a b c d

b

$ N = N_0 e^{-\lambda t}$ First case $ N = \frac{N_0}{3}$ $\Rightarrow N_0 e^{-\lambda t_1} = \frac{N_0}{3}$ $ t_1 = \frac{ln3}{\lambda}$ Second case $ N = \frac{2N_0}{3}$ $\Rightarrow N_0 e^{-\lambda t_2} = \frac{2N_0}{3}$ $ t_2 = \frac{ln3/2}{\lambda}$ $ t_1 - t_2 = \frac{ln2}{\lambda} = t_{\frac{1}{2}}=20 min$