Radioactive decay will occur as follows aligned & _ 86 ^ 220 Rn _...

Let $M_0$ be initial mass of $R n-220$$\begin{aligned} & \text{ Number of half-lives of } R n \text{ in } 5 \text{ minutes }=\frac{5 \mathrm{ min}}{55 \mathrm{ s}} \\ & =\frac{5 \times 60}{55} \approx 5.5 \end{aligned}$$\therefore$ Mass of $R n-220$ (left) $=\left(\frac{1}{2}\right)^{5.5} M_0=\frac{M_0}{45} \therefore$ The mass converted into Po -216 is (44/45) $M_0$, Number of half-lives of $P o-216$ in 5 minutes $=\frac{5 \mathrm{ min}}{0.66}=\frac{300}{0.66} \approx 455 \therefore$ Mass of $P o-216$ left $=\left(\frac{1}{2}\right)^{455} \times \frac{44}{45} M_0 \longrightarrow 0$ This means that $P o-216$ formed will soon decay to lead. Hence, lead will have maximum mass nearly \left(\frac{44}{45}\right)$ M_0