Question

An alpha particle and a deuterium ion are accelerated through the same potential difference. These are then directed towards a target nucleus to make a head-on collision. It is observed that their distance of closest approach is the same. Justify it theoretically.

Hint:

The distance of closest approach depends on the charge of the particle and the target. Although the alpha particle has a greater charge, the target nucleus can influence the effective distance at which the kinetic energy of the particle is fully converted into potential energy.

Answer 1

When an ion is accelerated through a potential difference \( V \), its acquired kinetic energy is \( K.E. = qV \), where \( q \) is the particle's charge. This kinetic energy is also given by \( K.E. = \frac{1}{2} mv^2 \). At the point of closest approach, all kinetic energy converts to electrostatic potential energy, \( U = \frac{1}{4 \pi \epsilon_0} \times \frac{q_1 q_2}{r} \), where \( q_1 \) and \( q_2 \) are the charges and \( r \) is the separation distance. Equating these at closest approach gives \( qV = \frac{1}{4 \pi \epsilon_0} \times \frac{q_1 q_2}{r} \). Here, \( q_1 \) is the incoming particle's charge, \( q_2 \) is the target nucleus's charge, and \( r \) is the closest approach distance. Consider two cases with the same potential difference \( V \): 1. Alpha particle (\( \alpha \)): Charge \( q_{\alpha} = 2e \). 2. Deuterium ion (\( D \)): Charge \( q_D = e \). The distance of closest approach for the alpha particle is \( r_{\alpha} = \frac{1}{4 \pi \epsilon_0} \times \frac{q_{\alpha} q_2}{qV} = \frac{1}{4 \pi \epsilon_0} \times \frac{2e q_2}{qV} \). For the deuterium ion, it is \( r_D = \frac{1}{4 \pi \epsilon_0} \times \frac{q_D q_2}{qV} = \frac{1}{4 \pi \epsilon_0} \times \frac{e q_2}{qV} \). Comparing these, \( r_{\alpha} = 2 \times r_D \). This implies the alpha particle's closest approach distance is twice that of the deuterium ion. However, the problem states these distances are the same. This suggests that the effective nuclear charge experienced by the alpha particle and the deuterium ion must differ in the presence of the target nucleus. This could occur if the target nucleus is more effective at slowing down the alpha particle than the deuterium ion, resulting in the same effective closest approach distance.