Atoms and Nuclei

Given:Initial amount of $A_1=40\ \mathrm{g}$Initial amount of $A_2=160\ \mathrm{g}$Half-life of $A_1,\ T_{A_1}=20\ \mathrm{s}$Half-life of $A_2,\ T_{A_2}=10\ \mathrm{s}$To determine when the amounts of $A_1$ and $A_2$ will be equal, we use the decay formula:$N=N_0 \left(\frac{1}{2}\right)^{t/T}$For $A_1$, the remaining amount is:$N_{A_1}=40 \left(\frac{1}{2}\right)^{t/20}$For $A_2$, the remaining amount is:$N_{A_2}=160 \left(\frac{1}{2}\right)^{t/10}$To find the time when these amounts are equal, set $N_{A_1}=N_{A_2}$ :$40 \left(\frac{1}{2}\right)^{t/20}\;=160 \left(\frac{1}{2}\right)^{t/10}$$2^{t/10}\;=4 \cdot 2^{t/20} \Rightarrow 2^{\frac{t}{10}-\frac{t}{20}}=4$$2^{t/20}\;=2^2 \Rightarrow \frac{t}{20}=2$$t\;=40\ \mathrm{s}$Therefore, the amounts of $A_1$ and $A_2$ in the mixture will become equal after 40 seconds.