Modern Physics 3 Question 28
29. 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
(2013 Adv.)
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Solution:
- Number of nuclei decayed in time $t$,
$$ \begin{aligned} N _d & =N _0\left(1-e^{-\lambda t}\right) \\ \therefore \quad % \text { decayed } & =\frac{N _d}{N _0} \times 100 \\ & =\left(1-e^{-\lambda t} d\right) \times 100 \end{aligned} $$
Here, $\lambda=\frac{0.693}{1386}=5 \times 10^{-4} s^{-1}$
$\therefore \quad %$ decayed $\approx(\lambda t) \times 100$
$$ =\left(5 \times 10^{-4}\right)(80)(100)=4 $$
