Nuclei of a radioactive element A are being produced at a constant rate `alpha`. The element has a decay constant `lambda`. At time `t=0`, there are `N_0` nuclei of the element.
(a) Calculate the number N of nuclei of A at time t.
(b) If `alpha=2N_0lambda`, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as `trarroo`.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Part (a): Calculate the number N of nuclei of A at time t.

1. **Understanding the Problem**:
We have a radioactive element A that is being produced at a constant rate \( \alpha \) and decaying at a rate characterized by the decay constant \( \lambda \). At time \( t = 0 \), the number of nuclei is \( N_0 \).

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

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

4. **Integrating Both Sides**:
We will integrate both sides. The left side will be integrated from \( N_0 \) to \( N \) and the right side from \( 0 \) to \( t \):
\[ \int_{N_0}^{N} \frac{dN}{\alpha - \lambda N} = \int_{0}^{t} dt \]

5. **Calculating the Left Integral**:
The integral on the left can be solved using the natural logarithm:
\[ -\frac{1}{\lambda} \ln |\alpha - \lambda N| \bigg|_{N_0}^{N} = t \]
This leads to:
\[ -\frac{1}{\lambda} \left( \ln |\alpha - \lambda N| - \ln |\alpha - \lambda N_0| \right) = t \]

6. **Simplifying the Equation**:
This simplifies to:
\[ \ln \left( \frac{\alpha - \lambda N_0}{\alpha - \lambda N} \right) = -\lambda t \]

7. **Exponentiating Both Sides**:
Exponentiating gives:
\[ \frac{\alpha - \lambda N}{\alpha - \lambda N_0} = e^{-\lambda t} \]

8. **Solving for N**:
Rearranging this equation leads to:
\[ \alpha - \lambda N = (\alpha - \lambda N_0)e^{-\lambda t} \]
Thus,
\[ \lambda N = \alpha - (\alpha - \lambda N_0)e^{-\lambda t} \]
Finally, we have:
\[ N = \frac{\alpha}{\lambda} - \frac{\alpha - \lambda N_0}{\lambda} e^{-\lambda t} \]

### Part (b): If \( \alpha = 2N_0\lambda \), calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as \( t \to \infty \).

1. **Substituting \( \alpha \)**:
Given \( \alpha = 2N_0\lambda \), we substitute this into the equation derived in part (a):
\[ N = \frac{2N_0\lambda}{\lambda} - \frac{2N_0\lambda - \lambda N_0}{\lambda} e^{-\lambda t} \]
Simplifying gives:
\[ N = 2N_0 - (2N_0 - N_0)e^{-\lambda t} = 2N_0 - N_0 e^{-\lambda t} \]

2. **Calculating After One Half-Life**:
The half-life \( t_{1/2} \) is given by:
\[ t_{1/2} = \frac{\ln 2}{\lambda} \]
Substituting \( t = t_{1/2} \):
\[ N = 2N_0 - N_0 e^{-\ln 2} = 2N_0 - N_0 \cdot \frac{1}{2} = 2N_0 - \frac{N_0}{2} = \frac{4N_0}{2} - \frac{N_0}{2} = \frac{3N_0}{2} \]

3. **Finding the Limiting Value as \( t \to \infty \)**:
As \( t \to \infty \), \( e^{-\lambda t} \to 0 \):
\[ N = 2N_0 - N_0 \cdot 0 = 2N_0 \]

### Final Answers:
- (a) The number of nuclei of A at time \( t \) is:
\[ N = \frac{2N_0}{\lambda} - \frac{(2N_0 - N_0)e^{-\lambda t}}{\lambda} \]
- (b) After one half-life, \( N = \frac{3N_0}{2} \) and as \( t \to \infty \), \( N = 2N_0 \).