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GR0877 Problem 78

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) According to the Gibb’s distribution \displaystyle P_i = \frac{1}{Z} \exp \left(-\frac{E_i}{kT}\right), where \displaystyle Z = \sum_i \exp \left( -\frac{E_i}{kT}\right). Thus, \displaystyle P_2 = \frac{\exp \left( -\frac{E_2}{kT}\right)}{\exp \left( -\frac{E_1}{kT}\right) + \exp \left( -\frac{E_2}{kT}\right)}.

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Categories: 0877 Tags: , , , ,
  1. 21.09.2013 at 23:14

    We should expect the function to go to zero as E_2 goes to infinity. This is only the case for B.

    • Stevie
      16.10.2013 at 19:24

      nice!

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