Question

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. $\dfrac {1}{{n}^{2}} (\dfrac {{h}^{2}}{8 {\pi}^{2}I})$
B. $\dfrac {1}{n}(\dfrac {{h}^{2}}{8 {\pi}^{2}I})$
C. $n(\dfrac {{h}^{2}}{8 {\pi}^{2}I})$
D. ${n}^{2}(\dfrac {{h}^{2}}{8 {\pi}^{2}I})$

Answer 1

Hint:To solve this problem, find the angular momentum of the electron in a molecule in the ${n}^{th}$ level. Then, from Bohr’s postulate, find the angular momentum of an orbiting electron in a molecule in ${n}^{th}$ level. Equate both the expressions for angular momentum and find the expression for angular velocity. Then, use the formula for rotational energy of a molecule ${n}^{th}$ level. Substitute the obtained angular velocity in that formula. Solve it and you will get the rotational energy in the ${n}^{th}$ level (n=0 is not allowed).
Formula used:
$ {L}_{n}= I{\omega}_{n}$
${L}_{n} = \dfrac {nh}{2\pi}$
$E = \dfrac {1}{2} I{{\omega}_{n}}^{2}$

Complete answer:
Let ${\omega}_{n}$ be the angular velocity of the molecule in ${n}^{th}$ level
Angular momentum of an orbiting electron in a molecule in ${n}^{th}$ level is given by,
$ {L}_{n}= I{\omega}_{n}$ …(1)
Where, I is the rotational inertia
From Bohr’s postulate we know,
${L}_{n} = \dfrac {nh}{2\pi}$ …(2)
Where, n is the principal quantum number of the molecule
From the equation. (1) and equation. (2) we get,
$ I{\omega}_{n}= \dfrac {nh}{2\pi}$
Rearranging above equation we get,
${\omega}_{n}=\dfrac {nh}{2\pi I}$ …(3)
Rotational energy of a molecule in ${n}^{th}$ level is given by,
$E = \dfrac {1}{2} I{{\omega}_{n}}^{2}$
Substituting equation. (3) in above equation we get,
$E = \dfrac {1}{2} I {\dfrac {nh}{2\pi I}}^{2}$
$\Rightarrow E = \dfrac {1}{2} I \dfrac {{n}^{2}{h}^{2}}{4 {\pi}^{2}{I}^{2}}$
$\Rightarrow E = \dfrac {{n}^{2}{h}^{2}}{8 {\pi}^{2}I}$
Hence, rotational energy in the ${n}^{th}$ level (n=0 is not allowed) is $\dfrac {{n}^{2}{h}^{2}}{8 {\pi}^{2}I}$.

So, the correct answer is “Option D”.

Note:
When an electron moves around a proton, angular momentum is conserved. This postulate is the key feature of Bohr’s quantization. According to Bohr, an electron can revolve only in certain discrete and non-orbiting orbits. Angular momentum of the revolving electron is the integral multiple of $\dfrac {h}{2\pi}$ where h is the Planck’s constant. Bohr’s theory is applicable only to hydrogen and hydrogen-like atoms. It does not give any information about the wave nature of an electron.