The half-life of a radioactive isotope $$X$$ is $$20\,yr.$$  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

Solution :
As we know that
$${N = {N_0}{{\left( {\frac{1}{2}} \right)}^n}}$$
$$\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{{1 + 7}} = \frac{1}{8}$$
So, number of half lifes = 3
$$\eqalign{ & \Rightarrow T = 20\,yr \cr & \therefore T = \frac{t}{n} \cr & \Rightarrow t = Tn = 20 \times 3\,yr = 60\,yr \cr} $$