Question

At some instant, a radioactive sample having activity $5\,\mu Ci$ has twice the number of nuclei as another sample $S_2$ which has an activity of $10\,\mu Ci$. The half-lives of $S_1$ and $S_2$ are
A. 20 years and 5 years, respectively
B. 20 years and 10 years, respectively
C. 10 year each${S_1}$
D. 5 year each

Answer 1

Use the relation between the activity of the element, decay constant and population of radioactive element at any time t. Also use the relation between the decay constant and half-life of the element. These two equations when combined give the relation between the activity and half-life of a radioactive element.

Formulae used:
The decay rate equation is given by
$A = \lambda N$ …… (1)
Here, $A$ is the activity of a radioactive element, $N$ is the population of the radioactive element at time $t$ and $\lambda$ is the decay constant.
The formula for the decay constant $\lambda$ is
$\lambda = \dfrac{{0.693}}{T}$ …… (2)
Here, $T$ is the half-life of the radioactive element.

Complete step by step answer:
It is given that there are two samples ${S_1}$ and ${S_2}$which have the activities $5\,\mu Ci$ and $10\,\mu Ci$ respectively.
${A_1} = 5\,\mu Ci$
${A_2} = 10\,\mu Ci$
The number of nuclei at any time for the sample ${S_1}$ are twice as that of the sample ${S_2}$.
${N_1} = 2{N_2}$
Substitute $\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ for $\lambda$ in equation (1).
$A = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}N$
Rewrite the above equation for the activities of the samples ${S_1}$ and ${S_2}$.
${A_1} = \dfrac{{0.693}}{{{T_1}}}{N_1}$
${A_2} = \dfrac{{0.693}}{{{T_2}}}{N_2}$
Divide the equation for ${A_1}$ by the equation for ${A_2}$.
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{{0.693}}{{{T_1}}}{N_1}}}{{\dfrac{{0.693}}{{{T_2}}}{N_2}}}$
$\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{N_1}}}{{{T_1}}}\dfrac{{{T_2}}}{{{N_2}}}$
Substitute $5\,\mu Ci$ for ${A_1}$, $10\,\mu Ci$ for ${A_2}$ and $2{N_2}$ for ${N_1}$ in the above equation.
$\Rightarrow \dfrac{{5\,\mu Ci}}{{10\,\mu Ci}} = \dfrac{{2{N_2}}}{{{T_1}}}\dfrac{{{T_2}}}{{{N_2}}}$s
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{4}{1}$
The ratio of the half-lives of the samples ${S_1}$ and ${S_2}$ are in the ratio $4:1$.
From the given options, the half-lives of the samples could only be 20 years and 5 years according to the obtained ratio.

So, the correct answer is “Option A”.

Note:
One may also solve the same question by another method using decay rate equation, activity formula and decay constant formula. The decay rate equation gives the relation between the population of the radioactive element at a particular time and the rate of change of this population with time gives the activity of the element. At last we get the relation between the activity of the radioactive element and its half-life.