In the original experiment, Geiger and Marsden calculated the distance of closest approach to the gold nucleus (Z=79)- of a 7.7MeV `alpha` particle before it comes momentarily to rest and reverses its direction. What is its value?

Explanation

To find the distance of closest approach of a 7.7 MeV alpha particle to a gold nucleus (Z=79), we can use the formula derived from the conservation of energy and Coulomb's law. Here’s a step-by-step solution:

### Step 1: Convert the energy of the alpha particle from MeV to Joules
The energy of the alpha particle is given as 7.7 MeV. We need to convert this energy into Joules using the conversion factor \(1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules}\).

\[ E = 7.7 \, \text{MeV} = 7.7 \times 1.6 \times 10^{-13} \, \text{J} = 1.232 \times 10^{-12} \, \text{J} \]
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