An `alpha`-particle with kinetic energy `K` is heading towards a stationary nucleus of atomic number `Z`. Find the distance of the closest approach.

Explanation

Between `alpha`-particle (charge `2e`) and nucleus , electric force is repulsive . In approaching nucleus, speed of `alpha`-particle is decreasing and at distance of the closest approach it comes to rest momentarily . Here kinetic energy of `alpha`-particle is converted into electric potential energy
`K = U = (1)/(4 pi in_(0)) .(Ze .Ze)/( r)`
`r = (1)/(4 pi in_(0)) .(2 Ze^(2))/(K)`