1. Quantization of rotational motion

For a rigid diatomic molecule, apply Bohr's quantization condition to angular momentum:

$L = n\frac{h}{2\pi} = n\hbar$

For rotational motion,

$L = I\omega$

So,

$I\omega = n\hbar$

The rotational kinetic energy is

$E = \frac{1}{2}I\omega^2$

Using $\omega = \dfrac{n\hbar}{I}$,

$E_n = \frac{1}{2}I\left(\frac{n\hbar}{I}\right)^2 = \frac{n^2\hbar^2}{2I}$

Since $\hbar = \dfrac{h}{2\pi}$,

$E_n = \frac{n^2 h^2}{8\pi^2 I}$

2. Energy difference between ground and first excited state

Ground state corresponds to $n=1$ and first excited state to $n=2$.

Thus,

$E_1 = \frac{h^2}{8\pi^2 I}$

$E_2 = \frac{4h^2}{8\pi^2 I}$

So the excitation energy is

$\Delta E = E_2 - E_1 = \frac{3h^2}{8\pi^2 I}$

This corresponds to absorption frequency $\nu$:

$\Delta E = h\nu$

Hence,

$h\nu = \frac{3h^2}{8\pi^2 I}$

$I = \frac{3h}{8\pi^2 \nu}$

3. Substitute the given values

Given:

$\nu = \frac{4}{\pi}\times 10^{11}\ \text{Hz}$

and

$h = 2\pi \times 10^{-34}\ \text{J s}$

Now,

$I = \frac{3(2\pi\times 10^{-34})}{8\pi^2 \left(\frac{4}{\pi}\times 10^{11}\right)}$

Simplify denominator:

$8\pi^2 \cdot \frac{4}{\pi} \times 10^{11} = 32\pi \times 10^{11}$

Therefore,

$I = \frac{6\pi\times 10^{-34}}{32\pi\times 10^{11}}$

$I = \frac{6}{32}\times 10^{-45}$

$I = 0.1875\times 10^{-45}$

$I = 1.875\times 10^{-46}\ \text{kg m}^2$

4. Match with options

$I \approx 1.87\times 10^{-46}\ \text{kg m}^2$

So the correct option is:

B

5. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.