At some instant, a radioactive sample `S_(1)` having an activity 5 `muCi` has twice the number of nuclei as another sample `S_(2)` which has an activity of 10 `muCi`. The half lives of `S_(1) and S_(2)` are :

Options

A. 20 years and 5 years, respectively
B. 20 years and 10 years, respectively
C. 10 years and 20 years, respectively
D. 5 years and 20 years, respectively

Answer

To solve the problem, we need to find the half-lives of two radioactive samples \( S_1 \) and \( S_2 \) based on their activities and the relationship between their number of nuclei.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Activity of sample \( S_1 \) (denoted as \( A_1 \)) = 5 µCi
- Activity of sample \( S_2 \) (denoted as \( A_2 \)) = 10 µCi
- Number of nuclei in sample \( S_1 \) (denoted as \( N_1 \)) is twice that of sample \( S_2 \) (denoted as \( N_2 \)):
\[ N_1 = 2N_2 \]

2. **Using the Relationship Between Activity and Decay Constant:**
- The activity \( A \) of a radioactive sample is given by:
\[ A = -\frac{dN}{dt} = \lambda N \]
- For sample \( S_1 \):
\[ A_1 = \lambda_1 N_1 \implies 5 = \lambda_1 N_1 \]
- For sample \( S_2 \):
\[ A_2 = \lambda_2 N_2 \implies 10 = \lambda_2 N_2 \]

3. **Substituting \( N_1 \) in Terms of \( N_2 \):**
- Since \( N_1 = 2N_2 \), we substitute this into the equation for \( A_1 \):
\[ 5 = \lambda_1 (2N_2) \implies \lambda_1 = \frac{5}{2N_2} \]

4. **Finding \( \lambda_2 \):**
- From the equation for \( A_2 \):
\[ 10 = \lambda_2 N_2 \implies \lambda_2 = \frac{10}{N_2} \]

5. **Finding the Ratio of the Decay Constants:**
- Now we can find the ratio of \( \lambda_1 \) to \( \lambda_2 \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{5}{2N_2}}{\frac{10}{N_2}} = \frac{5}{2} \cdot \frac{N_2}{10} = \frac{5}{20} = \frac{1}{4} \]

6. **Finding the Ratio of the Half-Lives:**
- The half-life \( T_{1/2} \) is related to the decay constant \( \lambda \) by:
\[ T_{1/2} = \frac{\ln 2}{\lambda} \]
- Therefore, the ratio of the half-lives is:
\[ \frac{T_{1/2,1}}{T_{1/2,2}} = \frac{\frac{\ln 2}{\lambda_1}}{\frac{\ln 2}{\lambda_2}} = \frac{\lambda_2}{\lambda_1} \]
- Since we found \( \frac{\lambda_1}{\lambda_2} = \frac{1}{4} \), we have:
\[ \frac{T_{1/2,1}}{T_{1/2,2}} = 4 \]

### Final Answer:
The ratio of the half-lives of \( S_1 \) to \( S_2 \) is 4.