In a radioactive decay process, the activity is defined as $$A = -\frac{dN}{dt}$$, where $$N(t)$$ is the number of radioactive nuclei at time $$t$$. Two radioactive sources, $$S_1$$ and $$S_2$$ have same activity at time $$t = 0$$. At a later time, the activities of $$S_1$$ and $$S_2$$ are $$A_1$$ and $$A_2$$, respectively. When $$S_1$$ and $$S_2$$ have just completed their 3rd and 7th half-lives, respectively, the ratio $$A_1/A_2$$ is ____.

The activity of a radioactive sample is given by $$A = -\frac{dN}{dt} = \lambda N$$, where $$\lambda$$ is the decay constant and $$N$$ is the number of undecayed nuclei at that instant.

At $$t = 0$$ both sources have the same activity:
$$A_0 = \lambda_1 N_{01} = \lambda_2 N_{02}$$ $$-(1)$$

After some time, source $$S_1$$ has just finished its 3rd half-life, while source $$S_2$$ has just finished its 7th half-life.

The number of nuclei remaining after $$n$$ half-lives is
$$N = N_0\left(\frac12\right)^n$$

Hence, after 3 half-lives,
$$N_1 = N_{01}\left(\frac12\right)^3$$
so the activity of $$S_1$$ is
$$A_1 = \lambda_1 N_{01}\left(\frac12\right)^3$$ $$-(2)$$

After 7 half-lives,
$$N_2 = N_{02}\left(\frac12\right)^7$$
so the activity of $$S_2$$ is
$$A_2 = \lambda_2 N_{02}\left(\frac12\right)^7$$ $$-(3)$$

Divide $$(2)$$ by $$(3)$$:

$$\frac{A_1}{A_2} = \frac{\lambda_1 N_{01}}{\lambda_2 N_{02}}\;\frac{\left(\frac12\right)^3}{\left(\frac12\right)^7} = \frac{\lambda_1 N_{01}}{\lambda_2 N_{02}}\;2^{7-3} = \frac{\lambda_1 N_{01}}{\lambda_2 N_{02}}\;2^{4}$$

From $$(1)$$, $$\lambda_1 N_{01} = \lambda_2 N_{02}$$, so their ratio is 1. Therefore

$$\frac{A_1}{A_2} = 2^{4} = 16$$

Thus, the required ratio is 16.