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 Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn.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?