How do I calculate the Sieverts/hour that a gamma meter would see at x distance from a radioactive sample?
I'm working on a little proof of concept program where I want to input some weight of a radioactive element, and then calculate what a meter would read at different distances. Ultimately I hope to also add some shielding to the calculation like a concrete wall or something. I've been reading and trying to learn how this calculation would work but I don't think I have it yet.
So far I understand I can calculate the activity if I understand the number of nuclei and the half life. As an example I take a 1gram sample of Cesium 137 with a half life of 2.6 min.
Cesium has an atomic weight of 136.9amu = 2.2733×10^-22 grams. A one gram sample should then have 4.398E21 nuclei in it. Although I also read that the gamma photon that comes from Cesium decay comes when it decays to boron 137 and it only does that 85% of the time. 617
So maybe I should reduce this by 15%?
Another formula tells me activity is equal to the decay constant times the number of nuclei:
lambda = ln(2)/half-life in seconds lambda = <PHONE_NUMBER> seconds lambda = 0.0044423
A(t) = lambda * Nt A(t) = 0.0044423 * 4.398E21 A(t) = 1.9537E19 Bq
Assuming my math is okay 1.9537E19 Bq is how radioactive my sample is. When I tried to search for dose at distance I just found a lot of formulas explaining the inverse square law. As in if you know how the dose at 1m here's how to calculate it at 4m. But I found one formula that said it calculated the dose from activity and range.
H = 1.5E-13 * A * E * n/r^2
Where: H = dose in Sievert/min A = activity E = photon energy 0.6617MeV for the Barium 137? n = # emitted per disintegration. 1 I think for Barium 137 r = distance in meters say 2 meters for this example
That would be: H = 1.5E-13 * 1.9537E19Bq * 0.6617E6 * 1/(2^2) H = 484786233750 Sv/hr
This seems pretty high to me but I don't know for sure. Am I on the right track here or way off base? Have I grossly over estimated how large a real sample of Cesium 137 would be?
Thank you 565