Two radioactive materials Y_1 and Y_2 initial | vedclass

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(A) Let $N_0$ be the initial number of nuclei for both materials.The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.Fo

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$\frac{1}{3 \lambda} \ s$

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(A) Let $N_0$ be the initial number of nuclei for both materials.The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For material $Y_1$,$N_1(t) = N_0 e^{-(9 \lambda) t}$.For material $Y_2$,$N_2(t) = N_0 e^{-(6 \lambda) t}$.The ratio of undecayed nuclei is $\frac{N_1(t)}{N_2(t)} = \frac{N_0 e^{-9 \lambda t}}{N_0 e^{-6 \lambda t}} = e^{-9 \lambda t + 6 \lambda t} = e^{-3 \lambda t}$.We are given that this ratio is $\frac{1}{e}$,which is $e^{-1}$.So,$e^{-3 \lambda t} = e^{-1}$.Equating the exponents: $-3 \lambda t = -1$.Therefore,$t = \frac{1}{3 \lambda} \ s$.

Two radioactive materials $Y_1$ and $Y_2$ initially contain the same number of nuclei. Their decay constants are $9 \lambda \ s^{-1}$ and $6 \lambda \ s^{-1}$ respectively. The time after which the ratio of the number of undecayed nuclei of $Y_1$ and $Y_2$ becomes $\frac{1}{e}$ is:

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