Solution to 1992 Problem 95

The relationship between the differential scattering cross section d \sigma/ d \Omega, the solid angle d \Omega, the number of incident particles per unit time L, the number of incident particles scattered into the solid angle d \Omega per unit time d N, and number of nuclei per unit area n is %(see Griffiths QM page 399 but note that the notation there is slightly different)
\begin{align*}\frac{d \sigma}{d \Omega} L n d \Omega = d N\end{align*}
We are given that
\begin{align*}L &= 10^{12}\mbox{ 1/s} \\n &= 10^{20} \;\mathrm{cm}^{-2} \\dN &= 10^2 \mbox{ 1/s} \\d \Omega &...
Plugging these values in, we find that
\begin{align*}\frac{d \sigma}{d \Omega} = \frac{d N}{L n d \Omega} = \boxed{10^{-26} \;\mathrm{cm}^2\mathrm{/}\mathrm{steradi...
Therefore, answer (C) is correct.

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