A particle of mass m, charge `q gt 0` and initial kinetic energy K is projected from infinity towards a heavy nucleus of charge Q assumed to have a fixed position.
If the aim is perfect, how close to the center of the nucleus is the particle when it comes instantaneously to rest ?

Answer

To solve the problem, we need to apply the principle of conservation of energy. The initial kinetic energy of the particle will be converted into electrostatic potential energy as it approaches the nucleus.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- A particle of mass \( m \) and charge \( q > 0 \) is projected from infinity towards a nucleus of charge \( Q \).
- The initial kinetic energy of the particle is \( K \).

2. **Identify Energy Conservation**:
- At infinity, the potential energy is zero, and the total energy is purely kinetic, given by \( K \).
- As the particle approaches the nucleus, it slows down due to the electrostatic force and will eventually come to rest at some distance \( R \) from the nucleus.

3. **Set Up the Energy Equation**:
- When the particle comes to rest at distance \( R \), all its initial kinetic energy \( K \) is converted into electrostatic potential energy \( U \).
- The electrostatic potential energy \( U \) between two point charges is given by:
\[ U = \frac{1}{4 \pi \epsilon_0} \frac{Qq}{R} \]

4. **Apply Conservation of Energy**:
- According to the conservation of energy:
\[ K = U \]
- Substituting the expression for potential energy:
\[ K = \frac{1}{4 \pi \epsilon_0} \frac{Qq}{R} \]

5. **Rearranging to Find \( R \)**:
- Rearranging the equation to solve for \( R \):
\[ R = \frac{1}{4 \pi \epsilon_0} \frac{Qq}{K} \]

6. **Final Expression**:
- Thus, the distance \( R \) at which the particle comes to rest is given by:
\[ R = \frac{Qq}{4 \pi \epsilon_0 K} \]

### Final Answer:
The closest distance to the center of the nucleus when the particle comes instantaneously to rest is:
\[ R = \frac{Qq}{4 \pi \epsilon_0 K} \]