Two radioactive substance A and B have decay constants $9 λ$ and $3 λ$ respectively. At t = 0 , they have the same number of nuclei. The ratio of number of nuclei of A to these of B will be $\frac{1}{e^{2}}$ after a time interval of

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$\frac{1}{2 λ}$

$\frac{1}{4 λ}$

$\frac{1}{3 λ}$

$\frac{1}{5 λ}$

answer is D.

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Detailed Solution

$D e c a y L a w, N = N_{o} e^{- λ t} C a s e 1 : N_{1} = N_{o} e^{- λ_{1} t} = N_{o} e^{- 9 λ t} C a s e 2 : N_{2} = N_{o} e^{- λ_{2} t} = N_{o} e^{- 3 λ t} D i v i d i n g t h e m, \frac{N_{1}}{N_{2}} = \frac{N_{o} e^{- 9 λ t}}{N_{o} e^{- 3 λ t}} = e^{- 6 λ t} B u t \frac{N_{1}}{N_{2}} = \frac{1}{e^{2}} \Rightarrow e^{- 6 λ t} = e^{- 2} \Rightarrow 6 λ t = + 2 \Rightarrow t = \frac{2}{6 λ} = \frac{1}{3 λ}$