The half line is `._(38)^(90)Sr` is 20 year. If the sample of this nuclide has an activity of `8,000` disintergrations `min^(-1)` today, what will be its activity after 80 year.

Explanation

Given, `(r_(0)) = 8000 dp m, t_(1//2) = 20` year
or `lambda = (0.693)/(20) yr^(-1)`
`r_(0) prop N_(0)`
`r prop N`
`:. (r_(0))/(r) = (N_(o))/(N)`
Now `t = (2.303)/(lambda)` log `(r_(0))/(r)`
`:. 80 = (2.303xx20)/(0.693)` log `(8000)/(r)`
`:. r = 500` dpm