18 karma · joined December 22, 2022
@defrost Great discussion and addition by the way. I only have one additional comment concerning radon. Radon is produced in a radioactive decay in the ground, and as it is gaseous it usually dissipates into the atmosphere rather quickly, so there is not too much of it near the detector at any given time. As you mention correctly, certain weather condition obviously can trap it near the ground. We usually study urban centers, with lots of asphalt covering the ground, affecting the amount released near the ground and near our detectors. We usually see it during rain, as then the radon in the atmosphere is washed down through the rain and accumulates on top of the ground for a while, which we often see as a strong uptick followed by an exponential decay. I have a coworker that just submitted a paper for review regards modeling this, sorry can't link the work yet. Overall, radon is quite tricky to understand and what you mention seem very reasonable. It would be quite interesting putting a detector for a while in these areas. Nevertheless, the measured amount of radon usually fluctuates on larger scales, maybe hundreds of meters, while the source will create a very localized signal, so it shouldn't cause too much troubles for a search like this.
[1] https://en.wikipedia.org/wiki/Western_Australian_radioactive...
For simplicity lets assume 50 Liter is a 1 meter x 1 meter x 5 centimeter thick panel. 5 centimeter is usually sufficient to give you a good change to measure gamma rays at most energies and you want to have a large surface area exposed to the source, so this seems a quite optimal configuration for 50 Liter of NaI.
A source emits gamma rays in all directions uniformly, so to calculate the fraction of gamma rays that end up in our detector we calculate in "solid angle space".
The surface area of such the detector is 1 meter^2. The solid angle coverage can be approximated by surface area / distance^2 [1], so at 40 meter that would be about 0.000625. The full solid sphere is 4PI ~ 12.5, which is the surface area of a ball with 1 meter radius.
The fraction of the full solid angle that is covered by the detector is 0.000625 / 12.5 = 0.00005.
You have 19 Becequerel = 19 Gamma rays per seconds leaving the source so 0.00005 * 19 Bequerel * 60 seconds = 0.057 gamma rays per minute reaching the detector. You won't see this over the background.
Putting 37 Million Becquerel into this equation, we will get roughly 111,000 counts / minute in the detector. Remembering correctly (and this is only a rough guess) you would expect about 700,000 gamma rays per minute from background (what we call naturally occurring radioactive materials or NORM) in that much detector volume. So that is about a 1:7 signal to background ratio. You should have a very good chance of seeing that.
I neglected here that not all gamma rays that reach the detector actually will leave a signal in the detector. But at 662 keV, the energy most of the gamma rays in Cs137 are emitted at, about half of them do so in a 3 cm thick detector. At 5 cm I would estimate about 75% or more doing so.
Hope that gives you an idea how you can estimate these types of questions.
[1] https://en.wikipedia.org/wiki/Solid_angle
Edit: Forgot to mention the radon. Usually you do not measure much radon unless it rains. It is mostly a dessert there so I don't think that should be an issue
I am working at a National Lab in the US and this is the main focus of my research. Please have a look at our website if you are interested in learning more about this type of things [1].
[1]: https://anp.lbl.gov/
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[1] https://www.honeywellstore.com/store/news/honeywell-warm-mis...
[2] https://www.amazon.com/gp/customer-reviews/R911WGTC1OHLL/ref...
[3] https://learn.kaiterra.com/en/air-academy/humidifiers-cause-...