Half-lives of two radioactive elements A and | vedclass

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(B) For element $A$,half-life $T_{1/2} = 20 \ min$. Total time $t = 80 \ min$. Number of half-lives $n_A = \frac{80}{20} = 4$. Remaining nuclei $N_A =

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(B) For element $A$,half-life $T_{1/2} = 20 \ min$. Total time $t = 80 \ min$. Number of half-lives $n_A = \frac{80}{20} = 4$. Remaining nuclei $N_A = \frac{N_0}{2^4} = \frac{N_0}{16}$. Decayed nuclei $N_{A, decayed} = N_0 - \frac{N_0}{16} = \frac{15N_0}{16}$. For element $B$,half-life $T_{1/2} = 40 \ min$. Total time $t = 80 \ min$. Number of half-lives $n_B = \frac{80}{40} = 2$. Remaining nuclei $N_B = \frac{N_0}{2^2} = \frac{N_0}{4}$. Decayed nuclei $N_{B, decayed} = N_0 - \frac{N_0}{4} = \frac{3N_0}{4}$. The ratio of decayed nuclei is $\frac{N_{A, decayed}}{N_{B, decayed}} = \frac{15N_0 / 16}{3N_0 / 4} = \frac{15}{16} 	imes \frac{4}{3} = \frac{5}{4}$.

Half-lives of two radioactive elements $A$ and $B$ are $20 \ min$ and $40 \ min$,respectively. Initially,the samples have equal number of nuclei. After $80 \ min$,the ratio of decayed number of $A$ and $B$ nuclei will be

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