MHT CET · Physics · Atomic Physics
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 \(]\)
- A \(\frac{1}{\mathrm{n}^2}\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right)\)
- B \(\frac{1}{n}\left(\frac{h^2}{8 \pi^2 \mathrm{I}}\right)\)
- C \(\mathrm{n}\left(\frac{\mathrm{h}^2}{8 \pi^2 \mathrm{I}}\right)\)
- D \(\mathrm{n}^2\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right)\)
Answer & Solution
Correct Answer
(D) \(\mathrm{n}^2\left(\frac{\mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\right)\)
Step-by-step Solution
Detailed explanation
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 \(=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{I}}\right)^2=\frac{\mathrm{n}^2 \mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\)
\(\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\mathrm{I} \omega \Rightarrow \omega=\frac{\mathrm{nh}}{2 \pi \mathrm{I}}\)
Rotational KE \(=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{I}}\right)^2=\frac{\mathrm{n}^2 \mathrm{~h}^2}{8 \pi^2 \mathrm{I}}\)
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