Stalin Posted February 26, 2020 Report Share Posted February 26, 2020 (edited) Any chance of you putting a radon released from a massive source tool? This calculates the amount of radon released from a pile of uranium ore. As radon is continually generated by the decay of radium, some fraction of radon (the emanation fraction) escapes. Grade, mass and emanation fraction are input into the program and the amount of radon generated per second is returned. Edited February 26, 2020 by Stalin Quote Link to comment Share on other sites More sharing options...
Florian Michelin Posted February 27, 2020 Report Share Posted February 27, 2020 That is something we can work on. Would you use the emanation rate from exposed surface J (pCi/m2s) or emanating powers B (pCi/m3s)? Or both? Quote Link to comment Share on other sites More sharing options...
Stalin Posted February 27, 2020 Author Report Share Posted February 27, 2020 The grade times the mass gives you the total amount of U. Ra will be in secular equilibrium with U. The specific activity of U is 12350 Bq/g. Each Bq of Ra makes 1 atom of Rn per second. That gives you the amount of Rn (in atoms) generated inside the ore. Multiply by the emanation fraction to get the amount released (in atoms). Multiplying that by 0.0000021 converts atoms of Rn to Bq. For the water, its very similar. The emanation fraction, in that case, will give you the fraction of the radon that stays in the ore vs what gets into the pore space. Quote Link to comment Share on other sites More sharing options...
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