Answer
$r=1.6\times 10^{-9}m$
Work Step by Step
From law of conservation of energy,
$K.E=P.E$
$\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=K\frac{e^2}{r}$
We plug in the known values to obtain:
$\frac{1}{2}(9.109\times 10^{-31})(2\times 3.2\times 10^{5})^2-\frac{1}{2}(9.109\times 10^{-31})(3.2\times 10^{5})^2=9\times 10^9\times \frac{(1.6\times 10^{-19})^2}{r}$
Solving this, we find that
$r=1.6\times 10^{-9}m$
