Interatomic Distance in CO Molecule from Reduced Mass and Inertia

To determine the interatomic distance in the $\text{CO}$ molecule, we use the result for the moment of inertia from the rotational excitation of the molecule.

Step 1: Moment of Inertia of the CO Molecule

From the preceding problem, the moment of inertia of the $\text{CO}$ molecule about its center of mass is:
$I \approx 1.87 \times 10^{-46}\text{ kg}\cdot\text{m}^2$

Step 2: Reduced Mass of the Diatomic Molecule

The $\text{CO}$ molecule consists of a carbon atom of mass $m_{\text{C}} = 12\text{ a.m.u.}$ and an oxygen atom of mass $m_{\text{O}} = 16\text{ a.m.u.}$, where:
$1\text{ a.m.u.} = \frac{5}{3} \times 10^{-27}\text{ kg}$

The reduced mass $\mu$ of the system is given by:
$\mu = \frac{m_{\text{C}} m_{\text{O}}}{m_{\text{C}} + m_{\text{O}}}$

Substituting the given values:
$\mu = \frac{12 \times 16}{12 + 16}\text{ a.m.u.} = \frac{192}{28}\text{ a.m.u.} = \frac{48}{7}\text{ a.m.u.}$

Converting $\mu$ to kilograms:
$\mu = \left(\frac{48}{7}\right) \times \left(\frac{5}{3} \times 10^{-27}\right)\text{ kg} = \frac{16 \times 5}{7} \times 10^{-27}\text{ kg} = \frac{80}{7} \times 10^{-27}\text{ kg} \approx 1.143 \times 10^{-26}\text{ kg}$

Step 3: Calculation of Interatomic Distance

The moment of inertia of a diatomic molecule about its center of mass is related to the bond length (interatomic distance) $d$ by:
$I = \mu d^2$

Rearranging for $d$:
$d = \sqrt{\frac{I}{\mu}}$

Substituting the values of $I$ and $\mu$:
$d^2 = \frac{1.87 \times 10^{-46}}{\frac{80}{7} \times 10^{-27}} = \frac{1.87 \times 7}{80} \times 10^{-19}\text{ m}^2$
$d^2 = \frac{13.09}{80} \times 10^{-19}\text{ m}^2 \approx 0.1636 \times 10^{-19}\text{ m}^2 = 1.636 \times 10^{-20}\text{ m}^2$

Taking the square root:
$d = \sqrt{1.636 \times 10^{-20}}\text{ m} \approx 1.28 \times 10^{-10}\text{ m} \approx 1.3 \times 10^{-10}\text{ m}$

Therefore, the distance between the carbon and oxygen atoms is close to $1.3\times 10^{-10}\text{ m}$.

Correct Option: C