At some instant, a radioactive sample S$$_1$$ having an activity 5$$\mu$$Ci has twice the number of nuclei as another sample S$$_2$$ which has an activity of 10$$\mu$$Ci. The half lives of S$$_1$$ and S$$_2$$ are:

We begin with the basic law of radioactivity which says that the activity $$A$$ of a sample is directly proportional to the decay constant $$\lambda$$ and to the number of undecayed nuclei $$N$$ present at that instant. Symbolically, the relation is written as

$$A = \lambda N.$$

For the two samples we have:

$$A_1 = \lambda_1 N_1,$$

$$A_2 = \lambda_2 N_2.$$

According to the statement of the problem, the first sample S1 has an activity of $$5\ \mu\text{Ci}$$ and contains twice as many nuclei as the second sample S2, which itself has an activity of $$10\ \mu\text{Ci}.$$ Translating this wording into equations we write

$$A_1 = 5\ \mu\text{Ci}, \qquad A_2 = 10\ \mu\text{Ci},$$

$$N_1 = 2\,N_2.$$

Substituting the values of the activities into the activity formula for each sample we get

$$5 = \lambda_1 N_1,$$

$$10 = \lambda_2 N_2.$$

Next we use the given relation $$N_1 = 2N_2$$ in the first equation. Replacing $$N_1$$ by $$2N_2$$ yields

$$5 = \lambda_1 (2N_2).$$

Simplifying, this becomes

$$5 = 2\lambda_1 N_2.$$

Now we have two simultaneous equations containing the common factor $$N_2$$:

$$5 = 2\lambda_1 N_2,$$

$$10 = \lambda_2 N_2.$$

To eliminate $$N_2$$ we divide the first equation by the second equation term by term:

$$\frac{5}{10} = \frac{2\lambda_1 N_2}{\lambda_2 N_2}.$$

The quantities $$N_2$$ cancel out immediately, giving

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

Multiplying both sides by $$\lambda_2$$ and then by $$2$$, we obtain

$$\lambda_1 = \frac{1}{4}\,\lambda_2.$$

Thus the ratio of the decay constants is

$$\lambda_1 : \lambda_2 = 1 : 4.$$

We now connect decay constant to half-life. The standard formula relating the two is

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

where $$T_{1/2}$$ is the half-life and $$\ln 2$$ is a constant. Because $$\ln 2$$ is the same for both samples, the half-life is inversely proportional to the decay constant. Therefore

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

Substituting the ratio $$\lambda_1 : \lambda_2 = 1 : 4$$ into this inverse relation we get

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

Hence the half-life of S1 is four times the half-life of S2, giving the required ratio

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

Hence, the correct answer is Option A.