由 A=A0e−λtA = A_0 e^{-\lambda t}A=A0e−λt 得 t=ln(200/2)/λ=ln100/λt = \ln(200/2)/\lambda = \ln 100/\lambdat=ln(200/2)/λ=ln100/λ。取 λ=3.209×10−5\lambda = 3.209\times 10^{-5}λ=3.209×10−5 s−1^{-1}−1,t=1.435×105t = 1.435\times 10^{5}t=1.435×105 s ≈39.9\approx 39.9≈39.9 h,约等于 ln100/ln2=6.64\ln 100/\ln 2 = 6.64ln100/ln2=6.64 个半衰期。