The decay constants of two radioactive elemen | vedclass

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(A) The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For the first radioactive element,$\lambda_1 = 15x$ and $t =

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$\frac{1}{e^2}$

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(A) The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.For the first radioactive element,$\lambda_1 = 15x$ and $t = \frac{1}{6x}$.$N_1 = N_0 e^{-(15x) 	imes \frac{1}{6x}} = N_0 e^{-\frac{15}{6}} = N_0 e^{-\frac{5}{2}}$.For the second radioactive element,$\lambda_2 = 3x$ and $t = \frac{1}{6x}$.$N_2 = N_0 e^{-(3x) 	imes \frac{1}{6x}} = N_0 e^{-\frac{3}{6}} = N_0 e^{-\frac{1}{2}}$.The ratio of the number of nuclei is $\frac{N_1}{N_2} = \frac{N_0 e^{-\frac{5}{2}}}{N_0 e^{-\frac{1}{2}}}$.$\frac{N_1}{N_2} = e^{-\frac{5}{2} + \frac{1}{2}} = e^{-\frac{4}{2}} = e^{-2} = \frac{1}{e^2}$.

The decay constants of two radioactive elements are $15x$ and $3x$ respectively. Initially,they have the same number of nuclei. After a time of $\frac{1}{6x}$,the ratio of the number of their nuclei will be ........

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