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Two radioactive substance 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 $\left(\frac{1}{e}\right)^2$ after a time interval:
Option: 1
$\frac{1}{4 \lambda}$

Option: 2
$4 \lambda$

Option: 3
$2 \lambda$

Option: 4
$\frac{1}{2 \lambda}$

Answers (1)

best_answer

Number of nuclei remained after time t can be written as

$\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$

Where, $\mathrm{N}_0$ is initial number of nuclei of both the substances

$\mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}}$ $(i)$

and $\mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$ $(ii)$

On dividing eq. (i) by eq. (ii), we get

$\frac{N_1}{N_2}=e^{(-5 \lambda+\lambda) t}=e^{-4 \lambda t}=\frac{1}{e^{4 \lambda t}}$

But, we have given

$\frac{\mathrm{N}_1}{\mathrm{~N}_2}=\left(\frac{1}{\mathrm{e}}\right)^2=\frac{1}{\mathrm{e}^2}$

Comparing the powers, we get

$2=4 \lambda \mathrm{t} \Rightarrow \mathrm{t}=\frac{2}{4 \pi}=\frac{1}{2 \lambda}$

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