Nuclear Fission and Fusion

Nuclear Fission

  • In nuclear fission an atomic nucleus splits into smaller atomic nuclei:
    • (1 neutron +1 Uranium-235) (1 Barium-144 + 3 neutrons + 1 Krypton-89)
  • The mass of the reacting particles is greater than the mass of the product particles, the lost mass converted into energy per Einstein’s E=mc2
  • Lost Mass = (Mass of uranium-235 + mass of 1 neutron) ➖ (mass of barium-144 + mass of krypton-89 + mass of 3 neutrons) = 173.3 MeV/c2
  • Mathematica Calculation
    • (UnitConvert[IsotopeData[“Uranium235”, “AtomicMass”], “MeV/c^2”] +ParticleData[“Neutron”, “Mass”]) – (UnitConvert[ IsotopeData[“Barium144”, “AtomicMass”], “MeV/c^2”] +UnitConvert[IsotopeData[“Krypton89”, “AtomicMass”], “MeV/c^2”] + 3*ParticleData[“Neutron”, “Mass”])
  • So 173.3 MeV of energy is released, taking the form of kinetic energy of the fission products and gamma-ray photons

Nuclear Fusion

  • In nuclear fusion atomic nuclei combine to form a larger atomic nucleus.
  • The mass of the reacting particles is greater than the mass of the product particles, the lost mass converted into energy per Einstein’s E=mc2 
  • Nuclear fusion in a thermonuclear bomb:
    • (1 Hydrogen-2 + 1 Hydrogen-3) → (1 Helium-4 + 1 neutron)
  • Nuclear fusion in the sun:
    • (6  Hydrogen-1) → (1 Helium-4 + 2 Hydrogen-1)
Nuclear Fusion in a Thermonuclear Bomb
  • (Mass of hydrogen-2 + mass of hydrogen-3) ➖(mass of 1 helium-4 + mass of 1 neutron) = 17.589 MeV/c2
  • Mathematica Calculation
    • (UnitConvert[IsotopeData[“Hydrogen2”, “AtomicMass”], “MeV/c^2”] + UnitConvert[IsotopeData[“Hydrogen3”, “AtomicMass”],  “MeV/c^2”]) – (UnitConvert[IsotopeData[“Helium4”, “AtomicMass”], “MeV/c^2”] + ParticleData[“Neutron”, “Mass”])
  • So 17.589 MeV of energy is released, taking the form of kinetic energy of the products
Nuclear Fusion in the Sun
  • Mass of 6 hydrogen-1 ➖ (mass of 1 helium-4 + mass of 2 hydrogen-1) = 26.731 MeV/c2
  • Mathematica Calculation
    • (6 * UnitConvert[IsotopeData[“Hydrogen1”, “AtomicMass”], “MeV/c^2”]) – (UnitConvert[IsotopeData[“Helium4”, “AtomicMass”], “MeV/c^2”] + 2 * UnitConvert[IsotopeData[“Hydrogen1”, “AtomicMass”], “MeV/c^2”])
  • So 26.731 MeV of energy is released, taking the form of kinetic energy of the products, gamma-ray photons, neutrinos, and positrons