Answer
$3.0\times10^{9}$
Work Step by Step
Decay constant $k=\frac{0.693}{8.0\,d}=0.086625\,d^{-1}$
Time $t=1\,month=30\,d$
Number of atoms at the beginning $N_{0}=4.0\times10^{10}$
Let the number of atoms left after time $t$ be $N$.
Recall that $\ln(\frac{N_{0}}{N})=kt$
$\implies \ln(\frac{4.0\times10^{10}}{N})=0.086625\,d^{-1} \times 30\,d=2.59875$
Taking the inverse $\ln$ of both sides, we have
$\frac{4.0\times10^{10}}{N}=e^{2.59875}=13.447$
Or $N=\frac{4.0\times10^{10}}{13.447}=3.0\times10^{9}$
