A diatomic molecule has moment of inertia 'I'. By applying Bohr's quantization condition, its rotational…

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 $]$
  1. $\frac{1}{\mathrm{n}^2}\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right)$
  2. $\frac{1}{n}\left(\frac{h^2}{8 \pi^2 \mathrm{I}}\right)$
  3. $\quad \mathrm{n}\left(\frac{\mathrm{h}^2}{8 \pi^2 \mathrm{I}}\right)$
  4. $\quad \mathrm{n}^2\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right)$

Solution

According to Bohr's quantization condition $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\mathrm{I} \omega \Rightarrow \omega=\frac{\mathrm{nh}}{2 \pi \mathrm{I}}$ Rotational KE $=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{I}}\right)^2=\frac{\mathrm{n}^2 \mathrm{~h}^2}{8 \pi^2 \mathrm{I}}$

Asked in: MHT CET 2024 (03 May Shift 2)

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