Physics - W.E.P. Question with Solution | TestHub

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, $$\begin{array}{rl}& \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{4\pi {\epsilon }_0}\frac{(2\mathrm{e})(\mathrm{Ze})}{{\mathrm{r}}_0}\\ & {\mathrm{r}}_0=\frac{1}{4\pi {\epsilon }_0}\frac{2{\mathrm{Ze}}^2}{\frac{1}{2}{\mathrm{mv}}^2}\end{array}$$