Question:

There are hydrogen molecules contained 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

\[ \frac{-\hbar^2}{2I}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right]\psi=E\psi  \]

where \(I\) is the moment of inertia of a hydrogen molecule, which is \(\mu r^2\). The rotational energy level is

\[  E_j = \frac{\hbar^2}{2\mu r^2}j(j+1)  \]

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 the Schroedinger equation. 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

\[  P = \exp\left[-\frac{E_1}{kT}\right] = \exp\left[-\frac{1.18287\times 10^{-21}}{4.14192 \times 10^{-21}}\right] = 0.752 = 75.2 \%  \]