A head-on collision takes place between an $\alpha$-particle of kinetic energy $5.5MeV$ and a gold nucleus $(Z=79)$. Calculate the distance of closest approach.

Since, the distance of closest approach is given by

$$\begin{gathered} r_0=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Z{e}^2}{m_{\alpha }{{\nu }^2}_{\alpha }}\right)=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Z{e}^2}{K_{\alpha }}\right) \\ \Rightarrow r_0=9\times {10}^9\times \frac{2\times 79\times {\left(1.6\times {10}^{-19}\right)}^2}{5.5\times \left(1.6\times {10}^{-13}\right)} \\ \Rightarrow r_0=4.13\times {10}^{-14}m \end{gathered}$$

The radius of gold nucleus must be smaller than $r_0$, so it may lie between ${10}^{-14}$ to ${10}^{-15}m$.