For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

Step by step solution

01

Conceptual Understanding

When radium-226 decays, it forms radon-222. At first, the number of radon-222 atoms increases because radon-222 starts accumulating as radium-226 decays. However, since radon-222 itself is radioactive and decays faster than radium-226 due to its shorter half-life, its concentration starts to decrease once it has reached a saturation point where the rate of its formation equals the rate of its decay.

02

Derive Rate Expression

The rate of change of radon-222 (dD/dt) can be expressed as the difference between its rate of production and its rate of decay. The rate of production is directly proportional to number of radium-226 atoms and their decay constant \(\lambda_{\mathrm{p}}\), and the rate of decay is directly proportional to the number of radon-222 atoms and its decay constant \(\lambda_{\mathrm{d}}\). This leads to the differential equation: \(dD/dt = \lambda_{p}(P - D) - \lambda_{d} D\).

03

Apply Calculus

To find the time when the radon-222 atom concentration is at its maximum, we need to find when the rate of change is zero (dD/dt = 0). Substituting the values for the decay constants \(\lambda_{p}\) and \(\lambda_{d}\) and solving the differential equation, we can obtain the time it takes for radon-222 to reach maximum concentration. Solving this, we get t = (\(ln(\lambda_{\mathrm{d}} / \lambda_{\mathrm{p}}) / (\lambda_{\mathrm{d}} - \lambda_{\mathrm{p}})\))

04

Compute Maximum Time

Substituting the values of \(\lambda_{\mathrm{p}}\) and \(\lambda_{\mathrm{d}}\) as 1.37 x \(-14\) yr\(-1\) and 0.181 day\(-1\) respectively (derived from their half-lifes), we gets that it takes approximately 4.8 days for the amount of radon222 to reach its maximum value

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