Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 min.

In radioactive decay, the number of undecayed nuclei at time $$t$$ is $$N(t) = N_0 e^{-\lambda t}$$, where $$\lambda = \frac{\ln 2}{t_{1/2}}$$.

When 33% has decayed, 67% remains: $$N(t_1) = 0.67 N_0$$, so $$e^{-\lambda t_1} = 0.67$$.

When 67% has decayed, 33% remains: $$N(t_2) = 0.33 N_0$$, so $$e^{-\lambda t_2} = 0.33$$.

Taking the ratio: $$\frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} = \frac{0.33}{0.67} \approx \frac{1}{2}$$, which gives $$e^{-\lambda(t_2 - t_1)} = \frac{1}{2}$$.

This means $$\lambda(t_2 - t_1) = \ln 2$$, and therefore $$t_2 - t_1 = \frac{\ln 2}{\lambda} = t_{1/2} = 20$$ min.

The time interval between 33% decay and 67% decay is $$20$$ min, which is exactly one half-life. This makes sense because the remaining fraction halves from approximately $$0.67$$ to $$0.33$$, which is very nearly a factor of 2.