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Maxwell-Boltzmann Distribution function and equipartition theorem

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A gas of electrons is contrained to lie on a two-dimensional surface. I.e. they
    have no movement in the z direction but may move freely in the x and y.

    a) From the equipartition theorem what is the expected average kinetic energy
    as a function of T?

    b) For T = 293K what speed would this correspond to?

    c) The equivalent of the Maxwell-Boltzman distribution for a two-dimensional
    gas is
    [tex]p(v) = Cve^{-\frac{mv^{2}}{kT}}[/tex]

    Define C such that,
    [tex]\int^{\infty}_{0} dvp(v)=N[/tex]

    3. The attempt at a solution

    a) As the particles are able to move in only two dimensions the equipartition theorem reduces to only have two velocity terms x and y. therefore;
    Where Ek is kinetic energy

    This would mean that the total kinetic energy in terms of T would be

    Would this be correct?

    Part b) will be a simple rearrangement and calculation. no real problems there

    c) I have rearranged the integration to;
    [tex]C\int^{\infty}_{0}ve^{-\frac{m}{kT}v^{2}} dv=N[/tex]

    But have been unable to process further. I think it is supposed to be a standard integral of some sort but have no real clue how to progress
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 9, 2010 #2


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    Staff Emeritus
    Science Advisor
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    a) Yes. Each quadratic term in the energy contributes 1/2 kT to the average energy.

    c) It's a simple substitution. Try something like [tex]u=v^2[/tex].
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