Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( h = Planck's constant, I = moment of inertia of diatomic molecule)

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According to Bohr's quantization condition, the angular momentum $L$ of the molecule is an integral multiple of $\frac{h}{2 π}$ .

$L = \frac{nh}{2 π}$

For the third orbit, $n = 3$ .

$L = \frac{3 h}{2 π}$

The rotational kinetic energy is given by:

$E = \frac{L ^{2}}{2 I}$

Substituting the value of $L$ :

$E = \frac{1}{2 I} (\frac{3 h}{2 π})^{2}$

$E = \frac{9 h ^{2}}{8 π ^{2} I}$

Answer: $\frac{9 h ^{2}}{8 π ^{2} I}$

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