Order of magnitude check: the energy released per fission event in a typical heavy nucleus (e.g., U-235 or Pu-239) is approximately how many MeV?

Introduction / Context:When a heavy nucleus such as U-235 undergoes fission after absorbing a neutron, it splits into two fission fragments plus neutrons and gamma rays. The mass defect between reactants and products is converted to energy according to E = m * c^2. Engineers frequently use a rounded order-of-magnitude value for rapid calculations.

Given Data / Assumptions:

• Thermal-neutron fission of U-235 or Pu-239.
• Energy includes kinetic energy of fission fragments, prompt/delayed neutrons, and gamma radiation.
• Interest is per fission event, not per mole or per kilogram.

Concept / Approach:Experimental tally of product energies gives a figure near 200 MeV per fission. Of this, roughly 165–170 MeV appears as kinetic energy of the two fragments; a few MeV appear as prompt neutrons; and the remainder is carried as gamma rays and beta decays of fission products (delayed energy).

Step-by-Step Solution:1) Recognise that fission fragments carry most kinetic energy (largest term).2) Add contributions from prompt neutrons and gamma rays to reach ~200 MeV total.3) Convert to macroscopic units if desired: 1 MeV ≈ 1.602e-13 J; therefore 200 MeV ≈ 3.2e-11 J per fission.4) Multiply by Avogadro-scale fission counts to estimate reactor thermal power from fission rate.

Verification / Alternative check:Back-calculating from 1 GW(th) requires ~3.1e19 fissions per second using 200 MeV per fission, which matches standard reactor physics back-of-the-envelope checks.

Why Other Options Are Wrong:20 MeV underestimates by an order of magnitude.1000 or 2000 MeV exceed typical fission energy and are closer to GeV-scale particle processes.

Common Pitfalls:Confusing per-fission energy with binding energy per nucleon; mixing MeV per fission with MeV per neutron capture in other reactions.

Final Answer:200