Rutherford α-Scattering

Most α particles go straight through the foil. Some deflect slightly. Even less deflect massively (over 90°). That works if the atom is mostly empty space with a teeny tiny nucleus.

most go straight

some large angles

rare backwards deflection

What Rutherford was actually counting

The detector is basically a screen where you see flashes when α particles hit. It shows how often particles hit the screen and where they hit.

He observed large scatterings every now and then, which couldn't be explained by previous models of the atom which did not have empty space.

How they used it to model the nucleus' size

A big deflection happens only if the α passes close enough to or collides with the nucleus.

If D is the atom diameter and d is the nucleus diameter, then for one layer of atoms the chance of a close pass is shown by the ratio of the cross-sectional areas of the nucleus to the atom:

P(1 layer) ≈ π(d/2)² / π(D/2)² = d² / D²

If the foil has n layers, the α gets n chances. So we multiply the probability of it passing an atom per layer by the amount of layers.

P(total) ≈ n · (d² / D²)

Where the 10000 comes from

Experimentally, only about 1 in 10,000 α particles show a huge deflection. So:

P(total) ≈ 1/10000

So:

n · (d² / D²) = 1/10000

Rearrange:

d² = D² / (10000 n)

This is a rough estimation.

Why this works conceptually

Since the nucleus is so small, the large deflections are rare as collisions with the nucleus are rare (remember every atom passes through at least one atom). The derived equation supports the idea that d is less than D. I intend on adding the distance of closest approach equation to this page soon enough.

the setup