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

At distance of closest approach 

$PE=K_{initail}$

Where $K_{initail}$ is initial kinetic energy.

$\begin{array}{l}\therefore \left(\frac{1}{4\pi {\epsilon }_0}\right)\frac{2Z{e}^2}{d}=6\times 1.6\times {10}^{-13}J\\ \Rightarrow d=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2\times 2\times 79}{6\times 1.6\times {10}^{-13}}m\\ =3.8\times {10}^{-14}m\end{array}$