Question

A radioactive isotope X with a half-life of $6.93\times {{10}^{9}}$ years decays to Y which is stable. A sample of rock from the moon was found to contain both the elements X and Y which were in the ratio of 1:7. The age of the rock is.
(A) $2.079\times {{10}^{10}}years$
(B) $1.94\times {{10}^{10}}years$
(C) $1.33\times {{10}^{10}}years$
(D) ${{10}^{10}}years$

Answer 1

Hint: To understand this question, we should note that X is producing Y which is stable. That means there is no further decay. And we should note that Y is producing only from X, so we can now find the answer by doing calculation.

Complete step by step answer:
In this question, it is given that there is a radioactive isotope X, which has a half-life of $6.93\times {{10}^{9}}$years.
$X={{t}_{\dfrac{1}{2}}}=6.93\times {{10}^{9}}years.$
And, in this question it is given that Y is producing only from X.
$X\to Y\to (No\,further\,decay)$
By radioactive decay law, we know that: $Y=X\left( 1-{{e}^{-\lambda t}} \right)$
And, in the question it is given that X and Y are found in moon rock in the ratio 1:7.
$\dfrac{X}{Y}=\dfrac{1}{7}$
The above ratio states that initially we had ${{X}_{0}}$ and from this some part got converted into Y and some part remained as X. then, we multiplied it by k.
$\begin{aligned} & {{X}_{0}}\to Y=7\times k \\\ & \downarrow \\\ & X=1\times k \\\ \end{aligned}$
After multiplication by k, we find that total ${{X}_{0}}$ was 8k.
$\begin{aligned} & Y=\dfrac{7}{8}{{X}_{0}} \\\ & X=\dfrac{1}{8}{{X}_{0}} \\\ \end{aligned}$
From the above equation we can say that the final value of X that remains is only$\dfrac{1}{8}{{X}_{0}}$. And this states that three half-lives are gone. Let us calculate this by calculating this after each half life:
${{X}_{0}}\xrightarrow{one\,half\,life}\dfrac{{{X}_{0}}}{2}\xrightarrow{\operatorname{Sec}ond\,half\,life}\dfrac{{{X}_{0}}}{4}\xrightarrow{Third\,half\,life}\dfrac{{{X}_{0}}}{8}$
So, three half-lives are gone. $(3{{t}_{\dfrac{1}{2}}})$
Now, we can easily solve our question.
$\begin{aligned} & 3{{t}_{\dfrac{1}{2}}} \\\ & 3\times (6.93\times {{10}^{9}}) \\\ & 20.79\times {{10}^{9}}years \\\ \end{aligned}$
Or $2.079\times {{10}^{10}}years$
So, from calculation we can say that answer of this question option A.

Note:We should know the radioactive decay. It is the phenomenon that is performed by the nuclei of an atom as a result of nuclear instability or we can say that it is a process by which the nucleus of an unstable atom loses energy by emitting radiation.
And one more concept that is important that is Half-life (symbol t1⁄2). It is the time required for a quantity to reduce to half of its initial value. We use this term in nuclear physics to describe how quickly unstable atoms undergo, or how long stable atoms survive.