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(A) The number of nuclei remaining after time $t$ is given by $N = N_0 (1/2)^n$, where $n = t / T_{1/2}$ is the number of half-lives.For substance $A$

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$1: 16$

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(A) The number of nuclei remaining after time $t$ is given by $N = N_0 (1/2)^n$, where $n = t / T_{1/2}$ is the number of half-lives.For substance $A$: $T_{1/2, A} = 1.5 \ days$, $t = 9 \ days$. Number of half-lives $n_A = 9 / 1.5 = 6$. Remaining nuclei $N_A = N_0 (1/2)^6 = N_0 / 64$.For substance $B$: $T_{1/2, B} = 4.5 \ days$, $t = 9 \ days$. Number of half-lives $n_B = 9 / 4.5 = 2$. Remaining nuclei $N_B = N_0 (1/2)^2 = N_0 / 4$.The ratio $N_A / N_B = (N_0 / 64) / (N_0 / 4) = 4 / 64 = 1 / 16$.Thus, the ratio is $1: 16$.

Two radioactive substances $A$ and $B$ have the same number of initial nuclei. If the half-lives of $A$ and $B$ are $1.5 \ days$ and $4.5 \ days$ respectively, then the ratio of the number of nuclei remaining in $A$ and $B$ after $9 \ days$ is

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