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Thermal Statistics - Microstate Probabilities

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data

    For a system in equilibrium at temperature T, the probability of finding it in a microstate m is:

    P(m) = ([itex]1/Z[/itex])exp([itex]-E/kT[/itex])
    where Z is the partition function.

    There are three accessible microstates, two with energy E[itex]_{a}[/itex] and one with energy E[itex]_{b}[/itex].

    Two identical and indistinguishable bosons can occupy these microstates. What's the probability that both these bosons have energy E[itex]_{b}[/itex]?

    2. Relevant equations

    I have the values for E[itex]_{a}[/itex] and E[itex]_{b}[/itex].
    It asked me to give a formula for Z which I put as Z = [itex]\Sigma[/itex]exp([itex]-E/kT[/itex])
    T = 300K
    I calculated Z to be 1.08

    3. The attempt at a solution

    What I've got at the moment is this:
    Let the number of particles in states a, a, b be n[itex]_{1}[/itex], n[itex]_{2}[/itex], n[itex]_{3}[/itex]:
    n[itex]_{1}[/itex] n[itex]_{2}[/itex] n[itex]_{3}[/itex]
    1 1 0
    1 0 1
    0 1 1
    2 0 0
    0 2 0
    0 0 2

    So the probability of both bosons being in state n[itex]_{3}[/itex] aka b would be [itex]1/6[/itex], but this seems way too obvious for 4 marks. I also tried to use the Bose-Einstein distribution but I don't know the chemical potential.

    Thanks
     
  2. jcsd
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