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.
$\therefore \frac{N_0}{3}=N_0{e}^{-{\lambda }_{\text{net }}}$
$\Rightarrow \ln \frac{1}{3}=-{\lambda }_{\text{net }}t$
$\Rightarrow 1.1=\frac{3\times 0.693}{1400}t$
$\Rightarrow t\approx 740\text{ years }$