When a particle is thrown towards the nucleus, its kinetic energy decreases, and potential energy increases as approaches the nucleus and comes to momentary rest at a particular distance to the nucleus. This distance is called the distance of the nearest approach. After that, it retraces its path, and its kinetic energy increases and potential energy decreases as going away from the nucleus.
Distance Of The Nearest Approach
Suppose an \alpha-particle is directed towards the center of the nucleus. It momentarily stopped at a point from the nucleus and then retraced its path.
The distance of the nearest approach is the minimum distance between \alpha-particle and center of the nucleus just before being reflected back by 180^0.
Let m be the mass of an \alpha-particle, v be the velocity with which it approaches the nucleus, and d is the distance of nearest approach.
The initial kinetic energy of the \alpha-particle is;
E = \dfrac{1}{2}mv^2
Since there is no external and non-conservative force acting on the system of the \alpha-particle and the nucleus so, the mechanical energy of the system will be conserved. And hence, as the \alpha-particle approaches the nucleus, its kinetic energy decreases and gradually is converted into electrostatic potential energy. At the point of nearest approach, the kinetic energy of the \alpha-particle is fully converted into electrostatic potential energy due to Coulomb’s force of repulsion of the nucleus on \alpha-particles.
That is,
\begin{aligned}E &= \dfrac{1}{4\pi\epsilon_0}[(2e)(Ze)/d]\\ \Rightarrow \dfrac{1}{2}mv^2 &= \dfrac{1}{4\pi\epsilon_0} \dfrac{(2e)(Ze)}{d}\\ \Rightarrow d &= \dfrac{Ze^2}{\pi\epsilon_0mv^2}\\ \end{aligned}