A stable nuclei C is formed from two radioactive nuclei A and B with decay constant of `lambda_1` and `lambda_2` respectively. Initially, the number of nuclei of A is `N_0` and that of B is zero. Nuclei B are produced at a constant rate of P. Find the number of the nuclei of C after time t.

Explanation

To find the number of nuclei of C after time \( t \), we will follow a systematic approach based on the information given in the problem.

### Step 1: Determine the number of nuclei A at time t
The number of radioactive nuclei A at time \( t \) can be expressed using the exponential decay formula:
\[ N_A(t) = N_0 e^{-\lambda_1 t} \]
where \( N_0 \) is the initial number of nuclei A and \( \lambda_1 \) is the decay constant of A.

**Hint:** Remember that the decay of a radioactive substance is exponential, and the number of nuclei decreases over time.

### Step 2: Set up the equation for the number of nuclei B
Nuclei B are produced at a constant rate \( P \) and also decay with a decay constant \( \lambda_2 \). The rate of change of nuclei B can be expressed as:
\[ \frac{dN_B}{dt} = P - \lambda_2 N_B \]
This is a first-order linear differential equation.

**Hint:** When dealing with production and decay, think of the balance between the two processes.

### Step 3: Solve the differential equation for nuclei B
To solve the equation, we can rearrange it:
\[ \frac{dN_B}{P - \lambda_2 N_B} = dt \]
Integrating both sides, we get:
\[ \int \frac{dN_B}{P - \lambda_2 N_B} = \int dt \]
This leads to:
\[ -\frac{1}{\lambda_2} \ln |P - \lambda_2 N_B| = t + C \]
Solving for \( N_B \) gives:
\[ N_B(t) = \frac{P}{\lambda_2} (1 - e^{-\lambda_2 t}) \]
where \( C \) is determined by the initial condition \( N_B(0) = 0 \).

**Hint:** Make sure to apply the initial conditions correctly when integrating.

### Step 4: Set up the equation for the number of nuclei C
The nuclei C are formed from both A and B. The rate of formation of C can be expressed as:
\[ \frac{dN_C}{dt} = \lambda_1 N_A + \lambda_2 N_B \]
Substituting the expressions for \( N_A \) and \( N_B \):
\[ \frac{dN_C}{dt} = \lambda_1 N_0 e^{-\lambda_1 t} + \lambda_2 \left(\frac{P}{\lambda_2} (1 - e^{-\lambda_2 t})\right) \]

**Hint:** The formation of C depends on the decay of A and the production of B.

### Step 5: Integrate to find the number of nuclei C
Now, we integrate \( \frac{dN_C}{dt} \):
\[ N_C(t) = \int \left( \lambda_1 N_0 e^{-\lambda_1 t} + P (1 - e^{-\lambda_2 t}) \right) dt \]
This results in:
\[ N_C(t) = -N_0 e^{-\lambda_1 t} + P t + \frac{P}{\lambda_2} e^{-\lambda_2 t} + C \]
Using the initial condition \( N_C(0) = 0 \), we can find \( C \).

**Hint:** Always check initial conditions to solve for constants of integration.

### Final Result
After evaluating the integral and applying the initial conditions, we find:
\[ N_C(t) = N_0 (1 - e^{-\lambda_1 t}) + \frac{P}{\lambda_2} (1 - e^{-\lambda_2 t}) + Pt \]

This is the number of nuclei C after time \( t \).