Question

Using Bohr's quantisation condition, what is the rotational energy in the second orbit for a diatomic molecule? ($I$ = moment of inertia and $h$ = Planck's constant) ______.

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

Hint:

Bohr's postulate $mvr = \frac{nh}{2\pi}$ isn't just for electrons! It dictates that angular momentum ($L$) is fundamentally quantized for any rotating system in quantum mechanics.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Bohr's quantization condition for angular momentum is $L = \frac{nh}{2\pi}$. Rotational kinetic energy is $E = \frac{L^2}{2I}$.

Step 2: Formula Application:
For the second orbit, $n = 2$.
$L = \frac{2h}{2\pi} = \frac{h}{\pi}$.

Step 3: Explanation:
$E = \frac{(h/\pi)^2}{2I} = \frac{h^2}{2I\pi^2}$.

Step 4: Final Answer:
The rotational energy is $\frac{h^2}{2I\pi^2}$.