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Q.

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

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answer is 4.

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Detailed Solution

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Since, $N = N_{0} e^{- λ t}$

$\Rightarrow \frac{N}{N_{0}} = e^{- λ t} \Rightarrow \frac{N}{N_{0}} = e^{- \frac{\log_{c} 2}{1386} \times 80} \Rightarrow \frac{N}{N_{0}} = e^{- \frac{0.693 \times 80}{1386}} \Rightarrow \frac{N}{N_{0}} = e^{- 0.04} \Rightarrow \frac{N}{N_{0}} = {(\frac{1}{e})}^{0.04} F r a c t i o n o f n u c l e i d e c a y e d = 1 - \frac{N}{N_{0}} = 1 - {(\frac{1}{e})}^{0.04} = 0.04 = 4 %$

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