Two radioactive materials A and B have decay | vedclass

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(C) Let $N_0$ be the initial number of nuclei for both materials.For material $A$,the number of nuclei at time $t$ is $N_A = N_0 e^{-10\lambda t}$.For

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$\frac{1}{9\lambda}$

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(C) Let $N_0$ be the initial number of nuclei for both materials.For material $A$,the number of nuclei at time $t$ is $N_A = N_0 e^{-10\lambda t}$.For material $B$,the number of nuclei at time $t$ is $N_B = N_0 e^{-\lambda t}$.We are given that the ratio $\frac{N_A}{N_B} = \frac{1}{e}$.Substituting the expressions,we get $\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$.This simplifies to $e^{-10\lambda t + \lambda t} = e^{-1}$,which means $e^{-9\lambda t} = e^{-1}$.Equating the exponents,we have $-9\lambda t = -1$.Therefore,$t = \frac{1}{9\lambda}$.

Two radioactive materials $A$ and $B$ have decay constants $10\lambda$ and $\lambda$,respectively. If initially they have the same number of nuclei,then the ratio of the number of nuclei of $A$ to that of $B$ will be $1/e$ after a time:

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