Determine the energies and degeneracies of the lowest four energy levels of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule freely rotating in three dimensions. What is the frequency of the transition between the lowest two rotational levels? The moment of inertia of an ${ }^1 \mathrm{H}^{35} \mathrm{Cl}$ molecule is $2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2$.

First, note that
$$
\frac{\hbar^2}{2 I}=\frac{\left(1.055 \times 10^{-34} \mathrm{Js}^2\right.}{2 \times\left(2.6422 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2\right)}=2.106 \ldots \times 10^{-22} \mathrm{~J}
$$
or $0.2106 \ldots$ zJ. We now draw up the following table, where the molar energies are obtained by multiplying the individual energies by Avogadro's constant:
\begin{tabular}{llll}
\hline$J$ & $E / z J$ & $E /\left(\mathrm{J} \mathrm{mol}^{-1}\right)$ & Degeneracy \\
\hline 0 & 0 & 0 & 1 \\
1 & 0.4212 & 253.6 & 3 \\
2 & 1.264 & 760.9 & 5 \\
3 & 2.527 & 1522 & 7 \\
\hline
\end{tabular}

The energy separation between the two lowest rotational energy levels $\left(J=0\right.$ and 1 ) is $4.212 \times 10^{-22} \mathrm{~J}$, which corresponds to a photon frequency of
$$
\nu=\frac{\Delta E}{h}=\frac{4.212 \times 10^{-22} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}=6.357 \times 10^{11} \mathrm{~s}^{-1}=635.7 \mathrm{GHz}
$$