The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition. A diatomic molecule has moment of inertia I. By Bohr's quantization condition, its rotational energy in the nth level (n = 0 is not allowed) is

1 \over n^2( h^2 \over 8\pi ^2I )

1 \over n( h^2 \over 8\pi ^2I )

Atoms and Nuclei PYQ — JEE Advanced 2010 · Shift 2 · Q53 with Solution

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Correct answer: D

For rotational motion, Bohr’s quantization says that the angular momentum is quantized:

$L = n\hbar = n\frac{h}{2\pi}, \qquad n=1,2,3,\dots$

Here, $n=0$ is not allowed as stated in the question.

For a rigid diatomic molecule of moment of inertia $I$ ,

$L = I\omega$

So,

$I\omega = n\frac{h}{2\pi}$

which gives

$\omega = \frac{nh}{2\pi I}$

The rotational energy of a rigid body is

$E = \frac{1}{2}I\omega^2$

Substitute the value of $\omega$ :

$E_n = \frac{1}{2}I\left(\frac{nh}{2\pi I}\right)^2$

$E_n = \frac{1}{2}I\cdot \frac{n^2h^2}{4\pi^2 I^2}$

$E_n = \frac{n^2h^2}{8\pi^2 I}$

Thus,

$E_n = n^2\left(\frac{h^2}{8\pi^2 I}\right)$

Therefore, the correct option is D.