Define the distance of closest approach. An `alpha`-particle of kinetic enegy `'K'` is bombarded on a thin gold foil. The distance of the closest approach is `'r'` . What will be the distance of closest approach for an `alpha`-particle of double the kinetic energy ?

Explanation

It `alpha` particle with KE = E is directed towards centre of atom (nucleus) , due to Columbian Repulsion, it won't hit the nucleus, rather stop at a distancs. For given K.E., distance of closest approach of `alpha`-particleis given by r. if `alpha`-particle has K.E, and the distance of closest approach is r,
then, `KE =1/2 mv^(2) = (2e(Ze))/(4piepsi_(o)r)`
`rArr r = (2Ze^(2))/(4piin_(0)K) , r prop 1/(K)`
`rArr (r_(1))/(r_(2)) = (K_(2))/(K_(1)) rArr r_(2) = (K)/(2K) r_(1) rArr r_(2) = (r_(1))/(2)`.
i.e., distance of closest approach is reduced by a factor of `1/2`.