The activity of a sample of radioactive material is `A_1` at time `t_(1)` and `A_(2)` at time `t_(2)(t_(2) le t_(1))` . Obtain an expression for its mean life.

Options

A. `A_1 t_1 = A_2 t_2`
B. `A_1 - A_2 = t_2 - t_1`
C. `A_2 = A_1 e^((t_1 -t_2)//T)`
D. `A_2 - A_1 e^((t_1//t_2)T)`

Answer

C

Explanation

( c) `A = A_0 e^(- lamda t) = A_0 e^(-t//tau)` , where `tau` = mean life
So `A_1 = A_0 e^(-t_1//T) rArr A_0 = (A_0)/(e^(-t_1//T)) = A_1 e^(t_1//T)`
`:. A_2 = A_0 e^(-t//T) = (A_1 e^(t_1//T)) e^(-t_2//T) rArr A_2 = A_1 e^((t_1 - t_2)//T)`.