Question

A radio isotope 'X' with a half life $1.4\times {10}^9$ years decays to 'Y' which is stable. A sample of the rock from a cave was found to contain '$X$' and '$Y$' in the ratio $1:7$. The age of the rock is

• A. $1.96\times {10}^{9}years$
• B. $3.92\times {10}^{9}years$
• C. $4.20\times {10}^{9}years$
• D. $8.40\times {10}^{9}years$

Answer 1

The correct option is C $4.20\times {10}^{9}years$

$\underset{\underset{attimet}{t=0}}{-}\underset{\underset{N_0-x}{N_0}}{X}\to \underset{\underset{x}{0}}{Y}$

$\frac{N_0-x}{x}=\frac{1}{7}\Rightarrow x=\frac{7N_0}{8}$

So, remaining nuclei of: $X=N_0-x=\frac{N_0}{8}=\frac{N_0}{2^3}$
So, three half life periods would have been passed:
$\Rightarrow t=3t_{1/2}=3\times 1.4\times {10}^9=4.2\times {10}^{9}year$