Solution to 1986 Problem 42

The relevant equation is
\begin{align*}\mbox{probability of scattering} = n t \sigma\end{align*}
where n is the number density of targets, t is the thickness of the target, and \sigma is the cross section.
\begin{align*}\sigma = \frac{\mbox{probability of scattering}}{n t} = \frac{10^{-6}}{10^{20} \;\mathrm{cm}^{-3} \cdot 0.1 \mb...
Therefore, answer (C) is correct.

return to the 1986 problem list

return to homepage

Please send questions or comments to where X = physgre.