In Rutherford\'s scattering experiments, alpha particles (charge =2e ) were fire

Question

In Rutherford's scattering experiments, alpha particles (charge =2e ) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy 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 = 3.5 MeV.

I keep getting: 6.4933 x 10^-14 and MP says its wrong....please help me.

Explanation / Answer

I believe you need to convert from m to fm. The answer is 65.

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