Question

Radioactive decay will occur as follows.
${}_{86}^{220}Rn\to {}_{84}^{216}PO+{}_2^{4}He$ Hald life $=55$s
${}_{84}^{216}Po\to {}_{82}^{212}Pb+{}_2^{4}He$ Half life$=0.66$s
${}_{82}^{822}Pb\to {}_{82}^{212}BL+{\lambda }^o$e Half life $=10.6$h
If a certain mass of radon(Rn$=220$) is allowed to decay in a certain container; then after $5$ minutes the element with the greater mass will be.

• A. Radon
• B. Polonium
• C. Lead
• D. Bismuth

Answer 1

The correct option is C Lead

Let $M_0$ be initial mass of $Rn-220$
Number of half-lives of Rn in $5$ minutes $=\frac{5min}{55s}$
$=\frac{5\times 60}{55}=5.5$
$\therefore$ Mass of Rri$-220$(left)$={\left(\frac{1}{2}\right)}^{5.6}{M}_0=\frac{M_0}{45}$
$\therefore$ The mass converted into $Po-216$ is $\left(44/45\right)$
$M_0$, Number of half-lives of Po$-216$ in $5$ minutes$=\frac{5min}{0.66}=\frac{300}{0.66}=455$
$\therefore$ Mass of $po-216$ left
$={\left(\frac{1}{2}\right)}^{455}\times \frac{44}{45}{M}_0\to 0$
This means that Po$-216$ formed will soon decay to lead. Hence, lead will have maximum mass nearly $\left(\frac{44}{45}\right)M_0$.