A radioactive material decays by simultaneous emissions

Given $\lambda_{1}=\frac{\ell \mathrm{n} 2}{700} /$ year,$\lambda_{2}=\frac{\ell \mathrm{n} 2}{1400} /$ year

$\therefore \lambda_{\text {net }}=\lambda_{1}+\lambda_{2}=\ell \operatorname{n} 2\left[\frac{1}{700}+\frac{1}{1400}\right]$

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

Now, Let initial no. of radioactive nuclei be

No.

$\therefore \frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\text {net }} \mathrm{t}}$

$\Rightarrow \ln \frac{1}{3}=-\lambda_{\text {net }} \mathrm{t}$

$\Rightarrow 1.1=\frac{3 \times 0.693}{1400} \mathrm{t} \Rightarrow \mathrm{t} \approx 740$ years

Hence option 4