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(C) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^n$,where $n = \frac{t}{T_{1/2}}$ is the number of ha

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$1 : 4$

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(C) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} ight)^n$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.For substance $A$: $T_{1/2, A} = 20 	ext{ min}$,$t = 80 	ext{ min}$. Number of half-lives $n_A = \frac{80}{20} = 4$.Remaining nuclei $N_A = N_0 \left( \frac{1}{2} ight)^4 = \frac{N_0}{16}$.For substance $B$: $T_{1/2, B} = 40 	ext{ min}$,$t = 80 	ext{ min}$. Number of half-lives $n_B = \frac{80}{40} = 2$.Remaining nuclei $N_B = N_0 \left( \frac{1}{2} ight)^2 = \frac{N_0}{4}$.The ratio of remaining nuclei is $\frac{N_A}{N_B} = \frac{N_0/16}{N_0/4} = \frac{4}{16} = \frac{1}{4}$.

Half lives of two radioactive substances $A$ and $B$ are respectively $20$ minutes and $40$ minutes. Initially,the sample of $A$ and $B$ have equal number of nuclei. After $80$ minutes,the ratio of the remaining numbers of $A$ and $B$ nuclei is

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