Discussion
The mean distance between the two atoms of HD molecule is \(r\), where H and D denote hydrogen and deuterium, respectively. The mass of the hydrogen atom is \(m_{H}\). The energy difference between two lowest lying rotational states of HD in multiples of \(\hbar^{2}/(m_{H}r^{2})\) is
Answer: A
The energy difference is \(\dfrac{3}{2}\) in units of \(\hbar^{2}/(m_{H}r^{2})\).
Rotational levels are EJ = ℏ2J(J + 1)/(2μr2). With mD = 2mH, the reduced mass of HD is
\[ \begin{aligned} \mu &= \frac{m_{H} \cdot 2m_{H}}{3m_{H}} = \tfrac{2}{3}m_{H} \\ E_{1} - E_{0} &= \frac{2\hbar^{2}}{2\mu r^{2}} = \frac{3}{2}\,\frac{\hbar^{2}}{m_{H}r^{2}} \end{aligned} \]
2/3 is the reduced mass ratio itself, not the energy. 6 and 4/3 come from using J(J + 1) = 6 for the J = 2 level, or from treating the molecule with the full mass instead of the reduced mass.
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