In the above problem if each decay produces `E_0` energy, then find
(a) power produced at time t
(b) total energy produced upto time t

Explanation

To solve the problem step by step, we will find the power produced at time \( t \) and the total energy produced up to time \( t \).

### Step 1: Understanding the problem
We know that each decay produces an energy \( E_0 \). The decay process is characterized by a decay constant \( \lambda \), and the rate of formation of the nuclei is given by \( \alpha \).

### Step 2: Finding the power produced at time \( t \)
The power produced at time \( t \) can be expressed as the product of the number of decays per second and the energy produced per decay.

1. The rate of decay at time \( t \) is given by:
\[ R(t) = \alpha (1 - e^{-\lambda t}) \]
where \( R(t) \) is the rate of formation of the nuclei.

2. The power \( P(t) \) produced at time \( t \) is given by:
\[ P(t) = R(t) \cdot E_0 \]
Substituting the expression for \( R(t) \):
\[ P(t) = \alpha (1 - e^{-\lambda t}) E_0 \]

### Step 3: Finding the total energy produced up to time \( t \)
To find the total energy produced up to time \( t \), we need to integrate the power over the time interval from \( 0 \) to \( t \).

1. The total energy \( E_{\text{total}} \) produced up to time \( t \) is given by:
\[ E_{\text{total}} = \int_0^t P(t') dt' \]
where \( P(t') = \alpha (1 - e^{-\lambda t'}) E_0 \).

2. Substituting for \( P(t') \):
\[ E_{\text{total}} = \int_0^t \alpha (1 - e^{-\lambda t'}) E_0 dt' \]

3. Since \( \alpha \) and \( E_0 \) are constants, we can factor them out of the integral:
\[ E_{\text{total}} = \alpha E_0 \int_0^t (1 - e^{-\lambda t'}) dt' \]

4. Now, we can evaluate the integral:
\[ \int_0^t (1 - e^{-\lambda t'}) dt' = \left[ t' + \frac{1}{\lambda} e^{-\lambda t'} \right]_0^t \]
Evaluating this from \( 0 \) to \( t \):
\[ = \left( t + \frac{1}{\lambda} e^{-\lambda t} \right) - \left( 0 + \frac{1}{\lambda} \cdot 1 \right) \]
\[ = t + \frac{1}{\lambda} e^{-\lambda t} - \frac{1}{\lambda} \]
\[ = t + \frac{1}{\lambda} (e^{-\lambda t} - 1) \]

5. Thus, the total energy produced up to time \( t \) is:
\[ E_{\text{total}} = \alpha E_0 \left( t + \frac{1}{\lambda} (e^{-\lambda t} - 1) \right) \]

### Final Results
- **Power produced at time \( t \)**:
\[ P(t) = \alpha (1 - e^{-\lambda t}) E_0 \]

- **Total energy produced up to time \( t \)**:
\[ E_{\text{total}} = \alpha E_0 \left( t + \frac{1}{\lambda} (e^{-\lambda t} - 1) \right) \]