# Distance of closest approach of an alpha particle to a gold nucleus

If an alpha particle with a kinetic energy Ea is fired directly towards a gold nucleus it will feel a repulsion which increases as it gets closer - climbing the potential hill surrounding the nucleus. When all the kinetic energy has been converted to potential energy the alpha particle (charge q) has reached its distance of closest approach (d0) and comes to rest. At that point :

Ea = qVE = (1/4peo)qQ/do    so therefore: do = (1/4peo)qQ/Ea

The diagram shows the paths of other alpha particles that were not travelling directly towards the centre of the nucleus. These are scattered.

## Distance of closest approach animation

For the schoolphysics animation showing the distance of closest approach please click the animation symbol:

## Alpha particle scattering animation

For the schoolphysics animation of alpha particle scattering please click the animation symbol:

Example problem
Calculate the distance of closest approach between an alpha particle of energy 5.5 MeV (5.5x106x1.6x10-19 = 8.8x10-13 J) if the charge on the gold nucleus is +79e = +79x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4peo)qQ/Ea we have
dc = [9x109x(2x1.6x10-19x79x1.6x10-19)]/8.8x10-13 = 4.1x10-14 m.
This is well inside the atom but some eight nuclear diameters from the centre of the gold nucleus.