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(D) Let $t$ be the time after which the amounts of $A_1$ and $A_2$ become equal.Using the radioactive decay law,the amount $N$ at time $t$ is given by

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$40$

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(D) Let $t$ be the time after which the amounts of $A_1$ and $A_2$ become equal.Using the radioactive decay law,the amount $N$ at time $t$ is given by $N = N_0 (1/2)^{t/T_{1/2}}$.For $A_1$: $N_1 = 40 	imes (1/2)^{t/20}$.For $A_2$: $N_2 = 160 	imes (1/2)^{t/10}$.Setting $N_1 = N_2$,we get:$40 	imes (1/2)^{t/20} = 160 	imes (1/2)^{t/10}$.Dividing both sides by $40$:$(1/2)^{t/20} = 4 	imes (1/2)^{t/10}$.Rearranging the terms:$(1/2)^{t/20} / (1/2)^{t/10} = 4$.$(1/2)^{(t/20 - t/10)} = 4$.$(1/2)^{-t/20} = 4$.$2^{t/20} = 2^2$.Comparing the exponents,$t/20 = 2$,which gives $t = 40 \, s$.

$A$ mixture consists of two radioactive materials $A_1$ and $A_2$ with half-lives of $20 \, s$ and $10 \, s$ respectively. Initially,the mixture has $40 \, g$ of $A_1$ and $160 \, g$ of $A_2$. The amount of the two in the mixture will become equal after ......... $s$.

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