Question

The half-life of a radioactive substance is 20mins. The time taken between 50% decay and 87.5% decay of the substance will be
\[ (a){\text{ 20mins}} \\
  (b){\text{ 30mins}} \\
  (c){\text{ 40mins}} \\
  (d){\text{ 25mins}} \\
  {\text{(e) 10mins}} \\
 \]

Answer 1

Hint – In this question use the direct formula for the reaction constant for the first order reaction that is $\lambda = \dfrac{{\ln 2}}{\tau }$ where $\tau $ is the half-life of the substance. The relationship the substance remaining after some time is given as $N = {N_o}{e^{\left( { - \lambda t} \right)}}$ where ${N_o}$ is the original amount of the substance. This will help to approach the solution of this problem.

Formula used: 1. $\lambda = \dfrac{{\ln 2}}{\tau }$
2. $N = {N_o}{e^{\left( { - \lambda t} \right)}}$ where ${N_o}$ is the original amount of the substance.

Complete step-by-step solution -
As we know a radioactive decay is a first order reaction and the reaction constant for the for the first order reaction is given as,
$\lambda = \dfrac{{\ln 2}}{\tau }$................. (1)
Where, $\tau $ is the half-life of the substance.
Now let it takes ${t_1}$ time to decay 50 % of the original substance
The substance remaining after ${t_1}$ time is given as
$ \Rightarrow N = {N_o}{e^{\left( { - \lambda t} \right)}}$
Where, N = amount of substance remaining
             ${N_o}$ = original amount of the substance
As 50% decay so amount of substance remained = 50%
Now substitute the value we have,
$ \Rightarrow \dfrac{{{N_o}}}{2} = {N_o}{e^{\left( { - \dfrac{{\ln 2}}{\tau }{t_1}} \right)}}$
Now cancel out the common terms we have,
$ \Rightarrow \dfrac{1}{2} = {e^{\left( { - \dfrac{{\ln 2}}{\tau }{t_1}} \right)}}$
Now take (ln) on both sides we have,
 $ \Rightarrow \ln \dfrac{1}{2} = - \dfrac{{\ln 2}}{\tau }{t_1}$
$ \Rightarrow - \ln 2 = - \dfrac{{\ln 2}}{\tau }{t_1}$
$ \Rightarrow {t_1} = \tau $
Now let it takes ${t_2}$ time to decay 87.5 % of the original substance
The substance remaining after ${t_2}$ time is given as
$ \Rightarrow N = {N_o}{e^{\left( { - \lambda t} \right)}}$
Where, N = amount of substance remaining
             ${N_o}$ = original amount of the substance
As 87.5% decay so amount of substance remained = 12.5%
Now substitute the value we have,
$ \Rightarrow \dfrac{{12.5{N_o}}}{{100}} = {N_o}{e^{\left( { - \dfrac{{\ln 2}}{\tau }{t_2}} \right)}}$
Now cancel out the common terms we have,
$ \Rightarrow 0.125 = {e^{\left( { - \dfrac{{\ln 2}}{\tau }{t_2}} \right)}}$
Now take (ln) on both sides we have,
 $ \Rightarrow \ln 0.125 = - \dfrac{{\ln 2}}{\tau }{t_2}$
$ \Rightarrow - 2.079 = - \dfrac{{\ln 2}}{\tau }{t_2}$
$ \Rightarrow {t_2} = \dfrac{{\tau \times 2.079}}{{\ln 2}} = \dfrac{{\tau \times 2.079}}{{0.693}} = 3\tau $
So the time taken between 50% decay and 87.5% decay of the substance is, t = ${t_2} - {t_1}$
$ \Rightarrow t = 3\tau - \tau = 2\tau $
Now it is given that the half-life of the substance is 20 minute.
$ \Rightarrow t = 2\tau = 2\left( {20} \right) = 40$min.
So this is the required answer.
Hence option (C) is the correct answer.

Note – A radioactive substance has the tendency to decompose thus the time after which the radioactive material is being decayed to half of its initial concentration or the radioactive substance is left only half of its initial concentration is termed as the half-life time of a radioactive material.