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

Answer

4

Explanation

Given `T_(1//2)=1386 sec`.
`|(dN)/(dt)|=10^(3)`
Fraction
`=(N_(0)-N)/N_(0)=(N_(0)-N_(0)e^(-lambdat))/N_(0)=1-e^(-lambdat)`
`=1-e^((-ln2)/1386xx80)`
`=1-e^((-4)/100)=1-[1-4/100]=4/100 equiv 4%`