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I Degeneracy of energy levels greater than no. of particles?

  1. Oct 26, 2016 #1
    From statistical mechanics in zeemansky's book . He states that it's easy to see that for a closed system the no. Of degenerate states ##g_i## for energy level ##E_i## is greater than the number of particles ##N_i##occupying that energy state. I can't find a mathematical proof for it. Can I get any hints on how to prove this? A proof would be even better.
     
  2. jcsd
  3. Oct 26, 2016 #2

    DrClaude

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    Staff: Mentor

    Could you give an exact quote?
     
  4. Oct 26, 2016 #3
    "However close they may be, there is still only a finite number of discrete energy levels for the atoms of an ideal gas. It is the fundamental problem of statistical mechanics to determine, at the equilibrium, the populations of these energy levels - that is, the number of particles ##N_1## having energy ##\epsilon_1##, the number of particles ##N_2## having energy ##\epsilon_2##, and so on. It is a simple matter to show (see Prob 13.4) that the number of quantum states ##g_i## corresponding to energy level ##\epsilon_i## (the degeneracy of this level) is very much larger than the number of particles ##N_i## occupying that the level at room temperature, Thus: ##g_i >> N_i##

    This is problem 13.4:

    Show that, when N ideal-gas atoms come to equilibrium,
    $$\frac{g_i}{N_i} = \frac{Z}{N}e^{\frac{\epsilon_i}{kT}}$$ and $$\frac{Z}{N}=\frac{(kT)^{5/2}}{P} \left( \frac{2\pi m}{h^2} \right)^{3/2}$$

    Z seems to be the partition function
     
  5. Oct 26, 2016 #4

    DrClaude

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    This has now become a homework-type problem. For further help with this precise statement, you will have have to open a new thread in a homework forum and have an attempt at a solution. And yes, ##Z## is the partition function.
     
  6. Oct 26, 2016 #5
    Okay.
     
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