A radioactive element X with half life 2 h decays giving a stable element Y. After a time t, ratio of X and Y atoms is 1:16 .Time t is

Options

A. 6 h
B. 4 h
C. 8 h
D. 16 h

Explanation

To solve the problem, we need to determine the time \( t \) when the ratio of the number of atoms of radioactive element \( X \) to the stable element \( Y \) is \( 1:16 \). We know that the half-life of element \( X \) is \( 2 \) hours.

### Step-by-Step Solution:

1. **Initial Setup**:
- Let the initial number of atoms of element \( X \) be \( N_0 \).
- Initially, the number of atoms of element \( Y \) is \( 0 \).

2. **Decay After One Half-Life**:
- After \( 2 \) hours (one half-life), the number of atoms of \( X \) will be:
\[ N_X = \frac{N_0}{2} \]
- The number of atoms of \( Y \) formed will be:
\[ N_Y = N_0 - N_X = N_0 - \frac{N_0}{2} = \frac{N_0}{2} \]
- The ratio \( \frac{N_X}{N_Y} \) is:
\[ \frac{N_X}{N_Y} = \frac{\frac{N_0}{2}}{\frac{N_0}{2}} = 1:1 \]

3. **Decay After Two Half-Lives**:
- After \( 4 \) hours (two half-lives), the number of atoms of \( X \) will be:
\[ N_X = \frac{N_0}{4} \]
- The number of atoms of \( Y \) formed will be:
\[ N_Y = N_0 - N_X = N_0 - \frac{N_0}{4} = \frac{3N_0}{4} \]
- The ratio \( \frac{N_X}{N_Y} \) is:
\[ \frac{N_X}{N_Y} = \frac{\frac{N_0}{4}}{\frac{3N_0}{4}} = \frac{1}{3} \quad \text{(which is } 1:3\text{)} \]

4. **Decay After Three Half-Lives**:
- After \( 6 \) hours (three half-lives), the number of atoms of \( X \) will be:
\[ N_X = \frac{N_0}{8} \]
- The number of atoms of \( Y \) formed will be:
\[ N_Y = N_0 - N_X = N_0 - \frac{N_0}{8} = \frac{7N_0}{8} \]
- The ratio \( \frac{N_X}{N_Y} \) is:
\[ \frac{N_X}{N_Y} = \frac{\frac{N_0}{8}}{\frac{7N_0}{8}} = \frac{1}{7} \quad \text{(which is } 1:7\text{)} \]

5. **Decay After Four Half-Lives**:
- After \( 8 \) hours (four half-lives), the number of atoms of \( X \) will be:
\[ N_X = \frac{N_0}{16} \]
- The number of atoms of \( Y \) formed will be:
\[ N_Y = N_0 - N_X = N_0 - \frac{N_0}{16} = \frac{15N_0}{16} \]
- The ratio \( \frac{N_X}{N_Y} \) is:
\[ \frac{N_X}{N_Y} = \frac{\frac{N_0}{16}}{\frac{15N_0}{16}} = \frac{1}{15} \quad \text{(which is } 1:15\text{)} \]

6. **Finding the Required Ratio**:
- We need the ratio \( \frac{N_X}{N_Y} = \frac{1}{16} \).
- This occurs after \( 8 \) hours, as we have established that after \( 8 \) hours the ratio is \( 1:15 \) and we can see that the next half-life would yield \( 1:16 \).

### Conclusion:
The time \( t \) when the ratio of \( X \) to \( Y \) is \( 1:16 \) is \( 8 \) hours.