A neutron of energy 1 MeV and mass `1.6 xx 10^(-27)` kg passes a proton at such a distance that the angular momentum of the neutron relative to the proton approximately equals `10^(-33) J s`. The distance of closest approach neglecting the interaction between particles is

Options

A. `0.44nm`
B. `0.44mm`
C. `0.44Å`
D. `44fm`

Explanation

To solve the problem, we need to find the distance of closest approach of a neutron with a proton given that the angular momentum of the neutron relative to the proton is \(10^{-33} \, \text{J s}\).

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \(L\) of the neutron relative to the proton can be expressed as:
\[ L = M V R \]
where:
- \(M\) is the mass of the neutron,
- \(V\) is the velocity of the neutron,
- \(R\) is the distance of closest approach.

2. **Given Values**:
- Mass of neutron \(M = 1.6 \times 10^{-27} \, \text{kg}\)
- Energy of neutron \(E = 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\)
- Angular momentum \(L = 10^{-33} \, \text{J s}\)

3. **Relating Energy to Velocity**:
The kinetic energy \(E\) of the neutron can be expressed in terms of its mass and velocity:
\[ E = \frac{1}{2} M V^2 \]
Rearranging this gives:
\[ V = \sqrt{\frac{2E}{M}} \]

4. **Calculating Velocity**:
Substituting the values of \(E\) and \(M\):
\[ V = \sqrt{\frac{2 \times 1.6 \times 10^{-13}}{1.6 \times 10^{-27}}} \]
\[ V = \sqrt{2 \times 10^{14}} = \sqrt{2} \times 10^{7} \, \text{m/s} \approx 1.41 \times 10^{7} \, \text{m/s} \]

5. **Substituting into Angular Momentum Equation**:
Now, substituting \(V\) back into the angular momentum equation:
\[ L = M V R \]
Rearranging for \(R\):
\[ R = \frac{L}{M V} \]

6. **Calculating Distance of Closest Approach**:
Substituting the known values:
\[ R = \frac{10^{-33}}{(1.6 \times 10^{-27})(1.41 \times 10^{7})} \]
Calculating \(M V\):
\[ M V = 1.6 \times 10^{-27} \times 1.41 \times 10^{7} \approx 2.26 \times 10^{-20} \, \text{kg m/s} \]
Now substituting this back into the equation for \(R\):
\[ R = \frac{10^{-33}}{2.26 \times 10^{-20}} \approx 4.42 \times 10^{-14} \, \text{m} \]

7. **Final Conversion**:
Converting \(R\) to femtometers:
\[ R \approx 44.2 \, \text{fm} \]

### Final Answer:
The distance of closest approach is approximately \(44 \, \text{fm}\) (femtometers).