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 $Ar$ 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.