The half life of a radioactive nuclide is 20 hours. What fraction of original activity will remain after 80 hours?
Answer & explanation
Correct answer: option 2
$ N = \frac{N_0}{2^n}$
$ n = \frac{t}{t_\frac{1}{2}} = \frac{80}{20} = 4$
$ N = \frac{N_0}{2^4} = \frac{N_0}{16}$
The correct answer is Option (2) → $\frac{1}{16}$