MHT CET · Physics · Atomic Physics
Using Bohr's quantization, what is the rotational energy in the \(2^{\text {nd }}\) orbit for a diatomic molecule?
( \(\mathrm{I}\) = moment of inertia of a diatomic molecule, \(\mathrm{h}=\) Planck's constant \()\)
- A \(\frac{h^2}{2 I \pi^2}\)
- B \(\frac{h^2}{I \pi^2}\)
- C \(\frac{h}{2 \pi}\)
- D \(\frac{h}{2 \mathrm{I} \pi^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{h^2}{2 I \pi^2}\)
Step-by-step Solution
Detailed explanation
According to Bohr's quantization condition:
\(\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{\pi}\) [For \(2^{\text {nd }}\) orbit]
The rotational \(\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}=\frac{\mathrm{h}^2}{2 \mathrm{I} \pi^2}\)
\(\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{\pi}\) [For \(2^{\text {nd }}\) orbit]
The rotational \(\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}=\frac{\mathrm{h}^2}{2 \mathrm{I} \pi^2}\)
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