Question

200 MeV of energy is obtained in the fission of one nucleus of \( ^{235}U \). A reactor is generating 1000 kW power. The rate of nuclear fission in the reactor is:

• A. 1000
• B. \( 2 \times 10^5 \)
• C. \( 3.125 \times 10^{16} \)
• D. 931

Hint:

Fission reactors produce energy by splitting heavy nuclei, releasing large amounts of energy.

Answer 1

The Correct Option is C

The number of fissions per second is given by:
\[ \frac{\text{Power output (W)}}{\text{Energy per fission (J)}} \] Given that 1 MeV = \( 1.6 \times 10^{-13} \) J, the energy per fission is:
\[ 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11} \text{ J} \] Power output is 1000 kW = \( 10^6 \) W. Thus,
\[ \frac{10^6}{3.2 \times 10^{-11}} = 3.125 \times 10^{16} \text{ fissions/s} \]