A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10 3 disintegrations per second. Given that In 2 = 0.693, the fraction of the initial number of nuclei (expressed in neares

Question

A freshly prepared sample of a radioisotope of half-life 1386 s has activity $10^{3}$ disintegrations per second. Given that In 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:

4.00

Solution:

$f = (1 - e^{- λ t}) = 1 - e^{- λ t} \approx 1 - (1 - λ t) = λ t$
f = 0.04
Hence $% decay \approx 4 %$

TestHub · Accepted Answer

f = 1 - e - λ t = 1 - e - λ t ≈ 1 - 1 - λ t = λ t f = 0.04 Hence % decay ≈ 4 %

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