Question

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, find 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.___

Answer 1

Number of nuclei decayed in time t.
$N_d=N_0(1-e^{-\lambda t})$
$\therefore$ % decayed $=\left(\frac{N_d}{N_0}\right)\times 100=(1-e^{-\lambda t})\times 100$ $\cdots \left(i\right)$
Here, $\lambda =\frac{0.693}{1386}=5\times {10}^{-4}{s}^{-1}$
$\therefore$ % decayed $=\left(\lambda t\right)\times 100=(5\times {10}^{-4})\left(80\right)\left(100\right)=4$