Question

The half life of ${}^{212}Pb$ is $10.6hour$. It undergoes decay to its daughter (unstable) element ${}^{212}Bi$ of half life $60.5$ minute. Calculate the time at which the daughter element will have maximum activity.

Answer 1

${\lambda }_{Pb}=\frac{0.693}{10.6\times 60}=1.0896\times {10}^{-3}{min}^{-1}$
${\lambda }_{Bi}=\frac{0.693}{60.5}=11.45\times {10}^{-3}{min}^{-1}$
$t_{max}=\frac{2.303}{{\lambda }_{Bi}-{\lambda }_{Pb}}\log \frac{{\lambda }_{Bi}}{{\lambda }_{Pb}}$

$=\frac{2.303}{(11.45\times {10}^{-3}-1.0896\times {10}^{-3})}\times \log \frac{11.45\times {10}^{-3}}{1.0896\times {10}^{-3}}=227.1min$