The activity of a radioactive sample is measured as N 0 counts per minute at t = 0 and N 0 e counts per minute at t = 5 min . The time (in minutes) at which the activity reduces to half its value is

Step-by-step solution

Fraction remains after n half-lives N N 0 = 1 2 n = 1 2 t T Given, N = N 0 e ⇒ N 0 e N 0 = 1 2 5 T ⇒   1 e = 1 2 5 T Taking log on both sides, we get log 1 - log e = 5 T log 1 2 ⇒   - 1 = 5 T - log 2 ⇒   T = 5 log e 2 Now, let t be the time after which activity reduces to half 1 2 = 1 2 t / 5 log e 2 ⇒   t ' = 5 log e 2