Nuclear Binding Energy - Example Values Deduced From Experimentally Measured Atom Nuclide Masses

Example Values Deduced From Experimentally Measured Atom Nuclide Masses

The following table lists some binding energies and mass defect values. Notice also that we use 1 u = 1 a.m.u = 931.494028(±0.000023) MeV. To calculate the "binding energy" we use the formula P*(m_p+m_e) + N * m_n - m_nuclide where P denotes the number of protons of the nuclides and N its number of neutrons. We take m_p = 938.2723 Mev, m_e = 0.5110 MeV and m_n = 939.5656 MeV. The letter A denotes the sum of P and N (number of nucleons in the nuclide). If we assume the reference nucleon has the mass of a neutron (so that all "total" binding energies calculated are maximal) we could define the total binding energy as the difference from the mass of the nucleus, and the mass of a collection of A free neutrons. In other words, it would be - m_nuclide. The "total binding energy per nucleon" would be this value divided by A.

56Fe has the lowest nucleon-specific mass of the four nuclides listed in this table, but this does not imply it is the strongest bound atom per hadron, unless the choice of beginning hadrons is completely free. Iron releases the largest energy if any 56 nucleons are allowed to build a nuclide—changing one to another if necessary, The highest binding energy per hadron, with the hadrons starting as the same number of protons Z and total nucleons A as in the bound nucleus, is 62Ni. Thus, the true absolute value of the total binding energy of a nucleus depends on what we are allowed to construct the nucleus out of. If all nuclei of mass number A were to be allowed to be constructed of A neutrons, then Fe-56 would release the most energy per nucleon, since it has a larger fraction of protons than Ni-62. However, if nucleons are required to be constructed of only the same number of protons and neutrons that they contain, then nickel-62 is the most tightly bound nucleus, per nucleon.

In the table above it can be seen that the decay of a neutron, as well as the transformation of tritium into helium-3, releases energy; hence, it manifests a stronger bound new state when measured against the mass of an equal number of neutrons (and also a lighter state per number of total hadrons). Such reactions are not driven by changes in binding energies as calculated from previously fixed N and Z numbers of neutrons and protons, but rather in decreases in the total mass of the nuclide/per nucleon, with the reaction.