Question

A radioactive isotope $X$ with a half-life of $1.4\times {{10}^{9}}$ years decays to $Y$ which is stable. A sample of the rock from a cake was found to contain $X$ and $Y$ in the ratio $1:7$. The age of the rock is
$A)\text{ }1.96\times {{10}^{6}}\text{ years}$
$B)\text{ 3}.92\times {{10}^{9}}\text{ years}$
$C)\text{ }4.20\times {{10}^{9}}\text{ years}$
$D)\text{ 8}.40\times {{10}^{9}}\text{ years}$

Answer 1

Hint: This problem can be solved by using the formula for the quantity of the original radioactive isotope present in terms of the number of half lives passed. After getting the number of half lives passed, we can get the age of the rock by multiplying the number with the half life of the radioactive isotope.

Formula used:
$A={{A}_{0}}{{\left( \dfrac{1}{2} \right)}^{n}}$

Complete step by step answer:
We will find the number of half lives passed by using the information of the quantity of the original radioactive isotope left in the quantity of the rock.
The quantity $A$ of radioactive isotope left in a sample after $n$ half lives have passed is given by
$A={{A}_{0}}{{\left( \dfrac{1}{2} \right)}^{n}}$ --(1)
Where ${{A}_{0}}$ is the quantity of the radioactive isotope in the original sample of the rock at the starting of the time period.
Hence, let the original quantity of the radioactive sample be ${{P}_{0}}$. It is given that $X$ decays to stable $Y$. Let the amount of $X$ left in the sample be$P$ and the amount of $Y$ be $Q$.
Let the number of half lives passed be $n$.
Now according to the question,
$P:Q=1:7$
$\therefore \dfrac{P}{Q}=\dfrac{1}{7}$
$\therefore Q=7P$ --(2)
Now, since $X$ decays to stable $Y$, it is obvious that the original quantity of $X$ will be equal to the total quantity of $X$ and $Y$ in the sample.
$\therefore {{P}_{0}}=P+Q$
Using (2), we get,
${{P}_{0}}=P+7P=8P$ --(3)
Now, using (1), we get,
$P={{P}_{0}}{{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow P=8P{{\left( \dfrac{1}{2} \right)}^{n}}$ [Using (3)]
$\Rightarrow \dfrac{P}{8P}={{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow \dfrac{1}{8}={{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow \dfrac{{{1}^{3}}}{{{2}^{3}}}={{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{n}}$
$\Rightarrow n=3$
Hence $n=3$ half lives have passed for the sample of the rock.
Now it is given that the half life of the isotope is $\lambda =1.4\times {{10}^{9}}\text{ years}$.
Now, since, the sample is 3 half lives old, its age will be
$\text{Age = Number of half lives passed}\times \text{Half life}$
$\therefore \text{Age = }1.4\times {{10}^{9}}\times 3=4.2\times {{10}^{9}}\text{years}$
Hence, the age of the rock is $4.2\times {{10}^{9}}\text{years}$.
Therefore, the correct option is $C)\text{ }4.20\times {{10}^{9}}\text{ years}$.

Note: By solving this problem, students must have got a feel for how the ages of rock samples and fossils are determined For determining the age of fossils and trees, usually carbon dating is used in which the radioactive element is the radioactive isotope of carbon, $C-14$ or $_{6}^{14}C$. This is a good choice for a radioactive element since it has a convenient half life and almost all living organisms contain some amount of this radioactive isotope within them.