Question

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

Answer 1

$\lambda t=ln\frac{N_o}{N}=\frac{0.693\times 80}{1386\times 1000}=0.04$