Discussion
A gas molecule moves equal distances \(\lambda\) between successive collisions with equal probability in any direction. After a total of N such collisions, the mean square displacement of the molecule \(\overline{d}^{2}\) will be :
Answer: B
The mean square displacement is \(N\lambda^{2}\).
Write the total displacement as the sum of N steps, \(\vec{d} = \sum \vec{\lambda}_{i}\), each of length \(\lambda\) in a random direction. Squaring and averaging:
\[ \begin{aligned} \overline{d^{2}} &= \sum_{i}\overline{\lambda_{i}^{2}} + \sum_{i \ne j}\overline{\vec{\lambda}_{i}\cdot\vec{\lambda}_{j}} \\ &= N\lambda^{2} + 0 \end{aligned} \]
The cross terms average to zero because successive directions are independent. So the root mean square distance grows only as \(\sqrt{N}\lambda\). \(N^{2}\lambda^{2}\) would be the result if all steps pointed the same way, and \(N\lambda\) is not even dimensionally a squared length.
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