A radioactive isotope is being produced at a constant rate `dN//dt=R` in an experiment. The isotope has a half-life `t_(1//2)`.Show that after a time `t gtgt t_(1//2)`,the number of active nuclei will become constant. Find the value of this constant.

Options

A. (a) AT
B. (b) `A/T1n2`
C. (c) AT 1n2
D. (d) `(AT)/(1n2)`

Explanation

To solve the problem, we need to analyze the behavior of a radioactive isotope being produced at a constant rate \( R \) and its decay over time. The goal is to show that after a time much greater than the half-life \( t_{1/2} \), the number of active nuclei becomes constant, and to find the value of this constant.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( N(t) \) be the number of active nuclei at time \( t \).
- The rate of production of the isotope is given as \( \frac{dN}{dt} = R \).
- The decay of the isotope is characterized by its decay constant \( \lambda \), which is related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]

2. **Set Up the Differential Equation**:
- The change in the number of active nuclei over time can be expressed as:
\[ \frac{dN}{dt} = R - \lambda N \]
- Here, \( R \) is the rate of production, and \( \lambda N \) is the rate of decay of the nuclei.

3. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ \frac{dN}{dt} + \lambda N = R \]

4. **Solve the Differential Equation**:
- This is a first-order linear ordinary differential equation. To solve it, we can use an integrating factor:
\[ \mu(t) = e^{\lambda t} \]
- Multiply the entire equation by the integrating factor:
\[ e^{\lambda t} \frac{dN}{dt} + \lambda e^{\lambda t} N = R e^{\lambda t} \]
- The left-hand side can be rewritten as:
\[ \frac{d}{dt}(e^{\lambda t} N) = R e^{\lambda t} \]

5. **Integrate Both Sides**:
- Integrating both sides with respect to \( t \):
\[ e^{\lambda t} N = \int R e^{\lambda t} dt + C \]
- The integral on the right side is:
\[ \int R e^{\lambda t} dt = \frac{R}{\lambda} e^{\lambda t} + C \]
- Thus, we have:
\[ e^{\lambda t} N = \frac{R}{\lambda} e^{\lambda t} + C \]

6. **Solve for \( N(t) \)**:
- Dividing through by \( e^{\lambda t} \):
\[ N(t) = \frac{R}{\lambda} + Ce^{-\lambda t} \]

7. **Determine the Constant \( C \)**:
- As \( t \to \infty \), the term \( Ce^{-\lambda t} \) approaches zero. Therefore, the number of active nuclei approaches:
\[ N(t) \to \frac{R}{\lambda} \]
- This shows that after a long time (i.e., \( t \gg t_{1/2} \)), the number of active nuclei becomes constant.

8. **Final Expression for the Constant**:
- Substituting \( \lambda = \frac{\ln(2)}{t_{1/2}} \):
\[ N_{\text{constant}} = \frac{R}{\lambda} = \frac{R t_{1/2}}{\ln(2)} \]

### Conclusion:
After a time much greater than the half-life \( t_{1/2} \), the number of active nuclei becomes constant and is given by:
\[ N_{\text{constant}} = \frac{R t_{1/2}}{\ln(2)} \]