Problema Solution
A certain radioactive isotope has leaked into a small stream. Four hundred days after the leak, 4% of the original amount of the substance remained. Determine the half-life of this radioactive isotope. Do not round until after final answer.
Answer provided by our tutors
fraction remaining, f = 0.5^(t/h) , where h is the half-life
0.08 = 0.5^(200/h)
(200/h)*ln 0.5 = ln 0.08
h = 200*ln 0.5 / ln 0.08 = 54.89, ≈ 55 days <-------