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An Explanation for the Binding Energies
of the Helium 4 (alpha particle), the
Helium 3, the Hydrogen 3 (triteron) and
the Hydrogen 2 (deuteron) nuclides

The binding energies of the heavier nuclides are very regular. A regression equation based upon a simple model of nuclide structure explains 99.968 percent of the variation. The big puzzle in nuclear physics is the relative binding energies of the small nuclides, particularly the deteron, the triteron and the alpha particle. The deuteron, the Hydrogen 2 nuclide, has binding energy of about 2.225 million electron volts (MeV). The triteron, the Hydrogen 3 nuclide, has binding energy of about 8.48 MeV. This is a very large increase, but the jump to the alpha particle, the Helium 4 nuclide, is even larger. The binding energy of the alpha particle is about 28.29 MeV. Note for later comparison that the Helium 3 nuclide has binding energy of about 7.72 MeV. The Helium 3 nuclide differs from the triteron in that it has two protons and one neutron rather than one proton and two neutrons.

This material is an attempt to explain the relative binding energies of the He4, He3, H3 and H2 nuclides. The results may not extend to the other small nuclides. The explanation for the relative values involves a number of steps.

It was found elsewhere that a two cluster system revolving about the center of mass has a potential energy proportional to the fourth power of the (geometric) mean of the numbers of nucleons in the clusters. This means that if the alpha particle is a deuteron of deuterons then it should have bond energy equal to 24=16 times that of the deuteron. Thus

V(He4) = 16V(H2)
which is equivalent to
23.846528 − Pnn − Ppp = 16(2.224573 − Pnp)
which reduces to
16Pnp − Pnn − Ppp = 11.74664

For the triteron (H3) considered as a deuteron and neutron revolving about their center of mass the cluster sizes are 2 and 1. The geometric mean of 2 and 1 is √2. The fourth power, (√2)4, is 4. Therefore the equation for the triteron is

V(H3) = 4V(H2)
6.257248 − Pnn − Pnp = 4(2.224573 − Pnp)
which reduces to
3Pnp − Pnn = 1.877281

Likewise

V(He3) = 4V(H2)
5.493485 − Ppp − Pnp = 4(2.224573 − Pnp)
which reduces to
3Pnp − Ppp = 3.404807

If the equations derived for V(H3) and V(He3) are added together the result is

6Pnp − (Pnn + Ppp) = 5.282088

If this last equation is subtracted from the equation for V(He4) the result is

10Pnp = 5.700789
and hence
Pnp = 0.5700789 MeV
and therefore
Pnn = −0.9308073 MeV
and
Ppp = −1.6945703

Thus

V(H2) = 2.224573 − 0.6464552 = 1.6544941 MeV
V(H3) = V(He3) = 6.6179764 MeV
V(He4) = 26.4719056 MeV

Thus the explanation for the binding energy of He4 is

B(He4) = 16V(H2) + Pnn + Ppp + 2B(H2)
= 28.295674 MeV

The binding energy of the triteron, H3, is explained as

B(H3) = 4V(H2) + Pnn + Pnp + B(H2)
= 8.481821 MeV

The binding energy of the He3nuclide is explained as

B(He3) = 4V(H2) + Ppp + Pnp + B(H2)
= 7.718058 MeV

The binding energy of the deuteron, H2, is explained as

B(H2) = V(H2) + Pnp
= 2.224573 MeV

The results give a separation of the binding energies of the He4, the He3, the H3 and the H2 nuclides into that which is due to the strong force bonding and that due to the pairing of the nucleons. The values appear to be reasonable.

Note that there is reason to believe that the conventional estimates of the mass deficits and hence binding energies of nuclides are in error because the mass of the neutron is underestimated. This underestimate arises because the mass deficit of the deuteron is taken to be exactly equal to the energy of the gamma ray involved in the formation or disassociation of the deuteron. Such an error in the estimates of binding energies would not affect the above analysis. The error in the binding energy for He4 is offset by the corresponding error in the binding energy of the two H2 nuclides which it is presumed to be composed of. Likewise for the H3 nuclide.


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