Question

A radioactive isotope $X$ with a half-life of $1.4\times {10}^9$ years decays to $Y$ 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. $1.96\times {10}^{6}years$
• 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 $4.20\times {10}^{9}years$

Initial amount $=1+7=8$
So, from $N=N_0(\frac{1}{2}{)}^n$
We have $\frac{1}{8}=(\frac{1}{2}{)}^n$
or $(\frac{1}{2}{)}^3=(\frac{1}{2}{)}^n$
or $n=3$
Age of rock $=nT)1/2$
$=3\times 1.4\times {10}^9$
$=4.2\times {10}^{9}years$