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Two radioactive substances A and B have decay constants $5\lambda$ and $\lambda$ respectively. At $t = 0$ they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $\left(\dfrac{1}{e}\right)^2$ after a time interval
A
$4\lambda$
B
$2\lambda$
C
$\dfrac{1}{2\lambda}$
D
$\dfrac{1}{4\lambda}$
Detailed Solution
$\lambda_A = 5\lambda$ and $\lambda_B = \lambda$. At $t = 0$, $(N_0)_A = (N_0)_B$.
According to the law of radioactive decay, $\dfrac{N}{N_0} = e^{-\lambda t}$
$\dfrac{N_A}{(N_0)_A} = e^{-\lambda_A t} = e^{-5\lambda t}$ ...(1)
$\dfrac{N_B}{(N_0)_B} = e^{-\lambda_B t} = e^{-\lambda t}$ ...(2)
Dividing (1) by (2): $\dfrac{N_A}{N_B} = e^{-(5\lambda - \lambda)t} = e^{-4\lambda t}$
Given $\dfrac{N_A}{N_B} = \left(\dfrac{1}{e}\right)^2 = e^{-2}$
$\Rightarrow e^{-4\lambda t} = e^{-2} \Rightarrow 4\lambda t = 2$
$\therefore t = \dfrac{1}{2\lambda}$
According to the law of radioactive decay, $\dfrac{N}{N_0} = e^{-\lambda t}$
$\dfrac{N_A}{(N_0)_A} = e^{-\lambda_A t} = e^{-5\lambda t}$ ...(1)
$\dfrac{N_B}{(N_0)_B} = e^{-\lambda_B t} = e^{-\lambda t}$ ...(2)
Dividing (1) by (2): $\dfrac{N_A}{N_B} = e^{-(5\lambda - \lambda)t} = e^{-4\lambda t}$
Given $\dfrac{N_A}{N_B} = \left(\dfrac{1}{e}\right)^2 = e^{-2}$
$\Rightarrow e^{-4\lambda t} = e^{-2} \Rightarrow 4\lambda t = 2$
$\therefore t = \dfrac{1}{2\lambda}$