`10^(14)` Fission per second are taking place in a nuclear reactor having efficiency `25 %.` The energy released per fission in `200 MeV.` The power output of the nuclear reactor

Options

A. 20KW
B. 40KW
C. 60KW
D. 80KW

Answer

To find the power output of the nuclear reactor, we can follow these steps:

### Step 1: Calculate the total energy released per second
The total number of fissions occurring per second is given as \(10^{14}\). The energy released per fission is \(200 \text{ MeV}\).

To find the total energy released per second, we multiply the number of fissions by the energy released per fission:

\[ \text{Total Energy Released per second} = \text{Number of fissions per second} \times \text{Energy per fission} \]

Substituting the values:

\[ \text{Total Energy Released per second} = 10^{14} \times 200 \text{ MeV} \]

### Step 2: Convert MeV to Joules
We need to convert the energy from MeV to Joules. The conversion factor is:

\[ 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ Joules} \]

So, we convert the total energy released per second:

\[ \text{Total Energy Released per second (in Joules)} = 10^{14} \times 200 \times 1.6 \times 10^{-13} \]

Calculating this gives:

\[ = 10^{14} \times 200 \times 1.6 \times 10^{-13} = 32000 \text{ Joules} \]

### Step 3: Calculate the output energy considering efficiency
The efficiency of the reactor is given as \(25\%\) or \(0.25\). The output energy can be calculated as:

\[ \text{Output Energy} = \text{Efficiency} \times \text{Total Energy Released per second} \]

Substituting the values:

\[ \text{Output Energy} = 0.25 \times 32000 \text{ Joules} = 8000 \text{ Joules} \]

### Step 4: Convert Joules to Watts
Power is defined as energy per unit time. Since we are calculating energy released per second, the power output in Watts is:

\[ \text{Power Output} = \text{Output Energy} = 8000 \text{ Watts} = 8 \text{ kW} \]

### Final Answer
The power output of the nuclear reactor is \(8 \text{ kW}\).

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