Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to Calculate the Number of $^{222}\text{Rn}$ Atoms in 1 m$^3$ of Air.
1 picocurie (pCi) is $ 3.7 \times 10^{-2} $ decays per second (Bq), so:
$$
4 \, \text{pCi/L} = 4 \times 3.7 \times 10^{-2} \, \text{Bq/L} = 0.148 \, \text{Bq/L}
$$
Thus, For 1 m$^3$ (1000 L):
$$
R= 0.148 \, \text{Bq/L} \times 1000 \, \text{L} = 148 \, \text{Bq}
$$
We know that the activity $ R $ is related to the number of atoms $ N $ by:
$$
R = \frac{\ln(2)}{t_{1/2}} N
$$
Solve for $ N $:
$$
N = \frac{t_{1/2}\;R }{\ln(2)}
$$
Plug the known;
$$
N = \frac{(3.82\times 24\times 3600)(148)}{\ln(2)} =\color{red}{\bf 7.05 \times 10^7 }\, \rm {atoms}
$$
$$\color{blue}{\bf [b]}$$
The effective exposure region around the person extends 3 cm out from their body. This means calculating the volume of a cylinder with an outer radius of $12.5 + 3 = 15.5$ cm and an inner radius of 12.5 cm, with a height of 1.83 m (180 cm+3 cm=183 cm).
Thus the volume of the air in the shell is given by
$$V=V_{\rm shell}-V_{\rm body}$$
$$
V = \pi (0.155)^2(1.83) -\pi (0.125)^2(1.8)
$$
$$
V =\bf 0.0498\, \rm{m}^3
$$
Now we need to calculate the number of $^{222}\rm{Rn}$ Atoms in This Shell.
From part (a) above, we know that the concentration of $^{222}\rm{Rn}$ atoms is $ 7.05 \times 10^7 \, \rm{atoms/m}^3 $.
So the number of atoms in the shell is given by
$$
N = (7.05 \times 10^7) V=(7.05 \times 10^7)(0.0498)$$
$$
N=\bf 3.51 \times 10^6 \, \rm{atoms}
$$
So, the total decays per second in the shell is given by
$$
R = rN = \dfrac{\ln (2)}{t_{1/2}}N$$
$$
R=\dfrac{\ln (2)}{(3.82\times 24\times 3600)} (3.51 \times 10^6) $$
$$
R =\bf 7.37\, \rm{decay/s}
$$
Only decays within 3 cm can cause exposure, and only 50% of these alpha particles are directed toward the person.
So the decays contributing to exposure per year is given by
$$
N_{\alpha} = \frac{1}{2} (7.37) (265\times 24\times 3600)$$
$$
N_{\alpha}=\bf 1.16 \times 10^8 \, \rm{paticle/yr}
$$
Now we need to calculate the Energy Absorbed per Year:
Each alpha particle has 5.50 MeV of energy, so
$$
E_{\rm yr}= N_{\alpha} E_\alpha= (1.16 \times 10^8) ( 5.5 \times 1.6 \times 10^{-13})$$
$$
E_{\rm yr}= \bf 1.02 \times 10^{-4} \, \rm{J}
$$
For a 65 kg person, a dose $ D $ in Gy is given by
$$
D = \dfrac{E_{\rm yr}}{m}=\frac{1.02 \times 10^{-4}}{65} =\bf 1.57 \times 10^{-6} \, \rm{Gy}
$$
Converting to mSv (using RBE of 20 for alpha particles):
$$
D = 1.57 \times 10^{-6} \times 20=3.138\times 10^{-5}\;\rm Sv/yr $$
$$
D =\color{red}{\bf 3.138 \times 10^{-2}} \, \rm{mSv/yr}
$$
$$\color{blue}{\bf [c]}$$
The calculated dose of 0.63 mSv/yr is too small compared to the average background radiation exposure, which is around 3 mSv/yr from all sources.
While the additional dose from Radon-222 exposure is small relative to natural background radiation, precautions are necessary due to the characteristics of alpha radiation and the higher potential impact on individuals with limited mobility. Setting a limit of 4 pCi/L helps to manage these risks effectively.
Therefore, the EPA's recommendation seems appropriate.
