Half-lives of two radioactive elements A and | 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 element $A$: $

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$4: 5$

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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 element $A$: $T_{1/2, A} = 30 	ext{ min}$,$t = 120 	ext{ min}$.Number of half-lives $n_A = 120 / 30 = 4$.Remaining nuclei $N_A = N_0 (1/2)^4 = N_0 / 16$.Decayed nuclei $N'_A = N_0 - N_A = N_0 - N_0 / 16 = (15/16) N_0$.For element $B$: $T_{1/2, B} = 60 	ext{ min}$,$t = 120 	ext{ min}$.Number of half-lives $n_B = 120 / 60 = 2$.Remaining nuclei $N_B = N_0 (1/2)^2 = N_0 / 4$.Decayed nuclei $N'_B = N_0 - N_B = N_0 - N_0 / 4 = (3/4) N_0 = (12/16) N_0$.The ratio of decayed nuclei of $B$ to $A$ is $N'_B / N'_A = (12/16) N_0 / (15/16) N_0 = 12 / 15 = 4 / 5$.

Half-lives of two radioactive elements $A$ and $B$ are $30 \text{ minute}$ and $60 \text{ minute}$ respectively. Initially,the samples have an equal number of nuclei. After $120 \text{ minute}$,the ratio of the number of decayed nuclei of $B$ to that of $A$ will be:

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