Average Energy - Solid State Physics - Solved Paper, Exams of Solid State Physics

Download Average Energy - Solid State Physics - Solved Paper and more Exams Solid State Physics in PDF only on Docsity! Physics 112 Spring 2010 Professor William Holzapfel Midterm 1 Solutions Problem 1 a) Since Z is a sum over microstates, not macrostates, we must include multiplic- ities (degeneracies). This gives Z = e−E0/τ (1 + 2e−/τ + e−2/τ ) = e−E0/τ (1 + e−/τ )2. b) The average energy is given by U = τ2 ∂ lnZ ∂τ = τ2 ∂ ∂τ [ −E0/τ + 2 ln(1 + e−/τ ) ] = E0 + 2e−/τ 1 + e−/τ = E0 + 2 1 + e/τ . c) The free energy is F = −τ lnZ = −τ [−E0/τ + 2 ln(1 + e−/τ )] = E0 − 2τ ln(1 + e−/τ ) and so the entropy is then σ = −∂F ∂τ = 2 ln(1 + e−/τ ) + 2 τ 1 1 + e/τ . d) Let’s take the τ →∞ limits first, since in this case the exponentials e±/τ can be Taylor expanded: lim τ→∞ U(τ) ≈ E0 + 2 1 + (1 + /τ) = E0 +  1 + /2τ = E0 + (1− /2τ). 1 This has limiting value E0 + , which is what we expect: in the large τ limit all microstates are equally probable, so the average energy should just be the average of all the possible values (weighted by multiplicity), which is E0 + . Similarly for the entropy (we carefully expand to second order here): lim τ→∞ σ(τ) ≈ 2 ln [ 2− /τ +  2 2τ2 ] + 2 τ 1 2 + /τ ≈ 2 ln 2 + 2 ln [ 1− ( /2τ −  2 4τ2 )] +  τ ( 1−  2τ ) ≈ ln 4 + 2 [ −/2τ +  2 4τ2 − 1 2 (/2τ)2 ] +  τ ( 1−  2τ ) = ln 4− 1 4 (/τ)2. Again, this has the limiting value that we expect at τ →∞ since there are four possible microstates, all of which are equally likely in this limit. For the low temperature limits, the exponentials cannot be Taylor expanded but the Boltzmann factor itself is small so that allows some simplifications: lim τ→0 U(τ) = E0 + 2e−/τ 1 + e−/τ ≈ E0 + 2e−/τ . Likewise, lim τ→0 σ(τ) = 2 ln(1 + e−/τ ) + 2 τ e−/τ 1 + e−/τ ≈ 2 τ e−/τ . Problem 2 a) Say we start with the first atom spin up. Then we’ll have all consecutive atoms spin up until we hit the first domain wall, where the next atom is spin down. The n domain walls are thus determined by choosing n atoms where the spin orientation flips. Thus if we start with the first domain spin up then choosing n out of N atoms completely specifies a microstate (we don’t choose out of N +1 because we can’t have a domain wall at the first atom). This then gives a multiplicity of ( N n ) . However, we could also have started with the first domain spin down, so the total multiplicity for n domains is g(N + 1, n) = 2 ( N n ) = 2 N ! N !(N − n)! . 2