Question

A radioactive material decays by simulataneous emission of two particles with half - lives 1620 and 810 years. The time after which one - fourth of the material remains, is

• A. 1080 years
• B. 2430 years
• C. 3240 years
• D. 4860 years

Answer 1

The correct option is A

1080 years

$\lambda ={\lambda }_1+{\lambda }_2=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{t_{\frac{1}{2}\left(1\right)}}+\frac{0.693}{t_{\frac{1}{2}\left(2\right)}}{t}_{\frac{1}{2}}=\frac{t_{\frac{1}{2}\left(1\right)}\times t_{\frac{1}{2}\left(2\right)}}{t_{\frac{1}{2}\left(1\right)}+t_{\frac{1}{2}\left(2\right)}}=\frac{1620\times 810}{1620+810}=540yearsFractionleft=\frac{1}{4}=(\frac{1}{2}{)}^2$ No. of half-lives = 2
Hence, time required $=2\times 540=1080$ year