Q103 / 150 · MHT CET 2022 (07 Aug Shift 1) · 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 $)$

  • Correct answer
    A
    $\frac{h^2}{2 I \pi^2}$
  • B
    $\frac{h^2}{I \pi^2}$
  • C
    $\frac{h}{2 \pi}$
  • D
    $\frac{h}{2 \mathrm{I} \pi^2}$

Solution

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 $\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}=\frac{\mathrm{h}^2}{2 \mathrm{I} \pi^2}$

Hence, the correct answer is option: 'A'.

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