Two radioactive substances A and B have decay constant 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 afte

Question

Two radioactive substances A and B have decay constant 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

The number of nuclei of radioactive substances $A$ and $B$ at time $t$ are given by:

$N_{A} = N_{0} e^{- 5 λ t}$

$N_{B} = N_{0} e^{- λ t}$

The ratio of the number of nuclei of $A$ to those of $B$ is:

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

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

$e^{- 4 λ t} = e^{- 2}$

Equating the exponents:

$4 λ t = 2$

$t = \frac{1}{2 λ}$

Answer: $\frac{1}{2 λ}$

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