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I want to estimate the heat capacity of a diatomic molecule whose movement has been constrained to a 2-dimensional plane.

I can assume that $$ k_\text{B} T ~\ll~ \hbar\sqrt{\frac{k_\text{Hooke}}{m}} \tag{1} \,,$$ where $k_\text{Hooke}$ is the Hooke's law constant.

The answer is $\frac{3}{2} k_\text{B} ,$ since there are 2 translational movements and one rotational, so: $$ U = \frac{f}{2}k_\text{B} T ~~\implies~~ C = \frac{3}{2} k_\text{B} \,.$$ Since the heat capacity, $C ,$ isn't dependent on $k_\text{Hooke} ,$ this seems to say that vibration can be neglected. However, how does the assumption in $\operatorname{Eq.}{\left(1\right)}$ end up resulting in this conclusion that vibration can be neglected?

Question: How does $k_\text{B} T \ll \hbar \sqrt{k_\text{Hooke}/m}$ imply that the vibrational motion is frozen out?

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    $\begingroup$ Just to note it, arbitrary symbols like $`` k "$ can make for decent variable-identifiers in specific contexts, but when symbols become degenerate due to merged contexts, it's best to add an additional discriminant to each. $\endgroup$ – Nat Aug 7 at 19:17
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To help understand what your assumption is saying, it might be easiest to start by defining what each side measures. What does $k_B T$ measure? What does $\hbar\sqrt{k/m}$ measure? What are the units of them?

And once you figure out the units and what each side represents, what does it mean that one side is much, much less than the other side? What would happen if the inequality is reversed? What happens when they are of similar order of magnitude?

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  • $\begingroup$ I belive, this should be a comment. It is giving clues, not answering the question. $\endgroup$ – J. Manuel Aug 7 at 19:21
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    $\begingroup$ @J.Manuel On questions tagged homework-and-excercises, we have a long-standing policy on not answering the question directly, but providing ways for the person to answer it themselves. If I had come out and answered it directly, I would be breaking that policy and my answer should be deleted. And since giving hints is the way to answer homework-and-excercises questions, it shouldn't be left as a comment. $\endgroup$ – tpg2114 Aug 7 at 19:24

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