find distance of closest approach when 5Mev proton approaches a gold nucleus(Z=79)?

the answer is 2.3X10^-14.........but im not able to get this answer

Question

find distance of closest approach when 5Mev proton approaches a
gold nucleus(Z=79)?
the answer is 2 3X10^-14 but im not able to get this answer

Vijay Yadav · Accepted Answer

At the distance of closest approach

Potential Energy = total KE

$= > k \frac{q_{1} q_{2}}{r} = K E$

Given that,

KE = 5 Mev = 5*10⁶ eV

q₁ = e

q₂ = 79e

k = 9*10⁹ Nm²/C²

$r = \frac{k q_{1} q_{2}}{K E} = > r = \frac{9 * 10^{9} * e * 79 e}{5 * e * 10^{6}} = > r = \frac{9 * 10^{3} * 79 * 1.6 * 10^{- 19}}{5}$

=> r = 2.28 *10^-14m