Question

A 20mg sample of thorium-233 decays to 17mg after 5 min. determine the half life of thorium-233?

Answer 1

Let r be the decay rate (time in minutes) 17 = 20 e^(5r) 0.85 = e^(5r) ln 0.85 = 5r similarly, ln 0.5 = hr ... where h is the half-life, so dividing the 2 equations, ln 0.85 / ln 0.50 = 5r / hr = 5/h h = 5 ln 0.50 / ln 0.85 h = 21.3 min b) 1/20 = 0.05, so ln 0.85 = 5r ln 0.05 = tr t = 5 ln 0.05 / ln 0.85 t = 92.2 min