We have a cubic room of side $10$ m, into which no fresh air has been allowed to flow for a week. We register a specific activity of radon ($Rn-222$) of $50$ Bqm$^{-3}$. Knowing that $Rn-222$ is a product in the $U-238$ chain, we would like to find the concentration of $U-238$ in the walls of this room. We take for granted that the radon diffuses through a $3$ cm thick layer of wall.

My attempt

I assume that the activity concentration of the Radon is the same as that of the Uranium from which it's coming. I compute the activity of the Radon using $A$ = ($50$ Bqm$^{-3}$)($10^3$m$^3$)$=$ $50 000$ Bq. Then taking one of the four walls, the concentration of Uranium is given by the activity per unit volume in the thin layer through which the Radon diffuses, that is, $C =$ ($50 000$ Bq $)$/($10$ x $10$ x $0.03$ m$^3$)

I think this approach is flawed. Any hint towards a more reasonable solution will be appreciated


One more assumption must be made regarding the $^{222}Rn$ concentration at time zero (week ago). If you assume radioactive equilibrium between $^{222}Rn$ and the parent $^{238}U$ you can proceed like you did. But instead of taking one wall, you should take into account all the walls made of the $^{238}U$ contaminated material.

On the other hand, if you assume that the room was well ventilated until week ago, i.e. the $^{222}Rn$ concentration in the air was zero at the beginning, you must take into account the activity build-up. Starting with zero activity, $^{222}Rn$ will gradually grow until the equilibrium with parent radionuclide is reached (it takes approximately one month in this case). Let's assume the radioactive equilibrium between the $^{238}U$ and its decay products exists until $^{226}Ra$ (which decays to the $^{222}Rn$). The decay-growth equations describing such situation are:

$ \frac{dN_1}{dt}=-\lambda_1N_1 $

$ \frac{dN_2}{dt}=-\lambda_2N_2+\lambda_1N_1 $

Where $N_1$, $N_2$ are numbers of atoms of $^{238}U$ and $^{222}Rn$ respectively (as a functions of time) and $\lambda_1$, $\lambda_2$ are decay constants of the $^{238}U$ and $^{222}Rn$ respectively.

Solution to this system of equations is

$ N_2=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1} (e^{-\lambda_1 t}-e^{-\lambda_2 t}) \Rightarrow A_2=A_1 \frac{\lambda_2}{\lambda_2-\lambda_1} (e^{-\lambda_1 t}-e^{-\lambda_2 t}) $

Since $\lambda_2= 2.1 \times 10^{-6} \ s^{-1} \gg \lambda_1= 4.9 \times 10^{-18} \ s^{-1}$ and $1 \ week \doteq 6 \times 10^{5} \ s$ we can approximate this solution as

$ A_2 \doteq A_1(1-e^{-\lambda_2 t}) $

where $t$ is a time of $^{222}Rn$ "building up" in seconds and $A_1$, $A_2$ are activities of $^{238}U$ and $^{222}Rn$ respectively and $N_0$ is the initial number of $^{238}U$ atoms. It means, that after a week, the $^{222}Rn$ will grow to the ~ 70 % of its equilibrium activity, see the figure:

Rn-22 buildup

Assuming, that $^{222}Rn$ emanates from all of the walls including the floor and the ceiling, we obtain the activity $A_1$ of $^{238}U$ per unit volume of a wall material as

$ A_1 =\frac{A_2 V_{air}}{V_{material}(1-e^{-\lambda_2 t})}=\frac{50 \times 10^3}{6 \times 10^2 \times 0.03 \times (1-e^{2.1 \times 10^{-6} \times 6 \times 10^5})} \doteq 3900 \ Bq/m^3 $

  • $\begingroup$ Thank you mira, this is a brilliant response. So, if the time of build-up (1 week in this case) had not been specified, I would have had no choice but to assume radioactive equilibrium between the uranium and radon from the beginning? $\endgroup$ – 2good4this Feb 4 '16 at 12:29
  • $\begingroup$ @2good4this Happy to help! If the time is not specified you must make an assumption or leave it as an insufficiently defined problem. Assuming the equilibrium seems as natural choice, but of course it depends on particular situation. Your results should be accompanied by a specification of your assumptions anyway... $\endgroup$ – mira Feb 4 '16 at 13:38
  • $\begingroup$ Just realised that your solution gives the activity per unit volume of the uranium, how would one obtain the concentration of uranium in the walls from this quantity? $\endgroup$ – 2good4this Feb 5 '16 at 2:24
  • $\begingroup$ It is, in fact, a concentration- "concentration is the abundance of a constituent divided by the total volume of a mixture". Abundance is expressed here in the terms of Becquerels. You can easily convert Bq to grams or mols-1 $m^3$ of the wall material contains $3900 \ Bq= 0.3 \ g = 1.3 \times 10^{-3} \ mol$ of $^{238}U$ (hint: specific activity). The most common way in this case would be in $Bq/kg$ but you don't know the material density... $\endgroup$ – mira Feb 5 '16 at 7:46

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