Question

Half-lives of two radioactive substances $A$ and $B$ are, respectively $20\min$ and $40\min$. Initially, the samples of $A$ and $B$ have equal numbers of nuclei. After $80\min$, the ratio of the remaining number of $A$ and $B$ nuclei is,
A. $1:16$
B. $4:1$
C. $1:4$
D. $1:1$

Answer 1

From the relation of half life and fraction of atoms at a time $t$ find the number of remaining nuclei.The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach $50\%$ of its initial concentration (i.e. the time taken for the reactant concentration to reach half of its initial value).

Formula used:
The relation between half life and fraction of atoms at a time $t$ is given by, $\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
Here $T$ is the half life, ${N_0}$ is the number of nuclei at the beginning. $N$ is the number of nuclei at a time $t$.

Complete step by step answer:
Here, we have two radioactive substances $A$ and $B$ with half life, $20\min$ and $40\min$ and the number of nuclei is the same for both the substances. Hence ${N_0}$ is the same for both of the substances.Hence, Half life of $A$ is ${T_A} = 20\min$ and $B$ is ${T_B} = 40\min$. Now, we have to find the number of nuclei after $t = 80\min$. So, Putting the values we get the number of remaining nuclei of the substance $A$ as,
${N_A} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_A}}}}}$.
$\Rightarrow{N_A} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{{80}}{{20}}}}$

Up on simplifying we get,
${N_A} = {N_0}{\left( {\dfrac{1}{2}} \right)^4}$
$\Rightarrow{N_A} = \dfrac{{{N_0}}}{{16}}$
Now, the half life of $B$ is ${T_B} = 40\min$ and we have to find the number of nuclei after $t = 80\min$. Hence, putting the values we get,
${N_B} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_A}}}}}$
$\Rightarrow{N_B} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{{80}}{{40}}}}$
On simplifying we get,
${N_B} = {N_0}{\left( {\dfrac{1}{2}} \right)^2}$
$\Rightarrow {N_B} = \dfrac{{{N_0}}}{4}$.
Now, ratio of the remaining nuclei for both of the substances is,
$\therefore \dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{\dfrac{{{N_0}}}{{16}}}}{{\dfrac{{{N_0}}}{4}}} = \dfrac{1}{4}$

Hence, the correct answer is option C.

Note: For a radioactive substance the half life is measured to find the number of nuclei remaining or decayed at a time $T,2T,3T,4T,5T,6T......$ easily. If we have the number of initial nuclei off the substance we can measure the number of remaining nuclei in terms of multiple of half life. For example, the number of remaining nuclei after $2T$ is $\dfrac{{{N_0}}}{{{2^2}}}$, after $3T$ it is $\dfrac{{{N_0}}}{{{2^3}}}$ and so on.