/τ vs. n/n Q is plotted for a Bose gas. (b) For fermions we have the same ideal gas behavior for n n Q 1, but for n n Q 1, we will have a degenerate Fermi gas with μ F . We can rewrite the expression for the Fermi energy in terms of n n Q , using n Q ( 2 π 2 ) 3 / 2 : f = 2 2 m ( 3 π 2 n ) 2 / 3 = τ 4 π ( 3 π 2 ) 2 / 3 n n Q 2 / 3 So we see that μ/τ n n Q 2 / 3 for a Fermi gas with n n Q 1. We’ve graphed this behavior in figure 2. The important thing to notice is that in a Bose gas, the chemical potential is always negative (as measured from the ground state energy) or close to zero (when Bose-Einstein condensation 1

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Figure 2: μ/τ vs. n/n Q for a Fermi gas. occurs), whereas in a Fermi gas, the chemical potential becomes positive in the quantum (degenerate) limit. Problem 2, Kittel 7.14: Two Orbital Boson System If the occupation of the ground state ( 0 = 0) is twice that of the first excited state ( 1 = ), then because there are only two states we have f ( 0 , τ ) + f ( 1 , τ ) = N and hence f ( 0 , τ ) = 2 3 N f ( 1 , τ ) = 1 3 N. The occupation of each state is given by the Bose-Einstein distribution function: f ( τ, 0) = 1 e - μ/τ - 1 = 2 3 N, (1) f ( τ, ) = 1 e ( - μ ) - 1 = 1 3 N. (2) Solving equation (1) for e - μ/τ gives e - μ/τ = 1 + 3 2 N .
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PS9Sol_f10 - Physics 112 Fall 2010 Professor William...

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