Question

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

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

Answer 1

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

$\begin{array}{c}\text{Given - * A Diatomic molecule-}\\ \text{let, the structure of the molecule be like-}\end{array}$

$\Rightarrow I=2m{r}^2$

$\begin{array}{c}\text{we howe Rotational encrgy-}\\ E=\frac{1}{2}I{\omega }^2=m{\omega }^2{r}^2-\left(1\right)\\ \text{By Bohr quantization of angular momentum}\\ 2{\text{mwr}}^2=\frac{nh}{2\pi }-\left(2\right)\\ \text{From}eq\left(1\right)\text{and (2) we get ,}\\ E=\frac{n^2{h}^2}{16{\pi }^{2}m{r}^2}=\frac{n^2{h}^2}{8{\pi }^{2}I}\end{array}$