Question

The half-life of ${}^{212}Ph$ is 10.6 h, that of its daughter ${}^{212}Bi$ is 60.5 min. How long will it take for a maximum daughter activity to grow in freshly separated ${}^{212}Pb$?
Plan We define time of maximum activity by equation
$t_{max}=\frac{2.303}{({\lambda }_d-{\lambda }_p)}log\frac{{\lambda }_d}{{\lambda }_p}$
where ${\lambda }_d$ is the disintegration constant of daughter element and ${\lambda }_p$ that of parent element.

• A. $227$ min
• B. $287$ min
• C. $247$ min
• D. None of these

Answer 1

The correct option is A $227$ min

${\lambda }_d=\frac{0.693}{T_{50}{(}^{212}Bi)}=\frac{0.693}{60.5min}$
$=0.01145mi{n}^{-1}$
${\lambda }_d=\frac{0.693}{T_{50}{(}^{212}Pb)}=\frac{0.693}{10.6\times 60}$
$=0.00109mi{n}^{-1}$
$\therefore$ $t_{max}=\frac{2.303}{(0.01145-0.00109)}log\left(\frac{0.01145}{0.00109}\right)$
$=227$ min