The activity of a radioactive sample is measured as N₀ counts per minute at time t = 0 and N₀ e counts per minute at time t = 3 minute. The time (in minute) in which the activity reduces to half the value, is

Step-by-step solution

The activity of a radioactive sample at time t is given by A = A₀ e^ - t . Given that at t = 0 , A = N₀ . At t = 3 minutes, A = N₀ e . Substituting these values into the decay equation: N₀ e = N₀ e^ -3 e⁻¹ = e^ -3 Comparing the powers of e , we get: 3 = 1 = 1 3 min ⁻¹ The time in which the activity reduces to half its initial value is the half-life T_ 1/2 , which is given by: T_ 1/2 = _e 2 Substituting the value of : T_ 1/2 = _e 2 1 3 = 3 _e 2 min Answer: 3 _e 2