Question

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, \( h \) = Planck’s constant)

• A. \( \frac{h^2}{2I^2 \pi^2} \)
• B. \( \frac{h}{2I^2 \pi} \)
• C. \( \frac{h}{2I^2 \pi^2} \)
• D. \( \frac{h^2}{2I^2 \pi} \)

Hint:

Bohr’s quantization condition for rotational energy is used to determine the energy levels in rotational spectra for molecules.

Answer 1

The Correct Option is A

Step 1: Bohr’s quantization condition.
Bohr's quantization condition for rotational energy gives: \[ E = \frac{n^2 h^2}{8 \pi^2 I} \] where \( n \) is the quantum number, \( h \) is Planck’s constant, and \( I \) is the moment of inertia.
Step 2: Energy in second orbit.
For the second orbit, \( n = 2 \). Substituting \( n = 2 \) into the equation gives the rotational energy in the second orbit as: \[ E = \frac{h^2}{2I^2 \pi^2} \]
Step 3: Conclusion.
Thus, the correct answer is (A) \( \frac{h^2}{2I^2 \pi^2} \).