Half Life of a certain radioactive substance is 69.3 days. Its disintegration constant is:

Options

A. `0.010 "day"^(-1)`
B. `0.100 "day"^(-1)`
C. `0.001 "day"^(-1)`
D. `1.00 "day"^(-1)`

Explanation

To find the disintegration constant (λ) of a radioactive substance given its half-life (t½), we can use the formula:

\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]

Where:
- \(\lambda\) is the disintegration constant,
- \(\ln(2)\) is the natural logarithm of 2 (approximately 0.693),
- \(t_{1/2}\) is the half-life of the substance.

### Step-by-Step Solution:

1. **Identify the half-life**:
The half-life \(t_{1/2}\) is given as 69.3 days.

2. **Use the formula for disintegration constant**:
Substitute the value of \(t_{1/2}\) into the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{69.3} \]

3. **Calculate the value**:
Now perform the division:
\[ \lambda = \frac{0.693}{69.3} \approx 0.010 \]

4. **Express in appropriate units**:
Since the half-life is in days, the disintegration constant will be in day\(^{-1}\):
\[ \lambda \approx 0.010 \text{ day}^{-1} \]

5. **Conclusion**:
The disintegration constant for the radioactive substance is approximately \(0.010 \text{ day}^{-1}\).

### Final Answer:
The disintegration constant (λ) is \(0.010 \text{ day}^{-1}\).

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