A radioactive material decays by simultaneous emissions of two particles with half life of $1400$ years and $700$ years respectively. What will be the time after which, one third of the material remains ?

Take $\ln 3\approx 1.1$

The correct option is B $740$ years

Given :

${\lambda }_1=\frac{\ln 2}{(T_{1/2}{)}_1}=\frac{\ln 2}{700}/\text{year}$

${\lambda }_2=\frac{\ln 2}{(T_{1/2}{)}_2}=\frac{\ln 2}{1400}/\text{year}$

$\therefore {\lambda }_{\text{net}}={\lambda }_1+{\lambda }_2=\ln 2[\frac{1}{700}+\frac{1}{1400}]$

$=\frac{3\ln 2}{1400}/\text{year}$

Now, let initial numbers of radioactive nuclei be $N_0$.

So,

$N=N_0{e}^{-{\lambda }_{\text{net}}t}$

$\Rightarrow \frac{N_0}{3}=N_0{e}^{-{\lambda }_{\text{net}}t}$

$\Rightarrow \ln \frac{1}{3}=-{\lambda }_{\text{net}}t$

$\Rightarrow \ln 1-\ln 3=-{\lambda }_{\text{net}}t$

$\Rightarrow 0-1.1=-\frac{3\times 0.693}{1400}\times t$

$\Rightarrow t\approx 740$ years

Hence, option $\left(D\right)$ is the correct answer.