Moment of Inertia of CO Molecule from Rotational Excitation Frequency

To find the moment of inertia of the $\text{CO}$ molecule, we apply Bohr's quantization condition to the rotational motion of the molecule.

According to Bohr's postulate, the angular momentum $L$ of a rotating system is quantized as:
$L = n\hbar = \frac{nh}{2\pi}, \quad \text{where } n = 1, 2, 3, \dots$

The rotational kinetic energy of the molecule with moment of inertia $I$ about its center of mass is given by:
$E_n = \frac{L^2}{2I} = \frac{n^2\hbar^2}{2I}$

The ground state corresponds to the quantum number $n = 1$:
$E_1 = \frac{\hbar^2}{2I}$

The first excited state of rotation corresponds to $n = 2$:
$E_2 = \frac{4\hbar^2}{2I} = \frac{2\hbar^2}{I}$

The excitation energy $\Delta E$ required for the transition from the ground state to the first excited state is:
$\Delta E = E_2 - E_1 = \frac{3\hbar^2}{2I}$

Using Planck's relation $\Delta E = h\nu$, where $\nu$ is the excitation frequency:
$h\nu = \frac{3\hbar^2}{2I}$

Since $\hbar = \frac{h}{2\pi}$, this can be written as:
$h\nu = \frac{3\left(\frac{h}{2\pi}\right)^2}{2I} = \frac{3h^2}{8\pi^2 I}$

Solving for the moment of inertia $I$:
$I = \frac{3h}{8\pi^2 \nu}$

Substitute the given values:

• $h = 2\pi \times 10^{-34}\text{ J}\cdot\text{s}$
• $\nu = \frac{4}{\pi} \times 10^{11}\text{ Hz}$

$I = \frac{3(2\pi \times 10^{-34})}{8\pi^2 \left(\frac{4}{\pi} \times 10^{11}\right)} = \frac{6\pi \times 10^{-34}}{32\pi \times 10^{11}} = \frac{3}{16} \times 10^{-45}\text{ kg}\cdot\text{m}^2$

Evaluating the fraction:
$I = 0.1875 \times 10^{-45}\text{ kg}\cdot\text{m}^2 \approx 1.87 \times 10^{-46}\text{ kg}\cdot\text{m}^2$

Therefore, the correct option is B.