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Firmi-Dirac Stats. - Calculate number of microstates.

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the number of microstates where two particles have have energy 2eV, three have 3eV, three have energy 4eV, and two have 5eV.


    10 indistinguishable Particles
    1 particle allowed per state.

    ε1 = 1eV ε2 = 2eV ε3 = 3eV ε4 = 4eV ε5 = 5eV ε6 = 6eV
    g1 = 1 g2 = 5 g3 = 10 g4 = 10 g5 = 5 g6 = 1


    2. Relevant equations[itex]\textit{}[/itex]

    possibly [itex]\overline{n}[/itex]i=[itex]\frac{1}{\textbf{e}^{(ε_{i}-μ)/kT}+1}[/itex]

    3. The attempt at a solution

    Since the particles are indistinguishable I thought that there would only be one microstate, but I don't think that this is right... can someone point me in the right direction on how to start this problem?
     
  2. jcsd
  3. Apr 23, 2012 #2
    figured it out, was missing a very relevant equation.
    [itex]
    w_{\text{FD}}\left(N_1,N_2,\text{...}N_n\right)=∏ \frac{g_j!}{N_j!\left(g_j-N_j\right)!}
    [/itex]
    [itex]
    w_{\text{FD}}\left(N_1,N_2,N_3,N_4,N_5,N_6\right)=\left(\frac{g_1!}{N_1!\left(g_1-N_1\right)!}\right)\left(\frac{g_2!}{N_2!\left(g_2-N_2\right)!}\right)\left(\frac{g_3!}{N_3!\left(g_3-N_3\right)!}\right)\left(\frac{g_4!}{N_4!\left(g_4-N_4\right)!}\right)\left(\frac{g_5!}{N_5!\left(g_5-N_5\right)!}\right)\left(\frac{g_6!}{N_6!\left(g_6-N_6\right)!}\right)
    [/itex]
    [itex]
    w_{\text{FD}}(0,2,3,3,2,0)=\left(\frac{1!}{0!(1-0)!}\right)\left(\frac{5!}{2!(5-2)!}\right)\left(\frac{10!}{3!(10-3)!}\right)\left(\frac{10!}{3!(10-3)!}\right)\left(\frac{5!}{2!(5-2)!}\right)\left(\frac{1!}{0!(1-0)!}\right)
    [/itex]
    evaluates to 1,440,000
     
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