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.
In a `CO` molecule, the distance between `C (mass = 12 a. m. u ) and O (mass = 16 a.m.u)` where `1 a.m.u = (5)/(3) xx 10^(-27) kg , `is close to

Options

A. a.`2.4 xx 10^(-10) m`
B. b.`1.9 xx 10^(-10) m`
C. c.`1.3 xx 10^(-10) m`
D. d.`4.4 xx 10^(-11) m`

Explanation

To find the distance between the carbon (C) and oxygen (O) atoms in a CO molecule, we will use the concept of the center of mass and the moment of inertia.

### Step-by-Step Solution:

1. **Identify the Masses:**
- Mass of Carbon (C), \( m_1 = 12 \, \text{amu} \)
- Mass of Oxygen (O), \( m_2 = 16 \, \text{amu} \)
- Convert amu to kg: \( 1 \, \text{amu} = \frac{5}{3} \times 10^{-27} \, \text{kg} \)

2. **Calculate the Center of Mass:**
- The center of mass (CM) of the system can be calculated using the formula:
\[ R_{CM} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \]
- We set the coordinate of the center of mass to be at the origin (0,0). Let \( r_1 \) be the distance from C to CM, and \( r_2 \) be the distance from O to CM.
- Since \( R_{CM} = 0 \):
\[ m_1 r_1 = m_2 r_2 \]
- From this, we can express \( r_2 \) in terms of \( r_1 \):
\[ r_2 = \frac{m_1}{m_2} r_1 \]

3. **Express Distances in Terms of Total Distance \( D \):**
- The total distance \( D \) between C and O is given by:
\[ D = r_1 + r_2 \]
- Substituting \( r_2 \):
\[ D = r_1 + \frac{m_1}{m_2} r_1 = r_1 \left(1 + \frac{m_1}{m_2}\right) \]
- Thus, we can write:
\[ D = r_1 \left(1 + \frac{12}{16}\right) = r_1 \left(1 + 0.75\right) = r_1 \cdot 1.75 \]

4. **Calculate the Moment of Inertia:**
- The moment of inertia \( I \) for a diatomic molecule is given by:
\[ I = m_1 r_1^2 + m_2 r_2^2 \]
- Substitute \( r_2 \):
\[ I = m_1 r_1^2 + m_2 \left(\frac{m_1}{m_2} r_1\right)^2 = m_1 r_1^2 + \frac{m_1^2}{m_2} r_1^2 \]
- Factor out \( r_1^2 \):
\[ I = r_1^2 \left(m_1 + \frac{m_1^2}{m_2}\right) \]

5. **Set Up the Equation for Moment of Inertia:**
- Given \( I = 1.87 \times 10^{-46} \, \text{kg m}^2 \):
\[ 1.87 \times 10^{-46} = r_1^2 \left(12 + \frac{12^2}{16}\right) \]
- Calculate the term in parentheses:
\[ 12 + \frac{144}{16} = 12 + 9 = 21 \]
- Thus:
\[ 1.87 \times 10^{-46} = 21 r_1^2 \]

6. **Solve for \( r_1^2 \):**
- Rearranging gives:
\[ r_1^2 = \frac{1.87 \times 10^{-46}}{21} \]

7. **Calculate the Distance \( D \):**
- Finally, substitute \( r_1 \) back into the equation for \( D \):
\[ D = 1.75 r_1 \]
- After calculating \( D \), we find that it is approximately \( 1.3 \times 10^{-10} \, \text{m} \).

### Final Answer:
The distance between the C and O atoms in a CO molecule is approximately \( 1.3 \times 10^{-10} \, \text{m} \).

---