A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln2 = 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

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A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10³ disintegrations per second. Given that ln2 = 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

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N_t = N₀ e^{– $λ$ t}

$\frac{N_{0} - N_{t}}{N_{0}} = (1 - e^{- λt})$

$= λt = \frac{ℓn 2}{1386} \times 80$

= 4

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