Question

${}_{19}{K}^{40}$ decays into $C{a}^{40}$ and $A{r}^{40}$ simultaneously as follows.
Ratio of atoms of $Ar$ to $K$ in a rock sample is $3:10$. Calculate age of rock sample assuming that source of $Ar$ is radioactive decay of ${}^{\prime }{K}^{40+}$ only and no $Ar$ was present at the time of formation of rock.
If your answer is $ln\left(x\right)\times {10}^8$ years, then the value of $x$ is:

Answer 1

${}_{19}{K}^{40}⟶{Ca}^{40}+{Ar}^{40}molesatt=0A_o--att=t{A}_o-xxx$

As radioactive decay follows first order and given that,

$\frac{\left[{Ar}^{40}\right]}{\left[{}_{19}{K}^{40}\right]}=\frac{3}{10}=\frac{x}{A_o-x}$

Using first order kinetics equation,

$t=\frac{1}{K}\ln \left(\frac{A_o}{A_o-x}\right)=\frac{1}{K}\ln \left(\frac{A_o-x+x}{A_o-x}\right)=\frac{1}{K}\ln (1+\frac{x}{A_o-x})t=\frac{1}{K}\ln (1+\frac{3}{10})=\frac{\ln \left(1.3\right)}{K}=\frac{0.26}{K}$

K value is not given in the question, this is the age of the rock.