This materials deals with the application of a model that successfully explains the ionization energies of electrons in atoms to the binding energies of nucleons in nuclei.

The ionization energy IE, or as it is usually called the ionization potential, for an electron in an atom or ion is the amount of energy required to dislodge it. The Bohr model of a hydrogen-like atom indicates that the energy required to remove an electron should follow the form

IE = RZ²/n²

where R is the Rydberg constant (approximately 13.6 electron Volts (eV)), Z is the net charge experienced by the electron and n is the principal quantum number, effectively given by shell number. The value of Z is the number of protons in the nucleus #p less the shielding ε by the other electrons. Thus the ionization potential would be

IE = (R/n²)(#p−ε)²

Application of the
Bohr Model to Nuclei

A nucleus consists of protons and neutrons linked together through spin pairing into shells. The filled shells form a nuclear core of equal numbers of protons and neutrons. But typically a nucleus has more neutrons than protons, The extra neutrons, called halo neutrons, are held in orbits around the nuclear core due the attraction between neutrons and protons through the nucleonic force.

The quantity for nucleons in a nucleus which would be comparable to the ionization energy for electrons in atoms is the incremental total binding energy. This would include the energy lost when the components of a nucleus come together. There is a loss of potential energy but a lesser gain in kinetic energy. In atoms that net loss of energy goes into the creation of a photon. For nuclei a similar thing happens but for nuclei there is another loss of energy which is manifested as a mass deficit; i.e., the mass of a nucleus is less than the masses of its components. These mass deficits converted into energy units are what are called the binding energies of nuclei, but they are not the total binding energies of nuclei.

Total binding energy is known only for one nuclide, the deuteron. When a proton and neutron come together to form a deuteron a gamma ray of about 2.25 million electron volts (MeV) of energy is emtted. When deuterons are subjected to gamma radiation of 2.25 MeV or higher in energy they dissociate into protons and neutrons.

The Isotope of Nickel with
the Maximum Number of Neutrons

The maximum number of neutrons in an isotope of Nickel is 50. Consider the shielding for that 50th neutron. The core consists of 28 protons and 28 neutrons. The nucleonic charge of a proton is 1 and that of a neutron is −2/3 so the core has a net nucleonic charge of 28/3.

The halo neutrons are all in the 29 through 50 shell. The number of halo neutrons in the same shell with the 50th neutron is 21. They have a collective nucleonic charge of −14. If there is a 50% ratio of shielding by neutrons in the same shell then 7 nucleonic charge units of the core are cancelled out leaving the net nucleonic charge units experienced by the 50th halo neutron being 28/3−7=7/3 nucleonic charge units.

Now consider the situation for a Nickel isotope with a 51st neutron. Such an isotope of Nickel does not exist. The core is still has 28 protons and 28 neutrons with a net nucleonic charge of 28/3. The 22 neutrons in the 29 through 50 shell have a collective charge of −(2/3)22=−44/3. They are in an inner shell with respect to a 51st neutron. The core charge is thus more than all cancelled out and such an isotope cannot exist.

The Isotope of Tin with
the Maximum Number of Neutrons

The maximum number of neutrons in an isotope of Tin is 87. Consider first the shielding for the 82nd neutron. The core consists of 50 protons and 50 neutrons. The nucleonic charge of that core is 50/3. The 31 neutrons in the same shell as the 82nd neutron have a collective nucleonic charge of 62/3. With a 0.5 shielding ratio the 82nd neutron would experience a nucleonic charge of 50/3−31/3=19/3.

Now consider the 87th neutron. With a 100% shielding ratio for the 64/3 charge of the inner shell neutrons that would more than cancel the core charge.

The work on ionization energies indicated that the best empirical estimates for the shielding ratios for inner shell and same shell electrons are about 0.8 and 0.4, respectively. The shielding ratios may vary with shell number.

Suppose those shielding ratios for Tin are 0.72 and 0.36. The the inner shell for the 87th neutron would cancel 46.08/3 units of the 50/3 units of charge. The 4 neutrons in the same shell as the 87th neutron would have −8/3 units of nucleonic charge. With a 0.36 shielding ratio they would cancel 0.96 units of charge. Together the inner and same shell halo neutrons cancel out 48.96/3 units of the 50/3 units of the core charge. That leaves about 1/3 unite of net nucleonic charge for the 87th neutron.

For an 88th neutron there woud be 50.4/3 units of 50/3 units of core charge cancelled so a Tin isotope with 88 neutrons would not exist.

The Isotope of Lead with
the Maximum Number of Neutrons

The maximum number of neutrons in an isotope of Lead is 132. Consider the situation for the 132nd neutron. The net nucleonic charge of the core is 82/3. The 83rd through 126th neutrons are in an inner shell with respect to the 132nd neutron. They have a collective nucleonic charge of (−2/3)44=(−88/3). The 127th through 131 neutrons are in the same shell. They have a collective nucleonic charge of (−2/3)5.

If the shielding ratio for inner shell neutrons is less than 0.88 there is a bit of the nucleonic charge to hold the 132nd neutron. But for a 133rd neutron and shielding ratios of 0.88 and 0.44 for the inner shell and same shell, respectively, the net nucleonic charge from the core is negative and hence an isotope of lead with 133 neutrons cannot exist.

Conclusions

The existence of the isotopes of Nickel, Tin and Lead with the maximum number of neutrons and the nonexistence of any isotopes beyond those numbers can be explained by reasonable estimates of the shielding ratios for inner shell neutrons and same shell neutrons.