To solve this problem, let's begin with some fundamental concepts of radioactive decay.

The activity of a radioactive sample, which is its decay rate, is given by the equation:

$$A = \lambda N$$

where:

• $$A$$ is the activity.
• $$\lambda$$ is the decay constant.
• $$N$$ is the number of radioactive nuclei present.

Given that sample S₁ has an activity of 5 $$\mu$$Ci and sample S₂ has an activity of 10 $$\mu$$Ci, and that sample S₁ has twice the number of nuclei as sample S₂, we have:

$$A_1 = \lambda_1 N_1$$

$$A_2 = \lambda_2 N_2$$

We are also given that:

$$N_1 = 2N_2$$

Substitute $$N_1 = 2N_2$$ into the first equation:

$$5 \, \mu\text{Ci} = \lambda_1 (2N_2)$$

And for the second equation:

$$10 \, \mu\text{Ci} = \lambda_2 N_2$$

Dividing the first equation by the second equation to eliminate $$N_2$$, we get:

$$\frac{5 \, \mu\text{Ci}}{10 \, \mu\text{Ci}} = \frac{\lambda_1 (2N_2)}{\lambda_2 N_2}$$

Simplifying this, we obtain:

$$\frac{1}{2} = \frac{2 \lambda_1}{\lambda_2}$$

Therefore:

$$\lambda_2 = 4 \lambda_1$$

Next, we can relate the decay constant to the half-life using the equation:

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

Using the relationship between the decay constants, we have:

$$\frac{\lambda_2}{\lambda_1} = \frac{T_{1/2,1}}{T_{1/2,2}}$$

Substituting $$\lambda_2 = 4 \lambda_1$$, we get:

$$4 = \frac{T_{1/2,1}}{T_{1/2,2}}$$

Therefore:

$$T_{1/2,1} = 4 T_{1/2,2}$$

Given the options, the half-lives that satisfy this relationship are:

Option A: 20 years and 5 years, respectively.

In this case:

$$\frac{20 \text{ years}}{5 \text{ years}} = 4$$

Thus, option A is correct. Therefore, the half-lives of S₁ and S₂ are 20 years and 5 years, respectively.