Activities of three radioactive substances $$A$$, $$B$$ and $$C$$ are represented by the curves $$A$$, $$B$$ and $$C$$, in the figure. Then their half-lives $$T_{1/2}(A) : T_{1/2}(B) : T_{1/2}(C)$$ are in the ratio:

$$R = R_0 e^{-\lambda t}$$

$$\ln R = \ln R_0 - \lambda t$$

$$\lambda = |\text{slope}| = \frac{\Delta y}{\Delta x}$$

$$\lambda_A = \frac{6 - 0}{10 - 0} = \frac{6}{10}$$

$$\lambda_B = \frac{6 - 0}{5 - 0} = \frac{6}{5}$$

$$\lambda_C = \frac{2 - 0}{5 - 0} = \frac{2}{5}$$

$$T_{1/2}(A) : T_{1/2}(B) : T_{1/2}(C) = \frac{1}{\lambda_A} : \frac{1}{\lambda_B} : \frac{1}{\lambda_C}$$

$$T_{1/2}(A) : T_{1/2}(B) : T_{1/2}(C) = \frac{10}{6} : \frac{5}{6} : \frac{5}{2} = 10 : 5 : 15 = 2 : 1 : 3$$