Model the diatomic molecule as a rigid rotator

Two masses $m_1$ and $m_2$ are separated by a fixed distance $r$.

For rotation about the center of mass, the moment of inertia is $I = \mu r^2$ where $\mu$ is the reduced mass: $\mu = \frac{m_1 m_2}{m_1 + m_2}$

Hence, $I = \frac{m_1 m_2}{m_1 + m_2} r^2$

Apply Bohr's angular momentum quantization

Bohr's rule gives $L = \frac{nh}{2\pi}$ where $n$ is an integer.

Write rotational energy in terms of angular momentum

Rotational kinetic energy is $E = \frac{L^2}{2I}$

Substituting $L = \dfrac{nh}{2\pi}$,

Substitute the moment of inertia

$E = \frac{n^2 h^2}{8\pi^2 \left(\dfrac{m_1 m_2}{m_1+m_2} r^2\right)}$

$E = \frac{(m_1+m_2)n^2 h^2}{8\pi^2 m_1 m_2 r^2}$

Compare with the options

The physically correct expression using Bohr quantization is $\boxed{E = \frac{(m_1+m_2)n^2 h^2}{8\pi^2 m_1 m_2 r^2}}$

None of the given options contains the factor $8\pi^2$.

However, if the question setter has used the simplified/incorrect quantization condition $L = nh$ instead of Bohr's actual rule $L = \dfrac{nh}{2\pi}$, then

Final judgment

• Using the stated phrase Bohr's rule of angular momentum quantization, the correct formula should include $8\pi^2$ in the denominator.
• Among the provided options, the intended answer is clearly D.