An alpha particle with kinetic energy `K_(i)=5.30` MeV happens, by chance,to be headed directly toward he nucleus of a neutral gold atom (fig). What is its distance of closest approach d (least center to center separation) to the nucleus? Assume that the atom remains stationary.

Explanation

To find the distance of closest approach \( d \) for an alpha particle with kinetic energy \( K_i = 5.30 \) MeV heading towards 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 \( K_i \) is given in MeV. We need to convert it to joules using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).

\[ K_i = 5.30 \, \text{MeV} = 5.30 \times 10^6 \, \text{eV} = 5.30 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} \]
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