A narrow beam of protons, each having $\displaystyle 4.1$ MeV energy is approaching a sheet of lead $\displaystyle (\mathrm{Z}=82)$. Calculate :

(i)

the speed of a proton in the beam, and

(ii)

the distance of its closest approach $\displaystyle 3$

$\displaystyle \frac{1}{2} m v^{2}=4 \cdot 1 \times 1 \cdot 6 \times 10^{-13} \mathrm{~J}$

$$v=\sqrt{\frac{$\displaystyle 2$ \times $\displaystyle 4$ \cdot $\displaystyle 1$ \times $\displaystyle 1.6$ \times $\displaystyle 10$^{-$\displaystyle 13$}}{$\displaystyle 1$ \cdot $\displaystyle 673$ \times $\displaystyle 10$^{-$\displaystyle 27$}}}

$$(ii) $\displaystyle d=\frac{Z e^{2}}{4 \pi \varepsilon_{0} \times E_{k}}$

$$\begin{aligned} & =\frac{$\displaystyle 9$ \times $\displaystyle 10$^{$\displaystyle 9$} \times $\displaystyle 82$ \times $\displaystyle 1$ \cdot $\displaystyle 6$ \times $\displaystyle 10$^{-$\displaystyle 19$} \times $\displaystyle 1$ \cdot $\displaystyle 6$ \times $\displaystyle 10$^{-$\displaystyle 19$}}{$\displaystyle 4$ \cdot $\displaystyle 1$ \times $\displaystyle 1$ \cdot $\displaystyle 6$ \times $\displaystyle 10$^{-$\displaystyle 13$}} & =$\displaystyle 2.88$ \times $\displaystyle 10$^{-$\displaystyle 14$} \mathrm{~m} \end{aligned}