Question

Calculate the nearest distance of approach of an α-particle of energy 2.5 MeV being scattered by a gold nucleus (Z = 79).

Answer 1

The electrostatic potential at a distance x due to nucleus is given by Ze/4πε₀ x, where Ze is the charge on the nucleus.

The P.E. of an -particle when it is at a distance x from the nucleus is given by

2e being the charge on -particle. Since the -particle momentarily stops when its initial K.E. is completely changed into P.E. here, hence

Substituting the values, we get