A radioactive sample `S_(1)` having the activity `A_(1)` has twice the number of nucleic as another sample `S_(2)` of activity `A_(2)`. If `A_(2)=2A_(1)`, then the ratio of half-life of `S_(1)` to the half-life of `S_(2)` is

Options

A. 4
B. 2
C. 0.25
D. 0.75

Answer

To solve the problem, we will use the relationship between activity, number of nuclei, and half-life of radioactive samples.

### Step-by-Step Solution:

1. **Understanding the relationship between activity and half-life:**
The activity \( A \) of a radioactive sample is given by the formula:
\[ A = \lambda N \]
where \( \lambda \) is the decay constant and \( N \) is the number of nuclei. The decay constant \( \lambda \) can also be related to the half-life \( t_{1/2} \) by:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Therefore, we can express activity in terms of half-life:
\[ A = \frac{0.693}{t_{1/2}} N \]

2. **Setting up the equations for the two samples:**
For sample \( S_1 \):
\[ A_1 = \frac{0.693}{t_{1/2,1}} N_1 \]
For sample \( S_2 \):
\[ A_2 = \frac{0.693}{t_{1/2,2}} N_2 \]

3. **Using the given information:**
We know that:
- \( N_1 = 2 N_2 \) (Sample \( S_1 \) has twice the number of nuclei as \( S_2 \))
- \( A_2 = 2 A_1 \)

4. **Substituting the values into the activity equations:**
From the second condition, substituting \( A_2 \):
\[ 2 A_1 = \frac{0.693}{t_{1/2,2}} N_2 \]
Now substituting \( A_1 \):
\[ 2 \left( \frac{0.693}{t_{1/2,1}} N_1 \right) = \frac{0.693}{t_{1/2,2}} N_2 \]

5. **Substituting \( N_1 = 2 N_2 \):**
\[ 2 \left( \frac{0.693}{t_{1/2,1}} (2 N_2) \right) = \frac{0.693}{t_{1/2,2}} N_2 \]
Simplifying gives:
\[ \frac{4 \cdot 0.693}{t_{1/2,1}} N_2 = \frac{0.693}{t_{1/2,2}} N_2 \]

6. **Cancelling \( 0.693 N_2 \) from both sides:**
\[ \frac{4}{t_{1/2,1}} = \frac{1}{t_{1/2,2}} \]

7. **Rearranging to find the ratio of half-lives:**
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = 4 \]

### Final Answer:
The ratio of the half-life of \( S_1 \) to the half-life of \( S_2 \) is:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = 4 \]