A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years, respectively. What will be the time after the which one third of the material remains? (Take ln 3 = 1.1)

When a radioactive material undergoes simultaneous decay by two processes with half-lives $$T_1 = 1400$$ years and $$T_2 = 700$$ years, the effective decay constant is $$\lambda = \lambda_1 + \lambda_2 = \frac{\ln 2}{T_1} + \frac{\ln 2}{T_2}$$.

$$\lambda = \ln 2\left(\frac{1}{1400} + \frac{1}{700}\right) = \ln 2 \cdot \frac{1+2}{1400} = \frac{3\ln 2}{1400}$$

For one-third of the material to remain: $$N = \frac{N_0}{3} = N_0 e^{-\lambda t}$$, so $$\lambda t = \ln 3$$.

$$t = \frac{\ln 3}{\lambda} = \frac{\ln 3 \times 1400}{3\ln 2} = \frac{1.1 \times 1400}{3 \times 0.693} = \frac{1540}{2.079} \approx 741 \approx 740$$ years.