A radio nuclide X is produced at constant rate `alpha`. At time `t=0`, number of nuclei of X are zero. Find
(a) the maximum number of nuclei of X.
the number of nuclei at time t.
Decay constant of X is `lambda`.

Explanation

To solve the problem step by step, we will break it down into two parts as requested: finding the maximum number of nuclei of X and finding the number of nuclei at time t.

### Step 1: Finding the Maximum Number of Nuclei (N_max)

1. **Understanding the Rates**:
- The production rate of nuclei X is constant and given by \( \alpha \).
- The decay rate of nuclei X is proportional to the number of nuclei present and is given by \( \lambda n \), where \( n \) is the number of nuclei at time \( t \).

2. **Setting Up the Equation**:
- At equilibrium (maximum number of nuclei), the rate of production equals the rate of decay:
\[ \alpha = \lambda n_{max} \]

3. **Solving for Maximum Nuclei**:
- Rearranging the equation gives:
\[ n_{max} = \frac{\alpha}{\lambda} \]

### Step 2: Finding the Number of Nuclei at Time t (n(t))

1. **Net Rate of Formation**:
- The net rate of formation of nuclei X can be expressed as:
\[ \frac{dn}{dt} = \alpha - \lambda n \]

2. **Rearranging the Equation**:
- We can rearrange this to separate variables:
\[ \frac{dn}{\alpha - \lambda n} = dt \]

3. **Integrating Both Sides**:
- We will integrate both sides. The left side will be integrated with respect to \( n \) from 0 to \( n \), and the right side will be integrated with respect to \( t \) from 0 to \( t \):
\[ \int_0^n \frac{1}{\alpha - \lambda n} dn = \int_0^t dt \]

4. **Performing the Integration**:
- The left side integrates to:
\[ -\frac{1}{\lambda} \ln|\alpha - \lambda n| \bigg|_0^n = t \]
- This leads to:
\[ -\frac{1}{\lambda} \left( \ln|\alpha - \lambda n| - \ln|\alpha| \right) = t \]

5. **Simplifying the Equation**:
- This can be simplified to:
\[ -\frac{1}{\lambda} \ln\left(\frac{\alpha - \lambda n}{\alpha}\right) = t \]
- Rearranging gives:
\[ \ln\left(\frac{\alpha - \lambda n}{\alpha}\right) = -\lambda t \]

6. **Exponentiating Both Sides**:
- Exponentiating both sides leads to:
\[ \frac{\alpha - \lambda n}{\alpha} = e^{-\lambda t} \]

7. **Solving for n**:
- Rearranging gives:
\[ \alpha - \lambda n = \alpha e^{-\lambda t} \]
- Thus:
\[ \lambda n = \alpha - \alpha e^{-\lambda t} \]
- Finally, solving for \( n \):
\[ n = \frac{\alpha}{\lambda} (1 - e^{-\lambda t}) \]

### Final Answers:
- (a) The maximum number of nuclei of X:
\[ n_{max} = \frac{\alpha}{\lambda} \]
- (b) The number of nuclei at time \( t \):
\[ n(t) = \frac{\alpha}{\lambda} (1 - e^{-\lambda t}) \]