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Statistical Mechanics

  1. Jan 17, 2014 #1
    1. The problem statement, all variables and given/known data

    A system of N particles has three possible energy levels namely; 0, E and 4E. How many particles does one expect in the second state at temperature T?

    2. Relevant equations

    It's a sample problem for our finals. Our Text book is Statistical Mechanics by Roger Bowley and Mariana Sanchez.

    3. The attempt at a solution

    Three Energy levels

    [itex]E_{1}=0[/itex], [itex]E_{2}=E[/itex], [itex]E_{3}=4E[/itex]

    Let us first fill the [itex]E_{1}[/itex] state with 3 particle.

    N distinguishable ways of selecting the first particle
    N-1 different ways to select second particle
    N-2 different ways to select third particle

    So the total number of ways to place first three particles in state [itex]E_{1}[/itex] is

    [itex]N(N-1)(N-2)=\frac{N!}{(N-3)!}[/itex]​

    Generally for [itex]n_{1}[/itex] particles placed in [itex]E_{1}[/itex] is,
    [itex]\frac{N!}{n_{1}!(N-1)!}[/itex]

    for [itex]E_{2}[/itex] state,

    [itex]\frac{(N-n_{1})!}{n_{2}!(N-n_{1}n_{2})!}[/itex]​

    for [itex]E_{3}[/itex] state,

    [itex]\frac{(N-n_{1}n_{2})!}{n_{3}!(N-n_{1}n_{2}n_{3})!}[/itex]​

    Total number of particles in all three state will be

    [itex]P=\frac{N!}{n_{1}!n_{2}!n_{3}!}[/itex]​

    Substituting values

    [itex]P=\frac{N!}{0!1!4!}[/itex]​


    Am I on right track?
     
  2. jcsd
  3. Jan 17, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    I don't think so. You haven't even invoked temperature in any way.

    If you had only one particle, what would be the probability of finding in in state 2 when the temperature is T?
     
  4. Jan 17, 2014 #3
    ok, I try to re attempt it after going through chapter 6 of the book.
     
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