Discussion
A particle passes through a medium and is deflected by a small angle \(\pm \delta_{0}\) upon each collision within the medium with equal probability in any direction. The mean square angular scattering after N such collision, \(\overline{\delta^{2}}\) is :
Answer: B
The mean square scattering angle is \(N\delta_{0}^{2}\).
Each collision adds an independent random deflection \(\pm\delta_{0}\) with equal probability, so each step has mean zero and mean square \(\delta_{0}^{2}\). For a sum of N independent steps the cross terms average to zero:
\[ \begin{aligned} \overline{\delta^{2}} &= \overline{\left(\sum \delta_{i}\right)^{2}} \\ &= \sum_{i} \overline{\delta_{i}^{2}} \\ &\quad + \sum_{i \ne j} \overline{\delta_{i}\delta_{j}} \\ &= N\delta_{0}^{2} + 0 \end{aligned} \]
This is the random-walk result: the root mean square angle grows as \(\sqrt{N}\). Zero is the mean angle, not the mean square; \(N^{2}\delta_{0}^{2}\) would hold only if all deflections were in the same direction, and \(\delta_{0}^{2}/N\) decreases with N, which is impossible.
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