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An $\alpha$-particle of energy $5 \mathrm{MeV}$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of
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The correct answer is:
$10^{-12} \mathrm{~cm}$
$10^{-12} \mathrm{~cm}$
At closest approach, all the kinetic energy of the $\alpha$-particle will converted into the potential energy of the system, K.E. = P.E.
$$
\begin{aligned}
& 5 \mathrm{MeV}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\
& 5 \times 10^6 \times \mathrm{e}=9 \times 10^9 \frac{\mathrm{Z}_1 \times \mathrm{Z}_2 \mathrm{e}^2}{\mathrm{r}} \\
& \mathrm{r}=\frac{9 \times 10^9 \times 92 \times 2 \times 1.6 \times 10^{-19}}{5 \times 10^6} \\
& \therefore \mathrm{r}=5.3 \times 10^{-14} \mathrm{~m}=5.3 \times 10^{-12} \mathrm{~cm}
\end{aligned}
$$
$$
\begin{aligned}
& 5 \mathrm{MeV}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\
& 5 \times 10^6 \times \mathrm{e}=9 \times 10^9 \frac{\mathrm{Z}_1 \times \mathrm{Z}_2 \mathrm{e}^2}{\mathrm{r}} \\
& \mathrm{r}=\frac{9 \times 10^9 \times 92 \times 2 \times 1.6 \times 10^{-19}}{5 \times 10^6} \\
& \therefore \mathrm{r}=5.3 \times 10^{-14} \mathrm{~m}=5.3 \times 10^{-12} \mathrm{~cm}
\end{aligned}
$$
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