Using Bohr's quantization condition, what is the rotational energy in the second orbit for a diatomic…
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)
$\frac{n^{2}}{2 I^2 \pi^{2}}$
$\frac{h}{2 I \pi^{2}}$
$\frac{h}{2 \mathrm{I}^{2} \pi}$
$\frac{n^{2}}{2 I^{2} \pi^{2}}$
Solution
According to Bohr's quantization condition $\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{\pi} \quad\left[\right.$ For $2^{\text {nd }}$ orbit $]$
Rotational $\mathrm{KE}=\frac{1}{2} \frac{\mathrm{L}^{2}}{\mathrm{I}}=\frac{1}{2} \frac{\mathrm{h}^{2}}{\mathrm{I} \pi^{2}}$