A radioactive substance is being consumed at a constant of `1 s^(-1)`. After what time will the number of radioactive nuclei becoem `100`. Initially, there were 200 nuceli present.

Options

A. 1 s
B. `(1)/(ln(2))s`
C. `ln (2) s`
D. 2 s

Answer

To solve the problem of determining the time it takes for the number of radioactive nuclei to decrease from 200 to 100, we can use the radioactive decay formula:

\[ n = n_0 e^{-\lambda t} \]

Where:
- \( n \) is the number of radioactive nuclei at time \( t \),
- \( n_0 \) is the initial number of radioactive nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial number of nuclei, \( n_0 = 200 \)
- Final number of nuclei, \( n = 100 \)
- Decay constant, \( \lambda = 1 \, \text{s}^{-1} \)

2. **Substitute the known values into the decay equation:**
\[ 100 = 200 e^{-1 \cdot t} \]

3. **Simplify the equation:**
- Divide both sides by 200:
\[ \frac{100}{200} = e^{-t} \]
- This simplifies to:
\[ \frac{1}{2} = e^{-t} \]

4. **Take the natural logarithm of both sides:**
\[ \ln\left(\frac{1}{2}\right) = -t \]

5. **Use the property of logarithms:**
- We know that \( \ln\left(\frac{1}{2}\right) = -\ln(2) \):
\[ -\ln(2) = -t \]
- Therefore:
\[ t = \ln(2) \]

6. **Convert to seconds:**
- Since we are using the natural logarithm, the time \( t \) is expressed in seconds:
\[ t = \log(2) \, \text{seconds} \]

### Final Answer:
The time after which the number of radioactive nuclei becomes 100 is \( t = \log(2) \) seconds.