A radioactive with decay constant `lambda` is being produced in a nuclear ractor at a rate `q_0` per second, where `q_(0)` is a positive constant and `t` is the time. During each decay, `E_(0)` energy is released. The production of radionuclide starts at time`t=0`.
Average power developed in time t due to the decay of the radionuclide is

Options

A. `((q_(0) t)/(2)-(q_(0))/(lambda) +(q_(0))/(lambda^(2)t) -(q_(0))/(lambda^(2)t) e^(-lambda t)E_(0))`
B. `((q_(0) t)/(2)+(q_(0))/(lambda) +(q_(0))/(lambda^(2)t) -(q_(0))/(lambda^(2)t) e^(-lambda t)E_(0))`
C. `((q_(0) t)/(2)-(q_(0))/(lambda) +(q_(0))/(lambda^(2)t) +(q_(0))/(lambda^(2)t) e^(-lambda t)E_(0))`
D. `((q_(0) t)/(2)+(q_(0))/(lambda) +(q_(0))/(lambda^(2)t) +(q_(0))/(lambda^(2)t) e^(-lambda t)E_(0))`

Explanation

To find the average power developed in time \( t \) due to the decay of the radionuclide, we can follow these steps:

### Step 1: Set up the differential equation
The rate of change of the number of radionuclides \( n \) can be expressed as:
\[ \frac{dn}{dt} = q_0 - \lambda n \]
where \( q_0 \) is the production rate and \( \lambda \) is the decay constant.

### Step 2: Rearranging the equation
Rearranging the equation gives us:
\[ \frac{dn}{dt} + \lambda n = q_0 \]

### Step 3: Find the integrating factor
The integrating factor \( \mu(t) \) is given by:
\[ \mu(t) = e^{\int \lambda dt} = e^{\lambda t} \]

### Step 4: Multiply through by the integrating factor
Multiplying the entire differential equation by the integrating factor:
\[ e^{\lambda t} \frac{dn}{dt} + \lambda e^{\lambda t} n = q_0 e^{\lambda t} \]

### Step 5: Integrate both sides
The left-hand side can be simplified to:
\[ \frac{d}{dt}(e^{\lambda t} n) = q_0 e^{\lambda t} \]
Integrating both sides gives:
\[ e^{\lambda t} n = \int q_0 e^{\lambda t} dt + C \]
where \( C \) is the constant of integration.

### Step 6: Solve the integral
The integral on the right-hand side can be computed as:
\[ \int q_0 e^{\lambda t} dt = \frac{q_0}{\lambda} e^{\lambda t} \]
Thus, we have:
\[ e^{\lambda t} n = \frac{q_0}{\lambda} e^{\lambda t} + C \]

### Step 7: Solve for \( n \)
Dividing through by \( e^{\lambda t} \):
\[ n = \frac{q_0}{\lambda} + Ce^{-\lambda t} \]

### Step 8: Apply initial conditions
At \( t = 0 \), \( n = 0 \):
\[ 0 = \frac{q_0}{\lambda} + C \implies C = -\frac{q_0}{\lambda} \]
Thus, we have:
\[ n = \frac{q_0}{\lambda} (1 - e^{-\lambda t}) \]

### Step 9: Calculate the average power
The average power \( P_{\text{avg}} \) developed due to the decay is given by:
\[ P_{\text{avg}} = \frac{E_0}{t} \int_0^t n \, dt \]
Substituting for \( n \):
\[ P_{\text{avg}} = \frac{E_0}{t} \int_0^t \frac{q_0}{\lambda} (1 - e^{-\lambda t}) \, dt \]

### Step 10: Evaluate the integral
Evaluating the integral:
\[ \int_0^t (1 - e^{-\lambda t}) dt = t - \frac{1}{\lambda}(1 - e^{-\lambda t}) \]
Thus,
\[ P_{\text{avg}} = \frac{E_0}{t} \left[ \frac{q_0}{\lambda} \left( t - \frac{1}{\lambda}(1 - e^{-\lambda t}) \right) \right] \]

### Step 11: Final expression for average power
After simplifying, we get:
\[ P_{\text{avg}} = \frac{E_0 q_0}{\lambda} \left( 1 - \frac{1 - e^{-\lambda t}}{\lambda t} \right) \]