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

Step-by-step solution

Since, from Rutherford - Soddy law, the number of atoms left after half-lives is given by N = N 0 1 2 n Where, N 0 is the original number of atoms. The number of half-lives, 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 λ = ln ⁡ 2 T ∴                                 λ 1 + λ 2 = ln ⁡ 2 T 1 + ln &