Discussion
An \(\alpha\) particle moves towards a fixed nucleus carrying charge \(Ze\), with initial speed \(v_{0}\) and impact parameter \(b\). Starting from a large distance from the nucleus, its distance of closest approach is \(r_{m}\) and its speed there is \(v_{m}\). Then which of the following options is/are correct?
(\(k = \frac{1}{4\pi\epsilon_{0}}\) and \(r_{0} = k\frac{Ze^{2}}{mv_{0}^{2}}\))
Answer: A
The correct statement is \(v_{0}b = v_{m}r_{m}\).
The Coulomb force is central, so angular momentum about the nucleus is conserved. Far away it is \(mv_{0}b\); at closest approach the velocity is perpendicular to the radius, so it is \(mv_{m}r_{m}\). Hence \(v_{0}b = v_{m}r_{m}\).
Energy conservation, with the alpha particle's charge 2e, gives \(v_{0}^{2} = v_{m}^{2} + 4r_{0}v_{0}^{2}/r_{m}\). Using \(v_{m} = v_{0}b/r_{m}\):
\[ \begin{aligned} r_{m}^{2} - 4r_{0}r_{m} - b^{2} &= 0 \\ r_{m} &= 2r_{0} + \sqrt{4r_{0}^{2} + b^{2}} \\ &\approx 4r_{0} + \frac{b^{2}}{4r_{0}} \end{aligned} \]
So the correction is \(b^{2}/4r_{0}\), not \(b^{2}/2r_{0}\) or \(b^{2}/8r_{0}\). And \(v_{0}b = 2v_{m}r_{0}\) would need \(r_{m} = 2r_{0}\), less than the head-on value \(4r_{0}\). Only the angular momentum statement holds.
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