Question

${}_{1}9K^{4}0$ isotope of potassium has a half life of $1.37\times {10}^9$ years and decays to an isotope of argon which is stable in a particular sample of moon rock, the ratio of potassium atoms to argon atoms was found to be $1:7$. The age of the rock, assuming that originally there was no argon present is

• A. $4.11\times {10}^{9}year$
• B. $2.74\times {10}^{9}years$
• C. $5.48\times {10}^{9}yeras$
• D. $1.37\times {10}^{9}year$

Answer 1

The correct option is A $4.11\times {10}^{9}year$

$\begin{array}{c}\text{Initially, there was No Argon .}\\ \text{Finally, ratio of}K\text{to}Ar,K/Ar=\frac{1}{7}\\ \text{Then, ratio of potassium initial to}\\ \text{potassium left is "}{8}^{\prime \prime }\left[\frac{k_0}{k}\right]\end{array}$

$\begin{array}{c}\text{Age of rock}=\frac{t_{1/2}}{\ln 2}\ln 8\\ =3t_{1/2}\\ =3(1.37\times {10}^{{}_9})\\ \therefore \text{Age of rock is}4.11\times {10}^9\text{years.}\end{array}$