
Answer & explanation
Correct answer: option 2
$ 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$