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Neutron and Its Implication for the Binding Energies of Nuclides |
The masses of charged particles can be determined by injecting them at high speed into a magnetic field. If the velocity of the particle is perpendicular to the direction of the magnetic field then the particle executes a circular path. The radius of the circular path depends upon the velocity of the particle, the intensity of the magnetic field and the ratio of the charge of the particle to its mass. If the charge of the particle is known then its mass can be computed.
Since the neutron has no net charge this method cannot be used to determine its mass. Instead its mass must be deduced. It is known that deuterons dissociate when subjected to gamma rays of energy of at least 2.22457 million electron volts (MeV). Likewise when protons and neutrons form deuterons there are given off gamma rays of energy 2.22457 MeV.
The masses of a deuteron and a proton, being charged particles, have been determined; i.e.,
When a deuteron is formed from a proton and a neutron previously separated at a great enough distance that the potential energy of the pair is essentially zero then the combined state of the proton and neutron would have a negative potential energy. This loss of potential energy could be transformed into kinetic energy or the emission of a gamma photon. However those gains in energy could also arise from the loss in mass of the system.
An energy balance requires that
For convenience of explanation let the above energy balance be represented as an equation; i.e.,
The conventional estimate of the mass of the neutron is based upon the assumption that D=B. (This would imply that A=C, as well.) Thus
However in electron transitions within an atom the changes in potential and kinetic energy can be computed and a different allocation of energy changes prevails. For electron transitions D=0 and A=B=C/2.
For nuclear transitions there is no universally accepted model for computing the potential and kinetic energies of nuclear states. The Yukawa model could be used for such a computation but generally it is not. There is no generally accepted theory explaining the mass deficits of nuclei. The phenomenon of mass deficits remains an enigma. Nevertheless the mass deficits of almost three thousand different nuclides have been computed and expressed as binding energy; i.e., the energy that would have to be supplied in order to break the nuclide up into its presumed constituent components of protons and neutrons. Of the nearly three thousand nuclides all but one have positive binding energies. That exception is the beryllium isotope with four protons and one neutron, Be5. Be5 has a binding energy of −0.75 MeV. This suggests that there might be an error in the estimated value of the mass of the neutron. The error is not empirical but rather conceptual; i.e., from assuming the mass deficit of the deuteron is equal to the energy of the gamma ray emitted or absorbed in its formation or breakup.
However it is quite plausible that for nuclear transitions that a more complex allocation of energy changes prevail than the assumed D=B and A=C. For example, it might be that C=A/2 and D=B+A/2. If the mass deficit of the deuteron were 3 MeV instead of the assumed 2.22457 MeV then the mass of the neutron would be 940.340776 MeV instead of 939.565346 MeV. This would make the mass deficit of Be5 a positive 0.02543 MeV instead of a negative 0.75 MeV.
The neutron has a mass surplus in the sense that its mass is greater than the mass of a proton and an electron which it disintegrates into. According to the conventional estimate of neutron mass this mass surplus is equal to about one and a half (1.531) electron masses (0.511 MeV). For a neutron mass of 940.340776 MeV the mass surplus would be about three (3.05) electron masses. The mass of a neutron with a mass surplus of exactly three electron masses (0.510998918 MeV) would be 940.316009 MeV. Such a neutron mass would make the mass deficit of Be5 equal to 0.000663 MeV.
According to the Theory of Relativity mass produces a curvature in space. Equally valid is the notion that mass is the curvature of space. This makes a nuclear particle something in the nature of a soliton or solitary wave. Putting two waves of the same curvature in close proximity creates a region of space of opposite curvature, in effect a negative mass. This explains how the mass deficit arises when two nucleons from a nuclei and how the mass deficit disappears when the nuclei is broken up. For more on this point see Nature of Mass and Mass Deficits.
The important conclusion is that the loss of potential energy and the mass deficit in the formation of nuclei are simply two ways of observing the same thing rather than competing processes. In terms of the previous terminology D=C and A+B=C. Therefore if the loss in potential energy involved in a nuclear transition can be computed then the mass deficits of nuclei can be explained.
As noted previously when an electron goes to a lower level of potential energy the loss in potential energy is divided equally between an increase in kinetic energy and the energy of the emitted photon. This means that the loss of potential energy is equal to twice the energy of the emitted photon.
If this same allocation prevailed for the formation of a deuteron then the loss in potential energy is 2(2.22457)=4.44914 MeV. And, if as maintained above the mass deficit is equal to the loss in potential energy then the mass deficit of the deuteron is 4.44914 MeV rather than 2.22457 MeV. This means that the mass of the neutron would be 941.789916 MeV instead of 939.565346 MeV. This in turn would make the binding energy of the Be5 nuclide 1.4745 MeV instead of −0.75 MeV.
For the conventional estimate of the mass of the neutron the binding energy of the alpha particle, He4, is 28.295674 MeV. There are 6 nucleon-nucleon pairings so the binding energy per bond in the alpha particle is 4.723918 MeV where for the deuteron having only one bond it is 2.22457 MeV. This is a puzzling difference. If the binding energy of the deuteron is 4.44914 MeV and that of the alpha particle is 28.295674+4.44914=32.744814 MeV then their binding energies per bond are 4.44914 MeV and 5.457469 MeV, respectively. Still a puzzle, but less of one.
The allocation of potential energy changes for the nuclear force is not necessarily the same as for the electrostatic force involved in electron transitions.
Gravitation and the electrostatic force are inversely proportional to the square of the distance. This is because the force is carried by particles and these particles are spread over an area which is proportional to the square of the distance. It is also the case that the particles which are carrying the force do not decay over time and hence over distance. The nuclear force is carried by pi mesons and these do decay. The proportion of surviving force-carrying particles is an exponential function of distance. Thus the nuclear force will have a formula of the form
where H and ρ are parameters. Values for H and ρ have been estimated based upon the binding energy of the deuteron and the dimensions of the deuteron. Based upon these estimates and a separation distance of 4 fermi between the proton and neutron in the deuteron loss in potential energy when a deuteron is formed is 4.81655 MeV, of which 2.59198 MeV goes into the kinetic energies of the proton and neutron within the deuteron. The other 2.22457 MeV goes into the energy of the emitted gamma ray. This means that 46.2 percent of the loss of potential energy goes into the emitted photon rather than 50.0 percent as in the case of electron transitions. The mass deficit (binding energy) of the deuteron is 4.81655 Mev rather 2.22457 MeV. This means the mass-energy of the neutron is 2.59198 MeV higher than the conventional estimate. It is 942.157326 MeV rather than 939.565346 MeV, 0.276 of 1 percent higher.
According to the above estimate of the error in the mass of the neutron the binding energy of Be5 is 1.84198 MeV, a comfortably positive value.
The conventional and corrected binding energies for the triteron and He3 are shown below.
Nuclide | Symbol | Conventional Binding Energy MeV | Bonds | Binding Energy per Bond MeV |
deuteron | H² | 2.224573 | 1 | 2.22457 |
triteron | H³ | 8.481821 | 3 | 2.827274 |
helium 3 | He³ | 7.718058 | 3 | 2.572689 |
alpha particle | He^{4} | 28.295674 | 6 | 4.715949 |
Nuclide | Symbol | Corrected Binding Energy MeV | Bonds | Binding Energy per Bond MeV |
deuteron | H² | 4.81655 | 1 | 4.81655 |
triteron | H³ | 13.66578 | 3 | 4.55526 |
helium 3 | He³ | 10.31004 | 3 | 3.43668 |
alpha particle | He^{4} | 33.47963 | 6 | 5.57994 |
Some part of the difference in binding energies of the triteron and He3 might attributable to the electostatic repulsion between the two protons in the He3 nuclide that is not there in the triteron. The conventional binding energies of the triteron and He3 imply a separation distance for the two protons in He3 of 1.884 fermi, which is surprisingly small given that the separation distance for the proton and neutron is measured at about 4 fermi. The conclusion to be drawn is that there is major difference in the structural arrangements of the nucleons for the two nuclides.
For the binding energies corrected for the error in neutron mass the differences in binding energies between the triteron and He3 are even greater which would correspond to an even smaller separation distance for the protons in He3. This re-enforces the importance of different structural arrangements for the two nuclides.
For the triteron there is likely an equilateral triangular arrangement. For He3, because of the mutual repulsion of the two proton, the arrangement is likely to be a linear one with the neutron situated between the two protons. In this arrangement there would be two bonds of equal strength and one of half or less strength. This would justify computing the binding energy per bond in He3 by dividing by a figure of 2.5 or less rather than 3. Using 2.5 the binding energy per bond for He3 is 4.124 MeV. This brings the figure for He3 significantly closer to the 4.555 figure for the triteron.
The diameter of the deuteron is conventionally given as 4.2 fermi. The diameters of the proton and neutron are conventionally given as 1 fermi each. These figures mean the separation of the centers of the nucleons is 3.2 fermi. For this separation distance the kinetic energy of kinetic energy of the nucleons in the deuteron is 4.054286 MeV which means the mass deficit of the deuteron would be 6.27886 MeV instead of 2.22457 MeV. The error in the neutron mass is then 4.054286 MeV.
With this error in the neutron mass corrected the binding energies of the small nuclides are
Nuclide | Symbol | Corrected Binding Energy MeV | Bonds | Binding Energy per Bond MeV |
deuteron | H² | 6.27886 | 1 | 6.27886 |
triteron | H³ | 16.59039 | 3 | 5.53013 |
helium 3 | He³ | 11.77234 | 3 | 3.92411 |
alpha particle | He^{4} | 36.40425 | 6 | 6.06737 |
The value of separation distance of 3.2 fermi makes the binding energy per bond greater for the deuteron than that of the alpha particle. This is as it should be because the alpha particle contains two repelling protons. The effect of the two repelling protons on the structure of the alpha particle is much less drastic than it is for He3.
Thus, if the centers separation distance of the nucleons in the deuteron is 3.2 fermi, then the mass of the neutron is 943.619632 instead of 939.565346 MeV, a 0.43 of 1 percent increase.
If the diameters of the proton and neutron are less than 1 fermi each the centers separation distance for the nucleons in the deuteron will be greater than 3.2 fermi and there will be smaller adjustments in the binding energies of the small nuclides and a small error in the mass of the neutron.
(To be continued.)
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