So I did an experiment in which the count rate of a disk source of Sr-90 is measured using a plane detector, varied from ~0.7cm to 100cm, multiplying the count rate by the distances squared (i.e. $\frac{N}{\Delta t}d^2$) I got the following:
So the dip at small distances is mainly due to the inverse square law (which only applies for a point source) failing as a result of the extendedness of the Sr-90 disk source, and the linearly falling region is due to the ionisation of beta particles with air (Sr-90 decays both gamma at 1.76MeV, and Beta at 2.27MeV & 0.57MeV).
I am trying come up with a geometric correction for the inverse square law accounting for the extendedness of the Sr-90 disk, and tried using the classical electric field at a point z from the centre of a disk charge:
that is since $E_z=2\pi k\sigma[1-\frac{z}{\sqrt{z^2+R^2}}]$
can I use this as the geometric correction to my plot? Like (excluding the ionisation of beta)
$$\frac{N}{\Delta t}(d) \propto [1-\frac{d}{\sqrt{d^2+R^2_{Sr-90-Source}}}]$$
I came across a paper https://www.sciencedirect.com/science/article/abs/pii/S0969804306002314 that uses something entirely different (equation (2) in the paper) $$\frac{N}{\Delta t}(d) \propto \frac{d^2}{R^2}ln(1+\frac{R^2}{d^2})$$ which I cannot derive from first principle, and seems very different to the electric field treament. Or is diffraction important?
Best Answer
If you ignore all attenuation and specifics of the detector, then yes. It's the same math. The solution of integrating a $1/r^2$ source over a disk is not anything specific to electrostatics.
The paper you've linked to is specifically interested in both this attenuation and specifics of the detector, so of course they will come up with a different result.
Note that your formula assumes that your detector is also a point source. If you want to do better, you should assume that both the source and the detector have some finite extent.