Question

At any instant the ratio of the two radioactive substances is $2:1$ if their half-lives are respectively, $12h$ and $16h$, then after two days, what will be the ratio of the substances?
(A) $1:1$
(B) $2:1$
(C) $1:2$
(D) $1:4$

Answer 1

Hint: In order to solve this question, we will first find the number of these two radioactive substance initially and then by using the general relation between number of radioactive substance at any instant and their half-lives, we will find ratio of these two radioactive substance after two days.

Formula Used:
If ${N_0},N$ be the number of radioactive substance at initial stage, at any instant of time t and $T$ denote for the half-live of radioactive substance then, we have
$\dfrac{N}{{{N_0}}} = {(\dfrac{1}{2})^{\dfrac{t}{T}}}$
and $1Day = 24h$ so $2Day = 48h$

Complete step by step answer:
According to the question, we have given the ratio of $2:1$ initially, so let ${N_0}$ be the number of second radioactive substance denoted by ${N_{\sec ond}}$ then we have,
$\dfrac{{{N_{first}}}}{{{N_{\sec ond}}}} = \dfrac{2}{1}$
$ \Rightarrow {N_{first}} = 2{N_0}$
Now for first radioactive substance let $N{'_{first}}$ be the number of radioactive substance after time period of $t = 2days = 48hr$ with half-live given as $T = 12hr$ so using formula, $\dfrac{N}{{{N_0}}} = {(\dfrac{1}{2})^{\dfrac{t}{T}}}$ we get,
$\dfrac{{N{'_{first}}}}{{2{N_0}}} = {(\dfrac{1}{2})^{\dfrac{{48}}{{12}}}}$
or
$\dfrac{{N{'_{first}}}}{{2{N_0}}} = {(\dfrac{1}{2})^4}$
$ \Rightarrow \dfrac{{N{'_{first}}}}{{2{N_0}}} = (\dfrac{1}{{16}}) \to (i)$
Now, for second radioactive substance let $N{'_{\sec ond}}$ be the number of radioactive substance afer time period of $t = 2days = 48hr$ with half-live given as $T = 16hr$ so again using formula, $\dfrac{N}{{{N_0}}} = {(\dfrac{1}{2})^{\dfrac{t}{T}}}$ we get,
$\dfrac{{N{'_{sec ond}} }}{{{N_0}}} = {(\dfrac{1}{2})^{\dfrac{{48}}{{16}}}}$

$\dfrac{{N{'_{second}} }}{{{N_0}}} = {(\dfrac{1}{2})^3}$

$ \Rightarrow \dfrac{{N{'_{second}} }}{{{N_0}}} = (\dfrac{1}{8}) \to (ii)$
Now, divide the equations (i) by (ii) we get,
$\dfrac{{\dfrac{{N{'_{first}}}}{{2{N_0}}}}}{{\dfrac{{N{'_{second}} }}{{{N_0}}}}} = \dfrac{{\dfrac{1}{{16}}}}{{\dfrac{1}{8}}}$

$\Rightarrow \dfrac{{N{'_{first}}}}{{N{'_{second}}}} = \dfrac{1}{1}$
Therefore the ratio of two radioactive substance after two days is $1:1$
Hence, the correct option is (A) $1:1$.

Note: It should be remembered that, while solving such questions always convert the time in the same unit like from days to hours or from hours to minute as per the need in questions, and the phenomenon of radioactivity was first discovered by scientist Marie Curie.