Maxwell-Boltzmann Distribution function and equipartition theorem

[redz]

Homework Statement

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

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 E_k 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

 

[vela]

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 $$u=v^2$$.

 

[kerimek]

further



Your approach to part a) and b) seems correct. The average kinetic energy for a two-dimensional gas of electrons would indeed be given by E_{k}=kT, and for T=293K, this corresponds to a speed of 1.38 x 10^5 m/s.

For part c), you are on the right track. The integral you have written is a standard integral known as the Gaussian integral. You can solve it by using the substitution u=v^2 and then using the standard form of the Gaussian integral, which is given by:

\int^{\infty}_{0} e^{-au^2} du = \frac{\sqrt{\pi}}{2a}

Substituting back for v, you should get:

C\frac{\sqrt{\pi}}{2}\frac{1}{\left(\frac{m}{2kT}\right)^{1/2}}=N

Solving for C, you should get:

C=\frac{2N}{\sqrt{\pi}}\left(\frac{m}{2kT}\right)^{1/2}

This is the value of C that satisfies the given condition. I hope this helps!