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Maxwell Boltzman Distribution, Please help!

  1. May 3, 2013 #1
    1. The problem statement, all variables and given/known data

    You will recall from our discussion of the Franck-Hertz experiment that the energy difference between the first excited state of mercury and the ground state is 4.86 eV. If a sample of mercury vaporized in a flame contains 1.02×10^20 atoms in thermal equilibrium at 1613 K, calculate the number of atoms in the first excited state. Assume that the Maxwell-Boltzmann distribution applies and that the n = 1 (ground) and n = 2 (first excited) states have equal statistical weights.

    2. Relevant equations

    n2/n1=g(E2)/g(E1)*e^((E1-E2)/(k T))
    k=boltzman constant = (1.3807*10^-23), divide by 1.602*10^-19 to get in eV

    3. The attempt at a solution

    So I felt like it was almost a plug and chug. The distinct states at n=2, first excited state, is 8. For n=1 g(E1)= 2. Sooo :
    n2/n1=(8/2)*e^((-4.86 eV)/(k*1613))
    I got 2.626*10^-15, so that's the ratio of atoms in the first excited state compared to the ground state? So I just multiply it by the number of atoms 1.02*10^20 and I get 267884 atoms. Answer isn't correct though and I cant figure out what I may be missing.
     
  2. jcsd
  3. May 3, 2013 #2

    DrClaude

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    Staff: Mentor

    Have you made use of the last part of this statement?
     
  4. May 3, 2013 #3
    I guess im not too sure of what that means.
     
  5. May 3, 2013 #4
    nmv I got it Thanks!
     
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