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Each alpha particle in a beam of alpha particles has a kinetic energy of 5.0 MeV....

Each alpha particle in a beam of alpha particles has a kinetic energy of 5.0 MeV. Through what potential difference would you have to accelerate these alpha particles in order that they would have enough energy so that if one is fired head-on at a gold nucleus it could reach a point 1.0x10^-14 m from the center of the nucleus?

I know that the answer is delta V = 9e6 Volts

I am struggling to get past KEi = 5.0 MeV; r = 1.0e-14 m; q1 = 79 electrons (gold particle); q2 = 2 electrons (alpha particle)

Finding Delta V

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