Question

The key features 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 $n^{th}$ level (n = 0 is not allowed) is
(A) $\frac{1}{n^{2} } \frac{h^{2} }{8\pi^{2}I }$(B) $\frac{1}{n} \frac{h^{2} }{8\pi^{2}I }$(C) $n\frac{h^{2} }{8\pi^{2}I }$(D) $n^{2} \frac{h^{2} }{8\pi^{2}I }$

Answer 1

Question:

A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the  level (n = 0 is not allowed) is ?

Solution:

Given,

The key features 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.

We know,

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

And, Rotational kinetic energy = $\dfrac{1}{2}I\omega^2 = \dfrac{1}{2} \dfrac{n^2h^2}{4\pi^2I} = \dfrac{n^2h^2}{8\pi^2I}$

Hence, D is the correct option.