Q 12-13-102JEE MainJEE Main 2021 (20 Jul, Shift 1)Medium
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 $\ln3 = 1.1$)
Answer: (D) 740 years
Decay constants add: $\lambda = \lambda_1 + \lambda_2 = \ln2\left(\frac{1}{1400} + \frac{1}{700}\right) = \frac{3\ln2}{1400}$ per year.
$N = \dfrac{N_0}{3} \Rightarrow \lambda t = \ln3$:
$$t = \frac{1400\ln3}{3\ln2} = \frac{1400\times1.1}{3\times0.693} \approx 740\ \text{years}$$
Solution by Sreeraj P, M.Sc Physics