A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is

Solution :
$$\eqalign{ & \frac{{ - dN}}{{dt}} = {\lambda _1}N + {\lambda _2}N \cr & \Rightarrow {\log _e}\frac{N}{{{N_o}}} = - \left( {{\lambda _1} + {\lambda _2}} \right)t \cr} $$
when $${N_o}$$ is initial number of atoms
$$\eqalign{ & {\text{Here }}{\lambda _1} = \frac{{0.693}}{{1620}}\,{\text{and}}\,{\lambda _2} = \frac{{0.693}}{{810}}; \cr & \frac{N}{{{N_o}}} = \frac{1}{4} \Rightarrow {\log _e}\frac{1}{4} = - \left( {\frac{{0.693}}{{1620}} + \frac{{0.693}}{{810}}} \right)t \cr & \Rightarrow t = 1080{\text{ years}} \cr} $$