The key feature of Bohr's spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid.The rule to be applied is Bohr's quantization condition.
it is found that the excitation from ground to the first excited state of rotation for the `CO` molecule is close to `(4)/(pi) xx 10^(11) Hz` then the moment of inertia of `CO` molecule about its center of mass is close to
`(Take h = 2 pi xx 10^(-34) J s )`

Options

A. `2.76 xx 10^(-46) kg m^(2)`
B. `1.87 xx 10^(-46) kg m^(2)`
C. `4.67 xx 10^(-47) kg m^(2)`
D. `1.17 xx 10^(-47) kg m^(2)`

Explanation

To find the moment of inertia of the CO molecule about its center of mass, we will follow these steps:

### Step 1: Understand the Problem
We are given that the excitation from the ground state to the first excited state of the CO molecule corresponds to a frequency of \( \nu = \frac{4}{\pi} \times 10^{11} \, \text{Hz} \). We need to apply Bohr's quantization condition to find the moment of inertia \( I \).

### Step 2: Use Bohr's Quantization Condition
According to Bohr's quantization condition for rotational motion, the angular momentum \( L \) is quantized and given by:
\[
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