The half life of a radioactive nuclide is 20 hours. What fraction of original activity will remain after 80 hours?

$\frac{1}{8}$ $\frac{1}{16}$ $\frac{1}{4}$ $\frac{1}{32}$

$\frac{1}{16}$

$ 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) → $\frac{1}{16}$