# D evaluate the linear magnetic susceptibility t and t

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Unformatted text preview: y. d) Evaluate the linear magnetic susceptibility T ε and T Hint: cosh x = (∂ 2F/∂H 2)H =0 in the limits ε. 1 + x2 /2, when x 1. Bonus question Suppose that we are now dealing with a large number of two-level systems and the level spacing is distributed with the probability distribution function P (ε) = 1 exp ( ε/ ). e) Evaluate approximately the mean value of the magnetic susceptibility ∞ 0 (ε) P (ε) dε in the limit T . Solution a) Z = 1 + 4 exp ( ε/T ) + exp ( 2ε/T ) F= T log Z = T log [1 + exp ( 2ε/T ) + 4 exp ( ε/T )] 1 = b) T (exp ( 2ε/T ) + 4 exp ( ε/T )) T ε T log 6 + ε F ε 4 (ε/T ) exp ( ε/T ) T ε log 6 ∂F ∂T S= T ε T 4ε exp ( ε/T ) T ε ε E = F + TS ε T c) Z = 1 + 2 exp ( ε/T ) + exp [( ε + 2 = 1 + 2 exp ( ε/T ) [1 + cosh (2 F= T log Z = B H ) /T ] B H/T )] + exp [( ε 2 B H ) /T ] + exp ( 2ε/T ) + exp ( 2ε/T ) T log {1 + 2 exp ( ε/T ) [1 + cosh (2 B H/T )] + exp ( 2ε/T )} d) F T log 1 + 2 exp ( ε/T ) 2 + 2 ( 2T exp ( ε/T ) 2 + 2 ( T log 2 + 2 2 + 2 ( 4T exp ( ε/T ) 1 + ( T log 6 = 2T ( B H/T ) H =0 2 2 T 2 + exp ( 2ε/T ) ε T ε T ε T /3 2 B = 2 2 B H/T ) B H/T ) ∂ 2F/∂H 2 B H/T ) B H/T ) ε 8T 1 exp ( ε/T ) T ε 4T 1 /3 T ε ε/ ) dε ∝ 1 e) ∞ = (ε) P (ε) dε = 0 2 B 8T 1 (4/3) T 1 ∞ T 1 exp ( ε/T 1 2 T 0 exp ( ε/ ) dε ∝ T ∝ 1 [1 exp ( T / )] ∝ 1 2 B...
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## This note was uploaded on 12/14/2013 for the course PHYS 713 taught by Professor Serota during the Spring '09 term at University of Cincinnati.

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