Question

${ }_{19}^{49} K$ isotope of potassium has a half-life of $1.4 \times 10^{9} yr$ and decays to form stable argon, ${ }_{18}^{40}$ Ar. A sample of rock has been taken which contains both potassium and argon in the ratio $1: 7$, i.e.,
$\frac{\text { Number of potassium- } 40 \text { atoms }}{\text { Number of argon }-40 \text { atoms }}=\frac{1}{7}$
Assuming that when the rock was formed no argon- 40 was present in the sample and none has escaped subsequently, determine the age of the rock.

• A. $4.2 \times 10^{9}$ years
• B. $9.8 \times 10^{9}$ years
• C. $1.4 \times 10^{9}$ years
• D. $10 \times 10^{9}$ years

Answer 1

Answer: (A)

At present,
$\frac{\text { Number of } K \text { atoms }}{\text { Number of Ar atoms }}=\frac{1}{7}$
Let age of rock be $n$ half-lives of $K$-nuclide. Then,
$\left(\frac{1}{2}\right)^{n}=\frac{\text { Number of } K \text {-atoms present now }}{\text { Number of } K \text {-atom present initially }}$
$=\frac{1}{1+7}$
where number of $K$ atoms present initially $=$ number of $K$ atoms + number of $A r$ atoms present now.
$\therefore n=3$
So, age of rock is $3$ half-lives of $K$ nuclides,
i.e., $4.2 \times 10^{9}$ years.