Question

a diatomic has a moment of inertia.By Bohr's quantization condition its rotational energy in the $n^{th}$ level ( n= 0 is not allowed ) is :

• A. $\frac{1}{n^2}\left(\frac{h^2}{8{\pi }^{2}I}\right)$
• B. $\frac{1}{n}\left(\frac{h^2}{8{\pi }^{2}I}\right)$
• C. $n\left(\frac{h^2}{8{\pi }^{2}I}\right)$
• D. $n^2\left(\frac{h^2}{8{\pi }^{2}I}\right)$

Answer 1

The correct option is D $n^2\left(\frac{h^2}{8{\pi }^{2}I}\right)$

$\begin{array}{c}V=R\omega \text{,}\omega \text{is Angular velocity.}\\ \text{Moment of inertia,}I=m{R}^2\\ \text{Bohr's Angular momentum,}\\ \begin{array}{c}mVR=\frac{nh}{2\pi }\\ m\omega R^2=\frac{nh}{2\pi }\end{array}\end{array}$

$\begin{array}{c}w=\frac{nh}{2\pi m{R}^2}\\ \text{Rotational energy}\\ \begin{array}{cc}R\cdot E& =\frac{1}{2}I{\omega }^2\\ & =\frac{1}{2}m{R}^2\times \frac{n^2{h}^2}{4{\pi }^2{\left(m{R}^2\right)}^2}\\ R\cdot E& =n^2\left(\frac{n^2}{8{\pi }^{2}m{Q}^2}\right)\\ \therefore & R\cdot E=n^2\left(\frac{h^2}{8{\pi }^{2}I}\right)\end{array}\end{array}$