Question

2. In 40 days, the number of radioactive nuclei decreases to one-sixteenth the number present initially. What is the half-life (in days) of the material?

Answer 1

The radioactive decay law says

$$N(t)=N_02^{-t/t_{1/2}}$$

Hence, the half-life period

$$t_{1/2}=-\frac{t}{\log_2(N/N_0)}=-\frac{40}{\log_2(1/16)}=10\:\rm days$$