In scattering experiment, find the distance of closest approach, if a $6MeV$ $\alpha -$ particle is used

The correct option is D $2\times {10}^{-14}m$
The distance of closest approach is given us in under :

$r_0=14\pi {ϵ}_0\times 2\times z\times e_{2}KE$

given $KE=6MeV$

and assuming the nucleus to be heat of copper, then $z=2q$ as $d.e=1.6\times {10}^{-19}C$.

Substituting the values in the above equation and calculating we have

$r_0=2\times {10}^{-14}m$.