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 $\ln 3=1.1$ )

• A. 1110 years
• B. 700 years
• C. 340 years
• D. 740 years

Answer 1

Answer: (D)

Given $\lambda_{1}=\frac{\ln 2}{700} /$ year,$\lambda_{2}=\frac{\ln 2}{1400} /$ year
$\therefore \lambda_{\text {net }}=\lambda_{1}+\lambda_{2}=\ln 2\left[\frac{1}{700}+\frac{1}{1400}\right]$ $=\frac{3 \ln 2}{1400} /$ year
Now, Let initial no. of radioactive nuclei be No.
$\therefore \frac{N_{0}}{3}=N_{0} e^{-\lambda_{\text {net }}} $
$\Rightarrow \ln \frac{1}{3}=-\lambda_{\text {net }} t$
$\Rightarrow 1.1=\frac{3 \times 0.693}{1400} t$
$ \Rightarrow t \approx 740 \text { years }$