The half life of a radioactive substance is   20   minutes. In how much time, the activity of substance drops to 1 16 t h of its initial value?

Step-by-step solution

Activity of substance is given by, A = A 0 2 n , where n is the number of half-lives. Now, A 0 16 = A 0 2 n ⇒ n = 4 Therefore, n = t T 1 2 ⇒ t = 4 T 1 2 = 4 × 20   min = 80   min