Calculate Ratio of Activities for Radioactive Decay Half Lives

The activity $A(t)$ of a radioactive sample as a function of time $t$ is given by the law of radioactive decay: $A(t) = A_0 e^{-\lambda t}$

After $n$ half-lives, the activity of a sample decreases to: $A(n) = A_0 \left(\frac{1}{2}\right)^n$ where $A_0$ is the initial activity at time $t = 0$, and $n$ is the number of half-lives completed.

Given that at $t = 0$, both radioactive sources $S_1$ and $S_2$ have the same initial activity, let: $A_{1,0} = A_{2,0} = A_0$

When source $S_1$ has completed $n_1 = 3$ half-lives, its activity $A_1$ is: $A_1 = A_0 \left(\frac{1}{2}\right)^3 = \frac{A_0}{8}$

When source $S_2$ has completed $n_2 = 7$ half-lives, its activity $A_2$ is: $A_2 = A_0 \left(\frac{1}{2}\right)^7 = \frac{A_0}{128}$

Therefore, the ratio of the activities $A_1 / A_2$ is: $\frac{A_1}{A_2} = \frac{\frac{A_0}{8}}{\frac{A_0}{128}} = \frac{128}{8} = 16$