how do you solve this physics problem?

Question

An alpha particle with a Kinetic E of 7.00 MeV makes a head on collision with a gold nucleus at rest. What is the distance of closest approach of the 2 particles ( assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is 79, and an alpha particle is a helium nucleus consisting of 2 protons and 2 electrons. )

Answer 1

The energy KE will equal 0 when the alpha reverses direction at the point of closest approach and the Coulomb repulsive potential energy will equal the initial KE