# Equipartition Theorem and Microscopic Motion

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1. Sep 5, 2012

### ghoultree

1. The problem statement, all variables and given/known data
What is the typical rotational frequency frot for a molecule like N2 at room temperature (25°C)? Assume that d for this molecule is 10-10m. Take the atomic mass of N2 to be mN2=4.65x10-26kg. You will need to account for rotations around two axes (not just one) to find the correct frequency.

2. Relevant equations
rms angular speed: ω=sqrt{(2kT)/(md2)}
frot=ω/(2$\pi$)
k=1.381x10-23 J/(molecule*K)

3. The attempt at a solution
For my first attempt, I plugged the given numbers into sqrt{(3kT)/(md2)}. I used 3 instead of 2 because I was accounting for the two different axes. I'm not sure if the was the correct way to go about it. Secondly, I divided the given mass by 2. Then, I divided my answer by 2pi.

2. Sep 6, 2012

### voko

The formula you have is derived from equating $\frac 1 2 kT$ to rotational energy, which is $\frac {J \omega^2} {2}$, where $J = \frac {md^2} {2}$ is the moment of inertia of two masses m on a weightless rod d. But you have two degrees of freedom, so you must equate $kT$. That means instead $2kT$ you have in the formula you must take $4kT$, not $3kT$.

As follows from the above, you should not divide m by two, because you are already given the atomic mass.