At some instant, a radioactive sample S 1 having an activity 5 μ Ci has twice the number of nuclei as another sample S 2 which has an activity of 10 μ Ci . The ratio of half lives of S 1 a n d S 2 is

Step-by-step solution

Given: N 1 = 2. N 2 Now, activity A 1 = λ 1 N 1 = 5 μ C i     . . . . . ( 1 ) and A 2 = λ 2 N 2 = 10 μ C i     . . . . . ( 2 ) Dividing ( 1 ) ( 2 ) we get λ 1 λ 2 . N 1 N 2 = 5 10 ⇒ t 2 t 1 .2 = 1 2 ⇒ t 1 t 2 = 4 1