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

Hint: 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}} $ .