# Why does $k_\text{B} T \ll \hbar\sqrt{k_\text{Hooke}/m}$ imply that vibrational motion is negligible?

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?

• 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. – Nat Aug 7 '19 at 19:17

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?