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 $\ln 3=1.1$ )
Correct Option: , 4
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