Question

The half-life of radioisotope is 4 hr. If the initial mass of the isotope was 200 g, then mass remaining after 24 hr. will be
A. 1.042 g
B. 2.084 g
C. 3.125 g
D. 4.167 g

Answer 1

- Hint: In this question use the concept that radioactive decay is a first order reaction. We know the formula to calculate the amount of substrate remaining after time T and with the help of it, we can easily calculate the mass remaining after a given amount of time. The equation is $N = {N_o}{e^{\left( { - \lambda t} \right)}}$ , where N is the amount of mass remaining after a given amount of time , $\lambda $is the reaction constant and is the time at which the remaining mass is to be calculated . We know the reaction constant for a first order reaction i.e. $\lambda = \dfrac{{\ln 2}}{\tau }$. Where $\tau $ is the half-life of the isotope. This will help approaching the problem.

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 it is given that the mass of the isotope is 200gm and the half-life of the isotope is 4 hr.
Then we have to find out the amount of mass remaining in 24hr.
So the time (t) = 24hr.
Now the amount of mass remaining in time t is given as
$ \Rightarrow N = {N_o}{e^{\left( { - \lambda t} \right)}}$................. (2)
Where, ${N_o}$ = actual mass of the substance
               N = amount of mass remaining after t.
Now substitute the value from equation (1) in equation (2) we have,
$ \Rightarrow N = {N_o}{e^{\left( { - \dfrac{{\ln 2}}{\tau }t} \right)}}$
Now substitute the values of half-life, time (t) and actual mass of the substance (${N_o}$) we have,
$ \Rightarrow N = 200{e^{\left( { - \dfrac{{\ln 2}}{4} \times 24} \right)}}$
Now simplify this we have,
$ \Rightarrow N = 200{e^{\left( { - 6\ln 2} \right)}}$
Now use the logarithmic property i.e. $a\ln b = \ln {b^a}$ and ${e^{\ln x}} = x$ so use this properties in the above equation we have,
 $ \Rightarrow N = 200{e^{\left( {\ln {2^{ - 6}}} \right)}}$
$ \Rightarrow N = 200\left( {{2^{ - 6}}} \right)$
$ \Rightarrow N = \dfrac{{200}}{{{2^6}}}$
Now we all know that ${2^6}$ = 64 so we have,
$ \Rightarrow N = \dfrac{{200}}{{64}} = 3.125$ gm.
So the mass remains after 24 hr. will be 3.125 gm.
So this is the required answer.
Hence option (C) is the correct answer.

Note – There can be an alternate approach to solve this problem statement involving half-life, we can use the below formula directly instead of the basic approach as discussed above. The formula is $N = \dfrac{{{N_0}}}{{{2^n}}}$ and $n = \dfrac{t}{\tau }$ ,where N is the amount of mass remaining after a given amount of time, ${N_0}$ is the initial mass of radio isotope,$\tau $is the half-life of the radioisotope and t is the time at which the remaining mass is to be calculated.