Question

The half life of ${}^{212}Pb$ is $10.6$ hours and that of its daughter element ${}^{212}Bi$ is $60.5$ minutes. After how much time will the daughter element have maximum activity?

Answer 1

${\lambda }_p=\frac{0.693}{10.6\times 60}=0.001089{min}^{-1}$
${\lambda }_d=\frac{0.693}{60.5}=0.01145{min}^{-1}$
$t_{max}=\frac{2.303}{({\lambda }_d-{\lambda }_p)}{\log }_{10}\frac{{\lambda }_d}{{\lambda }_p}$
$=\frac{2.303}{0.01145-0.001089}{\log }_{10}\left[\frac{0.01145}{0.001089}\right]$
$=222.2758{\log }_{10}\frac{0.01145}{0.001089}$
$=227.1min$
$=3.785$ hours