An `alpha`-particle accelerated through V volt is fired towards a nucleus . Its distance of closest approach is r. If a proton acceleration throught the same potential is fired towards the same nucleus, the distance of closest approach of proton will be xr. Find value of x.

Answer

To solve the problem, we need to analyze the concept of the distance of closest approach for charged particles when they are fired towards a nucleus.

### Step-by-Step Solution:

1. **Understanding the Distance of Closest Approach**:
The distance of closest approach (r) for a charged particle approaching a nucleus can be derived from the conservation of energy. The kinetic energy of the particle is converted into electric potential energy at the closest approach.

2. **Kinetic Energy of the Alpha Particle**:
The kinetic energy (KE) of an alpha particle accelerated through a potential V is given by:
\[ KE_{\alpha} = q_{\alpha} \cdot V \]
where \( q_{\alpha} \) is the charge of the alpha particle. An alpha particle consists of 2 protons and 2 neutrons, so its charge is \( q_{\alpha} = 2e \) (where \( e \) is the elementary charge).

3. **Potential Energy at Closest Approach**:
The potential energy (PE) at the distance of closest approach (r) is given by:
\[ PE = \frac{q_{\alpha} \cdot q_{nucleus}}{4 \pi \epsilon_0 r} \]
where \( q_{nucleus} \) is the charge of the nucleus.

4. **Setting Kinetic Energy Equal to Potential Energy**:
At the distance of closest approach, we set the kinetic energy equal to the potential energy:
\[ q_{\alpha} \cdot V = \frac{q_{\alpha} \cdot q_{nucleus}}{4 \pi \epsilon_0 r} \]
Simplifying this, we find:
\[ r = \frac{q_{nucleus}}{4 \pi \epsilon_0 V} \]

5. **Kinetic Energy of the Proton**:
Similarly, for a proton accelerated through the same potential V, the kinetic energy is:
\[ KE_{p} = q_{p} \cdot V \]
where \( q_{p} = e \) (the charge of the proton).

6. **Potential Energy for the Proton**:
The potential energy at the distance of closest approach (let's call it \( xr \)) is:
\[ PE = \frac{q_{p} \cdot q_{nucleus}}{4 \pi \epsilon_0 (xr)} \]

7. **Setting Kinetic Energy Equal to Potential Energy for Proton**:
Setting the kinetic energy equal to the potential energy for the proton gives:
\[ e \cdot V = \frac{e \cdot q_{nucleus}}{4 \pi \epsilon_0 (xr)} \]
Simplifying this, we find:
\[ xr = \frac{q_{nucleus}}{4 \pi \epsilon_0 V} \]

8. **Finding the Value of x**:
From the equations for r and xr, we can equate them:
\[ r = \frac{q_{nucleus}}{4 \pi \epsilon_0 V} \quad \text{and} \quad xr = \frac{q_{nucleus}}{4 \pi \epsilon_0 V} \]
This implies:
\[ x = 1 \]

### Final Answer:
The value of \( x \) is \( 1 \).