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What Determines the
Minimum Number of Neutrons
in the Isotopes of Each Element?

Nuclei are composed of protons and neutrons. The nuclides with the same number of protons are called isotopes. They differ in the number of neutrons. For example, the isotopes of the element Tin all have 50 protons but their number of neutrons ranges from a minimum of 50 to a maximum of 87.

A nuclide is unstable if a modification will produce a product or products with a higher binding energy. Such a modification is the ejection of a particle such as an alpha or beta particle. Here is the graph of the binding energies of the isotopes of Tin (p=50).

The above display does not pro‪vide any insight into why there are no isotopes with n<50.

A previous study demonstrated that the maximum number of neutrons is affected by the peripheral halo neutron being shielded from the nucleonic charge of the core nucleus by the halo neutrons in innner shells and subshells and those in the same subshell. That phenomenon cannot be involved in the determination of the minimum number of neutrons, therefore there must be some asymmetry between the maximum and minimum number of neutrons.

The graph of the maximum and minimum number of neutrons as a function of the number of protons does show that asymmetry.

The maximum number of neutrons is nearly a linear function of the number of protons. The minumum number function has a definite curvature indicating a quadratic dependence. So there is an asymmetry between the determination of the maximum and minimum number of neutrons in the isotopes of each element.

Consider the following graph showing the binding energies of a sequence of nuclides which would arise from beta transitions.

The binding energy reaches a maximum at p=56 and n=81. This is Ba137 which is a stable isotope of Barium. Below n=81 the nuclides convert a proton into a neutron and eject a positron. Above 81 a neutron decays into a proton and an electron, which is ejected.

Here is the data for the sequence.

The Binding Energies in a
Sequence of Beta Transitions
p n BE Incr
50 87 1117.2
51 86 1126.1 8.90
52 85 1134.65 8.55
53 84 1140.81 6.16
54 83 1145.903 5.093
55 82 1149.293 3.39
56 81 1149.686 0.393
57 80 1148.3 -1.386
58 79 1146.3 -2.00
59 78 1142.82 -3.48
60 77 1138.34 -4.48
61 76 1131.9 -6.44
62 75 1125.22 -6.68
63 74 1116.8 -8.42
64 73 1107.3 -9.50

The incremental binding energies, shown in the last column of the above table, indicate how strong of an impulse there is for the beta transition.

An extrapolation of the curve suggest that if there were a nuclide beyond the sequence a gain in binding energy of over 10 MeV could be achieved by ejecting a positron. This approximately the same that could be achieved at the other end of the sequence by ejecting an electron.

This suggest that the minimum number of neutrons occurs where the gain in binding energy from a positron ejection for a nuclide is so large that the nuclide does not last long enough to have its mass measured. That value can be roughly calculated from the the gain in binding energy found for the last transition in the beta sequence. Here are those gains for an arbitrary select of nuclides which have the minimum number of neutrons for various proton numbers.

p n ICBE
8 4 15.4923
14 8 14.77
24 18 14.97
39 39 11.308
64 73 9.5
75 86 10.6


The critical level for the minimum number of neutrons shows some regularity. It is above 15 MeV for the small nuclides and just above 10 MeV for the larger nuclides; i.e., the ones with proton numbers approaching 82. Beta positron emission is the mode of decay for nuclides with proton numbers 82 and below.

(To be continued.)

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