I need help for the below question

Question

|||A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm. If the proton is fired at a speed of 4.0  10⁷ m/s, what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest.

Answer 1

Using KE = PE

KE of incident proton = PE of Mercury and proton

1/2 mv² = K q1*q2/r

m = mass of proton = 1.67*10 ^-27 Kg

V= velocity of incident proton = 4* 10⁷ m/s

q1 = Charge of incident proton = +e = 1.6*10^-19 C

q2 = Charge of target Mercury nucleus= +80e = 80* 1.6*10^-19 C

r= distance of closest approach = ?

K = 9*10⁹ Nm² /C²

Gives ( Plug in values)

r = 2Kq1*q/ (m V² ) = 1.38 *10^-14 m = 13.8 fm = Distance from the center of Hg Nucleus , when proton reaches closest to it