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Equipartition Theorem and Microscopic Motion

  • #1
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Homework Statement


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.

Homework Equations


rms angular speed: ω=sqrt{(2kT)/(md2)}
frot=ω/(2[itex]\pi[/itex])
k=1.381x10-23 J/(molecule*K)

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.
 

Answers and Replies

  • #2
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The formula you have is derived from equating [itex]\frac 1 2 kT[/itex] to rotational energy, which is [itex]\frac {J \omega^2} {2}[/itex], where [itex]J = \frac {md^2} {2}[/itex] 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 [itex]kT[/itex]. That means instead [itex]2kT[/itex] you have in the formula you must take [itex]4kT[/itex], not [itex]3kT[/itex].

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

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