Ratio of Half Lives of Radioactive Samples S1 and S2

The activity $A$ of a radioactive sample is related to the decay constant $\lambda$ and the number of active nuclei $N$ by the relation:
$A = \lambda N$

Since the decay constant $\lambda$ is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}}$, the activity can be expressed as:
$A = \frac{\ln 2}{T_{1/2}} N$

For sample $S_1$:
$A_1 = \frac{\ln 2}{T_{1, 1/2}} N_1$

For sample $S_2$:
$A_2 = \frac{\ln 2}{T_{2, 1/2}} N_2$

Taking the ratio of the activities of $S_1$ and $S_2$:
$\frac{A_1}{A_2} = \left( \frac{T_{2, 1/2}}{T_{1, 1/2}} \right) \left( \frac{N_1}{N_2} \right)$

Given in the problem statement:

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

Substituting these values into the ratio equation:
$\frac{5}{10} = \left( \frac{T_{2, 1/2}}{T_{1, 1/2}} \right) \times 2$

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

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

Thus, the half-life of $S_1$ must be $4$ times the half-life of $S_2$.

Checking the given options for $T_{1, 1/2} = 4 T_{2, 1/2}$:

• (A) 20 years and 5 years: $20 = 4 \times 5$ (Valid)
• (B) 20 years and 10 years: $20 \neq 4 \times 10$ (Invalid)
• (C) 10 years each: $10 \neq 4 \times 10$ (Invalid)
• (D) 5 years each: $5 \neq 4 \times 5$ (Invalid)

Therefore, the half-lives of $S_1$ and $S_2$ can be 20 years and 5 years, respectively.

Correct Option: (A)