(a) Alpha particles having kinetic energy of `1.8` MeV each are incident on a thin gold foil , from a large distance. Applying the principle of conversation of energy, find the closest distance of approach of the alpha particle from the gold nucleus. (Atomic number of gold = `79`).

Explanation

To find the closest distance of approach of an alpha particle to a gold nucleus, we can use the principle of conservation of energy. The kinetic energy of the alpha particle will be converted into potential energy due to electrostatic repulsion when it gets close to the nucleus.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Kinetic Energy (KE) of the alpha particle: \( KE = 1.8 \, \text{MeV} = 1.8 \times 10^6 \, \text{eV} \)
- Atomic number of gold (Z): \( Z = 79 \)
- Charge of an electron (e): \( e = 1.6 \times 10^{-19} \, \text{C} \)
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