A diatomic molecule has moment of inertia I. | vedclass

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(D) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2 \pi}$.Since $L = I \omega$,we have $I \omega = \f

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$n^2\left(\frac{h^2}{8 \pi^2 I}\right)$

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(D) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2 \pi}$.Since $L = I \omega$,we have $I \omega = \frac{nh}{2 \pi}$,which implies $\omega = \frac{nh}{2 \pi I}$.The rotational kinetic energy $E_r$ is given by $E_r = \frac{1}{2} I \omega^2$.Substituting the value of $\omega$,we get $E_r = \frac{1}{2} I \left(\frac{nh}{2 \pi I}\right)^2$.Simplifying this expression,$E_r = \frac{1}{2} I \left(\frac{n^2 h^2}{4 \pi^2 I^2}\right) = \frac{n^2 h^2}{8 \pi^2 I}$.

$A$ diatomic molecule has moment of inertia $I$. By applying Bohr's quantization condition,its rotational energy in the $n^{\text{th}}$ level is $[n \geq 1]$ $(h = \text{Planck's constant})$

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