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)/(20 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` hours.