Half-life of a radioactive substance A is two | vedclass

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(B) Let $(T_H)_A$ and $(T_H)_B$ be the half-lives of substances $A$ and $B$ respectively.Given: $(T_H)_A = 2(T_H)_B$.Let $N_A$ and $N_B$ be the initia

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$\frac{1}{8}$

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(B) Let $(T_H)_A$ and $(T_H)_B$ be the half-lives of substances $A$ and $B$ respectively.Given: $(T_H)_A = 2(T_H)_B$.Let $N_A$ and $N_B$ be the initial number of nuclei.The number of nuclei remaining after time $t$ is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_H}$.After time $t = 3(T_H)_A$,the number of nuclei of $A$ is $N_A(t) = N_A \left(\frac{1}{2}\right)^3 = \frac{N_A}{8}$.Since $(T_H)_A = 2(T_H)_B$,we have $t = 3(2(T_H)_B) = 6(T_H)_B$.The number of nuclei of $B$ after time $t$ is $N_B(t) = N_B \left(\frac{1}{2}\right)^6 = \frac{N_B}{64}$.Given that $N_A(t) = N_B(t)$,we have $\frac{N_A}{8} = \frac{N_B}{64}$.Therefore,$\frac{N_A}{N_B} = \frac{8}{64} = \frac{1}{8}$.

Half-life of a radioactive substance $A$ is two times the half-life of another radioactive substance $B$. Initially,the number of nuclei of $A$ and $B$ are $N_A$ and $N_B$ respectively. After three half-lives of $A$,the number of nuclei of both are equal. Then the ratio $\frac{N_A}{N_B}$ is

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