San José State University

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Thayer Watkins
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 The Deuteron, the Virial Theorem, the Nuclear Strong Force and the Mass of the Neutron

In an atom when an electron drops from a state of higher potential energy to one of lower potential energy part of the loss in potential energy goes into a higher kinetic energy of the electron and the remainder is emitted as a photon. In this case the division between higher kinetic energy and the energy of the photon is exactly even. This is because the electrostatic force is exactly an inverse distance squared force.

In a nucleus something similar happens. For example, when a neutron and proton form a deuteron a gamma ray with 2.22457 million electron volts (MeV) of energy is emitted. Likewise it requires a gamma ray of 2.22457 MeV to disassociate a deuteron into its constitutent parts. The analysis below is an attempt to estimate the the other energy changes involved in the formation or disassociation of a deuteron.

## The Functional Form of the Nuclear Strong Force

The electrostatic force has an inverse distance squared dependence because it is carried by particles, photons, which are spread out over an area proportional to the distance squared. The photons last forever. The nuclear strong force are carried by particles, the π mesons, which decay over time and therefore with distance. The number of surviving force-carrying particles is thus a negative exponential function of distance. Therefore the nuclear force has an inverse distance squared dependence weighted by the survival factor which is a negative exponential function.

Thus the nuclear strong force between a neutron and a protron is given by the formula

#### F = −H*exp(-s/s0)/s²

where S is the separation distance of the centers of the nucleons and H and s0 are constants.

The value of the scale parameter s0 is given by the Yukawa Relation as a function of the mass of the force-carrying particle. For the π mesons this is 1.522 fermi. (A fermi is 10-15 meters.)

## Potential Energy

The potential energy, V(s), is defined such that

#### F = −(∂V/∂s)

This means that the potential energy function V(s) is given by

#### V(s) = H ∫s+∞(exp(-z/s0)/s²)ds

This expression can be be rearranged using the Integration-by-Parts formula ∫UdW=UW − ∫WdU with U=exp(-z/s0) and dW=(1/z²) so W=−(1/z). Thus

## The Virial Theorem

According to the Virial Theorem the time-averaged kinetic energy K is given by

#### K = −½sF(s)

Since F(s)=−H*exp(-s/s0)/s²

## The Allocation of a Decrease in Potential Energy to Kinetic Energy and Photon Emission

When one nucleon drops from +∞ and zero velocity, where the potential energy is zero and the kinetic energy is zero to a distance s away from the other nucleon some of the decrease in potential energy goes into kinetic energy and the rest goes for the energy of an emitted photon. The allocation between these two uses of the loss in potential energy is given by

#### K/|V| = (½H*exp(-s/s0)/s)/[H[exp(-s/s0)/s + (1/s0)∫s∞exp(-z/s0)(1/z)dz] which reduces toK/|V| = ½/[1 + (s/s0)exp(s/s0)∫s∞(exp(-z/s0)/z)dz]

The exponential term in the above equation can be brought inside the integration to yield

## The Physical Dimensions of a Deuteron

A group of physicists under the editorship of Savely G. Karshenboin published in 2008 a book devoted to the compilation of the best estimates of physical properties of particles, simple atoms and simple molecules (Precision Physics of Simple Atoms and Molecules, Springer-Verlag).

The best estimate of the root-mean-square-charge diameter of a deuteron from page 70 of the above mentioned work is 4.260 fermi with a margin of error of ±0.02 fermi. The recommended estimate of the rms-charge radius of the proton, given on page 49 of the above work, is 0.895 fermi. Precision Physics of Simple Atoms and Molecules does not give an estimate for the radius of the neutron. Another source gives the rms-radius of the neutron as 1.113 fermi.

Thus the separation distance of the centers of the nucleons is

#### s = 4.260−0.895−1.113=2.252 fermi.

The separation distance s1 in the physical deuteron is thus about 2.252 fermi (2.252×10-15 m), Thus the ratio s1/s0 is equal to 1.4796, a pure number.

## The Change in Kinetic Energy in the Formation of a Deuteron Relativeto the Change in Potential Energy

The integral ∫0(exp(-p)/(p+σ)dp can be evaluated numerically. For σ equal to 1.4796 its value is 0.67129888. Thus the ratio of the change in kinetic energy to the change in potential energy in the formation of a deuteron is

#### K/|V| = ½/[1 + 1.4796*0.67129888] = 0.299168513

The proportion of the loss in potential energy which goes into the emission of the gamma photon is therefore (1−0.299168513) or 0.700831487. Thus

## The Energy Budget of the Deuteron and the Mass of the Neutron

The values of the energies involved in the formation of a deuteron are of special significance because they are the basis for the estimation of the mass of the neutron. The masses of the proton and the deuteron as charged particles can be measured. Their values are

#### Mass of proton = 938.272013 MeV Mass of deuteron = 1875.613003 MeV

The mass of the neutron as a neutral particle cannot be directly measured. Its value is deduced.

It was perceived that there is a mass deficit for complex nuclei, the mass of the nucleus being less than the mass of its constituent particles. The mass deficit expressed in energy terms is called the binding energy. Conventionally it is presumed that the binding energy of the deuteron was equal to the energy of the gamma ray emitted in its formation, 2.22457 MeV.

To get an estimate of the mass of a neutron the presumed mass deficit in terms of binding energy is added to the measured mass of the deuteron to get what the mass of the constituents would be. Conventionally this is

#### 1875.613003 MeV + 2.22457 MeV = 1877.837573 MeV

From this is then subtracted the measured mass of the proton to get the mass of the neutron.

#### mass of neutron = 1877.837573 − 938.272013 = 939.56556 MeV

There is no theoretical justification for the assumption that the mass deficit of the deuteron is equal to the energy of the emitted gamma photon. The whole matter of mass deficits is such an enigma that there is little basis for saying anything about them. Mass does not just disappear when a nuclide like the deuteron is formed, it is there when the nuclide is disassembled. a mass deficit is a phenomenon associated with nucleons being brought into proximity just as the loss of potential energy is. It is more reasonable to identify the mass deficit with the loss of potential energy than the energy of the photon emission. Thus the binding energy of the deuteron should be 3.17418672 MeV rather than 2.22457 MeV.

Therefore the mass of the neutron is given by

#### 1875.613003 + 3.17418672 = 1878.78719 MeV and mass of neutron = 1878.78719 − 938.272013 = 940.5151767 MeV

With the mass of the neutron being 940.5151767 MeV rather than the conventional 939.56556 MeV, it is therefore 2.24316372 MeV more massive than the proton rather than only 1.293547 MeV more massive. This constitutes a 0.1 of 1 percent increase in the estimated mass of the neutron. The additional mass of the neutron is equivalent to 4.39 electron masses rather than the conventional 2.53 electron masses.

This means that the binding energies of all the other nuclides are likewise underestimated by an amount equal to 0.94961672 MeV times their number of neutrons. There is one troubling anomaly in the binding energies computed using the conventional neutron mass. The nuclide Berylium 5, with four protons and one neutron has a negative binding energy of 0.75 MeV in the conventional calculation. If the neutron mass is underestimated by 0.95 MeV then the binding energy of Be5 is underestimated by 0.95 MeV and thus should be a positive 0.2 MeV rather than the puzzling negative 0.75 MeV.

Since free neutrons decay into protons and electrons one can compute the mass deficit or surplus of a neutron by comparing its mass to the sum of the masses of the proton and electron. (The electron mass is 0.511 MeV) Accoring to the conventional estimate of the mass of the neutron it has a mass surpus of 1.293547 MeV; i.e., a binding energy of −1.293547 MeV. This would make neutrons highly unstable. As indicated, free neutrons do decay but the half life is about 15 minutes. This is not highly unstable as nuclear processes go. According to the estimate of neutron mass given above the neutron has an even more negative binding energy of 1.73 MeV, which would indicate some significant degree of instability.

## An Estimate of the Force Constant H for the Nuclear Force

The value of H cancelled out in finding the ratio of kinetic energy to potential energy. However the equation derived from the Virial Theorem

#### K = ½H*exp(-s/s0)/s or, equivalently H = 2K/(exp(-s/s0)/s) and hence H = 2Ks*exp(s/s0)

When the appropriate values are entered into the RHS of this equation the result is

#### H = 3.00879×10-27 kg*m3/sec2

This constant can be expressed as a multiple of hc, where h is Planck's constant divided by 2π and c is the speed of light; i.e.,

#### H = 0.095166hc = (1/10.508)hc

This is in analogy with the fine structure constant of 1/137.036 for the electrostatic force.

(To be continued.)