The activity of a radioactive material is 6.4 × 10^–4 curie. Its half life is 5 days. The activity will become 5 × 10^–6 curie after :

$\begin{array}{l}{\mathrm{A}}_0=6.4\times {10}^{-4}\text{ Curie }\\ {\mathrm{T}}_{1/2}=5\text{ days }=\frac{\ln 2}{\mathrm{\lambda }}\\ \mathrm{A}={\mathrm{A}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}\\ 5\times {10}^{-6}=6.4\times {10}^{-4}{\mathrm{e}}^{-\mathrm{\lambda t}}\\ \frac{5}{6.4}\times {10}^{-2}={\mathrm{e}}^{-\mathrm{\lambda t}}\\ 7.8\times {10}^{-3}={\mathrm{e}}^{-\mathrm{\lambda t}}\\ \ln \left(7.8\times {10}^{-3}\right)=-\mathrm{\lambda tln}\mathrm{e}\\ \ln \left(7.8\times {10}^{-3}\right)=-\frac{\mathrm{\lambda n}2}{5}\cdot \mathrm{t}\\ \therefore \frac{5\times 4.853}{0.693}=\mathrm{t}=35\text{ days }\end{array}$