# Maxwell Boltzmann translational energy distribution.

What is the Maxwell Boltzmann translational energy distribution in units of k T

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

The Maxwell Boltzmann speed distribution, P(v), is given by:

P(v) = 4 pi (m/(2 pi k T))^(3/2) Exp[-m v^2/(2 k T)] v^2

P(v)dv is the probability that a molecule of mass m has a speed between v and v + dv. Let's denote the energy distribution by Q(E). Then Q(E)dE is the probability that a molecule has an energy between E and E + dE. To find Q(E), eliminate v in P(v)dv in favor of E.

P(v) is proportional to Exp[-m v^2/(2 k T)] v^2 = Exp[-E/(kT)] v^2

P(v)dv = Exp[-E/(kT)]v^2dv = Exp[-E/kT]1/3 d(v^3)

v^3 is proportional to E^(3/2) and d E^(3/2) is proportional to

E^(1/2) dE. It thus follows that Q(E) is proportional to

E^(1/2) Exp[-E/kT].

It is convenient to measure E in units of kT. If you put E = U kT, then the distribution in terms of U is proportional to U^(1/2) Exp[-U]. The constant of proportionality can be found by demanding that this distribution is normalized to 1 (that's easier than starting from the correctly normalized p(v), keeping track of all the (temperature and mass dependent) proportionality constants while changing variables). If you do that then you find that the distribution in terms of U (let's call this W(U)) is:

W(U) = 2 pi^(-1/2)U^(1/2) Exp[-U]

So, the temperature and mass have completely dropped out of the distribution! This function is easy to draw. It has a maximum at U = 1/2. To see this immediately, take the logarithm of the function (you then get: constant + 1/2 Log[U] - U) and then take the derivative (you get 1/(2 U) - 1).

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