A newly prepared radioactive nuclide has a decay constant `lambda` of `10^(-6) s^(-1)`. What is the approximate half-life of the nuclide?

Options

A. `1 hour`
B. `1 day`
C. `1 week`
D. `1 month`

Explanation

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

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

Where:
- \( T_{1/2} \) is the half-life,
- \( \ln(2) \) is the natural logarithm of 2, approximately equal to 0.693,
- \( \lambda \) is the decay constant.

### Step-by-Step Solution:

1. **Identify the decay constant (λ)**:
Given that \( \lambda = 10^{-6} \, \text{s}^{-1} \).

2. **Use the half-life formula**:
Substitute the value of \( \lambda \) into the half-life formula:
\[ T_{1/2} = \frac{\ln(2)}{10^{-6}} \]

3. **Calculate \( \ln(2) \)**:
We know that \( \ln(2) \approx 0.693 \).

4. **Substitute \( \ln(2) \) into the formula**:
\[ T_{1/2} = \frac{0.693}{10^{-6}} = 0.693 \times 10^{6} \]

5. **Perform the multiplication**:
\[ T_{1/2} \approx 693000 \, \text{s} \]

6. **Convert seconds to hours**:
To convert seconds to hours, divide by 3600 (since there are 3600 seconds in an hour):
\[ T_{1/2} \approx \frac{693000}{3600} \approx 192.5 \, \text{hours} \]

7. **Convert hours to days**:
To convert hours to days, divide by 24 (since there are 24 hours in a day):
\[ T_{1/2} \approx \frac{192.5}{24} \approx 8.02 \, \text{days} \]

8. **Approximate to weeks**:
Since 8.02 days is approximately one week, we conclude that the half-life of the nuclide is about one week.

### Final Answer:
The approximate half-life of the nuclide is **one week**.

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