A fresbly prepared of a radioisotope of half - life `1386 s ` has activety `10^(3) ` disentegrations per second Given that ln `2 = 0.693` the fraction of the initial number of nuclei (expressedin nearest integer percentage ) that will decay in the first `80 s ` after preparation of the sample is
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D
For a radioactive decay ` N = N_(0) e^(-11)` `:. (N)/(N_(0)) = e^(-11) :. 1 - (N)/(N_(0)) = 1 - e^(-11)` `:. (N_(0) - N)/(N)= 1- e ^((0.693)/(t_(1//2)))e1 = 1 - e^(-0.04) = 1 - (1- 0.04)` `= 0.04 = 4% [:' e^(-5) = 1 - x x ltlt 1]`
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