Source: High school physics (Chinese)
Problem
An alpha particle (charge $q_\alpha = 2e$, mass $m = 6.7 \times 10^{-27}$ kg) is fired from a great distance with an initial speed $v_i = 1.6 \times 10^7$ m/s towards a stationary gold nucleus (charge $q_{Au} = 79e$).
- How should it be fired to get as close as possible to the gold nucleus, and why?
- What is this closest distance of approach?
[Q1] It should be fired directly towards the nucleus (head-on collision). This maximizes the conversion of kinetic energy to potential energy. [Q2] $r_{min} = 4.24 \times 10^{-14}$ m
[A1] To achieve the minimum separation distance, the alpha particle must be fired in a direct, head-on collision. This is because all its initial kinetic energy will be converted into electrostatic potential energy at the point of closest approach, where it momentarily stops. Any non-zero impact parameter would result in a larger distance of closest approach.
[A2] By the conservation of energy, the initial kinetic energy $K_i$ of the alpha particle is completely converted into electric potential energy $U_f$ at the distance of closest approach, $r_{min}$.
$$K_i = U_f$$ $$\frac{1}{2}mv_i^2 = k \frac{q_\alpha q_{Au}}{r_{min}}$$Solving for $r_{min}$:
$$r_{min} = \frac{2kq_\alpha q_{Au}}{mv_i^2} = \frac{2k(2e)(79e)}{mv_i^2} = \frac{316ke^2}{mv_i^2}$$