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, S₁ and S₂, have the 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 $\frac{A_1}{A_2}$ is .

Correct Answer: The ratio $\frac{A_1}{A_2}$ is 16.

Step-by-Step Explanation:

1. Understanding Radioactive Decay and Activity:

The activity $A$ of a radioactive sample at any time $t$ is proportional to the number of radioactive nuclei $N\left(t\right)$ remaining in the sample:

$A\left(t\right)=\lambda N\left(t\right)$

where $\lambda$ is the decay constant of the source.

2. Activity after $n$ Half-lives:

After a time corresponding to $n$ half-lives, the number of remaining nuclei decreases by a factor of $2^n$. Consequently, the activity of the sample also reduces by the same factor:

$A=\frac{A_0}{2^n}$

where $A_0$ is the initial activity at time $t=0$.

3. Applying to Sources S₁ and S₂:

It is given that both sources have the same initial activity at $t=0$. Let this initial activity be $A_0$.

For source S₁, which has completed $n_1=3$ half-lives:

$A_1=\frac{A_0}{2^3}=\frac{A_0}{8}$

For source S₂, which has completed $n_2=7$ half-lives:

$A_2=\frac{A_0}{2^7}=\frac{A_0}{128}$

4. Calculating the Ratio $\frac{A_1}{A_2}$:

Now, we find the ratio of the two activities:

$\frac{A_1}{A_2}=\frac{A_0/2^3}{A_0/2^7}=\frac{2^7}{2^3}=2^{7-3}=2^4$

$2^4=16$

Thus, the ratio $\frac{A_1}{A_2}$ is equal to 16.