A nuclear reactor starts producing a radionuclide of half - life T at a constant rate R starting at time t = 0 The activity of the radionuclide at t = T is found to the A. Then `(R )/(4)` is

Options

A. (a)`2:1`
B. (b)`3:2`
C. (c)`4:1`
D. (d)`2:3`

Answer

To solve the problem, we need to analyze the production and decay of a radionuclide in a nuclear reactor. The key points to consider are the half-life of the radionuclide, the constant production rate, and the activity at a specific time.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( R \) be the constant rate of production of the radionuclide.
- Let \( T \) be the half-life of the radionuclide.
- Let \( A \) be the activity of the radionuclide at time \( t = T \).
- The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T} \]

2. **Setting Up the Differential Equation**:
- The change in the number of radionuclide atoms \( n \) over time can be described by the equation:
\[ \frac{dn}{dt} = R - \lambda n \]
- This equation states that the rate of change of \( n \) is equal to the production rate \( R \) minus the decay rate \( \lambda n \).

3. **Integrating the Equation**:
- Rearranging the equation gives:
\[ \frac{dn}{R - \lambda n} = dt \]
- We will integrate both sides from \( n = 0 \) to \( n = N \) (the number of radionuclide atoms at time \( t = T \)) and from \( t = 0 \) to \( t = T \):
\[ \int_0^N \frac{dn}{R - \lambda n} = \int_0^T dt \]

4. **Solving the Integral**:
- The left-hand side integral can be solved using the formula for integrating a rational function:
\[ \int \frac{1}{a - bx} dx = -\frac{1}{b} \ln |a - bx| + C \]
- Applying this gives:
\[ -\frac{1}{\lambda} \ln |R - \lambda N| + \frac{1}{\lambda} \ln |R| = T \]
- Simplifying this leads to:
\[ \ln \left( \frac{R}{R - \lambda N} \right) = \lambda T \]

5. **Finding the Number of Atoms**:
- Exponentiating both sides gives:
\[ \frac{R}{R - \lambda N} = e^{\lambda T} \]
- Rearranging gives:
\[ R - \lambda N = \frac{R}{e^{\lambda T}} \]
- Thus, we can express \( N \) in terms of \( R \) and \( \lambda \):
\[ N = \frac{R}{\lambda} \left( 1 - \frac{1}{e^{\lambda T}} \right) \]

6. **Calculating the Activity**:
- The activity \( A \) is given by:
\[ A = \lambda N \]
- Substituting for \( N \) gives:
\[ A = \lambda \cdot \frac{R}{\lambda} \left( 1 - \frac{1}{e^{\lambda T}} \right) = R \left( 1 - \frac{1}{2} \right) = \frac{R}{2} \]

7. **Finding \( \frac{R}{A} \)**:
- Since \( A = \frac{R}{2} \), we can find \( \frac{R}{A} \):
\[ \frac{R}{A} = \frac{R}{\frac{R}{2}} = 2 \]

### Final Answer:
\[ \frac{R}{A} = 2 \]