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: In $10=2.303$]

• A. 1.15
• B. 9.2
• C. 2.3
• D. 4.6

Answer 1

The correct option is B 9.2

$\lambda ={\lambda }_1+{\lambda }_2=4.5\times {10}^{-10}+0.5\times {10}^{-10}$
$\lambda =5\times {10}^{-10}$ per year
$N=N_0{e}^{-\lambda t}$ .... (i)


In time $"t"$ 99% of $K$ decayed, so the undecayed part of the sample
$N$=$N_0-\left(\frac{99N_0}{100}\right)=N_0/100$,
and now putting the value of $N$ in equation (i)

$\frac{N_0}{100}=N_0{e}^{-\lambda t}$

$e^{\lambda t}=100$
$t=\frac{\left(2.303\right)\times 2}{\lambda }$
$\therefore$ $t=\frac{4.6}{\lambda }=\frac{4.6}{5}\times {10}^{10}$ years
$\frac{4.6}{5}\times {10}^{10}=t\times {10}^9$
$t=\frac{46}{5}=9.2$