What is the distance of closest approach when a `5.0 MeV` proton approaches a gold nucleus?

Explanation

To find the distance of closest approach of a 5.0 MeV proton approaching a gold nucleus, we can use the principle of conservation of energy. Here are the steps to solve the problem:

### Step 1: Convert Kinetic Energy to Joules
The kinetic energy (KE) of the proton is given as 5.0 MeV. We need to convert this energy into joules.
- 1 eV = \(1.6 \times 10^{-19}\) Joules
- Therefore, \(5.0 \text{ MeV} = 5.0 \times 10^6 \text{ eV} = 5.0 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}\)

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