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 )

Step-by-step solution

Given λ 1 = ℓ n 2 700 / year λ 2 = ℓ n 2 1400 / year ∴ λ net  = λ 1 + λ 2 = ln 2 1 700 + 1 1400 = 3 ln 2 1400 / year Now, let initial number of radioactive nuclei be N 0 , ∴ N 0 3 = N 0 e - λ mat t ⇒ ℓ n 1 3 = - λ net t ⇒ 1 . 1 = 3 × 0 . 693 1400 t ⇒ t ≈ 740 years Hence, option 4 .