An `alpha`-particle having `K.E. =7.7MeV` is scattered by gold `(z=79)` nucleus through `180^(@)` Find distance of closet approach.

Explanation

To find the distance of closest approach of an alpha particle scattered by a gold nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Kinetic Energy (K.E) of the alpha particle = 7.7 MeV
- Atomic number (Z) of gold = 79
- Charge of an electron (e) = \(1.6 \times 10^{-19}\) C
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