Question 418: Atoms & Nuclei

Question

A diatomic molecule has moment of inertia ' I ', By applying Bohr's quantization condition, its rotational energy in the $\mathrm{n}^{	ext {th }}$ level is $[\mathrm{n} \geq 1]$ [h= Planck's constant]

Accepted Answer

Bohr's quantization condition for the rotational motion of a diatomic molecule states that the angular momentum is quantized and given by:
$$ L = n \frac{h}{2\pi}, $$
where $ n $ is the quantum number (an integer), and $ h $ is Planck's constant.
The rotational energy $ E $ of a diatomic molecule in the $ n $-th level can be linked to its moment of inertia $ I $ and angular momentum $ L $ through the following expression:
$$ E = \frac{L^2}{2I}. $$
Substituting the quantized angular momentum into the expression for energy gives:
$$ E = \frac{\left(n \frac{h}{2\pi}\right)^2}{2I} = \frac{n^2 h^2}{8 \pi^2 I}. $$
Thus, the rotational energy of a diatomic molecule in the $ n $-th level is:
$$ \mathrm{n}^2\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right). $$
Therefore, the correct option is
Option D
.

Answer

$\frac{1}{\mathrm{n}^2}\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right)$

Answer

$\frac{1}{\mathrm{n}}\left(\frac{\mathrm{h}^2}{8 \pi^2 \mathrm{I}}\right)$

Answer

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

Answer

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

Atoms & Nuclei

A diatomic molecule has moment of inertia ' I ', By applying Bohr's quantization condition, its rotational energy in the \mathrm{n}^{\text {th }} level is [\mathrm{n} \geq 1] [h= Planck's constant]

Choose the correct answer: