A proton of mass `m` moving with a speed `v_(0)` apporoches a stationary proton that is free to move. Assuming impact parameter to be zero., i.e., head-on collision. How close will be incident proton go to other proton ?

Options

A. `(e^(3))/(pi epsilon_(0) m^(2) v_(0))`
B. `(e^(3))/(pi epsilon_(0) m v_(0))`
C. `(e^(2))/(pi epsilon_(0) m v_(0)^(2))`
D. None of these

Explanation

To solve the problem of how close a moving proton can get to a stationary proton during a head-on collision, we can use the principles of conservation of energy and conservation of momentum. Here’s a step-by-step solution:

### Step 1: Understand the System
We have two protons:
- Proton 1 (moving) with mass \( m \) and initial speed \( v_0 \).
- Proton 2 (stationary) with mass \( m \) and initial speed \( 0 \).

Since the impact parameter is zero, we can treat this as a head-on collision.

### Step 2: Conservation of Momentum
Before the collision, the total momentum of the system is:
\[ p_{\text{initial}} = mv_0 + 0 = mv_0 \]
After the collision, let \( V \) be the common velocity of both protons. Thus, the total momentum after the collision is:
\[ p_{\text{final}} = (m + m)V = 2mV \]
By conservation of momentum:
\[ mv_0 = 2mV \implies V = \frac{v_0}{2} \]

### Step 3: Conservation of Energy
The initial kinetic energy of the system is:
\[ KE_{\text{initial}} = \frac{1}{2}mv_0^2 \]
At the closest approach, the kinetic energy will be converted into potential energy due to the electrostatic repulsion between the two protons. The potential energy \( U \) at the closest distance \( R \) is given by:
\[ U = \frac{k e^2}{R} \]
where \( k \) is Coulomb's constant and \( e \) is the charge of a proton.

At the closest approach, the total energy is the sum of kinetic energy and potential energy:
\[ KE_{\text{final}} + U = \frac{1}{2}mv^2 + \frac{k e^2}{R} \]
Substituting \( V = \frac{v_0}{2} \):
\[ KE_{\text{final}} = 2 \cdot \frac{1}{2}m\left(\frac{v_0}{2}\right)^2 = \frac{1}{2}m\frac{v_0^2}{4} = \frac{mv_0^2}{8} \]

### Step 4: Set Up the Energy Conservation Equation
Equating initial and final energies:
\[ \frac{1}{2}mv_0^2 = \frac{mv_0^2}{8} + \frac{k e^2}{R} \]
Rearranging gives:
\[ \frac{1}{2}mv_0^2 - \frac{mv_0^2}{8} = \frac{k e^2}{R} \]
\[ \frac{4mv_0^2}{8} - \frac{mv_0^2}{8} = \frac{k e^2}{R} \]
\[ \frac{3mv_0^2}{8} = \frac{k e^2}{R} \]

### Step 5: Solve for \( R \)
Rearranging for \( R \):
\[ R = \frac{8k e^2}{3mv_0^2} \]

### Final Answer
The closest distance \( R \) that the incident proton can approach the stationary proton is:
\[ R = \frac{8k e^2}{3mv_0^2} \]