Distance of closest approach for alpha particle and gold nucleus in Rutherford scattering

[map7s]

In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy K heading directly for the nucleus of a gold atom (charge = +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus for the case K = 3.4 MeV.

I first set up the equation as -ke^2/r=1/2mv^2, with the first part of the equation as the electric potential energy and the second part as the initial kinetic energy set equal to each other. then I substituted the value given for K for the 1/2 mv^2 part. however when I solved, I got the radius and I am pretty sure that that value does not correspond to the distance between the two charges. also, I am not sure how to factor in the two charges - should I add them together and use that number?

 

[Mentz114]

Electrical potential energy is k.q1.q2/r.

 

[map7s]

oh...okay...so would I still be solving for r, even though that's supposed to be the radius or is there another equation that I am forgetting?

 

[hage567]

oh...okay...so would I still be solving for r, even though that's supposed to be the radius or is there another equation that I am forgetting?

Yes, you are solving for r. Are you remembering to convert your kinetic energy into Joules? (check the units)
I

am not sure how to factor in the two charges - should I add them together and use that number?

If you have a charge of +79e, that means 79*(1.602x10^-19 C).