Question

In a moon rock sample, the ratio of stable argon-40 atoms present to the number of radioactive potassium-40 atoms is 7:1. Assume all argon atoms were produced by the decay of potassium atoms, with a half-life of $2.5 \times {10^9}$ years, what is the age of the rock?
(A) $2.5 \times {10^9}$ years
(B) $5 \times {10^9}$ years
(C) $7.5 \times {10^9}$ years
(D) ${10^{10}}$ years

Answer 1

Hint
The rock will die or fully decay when all the potassium-40 is converted to argon-40 and the argon-40 gets decayed as well completely.

Complete step by step answer
Half-life of a substance is the time taken by the substance to decay up to half of its initial concentration.
The age of the rock will be equal to the time required for potassium-40 to decay into argon-40 and the decay of argon-40 is not considered.
The atomic reaction taking place in the decay of the rock is given by;
${}^{40}K \to {}^{40}Ar$
For the above reaction the kinetics can be represented as in the following table;

| ${}^{40}K$ | ${}^{40}Ar$
---|---|---
Number of atoms at t=0| ${N_0}$ | $0$
Number of atoms at t=t| $N$ | ${N_0} - N$

According to the kinetics table above we conclude that initially the number of ${}^{40}K$ is ${N_0}$ ant number of argon atoms is zero. After some time the number of atoms of ${}^{40}K$ is reduced to N so the number of argon atoms becomes, ${N_0} - N$ .
Also we know that ratio of stable argon to radioactive potassium is 7:1;
Hence $\dfrac{{{N_0} - N}}{N} = \dfrac{7}{1}$
Thus solving the above equation we get;
${N_0} - N = 7N$
$8N = {N_0}$
So, $N = \dfrac{{{N_0}}}{8}$ (equation: 1)
Here we also know that, the ratio of number of ${}^{40}K$ at time t to the number of ${}^{40}K$ initially is given by;
$\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^n}$ (here, n is the n ratio of age of substance to its half-life)
So we substitute the values from equation 1 to the above equation and w get;
${\left( {\dfrac{1}{2}} \right)^n} = \dfrac{1}{8}$
${\left( {\dfrac{1}{2}} \right)^n} = {\left( {\dfrac{1}{2}} \right)^3}$
Thus, $n = 3$
Also as mentioned earlier in the comments n is given by;
$n = \dfrac{t}{T}$ Here t is the age of the substance and T is the half-life of the substance.
So as T is given and is $2.5 \times {10^9}$ years, we get;
$t = n \times T = 3 \times 2.5 \times {10^9}$
Thus, age= $t = 7.5 \times {10^9}$ years.
Therefore, option (C) is correct.

Note
-The Argon-40 atoms are stable and so the rock’s age will be equal to the time taken by potassium-40 to decay into argon-40 completely.
-The half-life of a substance is the time required to decay to half of the initial concentration, but the time needed for a full decay is not the twice of half time.
-After decaying to half of the initial concentration this new concentration becomes the initial concentration.