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Q.

Two radioactive substances A and B have decay constants $5 λ and λ$ having same number of nuclei initially. The ratio of number of nuclei of A to those of B will be $(1 / e)^{2}$ after a time interval of

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a

$5 λ$

b

$1 / 2 λ$

c

$4 λ$

d

$2 λ$

answer is C.

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Detailed Solution

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$λ_{A} = 5 λ and λ_{B} = λ t = 0, {(N_{0})}_{A} = {(N_{0})}_{B} \frac{N_{A}}{N_{B}} = {(\frac{1}{e})}^{2}$

According to radioactive decay,

$\begin{array}{l} \frac{N}{N_{0}} = e^{- λ t} \\ \frac{N_{A}}{{(N_{o})}_{A}} = e^{- λ_{A} t} . . . . . . . . (1) \\ s i m i l a r l y, \frac{N_{B}}{{(N_{0})}_{B}} = e^{- λ_{B} t} \dots . . . (2) \\ From (1) and (2) \\ \frac{N_{A}}{N_{B}} = e^{- (5 λ - λ) t} \\ \Rightarrow {(\frac{1}{e})}^{2} = e^{- 4 λ t} = {(\frac{1}{e})}^{4 λ t} \\ \Rightarrow 4 λ t = 2 \\ t = \frac{1}{2 λ} . \end{array}$

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