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A freshly prepared sample of a radioisot...

A freshly prepared sample of a radioisotope of half - life `1386 s ` has activity `10^(3) ` disintegrations per second Given that `ln 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

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The correct Answer is:
4
`f=(1-e^(-lambda t))=1-e^(-lambda t)~~1-(1 -lambda t)=lambda t`
`f=0.04`
Hence` %` decay `~~ 40%`.
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