At a given instant, say t = 0 , two radioactive substance A and B have equal activities. The ratio R B R A of their activities after time t itself decays with time t as e - 3 t . If the half-life of A is l n 2 , the half-life of B is:

Step-by-step solution

Given, Half life of A = ln 2 Initial number of nucleus is, R 0 A = R 0 B = R 0 Half life is related with decay constant as, t 1 / 2 A = ln 2 λ λ A = 1 At t = 0   R A = R B From activity relation, we have, R 0 e - λ A T = R 0 E - λ B T at   T = t ,   ⇒ R B R A = R 0 e - λ B t R 0 e - λ A t As per given question, R B R A = e - 3 t , On comparing with above relation, we have ⇒ e - λ B - λ A t = e - t ⇒ λ B - λ A = 3 ⇒ λ B = 3