An alpha-particle accelerated through V volt is fired towards a nucleus . Lts distance of closest approach is r.lf a proton accelerated through the same potential is fired towards the same nucleus, then distance of closest approach of proton will be

Options

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

Explanation

To solve the problem of finding the distance of closest approach for a proton when it is accelerated through the same potential as an alpha particle, we can follow these steps:

### Step 1: Understand the concept of closest approach
The distance of closest approach occurs when the kinetic energy of the incoming particle is completely converted into potential energy due to the electrostatic force between the charged particle and the nucleus.

### Step 2: Write the expression for kinetic energy
For an alpha particle (which has a charge of +2e, where e is the charge of a proton), when it is accelerated through a potential V, its kinetic energy (KE) can be expressed as:
\[ KE_{\alpha} = Q_{\alpha} \cdot V = 2e \cdot V \]

### Step 3: Write the expression for potential energy
The potential energy (PE) at the distance of closest approach (R) is given by the formula:
\[ PE = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_{\alpha} \cdot Q_{nucleus}}{R} \]
Where \( Q_{nucleus} \) is the charge of the nucleus.

### Step 4: Set kinetic energy equal to potential energy
At the distance of closest approach, we set the kinetic energy equal to the potential energy:
\[ 2e \cdot V = \frac{1}{4\pi \epsilon_0} \cdot \frac{2e \cdot Q_{nucleus}}{R} \]

### Step 5: Solve for R
From the above equation, we can rearrange to find R:
\[ R = \frac{1}{4\pi \epsilon_0} \cdot \frac{2e \cdot Q_{nucleus}}{2e \cdot V} \]
This simplifies to:
\[ R = \frac{Q_{nucleus}}{4\pi \epsilon_0 \cdot V} \]

### Step 6: Repeat for the proton
For a proton (which has a charge of +e), the kinetic energy when accelerated through the same potential V is:
\[ KE_{p} = e \cdot V \]
And the potential energy at the distance of closest approach (let's denote it as \( R' \)) is:
\[ PE = \frac{1}{4\pi \epsilon_0} \cdot \frac{e \cdot Q_{nucleus}}{R'} \]

### Step 7: Set kinetic energy equal to potential energy for the proton
Setting the kinetic energy equal to the potential energy for the proton:
\[ e \cdot V = \frac{1}{4\pi \epsilon_0} \cdot \frac{e \cdot Q_{nucleus}}{R'} \]

### Step 8: Solve for R'
Rearranging gives:
\[ R' = \frac{Q_{nucleus}}{4\pi \epsilon_0 \cdot V} \]
Notice that this expression for \( R' \) is exactly half of the expression for R:
\[ R' = \frac{R}{2} \]

### Conclusion
Thus, the distance of closest approach for the proton is:
\[ R' = \frac{R}{2} \]

### Final Answer
The distance of closest approach for the proton is \( \frac{R}{2} \).

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