Question

A radioisotope $'X'$ with a half-life $1.4\times {{10}^{9}}$ years decays to $'Y'$ which is stable. A sample of the rock from a cave was found to obtain $'X'$ and $'Y'$ in the ratio $1:7$. The age of the rock is
$A.1.96\times {{10}^{9}}$ Years
$B.3.92\times {{10}^{9}}$ Years
$C.4.20\times {{10}^{9}}$ Years
$D.8.40\times {{10}^{9}}$ Years

Answer 1

We will use the concept of radioactive decay to solve this problem. Radioactivity is defined as the spontaneous disintegration without any external provocation of the heavy nucleus of an atom. Half-life of a substance is the time it takes for half of a given number of radioactive nuclei to decay. Relation of half-life with radioactive decay should be used here.

Formula used:
To solve this problem given above we are using the following mentioned formula:-
$\dfrac{N}{{{N}_{o}}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{T}}}$ .

Complete step-by-step answer:
From the problem given above we have following parameters with their respective meaning:-
Half-life of the radioisotope $'X'$, $T=1.4\times {{10}^{9}}$ years.
Ratio of $'X'$ and $'Y'$ is $1:7$, which is represented as $\dfrac{{{N}_{X}}}{{{N}_{Y}}}=\dfrac{1}{7}$ .
To solve the given problem we have $\dfrac{N}{{{N}_{o}}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{T}}}$ …………… $(i)$
Where $N$ represents the nuclei remaining at any time and ${{N}_{o}}$ represents the original number of nuclei, $T$ denotes half-life and $t$ represents the age of the rock.
Therefore, we can write
$\dfrac{N}{{{N}_{o}}}=\dfrac{{{N}_{X}}}{{{N}_{X}}+{{N}_{Y}}}$ ………………….. $(ii)$
On putting $(ii)$ in $(i)$ we get,
$\Rightarrow \dfrac{{{N}_{X}}}{{{N}_{X}}+{{N}_{Y}}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{T}}}$
$\Rightarrow \dfrac{1}{1+7}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{T}}}$
$\Rightarrow \dfrac{1}{8}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{T}}}$
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{T}}}$
Solving exponentially we get,
$\dfrac{t}{T}=3$
$\Rightarrow t=3T$ ……………. $(iii)$
Putting the value of $T$ in $(iii)$, we get
$\Rightarrow t=3\times 1.4\times {{10}^{9}}$ Years
$\Rightarrow t=4.2\times {{10}^{9}}$ Years
Therefore, option $(C)4.20\times {{10}^{9}}$ is correct among the given options.

So, the correct answer is “Option C”.

Additional Information: Three main types of radioactive radiations are $\alpha$ decay, $\beta$ decay and $\gamma$ decay. Radioactivity is a statistical process. Rutherford and Soddy analysed radioactive decay with a law which states that ‘’at given time the rate at which particular decay occurs in a radioactive substance is proportional to the number of radioactive nuclei present.’’ Decay rate is also known as Activity whose SI unit is Becquerel $(Bq)$ .

Note: Concept of half-life should be used correctly. We should not be confused between the terms half-life and mean life. Formula should be used correctly with proper units and symbols. Never be confused between $N$ and ${{N}_{o}}$ .