At low temperatures, a diatomic molecule can be well described as a
rigid rotor. The Hamiltonian of such a system is simply
proportional to the square of the angular momentum
\begin{align}
H &= \frac{1}{2I}L^2
\end{align}
and the energy eigenvalues are
\begin{align}
E_{\ell m} &= \hbar^2 \frac{\ell(\ell+1)}{2I}
\end{align}

What is the energy of the ground state and the first and
second excited states of the \(H_2\) molecule? i.e. the lowest three distinct energy eigenvalues.

At room temperature, what is the relative probability of
finding a hydrogen molecule in the \(\ell=0\) state versus finding it
in any one of the \(\ell=1\) states? i.e. what is
\(P_{\ell=0,m=0}/\left(P_{\ell=1,m=-1} + P_{\ell=1,m=0} + P_{\ell=1,m=1}\right)\)

At what temperature is the value of this ratio 1?

At room temperature, what is the probability of
finding a hydrogen molecule in any one of the \(\ell=2\) states versus
that of finding it in the ground state? i.e. what is
\(P_{\ell=0,m=0}/\left(P_{\ell=2,m=-2} + P_{\ell=2,m=-1} + \cdots + P_{\ell=2,m=2}\right)\)