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: enter image description here

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: enter image description here

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?


1 Answer 1


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.

  • $\begingroup$ I see , thanks. How can I incoporate beta ionisation into my correction? $\endgroup$ Mar 13, 2022 at 1:59
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    $\begingroup$ @ChernSimons The main things to consider there are the energy loss in air and the efficiency of your detector at various energies. A good source for the former is NIST's ESTAR. Keep in mind that these are averages, and you would also need to account for variations in energy loss. You would probably need to get some idea of the variation in energy losses and the efficiency of your detector at various energies by fitting to some of your data, and then you can extrapolate to other regions and see if it matches. $\endgroup$
    – Chris
    Mar 13, 2022 at 2:28

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