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.,
$\frac{\text{Number of potassium-40 atoms}}{\text{Number of argon-40 atoms}}=\frac{1}{7}$
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

The correct option is A $4.2\times {10}^{9}years$

$K\to Ar$
$N=N_o{e}^{-lambdat}$
If initial no. of $K$ atoms $N_o=8$
Then N(t) =1
$\lambda =\frac{ln2}{t_{1/2}}$
$1=8e^{-\lambda t}$
$t=\frac{3ln2}{\lambda }$
$t=3t_{1/2}=4.2\times {10}^{9}years$