Question

A freshly prepared sample of a radioisotope of half-life $1386\text{s}$ has activity ${10}^3$ disintegration per second. Given that $\ln 2=0.693$, the fraction of the initial number of nuclei (expressed in the nearest integer percentage) that will decay in the first $80\text{s}$ after preparation of the sample is ____.

Answer 1

From the given data,

$\lambda =\frac{0.693}{1386}=5\times {10}^{-4}{\text{s}}^{-1}$

$\text{Number of nuclei decayed}=N_0-N\left(t\right)$

$\text{\% of nuclei decayed}=\frac{N_0-N\left(t\right)}{N_0}\times 100$

Using law of radioactivity, $N\left(t\right)=N_0{e}^{-\lambda t}$

$\therefore \text{\% of nuclei decayed}=(1-e^{-\lambda t)}\times 100$

Using Taylor's series of exponential expansion, when $x<<1$,

$e^{-x}=1-x$

So, $\text{\% of nuclei decayed}\approx \lambda t\times 100$

$\Rightarrow \text{\% of nuclei decayed}=5\times {10}^{-4}\times 80\times 100$

$\therefore \text{fraction of initial number of nuclei}=4$

Hence, $4$ is the correct answer.