Question

The $_{19}^{40}{\text{K}}$ isotope of potassium has a half-life of $1.4 \times {10^9}\,{\text{yr}}$ and decay to form stable argon, $_{18}^{40}{\text{Ar}}$. A sample of rock has been taken which contains both potassium and argon in the ratio 1:7 i.e., $\dfrac{{{\text{Number of potassium}} - {\text{40 atoms}}}}{{{\text{Number of argon}} - {\text{40 atoms}}}} = \dfrac{1}{7}$. Assuming that when the rock was formed no argon-40 was present in the sample and none has escaped subsequently, determining the age of the rock.
A. $4.2 \times {10^9}\,{\text{yr}}$
B. $9.8 \times {10^9}\,{\text{yr}}$
C. $1.4 \times {10^9}\,{\text{yr}}$
D. $10 \times {10^9}\,{\text{yr}}$

Answer 1

We can use the formula for the decay constant of the radioactive element in terms of half-life of the radioactive element. Also the decay equation which gives the relation between the initial population of the radioactive element and population at time t. First we should determine the decay constant for the decay of potassium and then calculate the age of the rock using the decay equation.

Formulae used:
The decay constant $\lambda$ of a radioactive element is
$\lambda = \dfrac{{0.693}}{T}$ …… (1)
Here, $T$ is the half-life of the radioactive element.
The decay equation for the radioactive element is given by
$N = {N_0}{e^{ - \lambda t}}$ …… (2)
Here, ${N_0}$ is the initial population of the radioactive element, $N$ is the population of the radioactive element at any time $t$ and $\lambda$ is the decay constant for the decay of the radioactive element.

Complete step by step answer:
We have given that the half-life of the radioactive element is $1.4 \times {10^9}\,{\text{yr}}$.
$T = 1.4 \times {10^9}\,{\text{yr}}$
We have given that the ratio of the number of potassium atoms and argon atoms at the present age is given by
$\dfrac{{{\text{Number of potassium}} - {\text{40 atoms}}}}{{{\text{Number of argon}} - {\text{40 atoms}}}} = \dfrac{1}{7}$
We have asked to calculate the age of the rock.Let us first calculate the decay constant for the radioactive decay of potassium.Substitute $1.4 \times {10^9}\,{\text{yr}}$ for $T$ in equation (1).
$\lambda = \dfrac{{0.693}}{{1.4 \times {{10}^9}\,{\text{yr}}}}$
$\Rightarrow \lambda = 4.95 \times {10^{ - 10}}\,{\text{y}}{{\text{r}}^{ - 1}}$
Hence, the decay constant for the given radioactive decay of potassium is $4.95 \times {10^{ - 10}}\,{\text{y}}{{\text{r}}^{ - 1}}$.

Let us now calculate the age of the rock.Let $t$ be the age of the rock.Initially, no argon was formed in the rock sample. This means that the total number ${N_0}$ of potassium atoms in the rock when it has not started decaying is given by the sum of the number of potassium and argon atoms in the rock at the present time.
${N_0} = \left( {{\text{Number of potassium}} - {\text{40 atoms}}} \right) + \left( {{\text{Number of argon}} - {\text{40 atoms}}} \right)$
$\Rightarrow {N_0} = \left( 1 \right) + \left( 7 \right)$
$\Rightarrow {N_0} = 8$
Hence, the initial population of the potassium atoms in the rock is 8.
The population of potassium atoms in the rock at the present time is 1.
$\Rightarrow N = 1$
Substitute 1 for $N$, 8 for ${N_0}$ and $4.95 \times {10^{ - 10}}\,{\text{y}}{{\text{r}}^{ - 1}}$ for $\lambda$ in equation (2).
$1 = 8{e^{ - \left( {4.95 \times {{10}^{ - 10}}\,{\text{y}}{{\text{r}}^{ - 1}}} \right)t}}$
$\therefore t = 4.2 \times {10^9}\,{\text{yr}}$
Therefore, the age of the rock is $4.2 \times {10^9}\,{\text{yr}}$.

Hence, the correct option is A.

Note: One can also solve the same question by another method. One can consider that the age of the rock is equal to n half-lives of potassium and calculate the value of the n using the relation between initial and present population of potassium. Then multiply these numbers of half-lives by the half-life of the potassium atom which gives the age of the rock.