Question

The energy released by the fission of one uranium nucleus is 200 MeV. The number of fissions per second required to produce 128 W power is:

• A. \( 6 \times 10^{12} \)
• B. \( 8 \times 10^{12} \)
• C. \( 2 \times 10^{12} \)
• D. \( 4 \times 10^{12} \)

Hint:

Convert MeV to Joules using \( 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \).
- Use \( \text{Fission Rate} = \frac{P}{E} \) for power calculations in nuclear fission.

Answer 1

The Correct Option is D

The total power required is given as: \[ P = 128 \text{ W} \] The energy released per fission is: \[ E = 200 \text{ MeV} \] 1. Convert MeV to Joules: \[ 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \] \[ E = 200 \times 1.6 \times 10^{-13} \] \[ = 3.2 \times 10^{-11} \text{ J} \] 2. Find fission rate: \[ \text{Fission rate} = \frac{P}{E} \] \[ = \frac{128}{3.2 \times 10^{-11}} \] \[ = 4 \times 10^{12} \text{ fissions per second} \] Thus, the correct answer is \(\boxed{4 \times 10^{12}}\).