A radioactive nuclei with decay constant 0.5 nuclei/s is being produced at a constant rate of 100 nuclei/s. If at t = 0 there were no nuclei, the time when there are 50 nuclei is

Options

A. 1 s
B. `2 ln (4/3) s`
C. ln 2 s
D. `ln(4/3)s`

Explanation

To solve the problem, we need to derive the time when the number of radioactive nuclei reaches 50, given the production and decay rates.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a decay constant \( \lambda = 0.5 \) nuclei/s.
- The production rate of the nuclei is \( R_1 = 100 \) nuclei/s.
- Initially, at \( t = 0 \), there are no nuclei present.

2. **Set Up the Differential Equation**:
- The rate of change of the number of nuclei \( n \) can be expressed as:
\[ \frac{dn}{dt} = R_1 - \lambda n \]
- Substituting the values, we have:
\[ \frac{dn}{dt} = 100 - 0.5n \]

3. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{dn}{100 - 0.5n} = dt \]

4. **Integrate Both Sides**:
- We need to integrate from \( n = 0 \) to \( n = 50 \) and \( t = 0 \) to \( t = t \):
\[ \int_0^{50} \frac{dn}{100 - 0.5n} = \int_0^{t} dt \]

5. **Perform the Integration**:
- The left side integrates to:
\[ -\frac{2}{1} \ln |100 - 0.5n| \bigg|_0^{50} = t \]
- Evaluating the limits:
- When \( n = 50 \):
\[ 100 - 0.5 \times 50 = 100 - 25 = 75 \]
- When \( n = 0 \):
\[ 100 - 0.5 \times 0 = 100 \]
- Thus, we have:
\[ -2 \left( \ln(75) - \ln(100) \right) = t \]
- This simplifies to:
\[ -2 \ln \left( \frac{75}{100} \right) = t \]
- Which can be rewritten as:
\[ t = 2 \ln \left( \frac{100}{75} \right) = 2 \ln \left( \frac{4}{3} \right) \]

6. **Final Result**:
- Therefore, the time when there are 50 nuclei is:
\[ t = 2 \ln \left( \frac{4}{3} \right) \text{ seconds} \]