Although an ordinary H₂ molecule consists of two identical atoms, this is not the case for the molecule HD, with one atom of deuterium (i.e., heavy hydrogen, ²H). Because of its small moment of inertia, the HD molecule has a relatively large value of $ϵ:0.0057\mathrm{eV}$ At approximately what temperature would you expect the rotational heat capacity of a gas of HD molecules to "freeze out," that is, to fall significantly below the constant value predicted by the equipartition theorem?

An ordinary H₂ molecule consists of two identical atoms, this is not the case for the molecule HD, with one atom of deuterium (i.e., heavy hydrogen, ²H). Because of its small moment of inertia, the HD molecule has a relatively large value of$\epsilon :0.0057\mathrm{eV}$

The expression for heat capacity from previous problem is:

$C=\frac{k\sum (2j+1)e^{-j(j+1)/t}\sum j^2(j+1{)}^2(2j+1)e^{-j(j+1)/t}}{{\left(t\sum (2j+1)e^{-j(j+1)/t}\right)}^2}-\frac{k{\left(\sum j(j+1)(2j+1)e^{-j(j+1)/t}\right)}^2}{{\left(t\sum (2j+1)e^{-j(j+1)/t}\right)}^2}$

Using python, plot the function between t and $C/k$. The code is:

We can say that the heat capacity falls off steeply when:

$0.3<t<0.6$

Where,

$t=\frac{kT}{ϵ}$

Taking the average,

$0.4=\frac{kT}{ϵ}$

Solving for temperature,

$T=\frac{0.4ϵ}{k}$

Substituting the values of energy constant and Boltzmann's constant

$$\begin{gathered} T=\frac{0.4(0.0057\mathrm{eV})}{8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}} \\ T=26.45K \end{gathered}$$

The graph is: