A radioactive substance decays to (1/16)th of its initial activity in 80 days. The half-life of the radioactive substance expressed in days is …… .

According to question, a radioactive substance decays to $\left(\frac{1}{16}\right)$th of its initial activity in 80 days.

If N₀ is the number of atoms present initially, then in 80 days, the substance decays to $\frac{1}{16}\text{ th }$of N₀ as follows

$\begin{array}{l}{\mathrm{N}}_0\stackrel{{\mathrm{t}}_{1/2}}{\to }\frac{{\mathrm{N}}_0}{2}\stackrel{{\mathrm{t}}_{1/2}}{\to }\frac{{\mathrm{N}}_0}{4}\stackrel{{\mathrm{t}}_{1/2}}{\to }\frac{{\mathrm{N}}_0}{8}\stackrel{{\mathrm{t}}_{1/2}}{\to }\frac{{\mathrm{N}}_0}{16}\\ \Rightarrow 4\times {\mathrm{t}}_{1/2}=80\text{ days }\Rightarrow {\mathrm{t}}_{1/2}=\frac{80}{4}\\ \Rightarrow {\mathrm{t}}_{1/2}=20\text{ days }\end{array}$