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(D) Let the half-life of substance $B$ be $T_B = T$. Then the half-life of substance $A$ is $T_A = 2T$.After time $t$,the number of nuclei remaining i

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$\frac{1}{8}$

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(D) Let the half-life of substance $B$ be $T_B = T$. Then the half-life of substance $A$ is $T_A = 2T$.After time $t$,the number of nuclei remaining is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.The time elapsed is $t = 3 T_A = 3(2T) = 6T$.At this time,the number of nuclei of $A$ is $N_A(t) = N_A \left(\frac{1}{2}\right)^3$.At this time,the number of nuclei of $B$ is $N_B(t) = N_B \left(\frac{1}{2}\right)^{t/T_B} = N_B \left(\frac{1}{2}\right)^{6T/T} = N_B \left(\frac{1}{2}\right)^6$.Given that $N_A(t) = N_B(t)$,we have $N_A \left(\frac{1}{2}\right)^3 = N_B \left(\frac{1}{2}\right)^6$.Therefore,$\frac{N_A}{N_B} = \frac{(1/2)^6}{(1/2)^3} = \left(\frac{1}{2}\right)^{6-3} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

Half-life of a radioactive substance $A$ is two times the half-life of another radioactive substance $B$. Initially,the number of nuclei of $A$ and $B$ are $N_A$ and $N_B$ respectively. After three half-lives of $A$,the number of nuclei of both are equal. Then $\frac{N_A}{N_B}$ is

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