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Two radioactive materials $X_1$ and $X_2$ have decay constants $5\lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $\dfrac{1}{e}$ after a time -
A
$\dfrac{1}{4\lambda}$
B
$\dfrac{e}{\lambda}$
C
$\lambda$
D
$\dfrac{1}{2}\lambda$
Detailed Solution
Radioactive decay law: $N = N_0 e^{-\lambda t}$
For $X_1$: $N_1 = N_0 e^{-5\lambda t}$
For $X_2$: $N_2 = N_0 e^{-\lambda t}$ (same initial number $N_0$)
$\dfrac{N_1}{N_2} = \dfrac{e^{-5\lambda t}}{e^{-\lambda t}} = e^{-5\lambda t + \lambda t} = e^{-4\lambda t}$
Given $\dfrac{N_1}{N_2} = \dfrac{1}{e} = e^{-1}$
$e^{-1} = e^{-4\lambda t} \Rightarrow 4\lambda t = 1$
$t = \dfrac{1}{4\lambda}$
For $X_1$: $N_1 = N_0 e^{-5\lambda t}$
For $X_2$: $N_2 = N_0 e^{-\lambda t}$ (same initial number $N_0$)
$\dfrac{N_1}{N_2} = \dfrac{e^{-5\lambda t}}{e^{-\lambda t}} = e^{-5\lambda t + \lambda t} = e^{-4\lambda t}$
Given $\dfrac{N_1}{N_2} = \dfrac{1}{e} = e^{-1}$
$e^{-1} = e^{-4\lambda t} \Rightarrow 4\lambda t = 1$
$t = \dfrac{1}{4\lambda}$