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**1. Homework Statement**

In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle, very far from the gold foil, with an initial kinetic energy of 3.6 MeV heading directly for 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.

**2. Homework Equations**

Force:

mv^2=kZe^2/r

k=8.99x10^9 N^2/C^2

Z= # protons

e=1.6x10^-19 C

**3. The Attempt at a Solution**

From the question, I'm assuming that the gold and the alpha particle never collide. Thus, I figured that by Newton's third law, I can deduce the distance (r) between them. First, I did 3.6MeV*2 to find mv^2 (since KE=1/2mv^2 and I know that all KE is converted to PE). I got 7200 eV. Then, I plugged in 79 for Z on the other side of the equation and tried solving for r. I came up with

**However this answer is not correct. Had I missed an important step? Thanks in advance for your help!**

__2.525E-30 m.__