Question

A radioactive element X converts into another stable element Y. A radioactive element XX converts into another stable element YY. Half-life of XX is $3hrs$. Initially, only X nuclei are present. After time $'t'$, the ratio of the number of nuclei of X to that of Y is found to be $1:7$. Therefore
(A) \[t = 3hrs\]
(B) $t = 6hrs$
(C) $t = 9hrs$
(D) $t = 15hrs$

Answer 1

Hint: To solve this question, we need to evaluate the given ratio after each half life. The time when the required ratio will become equal to the given ratio, will be the correct answer.

Complete step-by-step solution:
Let the initial number of nuclei of X be ${X_0}$.
After one half life of $3hrs$, half of the original number of nuclei will disintegrate and convert to Y. So at \[t = 3hrs\], we have the number of X nuclei as
$X\left( 3 \right) = \dfrac{{{X_0}}}{2}$
And the number of nuclei of Y is
$Y\left( 3 \right) = \dfrac{{{X_0}}}{2}$
So the ratio of the number of nuclei of X to that of Y at \[t = 3hrs\] is
$\dfrac{{X\left( 3 \right)}}{{Y\left( 3 \right)}} = \dfrac{{{X_0}/2}}{{{X_0}/2}}$
 $ \Rightarrow \dfrac{{X\left( 3 \right)}}{{Y\left( 3 \right)}} = 1$
Therefore, at \[t = 3hrs\] the ratio is $1:1$.
So option A is incorrect.
Now, after the second half life, half of the remaining number of nuclei of X will convert to Y. So the nuclei of X remaining at $t = 6hrs$ is
$X\left( 6 \right) = \dfrac{{X\left( 3 \right)}}{2}$
$ \Rightarrow X\left( 6 \right) = \dfrac{{{X_0}}}{4}$
And the number of nuclei of Y at this time is
$Y\left( 6 \right) = Y\left( 3 \right) + \dfrac{{X\left( 3 \right)}}{2}$
$ \Rightarrow Y\left( 6 \right) = \dfrac{{{X_0}}}{2} + \dfrac{{X\left( 0 \right)/2}}{2}$
On simplifying we get
$Y\left( 6 \right) = \dfrac{{3{X_0}}}{4}$
So the ratio of the number of nuclei of X to that of Y at \[t = 6hrs\] is
$\dfrac{{X\left( 6 \right)}}{{Y\left( 6 \right)}} = \dfrac{{{X_0}/4}}{{3{X_0}/4}}$
$ \Rightarrow \dfrac{{X\left( 6 \right)}}{{Y\left( 6 \right)}} = \dfrac{1}{3}$
Therefore, at \[t = 6hrs\] the ratio is $1:3$.
Now, after the third half life, half of the remaining number of nuclei of X will convert to Y. So the nuclei of X remaining at $t = 9hrs$ is
$X\left( 9 \right) = \dfrac{{X\left( 6 \right)}}{2}$
$ \Rightarrow X\left( 9 \right) = \dfrac{{{X_0}}}{8}$
And the number of nuclei of Y at this time is
$Y\left( 9 \right) = Y\left( 6 \right) + \dfrac{{X\left( 6 \right)}}{2}$
$ \Rightarrow Y\left( 9 \right) = \dfrac{{3{X_0}}}{4} + \dfrac{{{X_0}/4}}{2}$
On simplifying we get
$Y\left( 9 \right) = \dfrac{{7{X_0}}}{8}$
So the ratio of the number of nuclei of X to that of Y at \[t = 9hrs\] is
$\dfrac{{X\left( 9 \right)}}{{Y\left( 9 \right)}} = \dfrac{{{X_0}/8}}{{7{X_0}/8}}$
$ \Rightarrow \dfrac{{X\left( 9 \right)}}{{Y\left( 9 \right)}} = \dfrac{1}{7}$
Therefore, at \[t = 9hrs\] the ratio is $1:7$.

Hence, the correct answer is option C.

Note: Do not forget to add the previously remaining nuclei of Y while calculating the number of nuclei of Y at a time. Otherwise, you will not be able to obtain the given ratio.