The activity of a radioactive sample is monitored over time. It is observed that its activity decreases by 87.5 % in 60 days. The half-life period of the radioactive sample is:

Step-by-step solution

The percentage of activity remaining after 60 days is 100 % - 87.5 % = 12.5 % . Converting this to a fraction, we get 12.5 100 = 1 8 . The activity of a radioactive sample is given by A = A₀ ( 1 2 )^n , where n is the number of half-lives. A A₀ = 1 8 = ( 1 2 )^3 Thus, n = 3 half-lives have elapsed. Since the total time t = 60 days, the half-life period is T_ 1/2 = t n = 60 3 = 20 days. Answer: 20 days