Half-lives of two radioactive substances A an | vedclass

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(C) 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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$1 : 4$

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(C) 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 $A$: $T_{1/2, A} = 20 \ min$. The number of half-lives in $80 \ min$ is $n_A = 80 / 20 = 4$.Remaining nuclei $N_A = N_0 (1/2)^4 = N_0 / 16$.For substance $B$: $T_{1/2, B} = 40 \ min$. The number of half-lives in $80 \ min$ is $n_B = 80 / 40 = 2$.Remaining nuclei $N_B = N_0 (1/2)^2 = N_0 / 4$.The ratio of remaining nuclei is $N_A / N_B = (N_0 / 16) / (N_0 / 4) = 4 / 16 = 1 / 4$.

Half-lives of two radioactive substances $A$ and $B$ are,respectively,$20 \ min$ and $40 \ min$. Initially,the samples of $A$ and $B$ have an equal number of nuclei. After $80 \ min$,the ratio of the remaining number of $A$ and $B$ nuclei is

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