Question

The half-life of a radioactive isotope $X$ is \[50\] years. It decays to another element $Y$ which is stable. The two elements $X$ and .. were found to be in the ratio \[1:15\] in a sample of a given rock. The age of the rock was estimated to be
A. \[150\] years
B. \[200\]years
C. \[250\]years
D. \[100\] years

Answer 1

Hint: Here we have calculated the age of the radioactive rock. Given that the ratio of the radioactive elements present in the rock is \[1:15\] . To find the age of the rock, we can use the radioactive law of decay by substituting the given data, as it relates the half-life and the ratio of decay.

Formula used:
$N=N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$

Complete step-by-step solution:
We know that the radioactive decay law states that the decay in the radioactive element $d\;N$ depends on the number of radioactive elements present initially$N$, over the time $d\;t$.
$\dfrac{-dN}{N}=\lambda dt$, where $\lambda$ is the radioactive decay constant.
The radioactivity is often measured in terms of curie or becquerel, depending on the nature of the element.
The decay constant $\lambda$ can be expressed in terms of the half-life as $\lambda=\dfrac{0.693}{T_{\dfrac{1}{2}}}$
From the above two statements, we get, $N=N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$, where $N$ is the amount of elements, with $N_0$ initial elements with half-life $T$, decayed at the given time $t$.
Given that $X$ and $Y$are in the ratio \[1:15\], then $N_0=X+Y=16$, and half-life $T=50 \;years$
Substituting, the given in $N=N_0\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$, we have
$\implies \dfrac{1}{16}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{50}}$
 $\implies\left(\dfrac{1}{2}\right)^{4}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{50}}$
Comparing the powers we have
$\implies 4=\dfrac{t}{50}$
$\therefore t=4\times 50=200$
Thus the age of the rock was estimated to be $200\;years$.
Hence the correct answer is option B. \[200\]years

Note: The radioactive decay law can be expressed in terms of fraction as shown above or in terms of exponential majorly. It can also be represented in terms of the half-life of the radioactive element as shown above. Knowing these conversions will be helpful in solving the questions. However, note that in all the expressions the common terms of the equation remain the same.