A radioactive material decays by simultaneous emission of two particles with half-lives 1620 yr and 810 yr respectively. The time in year after which one-fourth of the material remains, is

Step-by-step solution

From Rutherford-Soddy law, the number of atoms left after n half-lives is given by N = N 0 1 2 n Where, N 0 is original number of atoms. The number of half-life n = t i m e o f d e c a y e f f e c t i v e h a l f - l i f e Relation between effective disintegration constant ( λ ) and half-life ( T ) is λ = ln ⁡ 2 T ∴ λ 1 + λ 2 = ln ⁡ 2 T 1 + ln ⁡ 2 T 2 Effective half -life 1 T = 1 T 1 + 1 T 2 = 1 1620 + 1 810 1 T = 1 + 2 1620 ⇒ T ⇒ 540 y r ∴ n = t 540 ∴ N = N 0 1 2 t / 540 ⇒ N N 0 = 1 2 2 = 1 2 t / 540 ⇒ t 540 = 2 ⇒