Question

A freshly prepared sample of a radioisotopes of half-life $1386s$ has activity ${10}^3$ disintegrations per second. Given that $ln2=0.693$, the fraction of the initial number of nuclei (expressed in nearest percentage) that will decay in the first $80s$ after preparation of the sample is

Answer 1

$\frac{dN}{dt}=\lambda N_o{e}^{-\lambda t}$

At $t=0$: $\lambda N_o=1000$

Given: $t_{1/2}=1386s$
$\lambda =\frac{ln2}{t_{1/2}}=5\times {10}^{-4}$ s-1
$\frac{N\left(t\right)}{N_o}$ = exp$(-\lambda t)$

Put $t=80s$
$N\left(t\right)/N_o=0.96$
Decayed amount $=0.04=4$ %