The half-life of a radioactive nuclide is 20 hours. What fraction of original activity will remain after 40 hours?

Answer

4

Explanation

We have `t/t_(1/2) = ("40 hours")/("2 hours") = 2`
Thus, `A =(A_(0)/(2^(t//t) (1//2))) = A_(0)/2^(2) = A_(0)/4`
So, one-fourth of the original activity will remain after 40 hour.