The mean lives of a radioactive substance are 1620 years and 405 years for α and β emission respectively. If it is decaying by both α and β emission simultaneously, then the time during which three-fourths of the sample will decay is

Step-by-step solution

λ = λ α + λ β = 1 T α + 1 T β ∵ λ = 1 T = 1 1620 + 1 405 [given, T α = 1620 y r and T β = 405 y r ] = 5 1620 y r - 1 3 4 t h sample will decay, i.e. remaining 1 4 t h , N = N 0 1 2 n N 0 4 = N 0 1 2 n ⇒ n = 2 ∴ t = n T 1 2 = n ln ⁡ 2 λ = 2 × 0.693 5 1620 = 449 y r