Half-life of iodine (atomic mass 125) is 60 days. Calculate the present of radioactivity remain after 120 days.

Answer

C

Explanation

`t_(0.5) = 60` days t = 120 days `therefore n = (T)/(t_(0.5)) = (120)/(60) = 2`
% of radioactivity remain after 2half-lives, `N = N_(0) ((1)/(2))^(n) = 100 xx ((1)/(2))^(2) = (100)/(4) = 25%`