\(T _{1 / 2}=5 \text { days }=\frac{\ln 2}{\lambda}\)

\(A = A _{0} e ^{-\lambda t }\)

\(5 \times 10^{-6}=6.4 \times 10^{-4} e ^{-\lambda t }\)

\(\frac{5}{6.4} \times 10^{-2}=e^{-\lambda t}\)

\(7.8 \times 10^{-3}= e ^{-\lambda t}\)

\(\log \left(7.8 \times 10^{-3}\right)=-\lambda t \ln e\)

\(\ln \left(7.8 \times 10^{-3}\right)=-\frac{\lambda n 2}{5} \cdot t\)

\(\therefore \frac{5 \times 4.853}{0.693}= t =35 \text { days }\)