If the half of an isotope X is 10 years , its decay contant is

Options

A. `6.932 year^(-1)`
B. `0.6932 year^(-1)`
C. `0.06932 year^(-1)`
D. `0.006932 year^(-1)`

Answer

To find the decay constant (λ) of an isotope given its half-life (t₁/₂), we can use the relationship between half-life and decay constant for first-order reactions. The formula is:

\[ t_{1/2} = \frac{0.693}{\lambda} \]

### Step-by-Step Solution:

1. **Identify the half-life**:
The problem states that the half-life (t₁/₂) of isotope X is 10 years.

2. **Use the formula for half-life**:
We can rearrange the half-life formula to solve for the decay constant (λ):
\[ \lambda = \frac{0.693}{t_{1/2}} \]

3. **Substitute the half-life into the equation**:
Now, substitute the value of the half-life into the equation:
\[ \lambda = \frac{0.693}{10 \text{ years}} \]

4. **Calculate the decay constant**:
Perform the calculation:
\[ \lambda = 0.0693 \text{ year}^{-1} \]

5. **Final result**:
Therefore, the decay constant (λ) is approximately:
\[ \lambda \approx 0.0693 \text{ per year} \]

### Conclusion:
The decay constant for isotope X is approximately 0.0693 per year.

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