Answer
$r = 1.6\times 10^{-14}~m$
Work Step by Step
An alpha particle has 2 protons
A copper nucleus has 29 protons
We can express the kinetic energy in units of joules:
$K = (5.30~MeV)(\frac{1.6\times 10^{-19}~J}{1~eV}) = 8.48\times 10^{-13}~J$
To find the distance of closest approach, we can equate the final potential energy with the initial kinetic energy:
$K = U$
$K = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{r}$
$r = \frac{1}{4\pi~\epsilon_0}~\frac{q_1~q_2}{K}$
$r = \frac{1}{(4\pi)~(8.854\times 10^{-12}~F/m)}~\frac{(2)(1.6\times 10^{-19}~C)(29)(1.6\times 10^{-19}~C)}{8.48\times 10^{-13}~J}$
$r = 1.6\times 10^{-14}~m$
