A radioactive material decays by simultaneous emissions
Question:

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 $\ln 3=1.1$ )

1. 1110 years

2. 700 years

3. 340 years

4. 740 years

Correct Option: , 4

Solution:

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