Cool save but U-235 nucleus rarely splits exactly in half. The number of protons and neutrons that goes to each of the new nuclei after split is random and defined by probability function.
Reactants: H-2(2.0141017780u), H-3(3.0160492675u) Products: He-4(4.0026032497u), n(1.0086649233u). If TE is overcome, the change in mass is -0.0188818725u. To find energy in eV (amount of energy of electron across 1 volt), multiply the mass decrement (-delta m) by 931.5MeV/u. This fusion reaction (not as shown due to stated inaccuracies) would produce 17.58846MeV of energy.
The maximum TE is the energy needed, and the energy released is based on the change in mass (E=mc^2, literally). To calculate this, tables are needed giving the mass (in amu/u/(C-12/12)) of each and every nuclide to at least 6 places past the decimal. Multiplied by the speed of light squared, you take the mass of all the nuclides going in from the sum of the masses of all the products.
The energy needed to actually make contact with the other nucleus is what is being referred to (Nuclear Engineering Student here). Coulombic threshold energy (TE) and kinetic TE. Kinetic is used for all particles, while coulombic is only for those with charge
Also fusion creates He-4 and 1 neutron.
Deuterium has 1 neutrons (instead of 2). And tritium has 2 neutrons (instead of 3).
as a science geek, this is accurate +1
is he banned?
Cool save but U-235 nucleus rarely splits exactly in half. The number of protons and neutrons that goes to each of the new nuclei after split is random and defined by probability function.
Reactants: H-2(2.0141017780u), H-3(3.0160492675u) Products: He-4(4.0026032497u), n(1.0086649233u). If TE is overcome, the change in mass is -0.0188818725u. To find energy in eV (amount of energy of electron across 1 volt), multiply the mass decrement (-delta m) by 931.5MeV/u. This fusion reaction (not as shown due to stated inaccuracies) would produce 17.58846MeV of energy.
Also fusion creates He-4 and 1 neutron.
as a science geek, this is accurate +1