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

D

Explanation

Number of nuclei as a function of time can be written as follows:
`N=N_(0)e^(-lambdat)`
`rArr N/N_(0)=e^(-lambdat)`
`rArr 1-N/N_(0)=1-e^(-lambdat)`
`rArr (-DeltaN)/(N_(0))=(1-e^(-lambdat))`
`lambda t lt lt 1 rArr 1-e^(-lambdat) =lambdat`
`(-DeltaN)/N_(0)=(1-e^(-lambdat))`
`rArr (-lambdaN)/(Delta) xx 100 = 4 rArr |DeltaN|% = 4%`