An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^2$ bombards a heavy nuclear target of charge $Z e$. Then…
An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^2$ bombards a heavy nuclear target of charge $Z e$. Then the distance of closest approach for the alpha nucleus will be proportional to
$\frac{1}{\mathrm{Ze}}$
$v^2$
$\frac{1}{\mathrm{~m}}$
$\frac{1}{\mathrm{v}^4}$
Solution
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,
$\begin{aligned}
& \frac{1}{2} \mathrm{mv}^2=\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}_0} \\
& \mathrm{r}_0=\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{Ze}^2}{\frac{1}{2} \mathrm{mv}^2}
\end{aligned}$