﻿ A Correction of the Conventional Mass of the Neutron
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
U.S.A.

A Correction of the Conventional
Mass of the Neutron

## Nuclear Binding Energy

The binding energy of a nuclide is its mass deficit expressed in energy units via the Einstein equation E=mc². Its mass deficit is the difference between the total mass of its constituent nucleons and its measured mass,

## How the Mass of a Charged Particle is Determined

The mass of a particle is measured by injecting it into a cloud chamber pervaded by a magnetic field. The particle moves in a helix which can be observed because the particles create a track in the cloud chamber. The initial velocity of the particle can be determined from the length of the cloud track it forms. The radius of the helix can be measured. From the radius of the path, the velocity of the particle and the strength of the magnetic field the mass of the particle can be calculated from an equation based upon a balance of the force imposed upon the particle by the magnetic field and centrifugal force. The centrifugal force is mv²/r where m is the mass of the particle, v is its velocity perpendicular to the direction of the magnetic field and r the radius of path. The force exerted upon the particle by the magnetic field is e(Bv), where e is the charge of the particle, B the magnetic field intensity and v the velocity of the particle. Balance requires

#### mv²/r = eBv and thus m = eBr/v

The problem is that only charged particles can have their mass determined through injection into a magnetic field. The mass of the neutron has to be deduced.

## Photon Emission

When a deuteron is formed from a proton and a neutron a gamma ray of energy 2.22457 million electron volts (MeV) is emitted. Also 2.22457 MeV is the minimum energy required to disassociate a deuteron into a proton and neutron. The value of 2.22457 MeV is assumed to be the binding energy of the deuteron. The mass of the neutron is then deduced to be the mass of the deuteron less the mass of the proton and 2.22457 MeV expressed in mass units.

## Shortcomings of this Method for Estimating the Mass of a Neutron

The problem with this ploy is that in other physical systems, such as an electron in an atom, if the system drops into a lower potential energy state part of the energy loss goes into higher kinetic energy and part into the emission of a photon. If the force involved has an inverse distance squared law then the potential energy loss is equally divided between increased kinetic energy and the energy of the emitted photon. For other force laws the division of the potential energy loss between increased kinetic energy and the energy of the emitted photon will not be so simple but it will be there. Therefore taking the energy of the emitted photon to be the entire amount of the potential energy loss creates an error in the estimate of the mass of the neutron. Let us denote the error in the mass of the neutron expressed in energy units as Δ. Thus the binding energy of a nuclide with n neutrons and p protons which is reported as BE(n, p) should really be BE(n, p)+nΔ.

It is notable that the binding energy of one nuclide, beryllium 5, is reported as a negative value, −0.768 MeV. Beryllium 5 consists of 4 protons and one neutron, an highly unstable combination but it is stable enough to have its mass measured. An actual negative binding energy for a nuclide seems flat wrong. Undoubtedly this negative binding energy is an error and most likely due to an error in the mass of the neutron.

## The Deuteron

A deuteron is simply a neutron-proton spin pair. The binding energy associated with the formation of a neutron-proton spin pair can be determined. When the number of protons is less than the number of neutrons in a nuclide the addition of a proton will involve the formation of such a spin pair, but once the number of protons is equal to the number of neutrons the addition of another proton will not involve the formation of such a spin pair. Thus the change in the increment in binding energy when the number of one type of nucleon surpasses the number of nucleons of the other type gives the value of the binding energy associated with the formation of a neutron-proton spin pair.

Let BE(n, p) be the binding energy of a nuclide with n neutrons and p protons. The binding energy associated with the formation of a neutron-proton spin pair can thus be estimated by

#### FBEnpn = [BE(n+2, p) − BE(n, p)] − [BE(n+1, p) − BE(n-1, p)] and by FBEnpp = [BE(n, p+2) − BE(n, p)] − [BE(n, p+1) − BE(n, p-1)]

The first is the formation binding energy of a neutron-proton spin pair from the incremental binding energies of neutrons. The second is the formation binding energy of a neutron-proton spin pair from the incremental binding energies of protons. There are 42 cases in which FBEnpn can be computed and 36 where FEnpp can be computed.

It is important that FBEnpn and FBEnpp are independent of any error in the mass of the neutron. For the first formula each term would be in error by +2Δ and thus the effect of Δ is eliminated in their difference. In the second formula each term is independent of Δ.

Estimates were computed of the binding energy associated with the formation of neutron-proton pairs. These are given in np spin pairs neutrons and np spin pairs protons In particular the values found were 5.3226 MeV and 5.1264 MeV for nuclides with one proton and/or one neutron. The average of these two values is 5.2245 MeV. The difference between this value and the energy of the emitted photon is 5.2245−2.22457=2.99993 MeV. This means that the conventional mass of the neutron is about 3 MeV too low. The conventional mass of a neutron is 939.5654 MeV and should be 942.5654. This is only a 0.32 of 1 percent increase but it eliminates the anomaly of a negative binding energy for the Beryllium 5 nuclide. The binding energy of Beryllium 5 would be 2.232 MeV with the correction in the mass of the neutron.

## The Rotational Kinetic Energy of a Deuteron

According to the above analysis when a deuteron is formed the kinetic energy goes from zero to 3 MeV. One MeV is equal to 1.60217657×10-13 joules. Therefore the rotational kinetic energy of a deuteron should be 4.8×10-13 joules. The separation distance between the centers of mass of the two nucleons in a deuteron is about 3.2 fermi, which is 3.2×10-15 meters. The moment of inertia I of the two nucleons revolving about their center of mass is then (3.3475×10-27)(1.6×10-15)²=8.5696×10-57 kg-m². Since the kinetic energy of a rotating system is ½Iω²

#### ω² = (4.8×10-13/(½·8.5696×10-57) = 1.12×1044 and thus ω = 1.0584×1022 radians per sec which corresponds to ν = 1.684×1021 times per second.

(To be continued.)