A radioactive nuclide is produced at the constant rate of n per second (say, by bombarding a target with neutrons). The expected number `N` of nuclei in existence `t` s after the number is `N_(0)` is given by

Options

A. `N=N_(0)e^(-lambda t)`
B. `N=N_(0)e^(-lambda t)`
C. `N=n/lambda+(N_0-n/lambda)e^(-lambda t)`
D. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)`

Explanation

To solve the problem of determining the expected number \( N \) of radioactive nuclei in existence \( t \) seconds after the initial number \( N_0 \) is given, we can follow these steps:

### Step 1: Understand the Problem
We have a radioactive nuclide being produced at a constant rate \( n \) nuclei per second. Initially, we have \( N_0 \) nuclei. We need to derive the expression for the number of nuclei \( N \) after time \( t \).

### Step 2: Set Up the Rate Equation
The rate of change of the number of nuclei can be expressed as:
\[ \frac{dN}{dt} = n - \lambda N \]
where \( \lambda \) is the decay constant. The term \( n \) represents the constant production of new nuclei, while \( -\lambda N \) represents the decay of the existing nuclei.

### Step 3: Rearrange the Equation
Rearranging the equation gives:
\[ \frac{dN}{dt} + \lambda N = n \]

### Step 4: Solve the Differential Equation
This is a first-order linear differential equation. We can solve it using an integrating factor. The integrating factor \( \mu(t) \) is given by:
\[ \mu(t) = e^{\int \lambda dt} = e^{\lambda t} \]

### Step 5: Multiply Through by the Integrating Factor
Multiplying the entire equation by \( e^{\lambda t} \):
\[ e^{\lambda t} \frac{dN}{dt} + \lambda e^{\lambda t} N = n e^{\lambda t} \]

### Step 6: Recognize the Left Side as a Derivative
The left side can be recognized as the derivative of a product:
\[ \frac{d}{dt}(e^{\lambda t} N) = n e^{\lambda t} \]

### Step 7: Integrate Both Sides
Integrating both sides with respect to \( t \):
\[ e^{\lambda t} N = \int n e^{\lambda t} dt + C \]
where \( C \) is the constant of integration.

### Step 8: Solve the Integral
The integral on the right side can be solved as:
\[ \int n e^{\lambda t} dt = \frac{n}{\lambda} e^{\lambda t} + C \]

### Step 9: Substitute Back
So, we have:
\[ e^{\lambda t} N = \frac{n}{\lambda} e^{\lambda t} + C \]

### Step 10: Solve for \( N \)
Dividing through by \( e^{\lambda t} \):
\[ N = \frac{n}{\lambda} + Ce^{-\lambda t} \]

### Step 11: Determine the Constant \( C \)
To find \( C \), we use the initial condition \( N(0) = N_0 \):
\[ N_0 = \frac{n}{\lambda} + C \implies C = N_0 - \frac{n}{\lambda} \]

### Step 12: Final Expression for \( N \)
Substituting \( C \) back into the equation for \( N \):
\[ N = \frac{n}{\lambda} + \left(N_0 - \frac{n}{\lambda}\right)e^{-\lambda t} \]

### Final Answer
Thus, the expected number \( N \) of nuclei in existence \( t \) seconds after the initial number \( N_0 \) is:
\[ N = \frac{n}{\lambda} + \left(N_0 - \frac{n}{\lambda}\right)e^{-\lambda t} \]