The distance of closest approach of an `alpha`-particle fired at nucleus with momentum p is d. The distence of closest approach when the `alpha`-particle is fired at same nucleus with momentum 3p will be

Options

A. 3d
B. `(d)/(3)`
C. 9d
D. `(d)/(9)`

Explanation

To solve the problem of finding the distance of closest approach \( D_1 \) of an alpha particle fired at a nucleus with momentum \( 3p \), we can use the relationship between kinetic energy and potential energy.

### Step-by-Step Solution:

1. **Understand the Concept**: The distance of closest approach is determined by equating the kinetic energy of the alpha particle to the potential energy at that distance. The potential energy at the closest approach is given by the formula:
\[ U = \frac{k}{d} \]
where \( k \) is a constant related to the charge of the nucleus and \( d \) is the distance of closest approach.

2. **Kinetic Energy for Momentum \( p \)**: The kinetic energy \( K \) of the alpha particle when it has momentum \( p \) is given by:
\[ K = \frac{p^2}{2m} \]
where \( m \) is the mass of the alpha particle.

3. **Set Up the Equation for Momentum \( p \)**: For the initial momentum \( p \) and distance \( D \):
\[ \frac{p^2}{2m} = \frac{k}{D} \]
Rearranging gives:
\[ p^2 = \frac{2k}{D} m \quad \text{(Equation 1)} \]

4. **Kinetic Energy for Momentum \( 3p \)**: Now consider the case when the momentum is \( 3p \):
\[ K' = \frac{(3p)^2}{2m} = \frac{9p^2}{2m} \]
Setting this equal to the potential energy at distance \( D_1 \):
\[ \frac{9p^2}{2m} = \frac{k}{D_1} \]
Rearranging gives:
\[ 9p^2 = \frac{2k}{D_1} m \quad \text{(Equation 2)} \]

5. **Relate the Two Equations**: From Equation 1, we have \( p^2 = \frac{2k}{Dm} \). Substitute this into Equation 2:
\[ 9 \left(\frac{2k}{Dm}\right) = \frac{2k}{D_1} m \]
Simplifying gives:
\[ \frac{18k}{D} = \frac{2k}{D_1} \]

6. **Solve for \( D_1 \)**: Cross-multiplying yields:
\[ 18k D_1 = 2k D \]
Dividing both sides by \( 2k \) (assuming \( k \neq 0 \)):
\[ 9 D_1 = D \]
Thus:
\[ D_1 = \frac{D}{9} \]

### Final Answer:
The distance of closest approach when the alpha particle is fired with momentum \( 3p \) is:
\[ D_1 = \frac{D}{9} \]