# Maxwell-Boltzmann Distribution function and equipartition theorem

1. Jan 9, 2010

### redz

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
$$p(v) = Cve^{-\frac{mv^{2}}{kT}}$$

Define C such that,
$$\int^{\infty}_{0} dvp(v)=N$$

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;
$$E_{k}=\frac{1}{2}m(v^{2}_{x}+v^{2}_{y})$$
Where Ek is kinetic energy

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

$$E_{k}=kT$$
Would this be correct?

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

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

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. Jan 9, 2010

### vela

Staff Emeritus
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 $$u=v^2$$.