A radioisotope has `t_(1//2) = 3` years. After a given amount decays for 12 years, what fraction of the original isotope remains?

Explanation

Given half life `(t_(1//2)) = 3` years
Time for decay `(T) = 12` years
We know that `T = n xx t_(1//2)`
`12 = n xx 3`
`:.n = (12)/(3) = 4`
Let the original amount `N_(0)`
Let the amount left after three half-life periods be `N`.
we know that `N = ((1)/(2))^(n) N_(0)`
or `(N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(4) = (1)/(16)`