A radioactive element disintegrated into half after 18 hours what will be the %of radioactive element after 42.5 hours

Answer 1

We use exponential delay model: A = A₀e^rt, where r is rate of dilay. If t = 18 h, then A = 1/2A₀ = A₀e^18r,

18r = ln(1/2), r = ln(1/2)/18 ≈ - 0.0385. So A = A₀e^{- 0.0385t}. Now A/A₀ = e^{- 0.0385t.} For t = 42.5 h we have

A/A₀ = e^{- 0.0385·42.5} ≈ 0.1947 = 19.47%