Using Bohr's quantization, what is the rotational energy in the $2^{\text {nd }}$ orbit for a diatomic…
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 $)$
$\frac{h^2}{2 I \pi^2}$
$\frac{h^2}{I \pi^2}$
$\frac{h}{2 \pi}$
$\frac{h}{2 \mathrm{I} \pi^2}$
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 $\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}=\frac{\mathrm{h}^2}{2 \mathrm{I} \pi^2}$