Two radioactive samples A and B have same number of atoms initially. After 6 hours, I/8th of sample A and l/32th of sample B remains. The number of half-lives of A and B are `n_(1)` and `n_(2)` The ratio of half-lives of A and B is

Options

A. `5/7`
B. `n_(2)/n_(1)`
C. `n_(1)/n_(2)`
D. `5/3`

Explanation

To solve the problem, we need to determine the ratio of the half-lives of two radioactive samples A and B based on the information given about their decay over a period of 6 hours.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two samples A and B with the same initial number of atoms (let's denote it as \( N_0 \)).
- After 6 hours, \( \frac{1}{8} \) of sample A remains, and \( \frac{1}{32} \) of sample B remains.

2. **Setting Up the Equations**:
- For sample A:
\[ N_A = N_0 \cdot \left(\frac{1}{2}\right)^{n_1} \]
Given that \( N_A = \frac{N_0}{8} \), we can set up the equation:
\[ \frac{N_0}{8} = N_0 \cdot \left(\frac{1}{2}\right)^{n_1} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{8} = \left(\frac{1}{2}\right)^{n_1} \]

- For sample B:
\[ N_B = N_0 \cdot \left(\frac{1}{2}\right)^{n_2} \]
Given that \( N_B = \frac{N_0}{32} \), we can set up the equation:
\[ \frac{N_0}{32} = N_0 \cdot \left(\frac{1}{2}\right)^{n_2} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{32} = \left(\frac{1}{2}\right)^{n_2} \]

3. **Solving for \( n_1 \) and \( n_2 \)**:
- From \( \frac{1}{8} = \left(\frac{1}{2}\right)^{n_1} \):
\[ 8 = 2^{n_1} \implies n_1 = 3 \]
- From \( \frac{1}{32} = \left(\frac{1}{2}\right)^{n_2} \):
\[ 32 = 2^{n_2} \implies n_2 = 5 \]

4. **Finding the Ratio of Half-Lives**:
- The half-lives are related to the number of half-lives and the total time elapsed. We know:
\[ n_1 \cdot t_1 = n_2 \cdot t_2 \]
- To find the ratio of half-lives \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{n_2}{n_1} = \frac{5}{3} \]

5. **Conclusion**:
- The ratio of the half-lives of samples A and B is:
\[ \frac{t_1}{t_2} = \frac{5}{3} \]