Question

In a nuclear reactor of efficiency 25%, the number of fissions taking place per second is \( 5 \times 10^{13} \). If 200 MeV energy is released per fission, then the output power of the reactor is:

• A. \( 6400 \, W \)
• B. \( 1600 \, W \)
• C. \( 800 \, W \)
• D. \( 400 \, W \)

Hint:

For efficiency calculations in nuclear reactors, remember to use the conversion factor for MeV to joules: \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \).

Answer 1

The Correct Option is D

Step 1:
The total energy released per second can be calculated by the formula: \[ \text{Power} = \text{Number of fissions per second} \times \text{Energy released per fission} \] \[ P = 5 \times 10^{13} \times 200 \, \text{MeV} \] Since 1 MeV = \( 1.6 \times 10^{-13} \) J, we convert the energy into joules: \[ P = 5 \times 10^{13} \times 200 \times 1.6 \times 10^{-13} = 400 \, \text{W} \] Thus, the output power is \( 400 \, W \).