Question

The activity of a radioactive isotope falls to 12.5% in 90 days. The half-life and decay constant of isotope are:
A.$\mathop {2.31 \times 10}\nolimits^{ - 2} \mathop {day}\nolimits^{ - 1} ;\mathop {30days}\nolimits^{}$
B.$\mathop {2.65 \times 10}\nolimits^{ - 2} \mathop {day}\nolimits^{ - 1} ;\mathop {30days}\nolimits^{}$
C.$\mathop {2.12 \times 10}\nolimits^{ - 2} \mathop {day}\nolimits^{ - 1} ;\mathop {10days}\nolimits^{}$
D.None of these

Answer 1

The half-life of a reaction is the time in which the concentration of reactant is reduced to one half of its concentration. It is represented by $t_1/2$. The reaction in which rate of reaction is directly proportional to the first power of the concentration of the reactant is called first order reaction.

Complete step by step answer:
Now we have to find rate of the reaction first to calculate half-life so
$\mathop {k = \dfrac{{2.303}}{t}}\nolimits^{} \log \dfrac{{\mathop A\nolimits^0 }}{A}$

Where $A^\circ$ is the initial concentration of reactant = 100
A is the concentration of reactants at given time = 12.5
Time is given by $t{\text{ }} = {\text{ }}90{\text{ }}days\;$
Putting the value in
$\mathop {k = \dfrac{{2.303}}{t}}\nolimits^{} \log \dfrac{{\mathop A\nolimits^0 }}{A}$
$\mathop {k = \dfrac{{2.303}}{{90}}\log \dfrac{{100}}{{12.5}}}\nolimits^{}$

Now, $\mathop {k = 2.31 \times 10}\nolimits^{ - 2} \mathop {day}\nolimits^{ - 1}$
Let us write formula to find half life
$\mathop t\nolimits_{1/2} = \dfrac{{0.693}}{k}$
$\mathop t\nolimits_{1/2} = \dfrac{{0.693}}{{2.31 \times \mathop {10}\nolimits^{ - 2} }}$
$\mathop t\nolimits_{1/2} = 30days$

Our required answer is option A that is $\mathop {2.31 \times 10}\nolimits^{ - 2} \mathop {day}\nolimits^{ - 1} ;\mathop {30days}\nolimits^{}$

Note:
Zero order reaction is defined as the reaction in which rate of reaction is proportional to zero power of the concentration of reactant. So first of all we will find the order of reaction and then we can find the half-life of the reaction.