Using Bohr's quantisation condition, what is the rotational energy in the second orbit for a diatomic molecule? ( $I=$ moment of inertia of diatomic molecule and $h=$ Planck's constant)

Rotational energy quantization for a diatomic molecule

The angular momentum $L$ is quantized by Bohr's condition: $L = J\frac{h}{2\pi}$, where $J$ is the rotational quantum number.

For the second orbit, $J = 2$, yielding $L = 2\frac{h}{2\pi} = \frac{h}{\pi}$.

Rotational energy $E_R$ relates to angular momentum through $E_R = \frac{1}{2}I\omega^2$ with $L = I\omega$, giving $\omega = \frac{L}{I}$.

Substitution results in $E_R = \frac{1}{2}I\left(\frac{L}{I}\right)^2 = \frac{L^2}{2I}$.

Inserting the quantized $L$ value, $E_R = \frac{\left(\frac{h}{\pi}\right)^2}{2I} = \frac{h^2}{2I\pi^2}$.

This matches option B: $\frac{h^2}{2I\pi^2}$.

The correct answer is $\boxed{\text{B}}$.