The distance oif 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. `2r`
B. `4r`
C. `r//2`
D. `r//4`

Explanation

To solve the problem, we need to understand the relationship between the distance of closest approach (r) of an alpha particle to a nucleus and its momentum (p).

### Step-by-Step Solution:

1. **Understanding the Concept of Closest Approach**:
The distance of closest approach occurs when the kinetic energy of the alpha particle is completely converted into potential energy due to the electrostatic force between the alpha particle and the nucleus.

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

3. **Potential Energy at Closest Approach**:
The potential energy (PE) at the distance of closest approach (r) is given by the formula:
\[ PE = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} \]
where \( q_1 \) and \( q_2 \) are the charges of the alpha particle and the nucleus, respectively.

4. **Setting Kinetic Energy Equal to Potential Energy**:
At the closest approach, we set the kinetic energy equal to the potential energy:
\[ \frac{p^2}{2m} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} \]

5. **Finding the Relationship Between r and p**:
Rearranging the equation gives us:
\[ r = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2 \cdot 2m}{p^2} \]
This shows that the distance of closest approach (r) is inversely proportional to the square of the momentum (p):
\[ r \propto \frac{1}{p^2} \]

6. **Calculating the New Distance of Closest Approach**:
If the momentum of the alpha particle is doubled (i.e., \( p \) becomes \( 2p \)), we can substitute this into our relationship:
\[ r' = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2 \cdot 2m}{(2p)^2} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2 \cdot 2m}{4p^2} = \frac{1}{2} \cdot \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2 \cdot 2m}{p^2} \]
Thus, we find that:
\[ r' = \frac{r}{4} \]

### Conclusion:
When the momentum of the alpha particle is doubled, the distance of closest approach becomes \( \frac{r}{4} \).