Question

A freshly prepared sample of a radio isotope 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 first $80$ $sec$ after preparation of the sample is:

Answer 1

$4$
We know
$\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log \left(\frac{N_0}{N}\right)$
Given $t_{1/2}=1386$ $sec$, $t=80$ $sec$, $N_0=1$,
$N=$ remaining fraction after $80$ $sec$
$\frac{0.693}{1386}=\frac{2.303}{80}\log \left(\frac{1}{N}\right)$
On solving,
$N=0.96$
Fraction decayed $=1-0.96=0.04$
Percentage decay $=0.04\times 100=4$