Question

${}^{40}\text{K}$ 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. $4.11\times {10}^9\text{years}$
• B. $2.74\times {10}^9\text{years}$
• C. $5.48\times {10}^9\text{years}$
• D. $1.37\times {10}^9\text{years}$

Answer 1

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

Given that, $T_{1/2}=1.37\times {10}^9\text{years}$

At present,

$\frac{\text{Number of K atoms}}{\text{Number of Ar atoms}}=\frac{1}{7}$

As there were no argon present initially, so

Number of $\text{K}$ atoms present initially = number of $\text{K}$ atoms + number of $\text{Ar}$ atoms present now.

Let,

$\frac{\text{Number of K-atoms present now}}{\text{Number of K-atom present initially}}=\frac{N}{N_0}$

$\Rightarrow \frac{N}{N_0}=\frac{1}{1+7}=\frac{1}{8}$

Let age of rock be $n$ half-lives of $\text{K}-$nuclide. Then,

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$

$\Rightarrow \frac{1}{8}={\left(\frac{1}{2}\right)}^n$

$\Rightarrow n=3$

Therefore, the age of the rock is,

$t=n{T}_{1/2}$

$\Rightarrow t=3\times 1.37\times {10}^9$

$\Rightarrow t=4.11\times {10}^9\text{years}$

Hence, option $\left(A\right)$ is correct.