Two radioactive materials A and B have decay | vedclass

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(A) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Given that initially both have the same number of nuclei,$N_{0A

Vedclass · Accepted Answer

$9$

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(A) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Given that initially both have the same number of nuclei,$N_{0A} = N_{0B} = N_0$.The number of nuclei of $A$ at time $t$ is $N_A = N_0 e^{-25 \lambda t}$.The number of nuclei of $B$ at time $t$ is $N_B = N_0 e^{-16 \lambda t}$.We are given that the ratio $\frac{N_B}{N_A} = e$ at time $t = \frac{1}{a \lambda}$.$\frac{N_B}{N_A} = \frac{N_0 e^{-16 \lambda t}}{N_0 e^{-25 \lambda t}} = e^{(-16 \lambda + 25 \lambda) t} = e^{9 \lambda t}$.Setting this equal to $e^1$,we get $9 \lambda t = 1$,which implies $t = \frac{1}{9 \lambda}$.Comparing this with $t = \frac{1}{a \lambda}$,we find $a = 9$.

Two radioactive materials $A$ and $B$ have decay constants $25 \lambda$ and $16 \lambda$ respectively. If initially they have the same number of nuclei,then the ratio of the number of nuclei of $B$ to that of $A$ will be $e$ after a time $t = \frac{1}{a \lambda}$. The value of $a$ is $......$

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