A radioactive element` X` converts into another stable elemnet `Y`. Half-life of `X` is `2h`. Initally, only `X` is present. After time `t`, the ratio of atoms of `X` and `Y` is found to be `1:4` Then `t` in hours is .

Options

A. 2
B. 4
C. between 4 and 6
D. 4

Answer

To solve the problem, we need to analyze the decay of the radioactive element \( X \) and its conversion into the stable element \( Y \).

### Step-by-step Solution:

1. **Understanding the Half-life**:
The half-life of element \( X \) is given as 2 hours. This means that after every 2 hours, half of the remaining atoms of \( X \) will have decayed into \( Y \).

2. **Initial Conditions**:
Initially, we have \( n \) atoms of \( X \) and 0 atoms of \( Y \).

3. **After 1 Half-life (2 hours)**:
- Remaining atoms of \( X \):
\[ X = \frac{n}{2} \]
- Atoms of \( Y \) produced:
\[ Y = n - X = n - \frac{n}{2} = \frac{n}{2} \]
- Ratio of \( X \) to \( Y \):
\[ \text{Ratio} = \frac{X}{Y} = \frac{\frac{n}{2}}{\frac{n}{2}} = 1:1 \]

4. **After 2 Half-lives (4 hours)**:
- Remaining atoms of \( X \):
\[ X = \frac{n}{4} \]
- Atoms of \( Y \) produced:
\[ Y = n - X = n - \frac{n}{4} = \frac{3n}{4} \]
- Ratio of \( X \) to \( Y \):
\[ \text{Ratio} = \frac{X}{Y} = \frac{\frac{n}{4}}{\frac{3n}{4}} = 1:3 \]

5. **After 3 Half-lives (6 hours)**:
- Remaining atoms of \( X \):
\[ X = \frac{n}{8} \]
- Atoms of \( Y \) produced:
\[ Y = n - X = n - \frac{n}{8} = \frac{7n}{8} \]
- Ratio of \( X \) to \( Y \):
\[ \text{Ratio} = \frac{X}{Y} = \frac{\frac{n}{8}}{\frac{7n}{8}} = 1:7 \]

6. **Finding the Time for Ratio 1:4**:
- We have found that at 4 hours the ratio is \( 1:3 \) and at 6 hours the ratio is \( 1:7 \).
- Since the desired ratio \( 1:4 \) lies between these two ratios, we can conclude that the time \( t \) must be between 4 hours and 6 hours.

### Conclusion:
Thus, the time \( t \) in hours when the ratio of atoms of \( X \) to \( Y \) is \( 1:4 \) is between 4 and 6 hours.