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The half life of a radioactive isotope $'X'$ is $20$ years, 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 simple of a given rock . The age of the rock is estimated to be............$years$
$60$
$80$
$100$
$40 $
Solution
$X$ $\to $ $Y$
Initial number of atoms, $N_0$ $N$
Number of atoms after time $t$ $N$ $N_0-N$
As per question
$\frac{N}{{{N_0} – N}} = \frac{1}{7} \Rightarrow 7N = {N_0} – N$ or $8 N=N_{0}$
${\frac{N}{N_{0}}=\frac{1}{8}}$
As $\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}$ where $n$ is the no. of half lives
$\therefore \quad \frac{1}{8} = {\left( {\frac{1}{2}} \right)^n}$ or ${\left( {\frac{1}{2}} \right)^3} = {\left( {\frac{1}{2}} \right)^n}$
$\therefore n=3$
$n=\frac{t}{T_{1 / 2}} \text { or } t=n T_{1 / 2}=3 \times 20 \text { years }=60 \text { years }$
Hence, the age of rock is $60$ years.
