Question

A radioactive substance of half-life 69.3 days is kept in a container. The time in which 80% of the substance will disintegrate will be
A. 1.61 days
B. 16.1 days
C. 161 days
D. 1610 days

Answer 1

Throughout the course of the reactive disintegration the disintegration constant ($\lambda$) remains constant. Any radioactive decay follows the first-order kinetics. And is related to the half-life as given in the formula below. That $\lambda$ then can be used to find the time and the relation to time is also given below.

Formulas Used:
${{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\lambda }$
$t=\dfrac{2.303}{\lambda }\log \dfrac{a}{a-x}$

Complete step by step answer:
Through the course of any radioactive reaction, the disintegration constant ($\lambda$) remains constant. So let us start off by finding it out. In the question, it is given that the half-life of the substance is 69.3 days. And we also know that half-life and disintegration constant ($\lambda$) is related as
${{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\lambda }$
Or, $\lambda =\dfrac{\ln 2}{{{t}_{\dfrac{1}{2}}}}$
We also know that $\ln 2=0.693$, and it is given that ${{t}_{\dfrac{1}{2}}}=69.3days$
Plugging in all the values we get

& \lambda =\dfrac{\ln 2}{69.3} \\\ & \lambda =\dfrac{0.693}{69.3} \\\ & \lambda =0.01day{{s}^{-1}} \\\ \end{aligned}$$ So the disintegration constant ($\lambda$) is $$0.01day{{s}^{-1}}$$. Now coming to the second part of the question we have to find out the time in which 80% of the substance will disintegrate. We know that $$t=\dfrac{2.303}{\lambda }\log \dfrac{a}{a-x}$$ Where a is the original amount of substance let that be 100. x is the amount of substance left since 80% of the substance has disintegrated 20% remains. So x=20 Plugging in these values we get $$\begin{aligned} & t=\dfrac{2.303}{\lambda }\log \dfrac{a}{a-x} \\\ & t=\dfrac{2.303}{0.01}\log \dfrac{100}{20} \\\ & t\approx 161days \\\ \end{aligned}$$ So, the time in which 80% of the substance will disintegrate will be 161 days. **Option C is the correct one.** **Note:** The relation between time and amount of radioactive substance remaining differs with the order of the radioactive reaction. But it should be remembered that radioactive decay is a first-order process and thus the time required for the amount of substance to get halved is constant. The relation between time and amount remaining is given by $$t=\dfrac{2.303}{\lambda }\log \dfrac{a}{a-x}$$