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Quick statistical mechanics problem

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    In a large system of distinguishable harmonic oscillators, how high does the temperature have to be for the probable number of particles occupying the ground state to be less than 1?


    2. Relevant equations

    [itex]N(E_{n})=NAe^{\frac{-En}{K_{B}T}}[/itex]
    [itex]E_{0}=0[/itex] (the textbook shifted the harmonic oscillator energies down by -[itex]\frac{1}{2}hw_{o}[/itex])

    Answer: [itex]\frac{N\hbar\omega}{k_{B}}[/itex]

    3. The attempt at a solution

    Since this question is asking about the probable number of particles occupying a given state with energy 0, I decided to use the occupation number equation. I set the equation equal to one (I figured once I found out what temperature made it equal to one, I could see how the expression would have to change in order for the occupation number to be below 1) and set the energy equal to zero. Each time I try to solve the resulting equation, I end up with a natural logarithm (due to solving for T in the exponential) which isn't in the answer in the back of the book. I'm guessing I'm over complicating this problem but it is bothering me that I can't figure it out. Any help would be much appreciated. Thanks!
     
  2. jcsd
  3. May 21, 2013 #2

    BruceW

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    um. so you set this equation:
    [tex]N(E_{n})=NAe^{\frac{-En}{K_{B}T}}[/tex]
    equal to one? good. that's the right direction. And you set the energy to zero. So then your equation is:
    [tex]1=NA [/tex]
    right? I agree. So now, you 'solve the resulting equation' by solving for 'A' in the original equation, right? So what did you get for A ? please write it out, so I can see where you got to. It looks like you are headed in the right direction.
     
  4. May 21, 2013 #3
    Okay, so if I solve the original equation for A, I get [itex]A=\frac{e^{\frac{E_{n}}{K_{B}T}}}{N}[/itex].

    Plugging that in to the equation [itex]1=NA[/itex], I get [itex]1=e^{\frac{E_{n}}{K_{B}T}}[/itex].

    Taking the natural log of both sides gives me zero on the left hand side. What did I do wrong?
     
  5. May 21, 2013 #4
    sorry wrong post
     
  6. May 21, 2013 #5

    BruceW

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    you've just gone in a circle, I'm afraid. You need to go even further back, to calculate A. To calculate A, you need to use your original equation:
    [tex]N(E_{n})=NAe^{\frac{-En}{K_{B}T}}[/tex]
    So here, try doing a sum over all possible values of En. what do you get on the left-hand-side, and what do you get on the right-hand-side. This will give you the value of A.
     
  7. May 21, 2013 #6
    [itex]A=\frac{1}{Ʃe^{\frac{-E_{n}}{K_{B}T}}}[/itex]?
     
  8. May 21, 2013 #7

    BruceW

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    yep! that's it. Now, you can use the form of En, and the infinite sum just happens to have a nice answer.
     
  9. May 21, 2013 #8
    So [itex]1=\frac{N}{Ʃe^{\frac{-E_{n}}{K_{B}T}}}[/itex].

    I'm not sure what this infinite sum equals. However, I found this in my textbook...

    [itex]Ʃx^{n}=\frac{1}{1-x}[/itex]. So, could I use that and let [itex]x=e^{\frac{-1}{K_{B}T}}[/itex]?
     
  10. May 22, 2013 #9

    BruceW

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    that is the right idea. although En is not just equal to n. Remember the constant of proportionality.
     
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