In a radioactive decay process, the activity is defined as 𝐴 = − 𝑑𝑁 𝑑𝑡 , where 𝑁(𝑡) is the number of radioactive nuclei at time 𝑡. Two radioactive sources, 𝑆1 and 𝑆2 have same activity at time 𝑡 = 0. At a later time, the activities of 𝑆1 and 𝑆2 are 𝐴1 and 𝐴2, respectively. When 𝑆1 and 𝑆2 have just completed their 3rd and 7th half-lives, respectively, the ratio 𝐴1/𝐴2 is __________.

 To determine the ratio 

$\frac{A_1}{A_2}$ of the activities of two radioactive sources $S_1$ and $S_2$ after they have completed their 3rd and 7th half-lives respectively, we can follow these steps:

1. Understand the relationship between activity and half-life:

   The activity $A$ of a radioactive substance is related to the number of radioactive nuclei $N$ and the decay constant $\lambda$ by the equation:

   $$A=\lambda N$$

   The decay constant $\lambda$ is related to the half-life $T_{1\mathrm{/}2}$ by:

   $$\lambda =\frac{\ln (2)}{T_{1\mathrm{/}2}}$$

2. Express the number of nuclei after a certain number of half-lives:

   After $n$ half-lives, the number of radioactive nuclei remaining is:

   $$N(n)=N_0{\left(\frac{1}{2}\right)}^n$$

   where $N_0$ is the initial number of nuclei.

3. Calculate the activities $A_1$ and $A_2$:

   For source $S_1$, which has completed 3 half-lives:

   $$N_1=N_0{\left(\frac{1}{2}\right)}^3=\frac{N_0}{8}$$$$A_1={\lambda }_1{N}_1={\lambda }_1\frac{N_0}{8}$$

   For source $S_2$, which has completed 7 half-lives:

   $$N_2=N_0{\left(\frac{1}{2}\right)}^7=\frac{N_0}{128}$$$$A_2={\lambda }_2{N}_2={\lambda }_2\frac{N_0}{128}$$

4. Use the initial condition to relate ${\lambda }_1$ and ${\lambda }_2$:

   At time $t=0$, both sources have the same activity:

   $$A_{1,0}=A_{2,0}⟹{\lambda }_1{N}_0={\lambda }_2{N}_0⟹{\lambda }_1={\lambda }_2$$

   Therefore, ${\lambda }_1={\lambda }_2=\lambda$.

5. Compute the ratio $\frac{A_1}{A_2}$:

   Substituting the expressions for $A_1$ and $A_2$:

   $$\frac{A_1}{A_2}=\frac{\lambda \frac{N_0}{8}}{\lambda \frac{N_0}{128}}=\frac{\frac{1}{8}}{\frac{1}{128}}=\frac{128}{8}=16$$

So, the ratio $\frac{A_1}{A_2}$ is:

$$\boxed{{\displaystyle 16}}$$