A radioactive material decays by simultaneous emissions of two...

Given ${\lambda}_{1} = \dfrac{ℓn2}{700} /$year ${\lambda}_{2} = \dfrac{ℓn2}{1400} /$year
$∴ {\lambda}_{\text{net}} = {\lambda}_{1} + {\lambda}_{2} = \ln 2\left[\dfrac{1}{700} + \dfrac{1}{1400}\right]$
$= \dfrac{3\ln 2}{1400} /$year
Now, let initial number of radioactive nuclei be
${N}_{0}$,
$∴ \dfrac{{N}_{0}}{3} = {N}_{0}{e}^{- {\lambda}_{mat}t}$

$⇒ ℓn\dfrac{1}{3} = - {\lambda}_{net}t$
$⇒ 1 . 1 = \dfrac{3 \times 0 . 693}{1400}t ⇒ t \approx 740$ years
Hence, option $4$.