MHT CET · Physics · Atomic Physics
Using Bohr's quantization condition, what is the rotational energy in the second orbit for a diatomic molecule. (I = moment of inertia of diatomic molecule, \(\mathrm{h}=\) Planck's constant)
- A \(\frac{n^{2}}{2 \mathrm{~T}^{2}}\)
- B \(\frac{h}{2 I \pi^{2}}\)
- C \(\frac{h}{2 \mathrm{I}^{2} \pi}\)
- D \(\frac{n^{2}}{2 I^{2} \pi^{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{n^{2}}{2 \mathrm{~T}^{2}}\)
Step-by-step Solution
Detailed explanation
According to Bohr's quantization condition \(\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{\pi} \quad\left[\right.\) For \(2^{\text {nd }}\) orbit \(]\)
Rotational \(\mathrm{KE}=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}=\frac{1}{2} \frac{\mathrm{h}^{2}}{\mathrm{I} \pi^{2}}\)
Rotational \(\mathrm{KE}=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}=\frac{1}{2} \frac{\mathrm{h}^{2}}{\mathrm{I} \pi^{2}}\)
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