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Standard XI

Chemistry

Law of Radioactivity

Two radioacti...

Question

Two radioactive elements X and Y have decay constants $λ$ and $10 λ$ respectively. If decay begins with the same number of atoms of them, the ratio of atoms of X to those of Y after time $\frac{1}{9 λ}$ will be

$e$

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$e^{- 1}$

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$e^{2}$

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$e^{- 2}$

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Solution

The correct option is A
$e$

$N_{x} = N_{0} e^{- λ t} = N_{0} e^{\frac{- λ}{9 λ}} = N_{0} e^{\frac{- 1}{9}} N_{y} = N_{0} e^{- 10 λ t} = N_{0} e^{\frac{- 10 λ}{9 λ}} = N_{0} e^{\frac{- 10}{9}} H e n c e, \frac{N_{x}}{N_{y}} = \frac{e^{\frac{- 1}{9}}}{\frac{- 10}{9}} = e$

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$N_{x} = N_{0} e^{- λ t} = N_{0} e^{\frac{- λ}{9 λ}} = N_{0} e^{\frac{- 1}{9}} N_{y} = N_{0} e^{- 10 λ t} = N_{0} e^{\frac{- 10 λ}{9 λ}} = N_{0} e^{\frac{- 10}{9}} H e n c e, \frac{N_{x}}{N_{y}} = \frac{e^{\frac{- 1}{9}}}{\frac{- 10}{9}} = e$