The distance of closest approach of an alpha-particle fired towards a nucleus with momentum p is r. What will be the distance of closest approach when the momentum of alpha-particle is `2p`?

Options

A. r/2
B. 2r
C. 4r
D. r/4

Explanation

To solve the problem, we need to determine the distance of closest approach \( r' \) when the momentum of the alpha particle is doubled from \( p \) to \( 2p \).

### Step-by-Step Solution:

1. **Understanding the Distance of Closest Approach**:
The distance of closest approach \( r \) for an alpha particle with momentum \( p \) can be expressed in terms of the charges involved and the kinetic energy. The formula for the distance of closest approach is given by:
\[ r = \frac{2Z e^2}{k} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

2. **Kinetic Energy and Momentum Relation**:
The kinetic energy \( K \) of the alpha particle can be expressed in terms of momentum \( p \):
\[ K = \frac{p^2}{2m} \]
where \( m \) is the mass of the alpha particle.

3. **Relating Kinetic Energy to Distance of Closest Approach**:
At the distance of closest approach, the kinetic energy of the alpha particle is equal to the potential energy due to electrostatic interaction:
\[ \frac{p^2}{2m} = \frac{2Z e^2}{r} \]
Rearranging gives:
\[ r = \frac{2Z e^2}{\frac{p^2}{2m}} = \frac{4Z e^2 m}{p^2} \]

4. **Finding the New Distance of Closest Approach**:
When the momentum is doubled to \( 2p \), we can find the new distance of closest approach \( r' \):
\[ r' = \frac{4Z e^2 m}{(2p)^2} = \frac{4Z e^2 m}{4p^2} = \frac{Z e^2 m}{p^2} \]

5. **Relating the New Distance to the Original Distance**:
Now we can relate \( r' \) to \( r \):
\[ r' = \frac{1}{4} r \]

### Conclusion:
Thus, the distance of closest approach when the momentum of the alpha particle is \( 2p \) is:
\[ r' = \frac{r}{4} \]