The mass of charged particles can be determined by injecting them a 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 is 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.,

Proton mass = 938.272013 MeV
Deuteron mass = 1875.613789 MeV

When a deuteron is formed from a proton and a neutron previously separated at a great enough distance that the potential energy 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 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

the gain in kinetic energy of the proton and neutron plus
the energy of the emitted photon
equal
the loss of potential energy plus the loss of mass-energy

For convenience of explanation let the above energy balance be represented as an equation; i.e.,

A + B = C + D

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

mass of neutron = mass of deutron
− mass of proton
+ energy of gamma ray
or
mass of neutron = 1875.613789 − 938.272019 + 2.22457
= 939.565346 MeV

However in electron transitions within the atom the changes in potential and kinetic energy can be computed and a different allocation of energy changes prevails. For electron transitions B=0 and C=D=A/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. This phenomenon 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 a positive binding energy. 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., 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 postive 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 disintergrates 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.

(To be continued.)