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 \( V \), the kinetic energy acquired by the particle is given by: \[ K.E. = qV \] Where \( q \) is the charge of the particle. The kinetic energy is also related to the speed of the particle: \[ K.E. = \frac{1}{2} mv^2 \] At the point of closest approach, all the kinetic energy is converted into electrostatic potential energy. The electrostatic potential energy between two charges \( q_1 \) and \( q_2 \), separated by a distance \( r \), is given by: \[ U = \frac{1}{4 \pi \epsilon_0} \times \frac{q_1 q_2}{r} \] At the closest approach, this potential energy equals the kinetic energy: \[ qV = \frac{1}{4 \pi \epsilon_0} \times \frac{q_1 q_2}{r} \] Where: - \( q_1 \) is the charge of the incoming particle - \( q_2 \) is the charge of the target nucleus - \( r \) is the distance of closest approach Now, consider the two cases: 1. Alpha particle (\( \alpha \)): The charge of the alpha particle is \( q_{\alpha} = 2e \), where \( e \) is the elementary charge. 2. Deuterium ion (\( D \)): The charge of the deuterium ion is \( q_D = e \). Since the potential difference \( V \) is the same for both particles, we can write the equation for the distance of closest approach for each particle: \[ r_{\alpha} = \frac{1}{4 \pi \epsilon_0} \times \frac{q_{\alpha} q_2}{qV} = \frac{1}{4 \pi \epsilon_0} \times \frac{2e q_2}{qV} \] For the deuterium ion: \[ r_D = \frac{1}{4 \pi \epsilon_0} \times \frac{q_D q_2}{qV} = \frac{1}{4 \pi \epsilon_0} \times \frac{e q_2}{qV} \] Now, comparing the two expressions for \( r_{\alpha} \) and \( r_D \): \[ r_{\alpha} = 2 \times r_D \] Thus, the distance of closest approach for the alpha particle is twice that of the deuterium ion. However, the problem states that both distances of closest approach are observed to be the same. This means that in the presence of the target nucleus, the effective nuclear charge experienced by the alpha particle and the deuterium ion must be different. This can be explained by the fact that the target nucleus could be more effective in slowing down the alpha particle than the deuterium ion, leading to the same effective closest approach distance.