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

`lambda=(0.693)/(1386)=5xx10^(-4)`
Number decayed `=N_(0)-N(t)`
`%` age Decayed `=(N_(0)-N(t))/(N_(0))xx100`
`=(1-e^(-lambdat))xx100`
`~~lambda txx100`
`5xx10^(-4)xx80xx100`
`4`