If an `alpha`-particle is incident directly towards a nucleus then it will stop when the distance between them is the distance of closest approach. Which of these is correct?

Options

A. Its kinetic energy is converted into released radiation
B. When its kinetic energy becomes zero then it converts into electrostatic potential energy
C. At distance of closest approach the potential energy of the `alpha`-particle becomes zero
D. A `beta`-particle will show the same behaviour as an `alpha`-particle

Explanation

To solve the problem regarding the interaction between an alpha particle and a nucleus, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Interaction**:
- An alpha particle is a helium nucleus, which carries a charge of +2e (where e is the elementary charge).
- A nucleus also has a positive charge, typically represented as +Ze, where Z is the atomic number of the nucleus.

2. **Identifying Forces at Play**:
- When an alpha particle approaches a nucleus, there is a repulsive electrostatic force between the positively charged alpha particle and the positively charged nucleus.

3. **Distance of Closest Approach**:
- The distance of closest approach (r) is the point at which the alpha particle stops moving towards the nucleus due to the repulsive force overcoming its kinetic energy.

4. **Conservation of Energy**:
- According to the law of conservation of energy, the kinetic energy (KE) of the alpha particle is converted into electrostatic potential energy (PE) at the distance of closest approach.
- The kinetic energy of the alpha particle can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
- The electrostatic potential energy at the distance of closest approach is given by:
\[ PE = \frac{k \cdot Z \cdot 2e^2}{r} \]
where \( k \) is Coulomb's constant.

5. **Setting Up the Equation**:
- At the distance of closest approach, we set the kinetic energy equal to the potential energy:
\[ \frac{1}{2} mv^2 = \frac{k \cdot Z \cdot 2e^2}{r} \]

6. **Solving for the Distance of Closest Approach (r)**:
- Rearranging the equation to solve for r gives:
\[ r = \frac{2kZe^2}{KE} \]
- This equation allows us to calculate the distance of closest approach based on the kinetic energy of the alpha particle and the charge of the nucleus.

7. **Evaluating the Options**:
- Based on the understanding of the interaction and the conservation of energy, we can evaluate the provided options:
- **Option 1**: Incorrect - Kinetic energy is not converted into released radiation.
- **Option 2**: Correct - Kinetic energy is converted into electrostatic potential energy.
- **Option 3**: Incorrect - Potential energy is not zero at the distance of closest approach.
- **Option 4**: Incorrect - Beta particles behave differently due to their different charge.

### Final Answer:
The correct statement is **Option 2**: The kinetic energy of the alpha particle is converted into electrostatic potential energy.