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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Two radioactive elements X and Y have decay constantsλ and 10λ respectively. If decay begins with the same number of atoms of them, the ratio of atoms of X to those of Y after time 19λ will be

The correct option is A e Nx=N0e−λt=N0e−λ9λ=N0e−19Ny=N0e−10λt=N0e−10λ9λ=N0e−109Hence,NxNy=e−19−109=e

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

The correct option is A
$e$

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