nucleon in 12C has lost, on average, about 0.8% of its mass in the form of binding energy. Semiempirical formula for nuclear binding energy For a nucleus with A nucleons, including Z protons and N neutrons, a semi-empirical formula for the binding energy (EB) per nucleon is: where the coefficients are given by: ; ; ; ; . The first term is called the saturation contribution and ensures that the binding energy per nucleon is the same for all nuclei to a first approximation. The term is a surface tension effect and is proportional to the number of nucleons that are situated on the nuclear surface; it is largest for light nuclei. The term is the Coulomb electrostatic repulsion; this becomes more important as increases. The symmetry correction term takes into account the fact that in the absence of other effects the most stable arrangement has equal numbers of protons and neutrons; this is because the n–p interaction in a nucleus is stronger than either the n−n or p−p interaction. The pairing term is purely empirical; it is + for even–even nuclei and − for odd–odd nuclei. When A is odd, the pairing term is identically zero. 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 Da = . To calculate the binding energy we use the formula Z (mp + me) + N mn − mnuclide where Z denotes the number of protons in the nuclides and N their number of neutrons. We take , and . The letter A denotes the sum of Z 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 (Z + N) mn − mnuclide. 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