Chemistry - Chemical Kinetics Question with Solution | TestHub

Let, initial concentration $(a)=100$ Given, half-life $\left(t_{1/2}\right)=60$ days To find, radioactivity, i.e. $(a-x)$ after time $T$ ( 180 days) $$\because T=n\times t_{1/2}$$ where, $n=$ no. of half-lives $$\begin{array}{rl}180& =n\times 60\\ n& =\frac{180}{60}=3\end{array}$$ and $(a-x)=\frac{a}{2^n}=\frac{100}{2^3}=\frac{100}{8}$ $$(a-x)=12.5\%$$ Hence, (b) is the correct answer.