Physics - Modern Physics Question with Solution | TestHub

Given that,${ }_{19}^{49}K$decays in two stable${ }_{20}^{40}Ca of { }_{18}^{40}Ar$Nuclei. Let there be$N_0$active nuclei of${ }_{19}^{40}K$present at$t=0$and none of${ }_{20}^{40}Ca$and${ }_{18}^{40}Ar$present of$t=0$.
Where${\lambda }_1$(decay constant for$K\to Ca$)$=4.5\times \frac{{10}^{-10}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}$and
${\lambda }_2$(decay constant for$K\to Ar$)$=0.5\times \frac{{10}^{-10}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}$
These form two parallel nuclear reactions.
$\therefore$Equivalent decay consent,$\lambda ={\lambda }_1+{\lambda }_2=5\times \frac{{10}^{-10}}{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}}$
Now, number of active nuclei of${ }_{19}^{40}K$left at$\left(t=t\right)=N$
And according to Law of Radioactivity,
$N=N_0{e}^{-\lambda t}$........(i)
Now, according to question,
$\frac{N_0-N}{N}=99\Rightarrow N_0=100N\Rightarrow N=\frac{N_0}{100}$
Then using equation (i)
$\frac{N_0}{100}=N_0{e}^{-\lambda t}$
$\Rightarrow \mathrm{l}\mathrm{n}100=\lambda t\Rightarrow t=\frac{2\mathrm{l}\mathrm{n}10}{\lambda }=\frac{2\times 2.3}{5\times {10}^{-10}}=9.2\times {10}^{10}\mathrm{ }\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}$