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

We know
`(0.693)/(t_(1//2)) = (2.303)/(t) log ((N_(0))/(N))`
Given `t_(1//2) = 1386` sec t = 80 sec, `N_(0) = 1`
N = remaining fraction after 80 sec
`(0.693)/(1386) = (2.303)/(80) log ((1)/(N))`
On solving
N = 0.96
Fration decayed = 1 - 0.96 = 0.04
precentage decay `= 0.04 xx 100 = 4`