# Talking Radon: Activity or Concentration

Currently I'm working on my PhD in environmental sciences and we are measuring Radon-222 in air using a commercially available detector. The results from the detector are given in units Becquerel per cubic meter (Bq/m³).

For me it sounds more natural to talk about radon concentration while interpreting the results but my supervisor suggested to use activity instead. I'm not sure how to argue with that. Of course, Becquerel is defined as the unit of specific activity. But then this also describes an ammount of radioactive material in which one nucles decays per second. So in the end we have a ammount of radioactive material divided by the volume of the mixture? Which then, per definition, would be called a concentration?

In the literature regarding the topic it seems that somehow the two words are used synonymous. Some authors tend to prefer one above the other but I couldn't devise a rule from that.

• Radon has a specific half-life. A given quantity (concentration) of Radon in the air will have a number of decays per unit volume per unit time directly related to that half-life. Once it decays, it isn't Radon anymore. So, yes, one can go back and forth between number of decays per unit time and how much Radon per unit volume there are in the air. – Jon Custer Feb 5 '19 at 13:58
• @JonCuster Sounds like an answer to me – BioPhysicist Feb 5 '19 at 15:48
• So if I understood you correctly. you would say that the usage of both names, radon activity and radon concentration, is legit when interpreting the results? – sdb Feb 6 '19 at 7:59

As an example, $$3~\mu$$g/m$$^3$$ of radon-222 versus 3 $$\mu$$g/m$$^3$$ of hydrogen-3 (tritium) have an order of magnitude different activity density (along with a different type of radiation and energies): $$3~\frac{\mu\mathrm{g}}{\mathrm{m}^3}\text{ of } ^{222}\mathrm{Rn} \to 1.7\times 10^{10} \mathrm{Bq/m^3}$$ $$3~\frac{\mu\mathrm{g}}{\mathrm{m}^3}\text{ of } ^{3}\mathrm{H} \to 1.1\times 10^{9} \mathrm{Bq/m^3}$$
Edit to add carbon-14: $$3~\frac{\mu\mathrm{g}}{\mathrm{m}^3}\text{ of } ^{14}\mathrm{C} \to 4.9\times 10^{5} \mathrm{Bq/m^3}$$