The half-life of a radioactive isotope $X$ is $20 \mathrm{yr}$. It decays to another element $Y$ which is…
The half-life of a radioactive isotope $X$ is $20 \mathrm{yr}$. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio $1: 7$ in a sample of a given rock. The age of the rock is estimated to be
$40 \mathrm{yr}$
$60 y \mathrm{r}$
$80 \mathrm{yr}$
$100 \mathrm{yr}$
Solution
$\begin{aligned}
\text {(As } \frac{N}{N_0}= & \left(\frac{1}{2}\right)^n \\
& \frac{N}{N_0}=\left(\frac{1}{2}\right)^3=\frac{1}{8}
\end{aligned}$
Number of half lives $=3$
$\begin{aligned}
\Rightarrow & T & =20 \mathrm{yr} \\
\therefore & T & =\frac{t}{n}
\end{aligned}$
or $\quad t=T \times n=20 \times 3 \mathrm{yr}=60 \mathrm{yr}$