An alpha nucleus of energy (1)/(2) mv^2 bombards a heavy nuclear...

An $\alpha$-particle of mass $m$ possesses initial velocity $\mathrm{v}$, when it is at a large distance from the nucleus of an atom having atomic number $Z$. At the distance of closest approach, the kinetic energy of $\alpha$-particle is completely converted into potential energy. Mathematically, \begin{aligned}& \frac{1}{2} \mathrm{mv}^2=\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}_0}\\ & \mmathrm{r}}_{0}=\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{Ze}^2}{\frac{1}{2} \mathrm{mv}^2} \end{aligned}