The most significant source of natural radiation is radon- 222 . \({ }^{222} \mathrm{Rn}\), a decay product of \({ }^{238} \mathrm{U}\), is continuously generated in the earth's crust, allowing gaseous \(\mathrm{Rn}\) to seep into the basements of buildings. Because \({ }^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of \(3.82\) days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \({ }^{238} \mathrm{U}\) decays to \({ }^{222} \mathrm{Rn}\) ? What nuclei are produced when 222 \(\mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \({ }^{222} \mathrm{Rn}\) ? c. Another problem associated with \({ }^{222} \mathrm{Rn}\) is that the decay of \({ }^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) \(3.11 \mathrm{~min}\) ) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) -particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \({ }^{222} \mathrm{Rn}\) levels not exceed 4 pCi per liter of air \(\left(1 \mathrm{Ci}=1\right.\) curie \(=3.7 \times 10^{10}\) decay events per second; \(\left.1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\right) .\) Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \({ }^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(222 \mathrm{Rn}\) per liter of air.

Step by step solution

01

a. Calculate α and β particles from 238U to 222Rn and decay products of 222Rn

To find the number of α and β particles emitted, we must examine the decay series of \({ }^{238} \mathrm{U}\) to \({ }^{222} \mathrm{Rn}\). The decay series can be obtained either from the Decay Series Chart or via several online sources. The decay series is: \({ }^{238} \mathrm{U} \rightarrow { }^{234} \mathrm{Th} \rightarrow { }^{234} \mathrm{Pa} \rightarrow { }^{234} \mathrm{U} \rightarrow { }^{230} \mathrm{Th} \rightarrow { }^{226} \mathrm{Ra} \rightarrow { }^{222} \mathrm{Rn}.\) Counting the particles emitted in each decay, we get: - 6 alpha particles (from \({ }^{238}\mathrm{U}\), \({ }^{234}\mathrm{Th}\), \({ }^{234}\mathrm{U}\), \({ }^{230}\mathrm{Th}\), \({ }^{226}\mathrm{Ra}\)) - 2 beta particles (from \({ }^{234}\mathrm{Pa}\)) Thus, a total of 6 α particles and 2 β particles are produced when \({ }^{238}\mathrm{U}\) decays to \({ }^{222}\mathrm{Rn}\). As for the decay products of \({ }^{222}\mathrm{Rn}\), according to the decay series: \({ }^{222}\mathrm{Rn} \rightarrow {}^{218}\mathrm{Po}\). So, polonium-218 (218Po) is produced.

02

b. Concerns of inhaling 222Rn

Despite being a noble gas, inhaling \({ }^{222}\mathrm{Rn}\) is a concern because it decays into \({ }^{218}\mathrm{Po}\) which is a solid alpha emitter. These solid decay products can get lodged in the lung tissue, causing an increased risk of lung cancer over time as a result of alpha-particle exposure.

03

c. Identity of solid product and its α-particle production

The solid product of \({ }^{222}\mathrm{Rn}\) decay is \({ }^{218}\mathrm{Po}\) (polonium-218) which was determined in the previous step. To determine the balanced equation for this decay, we must consider that it is an alpha-particle decay, meaning an \({ }^{4}\mathrm{He}\) (Helium-4) nucleus is released. The equation for the decay is: \({ }^{222}\mathrm{Rn} \rightarrow { }^{218}\mathrm{Po} + { }^{4}\mathrm{He}\) The \({ }^{218}\mathrm{Po}\) is a more potent alpha-particle producer than \({ }^{222}\mathrm{Rn}\) due to its shorter half-life (3.11 minutes compared to 3.82 days). A shorter half-life means that the isotope decays more rapidly, releasing alpha particles at a faster rate which increases the potential for biological damage.

04

d. Conversion of radon concentration units

Firstly, we will convert pCi per liter to decay events per liter: Given that 1 pCi = \(1 \times 10^{-12}\) Ci and 1 Ci = \(3.7 \times 10^{10}\) decay events per second, \(4.0 \ \mathrm{pCi} \cdot \frac{1 \times 10^{-12} \ \mathrm{Ci}}{1 \ \mathrm{pCi}} \cdot \frac{3.7 \times 10^{10} \ \mathrm{decays/s}}{1 \ \mathrm{Ci}} = 1.48 \times 10^{-2} \ \mathrm{decays/s}\) per liter of air Now, we will convert decays per second into \({ }^{222}\mathrm{Rn}\) atoms per liter: Given that the half-life of \({ }^{222}\mathrm{Rn}\) is 3.82 days, the decay constant (λ) can be calculated as follows: \(λ = \frac{0.693}{t_{1/2}} = \frac{0.693}{3.82 \times 24 \times 3600\ \mathrm{s}} = 2.098 \times 10^{-6}\ \mathrm{s}^{-1}\) So, the number of \({ }^{222}\mathrm{Rn}\) atoms per liter is: \(\frac{1.48 \times 10^{-2} \ \mathrm{decays/s}}{2.098 \times 10^{-6}\ \mathrm{s}^{-1}} = 7.057 \times 10^{3}\ \mathrm{atoms}\) per liter of air Finally, we will convert \({ }^{222}\mathrm{Rn}\) atoms into moles per liter: Given that the Avogadro constant is approximately \(6.022 \times 10^{23}\) particles per mole, \(\frac{7.057 \times 10^{3}\ \mathrm{atoms}}{6.022 \times 10^{23} \ \mathrm{particles/mol}} = 1.17 \times 10^{-20}\ \mathrm{mol}\) per liter of air Thus, 4.0 pCi of \({ }^{222}\mathrm{Rn}\) per liter of air is equivalent to 1.17 \(\times 10^{-20}\) moles of \({ }^{222}\mathrm{Rn}\) per liter of air.

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