Two radioactive substances A and B have decay constants ' 5 λ ' and ' λ ' respectively. At t = 0 , they have the same number of nuclei. The ratio of number of nuclei of A to those

Question

Two radioactive substances A and B have decay constants ' 5 λ ' and ' λ ' 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 ( e 1 ​ ) 2 after a time interval

TestHub · Accepted Answer

Let the initial number of nuclei of A and B be $N_{0}$ .

The number of nuclei of A at time $t$ is $N_{A} = N_{0} e^{- 5 λ t}$ .

The number of nuclei of B at time $t$ is $N_{B} = N_{0} e^{- λ t}$ .

The ratio of the number of nuclei of A to B is given by:

$\frac{N _{A}}{N _{B}} = \frac{N _{0} e ^{- 5 λ t}}{N _{0} e ^{- λ t}} = e^{- 4 λ t}$

Given that the ratio is $(\frac{1}{e})^{2} = e^{- 2}$ , we have:

$e^{- 4 λ t} = e^{- 2}$

Equating the powers of $e$ , we get:

$4 λ t = 2$

$t = \frac{2}{4 λ} = \frac{1}{2 λ}$

Answer: $\frac{1}{2 λ}$

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