Solution to 1996 Problem 19

A scattering angle of 180^{\circ} means that an alpha particle will travels directly toward a nucleus, comes to a complete halt as it approaches the nucleus, and then turns around and starts moving directly away from the nucleus. At the time when the alpha particle comes to a complete halt, all of its kinetic energy will have been converted into potential energy. Therefore, fi we let r denote the distance between the alpha particle and nucleus at this moment, then
\begin{align*}5\cdot 10^6 \mbox{ eV} = \frac{2\cdot 50 e^2}{4 \pi \epsilon_0 r}\end{align*}
This can be easily solved for r, yielding
\begin{align*}r = \boxed{2.877 \cdot 10^{-14} \mbox{ m}}\end{align*}
Therefore, answer (B) is correct.

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