A radioactive substance decays to $$\left(\frac{1}{16}\right)^{th}$$ of its initial activity in 80 days. The half life of the radioactive substance expressed in days is ___.

The activity of a radioactive substance decreases according to $$A = A_0 \left(\frac{1}{2}\right)^{n}$$, where $$n$$ is the number of half-lives elapsed.

Given that the activity reduces to $$\frac{1}{16}$$ of its initial value:

$$\frac{A}{A_0} = \frac{1}{16} = \left(\frac{1}{2}\right)^4$$

This means 4 half-lives have passed in 80 days.

Therefore, the half-life is:

$$t_{1/2} = \frac{80}{4} = 20 \text{ days}$$