Question

In a radioactive sample, ${}_{19}^{40}K$ nuclei either decay into stable ${}_{20}^{40}Ca$ nuclei with decay constant $4.5\times 10^{-10}$ per year or into stable ${}_{18}^{40}Ar$ nuclei with decay constant $0.5\times 10^{-10}$ per year. Given that in this sample all the stable ${}_{20}^{40}Ca$ and ${}_{18}^{40}$ $Ar$ nuclei are produced by the ${}_{19}^{40}K$ nuclei only. In time $t\times 10^9$ years, if the ratio of the sum of stable ${}_{20}^{40}Ca$ and ${}_{18}^{40}Ar$ nuclei to the radioactive ${}_{19}^{40}K$ nuclei is $99$, the value of t will be : [Given $ln10=2.3$]

• A. 9.2
• B. 1.15
• C. 4.6
• D. 2.3

Answer 1

Answer: (A)

$\therefore \frac{dN}{dt}=-{\lambda }_{1}N-{\lambda }_{2}N$
$\therefore \frac{dN}{dT}=-\left({\lambda }_1+{\lambda }_2\right)dt$
$\Rightarrow N=N_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)t}$
For $N=N_0-99$ of $N_0=0.01N_0$
We get
$t=\frac{In100}{{\lambda }_1+{\lambda }_2}=\frac{2.3\times 2}{5\times 10^{-10}}$
$t=9.2\times 10^9$ year