A radioisotope has `t_(1//2) = 5` years. After a given amount decays for 15 years, what fraction of the original isotope remains?

Explanation

Half life `(t_(1//2) = 5` years
Time for decay `(T) = 15` years
We know that, `T = n xx t_(1//2)`
So, `15 = n xx 5`
or n = 3
We know that, `N = ((1)/(2))^(n) N_(0)`
or `(N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(3) = (1)/(8)`
Thus, after 15 years `(1)/(8)th` of the original amount remains.