The probability of survival of a radioactive nucleus for one mean life is

\frac{1}{e}

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1 - \frac{1}{e}

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\frac{l n 2}{e}

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1 - \frac{l n 2}{e}

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Solution

The correct option is A $\frac{1}{e}$
$P_{s u r v i v a l} = \frac{N}{N_{o}} = \frac{N_{o} e^{- λ t}}{N_{o}} = e^{- λ t}$

For, $t = \frac{1}{λ}, \to P_{s u r v i v a l} = \frac{1}{e}$

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Similar questions

Q. The probability of a radioactive nucleus to survive one mean life is

\frac{1}{e + x}

, then

x

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Q. The probability

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A nucleus of mass $M + △ m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. Speed of light is $c$ .
The binding energy per nucleon for the parent nucleus is $E_{1}$ and that for the daughter nuclei is $E_{2}$ . Then

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The probability of survival of a radioactive nucleus for one mean life is

The correct option is A 1ePsurvival=NNo=Noe−λtNo=e−λtFor, t=1λ,→Psurvival=1e

The correct option is A $\frac{1}{e}$
$P_{s u r v i v a l} = \frac{N}{N_{o}} = \frac{N_{o} e^{- λ t}}{N_{o}} = e^{- λ t}$

For, $t = \frac{1}{λ}, \to P_{s u r v i v a l} = \frac{1}{e}$