HW5 solutions - Physics 4120 Spring 2008 Homework #5...

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Unformatted text preview: Physics 4120 Spring 2008 Homework #5 solutions 1) Reif 3.3 (a) ( 29 ( 29 = = - + = - = 2 2 2 2 ln 2 2 2 ln ln ln 2 exp 2 2 ) ( H N E E N H E N H E N H E N H E E N N At equilibrium (where E is equal to its mean value), = , so 2 2 2 2 H N E H N E = But from Reif eq. 3.7.12, the mean value is equal to the most probable value, E E ~ = , so 2 2 2 2 ~ ~ H N E H N E = Furthermore, both systems are in the same field, so H=H and 2 2 ~ ~ = N E N E (b) The initial total energy of the system is given by H N b H bN + , and the final energy of the system is E E + ~ ~ . Energy conservation applies, so H N b H bN E E + = + ~ ~ We can rewrite E ~ as E N N E ~ ~ 2 2 = Which gives us 2 2 2 2 2 2 2 2 2 2 ~ ~ 1 ~ ~ ~ + + = + = + + = + + = + N N N b bN H N E H N b H bN N N N E H N b H bN N N E H N b H bN E N N E (c) The only change in energy of the system must come from heat, so...
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This note was uploaded on 04/17/2008 for the course PHYS 4120 taught by Professor Vajk during the Spring '08 term at Missouri (Mizzou).

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HW5 solutions - Physics 4120 Spring 2008 Homework #5...

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