Hydrogen molecules: Boltzmann factor

Question:
There are hydrogen molecules in a closed system by contacting a heat bath at $T=300$ K. Find the ratio of hydrogen molecules in the first rotational energy level relative to the ground state. (The molecular distance is given as $r=1.06$ angstroms.)

Answer:
This is a canonical ensemble in the sense of statistical mechanics. This gives a probability to each distinct state in terms of the Boltzmann constant.
\[
P=e^{-\frac{E_i}{kT}}
\]
The rotational part of the Schroedinger equation is
\begin{equation}
\frac{-\hbar^2}{2I}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right]\psi=E\psi
\end{equation}
where $I$ is the moment of inertia, $\mu r^2$. The rotational energy level is
\begin{equation}
E_j = \frac{\hbar^2}{2\mu r^2}j(j+1)
\end{equation}
The hydrogen atom has a proton and an electron. Thus, we have the mass.
\[
m_H = 1.67262\times 10^{-27} + 9.10938\times 10^{-31} = 1.67353\times 10^{-27} \ \mathrm{kg}
\]
The reduced mass for the molecule becomes
\[
\mu = \frac{m_{H1} m_{H2}}{m_{H1}+m_{H2}}=\frac{m_H}{2}=8.36765\times 10^{-28} \ \mathrm{kg}
\]
because $m_{H1}=m_{H2}$. The relative distance between two molecules is $1.06 \times 10^{-10}$ m. The ground state of energy is obviously $E_0 = 0$ from (1). Then, calculate the first level.
\[
E_1=\frac{\hbar^2}{2\mu r^2}2=\frac{(1.05457\times 10^{-34})^2}{8.36765\times 10^{-28} (1.06 \times 10^{-10})^2}=1.18287\times 10^{-21} \ \mathrm{J}
\]
Now, calculate the following:
\[
kT = 1.38064 \times 10^{-23} \times 300 = 4.14192 \times 10^{-21} \ \mathrm{J}
\]
Therefore, the probability is
\[
P = \exp\left[-\frac{E_1}{kT}\right] = \exp\left[-\frac{1.18287\times 10^{-21}}{4.14192 \times 10^{-21}}\right] = 0.752 = 75.2 \%
\]

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