An alpha-particle accelerated through `V` volt is fired towards a nucleus . Its distance of closest approach is . `r` If a protons is accelerated through the same potential and fired towards the same nucleus , the distance of closest approach of proton will be .

Options

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

Explanation

To solve the problem, we need to find the distance of closest approach for a proton when it is accelerated through the same potential \( V \) as an alpha particle.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two charged particles: an alpha particle and a proton. Both are accelerated through a potential difference \( V \) and are fired towards the same nucleus. The distance of closest approach for the alpha particle is given as \( r \).

2. **Kinetic Energy of the Particles**:
When a charged particle is accelerated through a potential \( V \), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy \( K \) can be expressed as:
\[ K = qV \]
where \( q \) is the charge of the particle.

3. **Potential Energy at Closest Approach**:
At the distance of closest approach, all the kinetic energy is converted into potential energy due to the electrostatic force between the charged particle and the nucleus. The potential energy \( U \) at a distance \( r \) is given by:
\[ U = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r} \]
where \( q_1 \) is the charge of the particle (alpha or proton) and \( q_2 \) is the charge of the nucleus.

4. **Equating Kinetic and Potential Energy**:
For the alpha particle, we can write:
\[ q_{\alpha} V = \frac{1}{4\pi \epsilon_0} \frac{q_{\alpha} q_2}{r} \]
Here, \( q_{\alpha} \) is the charge of the alpha particle (which is \( 2e \), where \( e \) is the elementary charge).

5. **Finding the Distance of Closest Approach for the Proton**:
For the proton, we can write a similar equation:
\[ q_{p} V = \frac{1}{4\pi \epsilon_0} \frac{q_{p} q_2}{r_p} \]
where \( q_{p} \) is the charge of the proton (which is \( e \)) and \( r_p \) is the distance of closest approach for the proton.

6. **Relating the Two Distances**:
From the equations for the alpha particle and the proton, we can rearrange them:
\[ r = \frac{1}{4\pi \epsilon_0} \frac{q_{\alpha} q_2}{q_{\alpha} V} = \frac{1}{4\pi \epsilon_0} \frac{q_2}{V/q_{\alpha}} \]
\[ r_p = \frac{1}{4\pi \epsilon_0} \frac{q_{p} q_2}{q_{p} V} = \frac{1}{4\pi \epsilon_0} \frac{q_2}{V/q_{p}} \]

7. **Comparing the Two Distances**:
Since \( q_{\alpha} = 2e \) and \( q_{p} = e \):
\[ r_p = \frac{1}{4\pi \epsilon_0} \frac{q_2}{V/e} = \frac{1}{4\pi \epsilon_0} \frac{q_2 e}{V} \]
The ratio of the distances of closest approach is:
\[ \frac{r_p}{r} = \frac{q_{p}}{q_{\alpha}} = \frac{e}{2e} = \frac{1}{2} \]

8. **Final Result**:
Therefore, the distance of closest approach for the proton is:
\[ r_p = \frac{r}{2} \]

### Conclusion:
The distance of closest approach of the proton is \( \frac{r}{2} \).