Step-by-Step Solution

1. Identify Given Information: We are given two radioactive samples, S1 and S2, with the following properties:

   • Activity of S1, $A_1 = 5 \, \mu\text{Ci}$
   • Activity of S2, $A_2 = 10 \, \mu\text{Ci}$
   • Let $N_1$ and $N_2$ be the number of nuclei in samples S1 and S2, respectively. The problem states that S1 has twice the number of nuclei as S2, so $N_1 = 2N_2$.

2. Recall the Formula for Radioactive Activity: The activity (A) of a radioactive sample is related to its decay constant ($\lambda$) and the number of radioactive nuclei (N) by the formula: $A = \lambda N$ The decay constant is related to the half-life ($T_{1/2}$) by: $\lambda = \frac{\ln 2}{T_{1/2}}$ Combining these two equations, we get the activity in terms of the half-life: $A = \frac{\ln 2}{T_{1/2}} N$

3. Set up Equations for Both Samples: For sample S1, with half-life $T_1$: $A_1 = \frac{\ln 2}{T_1} N_1 \quad \text{(Equation 1)}$ For sample S2, with half-life $T_2$: $A_2 = \frac{\ln 2}{T_2} N_2 \quad \text{(Equation 2)}$

4. Find the Relationship Between the Half-Lives: To find the relationship between $T_1$ and $T_2$, we can take the ratio of the two activity equations (Equation 1 / Equation 2): $\frac{A_1}{A_2} = \frac{\frac{\ln 2}{T_1} N_1}{\frac{\ln 2}{T_2} N_2}$ The $\ln 2$ terms cancel out: $\frac{A_1}{A_2} = \frac{N_1}{N_2} \times \frac{T_2}{T_1}$

5. Substitute the Given Values: Now, we substitute the known values into the ratio equation:

   • $A_1 = 5 \, \mu\text{Ci}$
   • $A_2 = 10 \, \mu\text{Ci}$
   • $N_1 = 2N_2$, which means $\frac{N_1}{N_2} = 2$

   $\frac{5}{10} = 2 \times \frac{T_2}{T_1}$

6. Solve for the Ratio of Half-Lives: Simplify the equation: $\frac{1}{2} = 2 \frac{T_2}{T_1}$ Rearrange to solve for the relationship between $T_1$ and $T_2$: $T_1 = 2 \times 2 T_2$ $T_1 = 4 T_2$ This means the half-life of sample S1 must be four times the half-life of sample S2.

7. Evaluate the Options: Let's check the given options to see which one satisfies the condition $T_1 = 4 T_2$.

   • A: 20 years and 5 years, respectively $T_1 = 20$ years, $T_2 = 5$ years. $4 \times T_2 = 4 \times 5 = 20$ years. Since $T_1 = 20$ years, this option is correct ($T_1 = 4T_2$).
   • B: 20 years and 10 years, respectively $T_1 = 20$ years, $T_2 = 10$ years. $4 \times T_2 = 4 \times 10 = 40$ years. This is not equal to $T_1$. Incorrect.
   • C: 10 years each $T_1 = 10$ years, $T_2 = 10$ years. $4 \times T_2 = 4 \times 10 = 40$ years. This is not equal to $T_1$. Incorrect.
   • D: 5 years each $T_1 = 5$ years, $T_2 = 5$ years. $4 \times T_2 = 4 \times 5 = 20$ years. This is not equal to $T_1$. Incorrect.

Conclusion

The only option that satisfies the derived relationship $T_1 = 4T_2$ is A.