The activity of a radioactive sample falls from `600 s^(-1)` to `500 s^(-1) `in 40 minutes. Calculate its half-life.

Explanation

We have,
A= A_0 e^(-M)`
or, `500 s^(-1) = (600 s^(-1) e^(-M)`
or, `e^(-M) = 5/6`
or, `lambdat = 1n(6/5)`
or, `lambda = (1n(6/5))/t` = (1n(6/5))/(40 min)` The half-life is `t_(1/2) = (1n 2)/(lambda)`
`= (1n 2)/(1n (6/5)` xx 40 min`
`= 152 min`.