Question

A radioisotope 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

Number of nuclei at t = 0
Number of nuclei after time t
(As per question) $\quad\begin{matrix}X&\to&Y\\ N_{0}&&0\\ N _{0}-x&&x\end{matrix}$
$\quad\frac{N_{0}-x}{x}=\frac{1}{7}$
$7N_{0}-7x=x$ or $x=\frac{7}{8} N_{0}$
$\therefore\quad$ Remaining nuclei of isotope $X$
$=N_{0}-x=N_{0}-\frac{7}{8}N_{0}=\frac{1}{8}N_{0}=\left(\frac{1}{2}\right)^{3}N_{0} $
So three half lives would have been passed.
$\therefore\quad$ $t=nT_{1 /2}=3\times1.4\times10^{9}$ Years
$=4.2 \times10^{9}$ Years
Hence, the age of the rock is $4.2\times10^{9}$ Years.