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

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`
or, `A/A_0 = 1/4`
So one fourth of the original activity will remain after 40 hours.