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 Question:
We must calculate the rotational kinetic energy of a molecule in its second allowed quantum orbit ($n=2$), applying Bohr's postulate of angular momentum quantization to a rotating rigid body.

Step 2: Key Formula or Approach:
1. Bohr's Quantization Condition: Angular momentum $L$ is an integral multiple of $\frac{h}{2\pi}$.
$$L = n \frac{h}{2\pi}$$
2. Rotational Kinetic Energy: The relationship between rotational energy $E_r$, angular momentum $L$, and moment of inertia $I$ is analogous to linear kinetic energy ($p^2/2m$):
$$E_r = \frac{L^2}{2I}$$

Step 3: Detailed Explanation:
The problem asks for the energy in the "second orbit", which corresponds to the principal quantum number $n = 2$.
First, find the quantized angular momentum $L$ for $n=2$:
$$L = 2 \left( \frac{h}{2\pi} \right) = \frac{h}{\pi}$$
Now, substitute this angular momentum $L$ into the rotational energy formula:
$$E_r = \frac{L^2}{2I}$$
$$E_r = \frac{\left( \frac{h}{\pi} \right)^2}{2I}$$
$$E_r = \frac{\frac{h^2}{\pi^2}}{2I}$$
$$E_r = \frac{h^2}{2I\pi^2}$$

Step 4: Final Answer:
The rotational energy is $\frac{h^2}{2I\pi^2}$, matching option (b).