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

Answer: (C)

Number of nuclei at t = 0
Number of nuclei after time t (As per question) $\begin{array}{ccc}X& \to & Y\\ N_0& & 0\\ N_0-x& & x\end{array}$
$\frac{N_0-x}{x}=\frac{1}{7}$
$7N_0-7x=x$ or $x=\frac{7}{8}{N}_0$
$\therefore$ 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.
$\therefore$ $t=n{T}_{1/2}=3\times 1.4\times 10^9$ Years
$=4.2\times 10^9$ Years
Hence, the age of the rock is $4.2\times 10^9$ Years.