A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10^3 disintegrations pe

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13.Nuclei

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A freshly prepared sample of a radioisotope of half-life $1386 \ s$ has activity $10^3$ disintegrations per second. Given that In $2=0.693$, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first $80 \ s$ after preparation of the sample is :

$4$

$5$

$6$

$7$

(IIT-2013)

$\lambda=\frac{0.693}{1386}=5 \times 10^{-4} $

$\text { Number decayed }= N _0- N ( t ) $

$\% \text { age Decayed } =\frac{ N _0- N ( t )}{ N _0} \times 100 $

$ =\left(1- e ^{-\lambda t}\right) \times 100 $

$ \approx \lambda t \times 100 $

$ =5 \times 10^{-4} \times 80 \times 100 $

$ =4$

Standard 12

Physics

easy

easy

medium

medium

medium