An `alpha` -particle moving with initial kinetic energy 'K' towards a nucleus of atomic number Z approaches a distance '`r_(0)`' at which it reverses its direction. Obtain the expression for the distance of closest approach '`r_(0)`' in terms of kinetic energy of a particle.

Explanation

Consider an o-particle of mass m, charge + 2e and an initial speed v (or initial kinetic energy K) moving straight along the central line of a nucleus of atomic number Z.
As `alpha` particle approaches the nucleus, on account of electrostatic repulsion, its speed goes on decreasing and at a point situated at a distance `r_(0)`, the `alpha`-particle momentarily comes to rest. Obviously in this position the initial kinetic energy of `alpha`-particle has been completely converted into electrostatic potential energy i.e.,
`K.E ` at `A = P.E` at B
`therefore " " k = (1)/(2) mv^(2) = (1)/(4 pi in_(0)) .((+2e)(+Ze))/(r_(0))`
`rArr " " r_(0) = (1)/(4 pi in_(0)) = (2Ze^(2))/(((1)/(2) mv^(2))) =(1)/(4pi in_(0)) .(2Ze^(2))/(K)`