In Ernest Rutherford’s Î±-particle scattering experiment, it is well-known that the *solid angle* density of Î±-particle flux

$$frac{ddot{N}}{dOmega}$$

on the inner surface of a sphere of which the centre is the point of impingement of a beam of Î±-particles on a (extremely) thin gold foil is proportional to

$$csc^4frac{phi}{2}$$

where $phi$ is the angle away from the *pole* of the apparatus – the point on the sphere upon which an udeflected beam would impinge. This proportionality relation incurs the problem of having no *normalisation*: integration of it over even the smallest non-zero region of the sphere comprising the pole will yield an infinite flux.

But there are various mechanisms operating to *blur* the distribution of flux over the sphere and ‘wash-out’ the singularity at the pole, such as the non-zero width of the beam and the non-zero spread of directions in the beam. Clearly, at the pole flux-density is not going to exceed the flux density of the beam, and

$$frac{ddot{N}}{dOmega}leq frac{a^2dot{N_0}}{pi b^2}$$

where $dot{N_0}$ is the flux in the beam, $a$ is the radius of the sphere, & $b$ the radius of the beam.

I do not find in Rutherford’s paper that there is any attempt to *hardwire* this blurring mathematically into the propotionality relation – it probably wasn’t necessary to do so in order sufficiently to evince that the nucleus is an extreme concentraction of charge and mass; and besides, the detection medium at the point on the sphere antipodal to the Î±-particle source was almost certainly *utterly saturated*: but I am curious anyway as to how it *would be* hardwired in. I fairly sure the greatest contribution to the blurring would be the *finite width* of the beam, followed by the *spread of directions* in the beam. I *think* the contribution from the finity of *cross-section* about a nucleus, in that more than a certain distance away another nucleus would be encountered, would be a small one.

So the question is *primarily* “how would the finite cross-sectional area of the beam be hardwired into the expression for solid-angle-density of flux (the $csc^4frac{phi}{2}$ expression) so that it can be normalised?”. It’s almost certainly going to be an integration over a disc centred on point at polar-angle $phi$ (I know $phi$ normally denotes *azimuth* these days, but Rutherford used it for polar-angle); but I can’t quite catch the particular details of how to do it.