﻿ The Statistical Explanation of the Masses of Nuclides
San José State University

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Thayer Watkins
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The Statistical Explanation
of the Masses of Nuclides

The mass of any nucleus is generally believed to be less than the sum of the masses of its constituent nucleons. This is called its mass deficit. This mass deficit expressed in energy units through the Einstein formula E=mc² it is called the binding energy of the nuclide.

The mass of a charged partcle such as an ionized atom can be determined by injecting it into a magnetic field and measuring the curvature of its path in relation to its velocity. To determine the mass of the neutral neutron another method must be used.

When a neutron and a proton come together to form a deuteron a gamma ray with energy 2.225 million electron volts (MeV) is emitted. Likewise when a deuteron absorbs a gamma ray of 2.225 MeV energy it breaks apart into a neutron and a proton. That energy is taken to be the binding energy of a deuteron and the mass of a neutron is computed from that value and the measured masses of the deuteron and the proton.

But consider the situation of an electron being absorbed by an ion. The electron would lose 27.2 electron volts (eV) of potential energy; 13.6 eV of that 27.2 eV would go into the emission of a photon of 13.6 eV energy. The other 13.6 eV would go into the kinetic energy of the electron in the atom into which it is absorbed. The potential energy loss is divided exactly equally between the emitted photon and the gain in kinetic energy because the electrostatic force is exactly a function of inverse distance squared. If one estimated the potential energy loss using only the photon energy there would be an erroneous underestimate. This may be what is involved with using the energy of the gamma ray emitted when a neutron and a proton form a deuteron. The mass deficit involved in the formation of the deuteron is equivalent to a loss of potential energy. Unfortunately, because the nuclear force is probably not an inverse distance squared force the division of the mass loss between increased kinetic energy and the energy of the emitted gamma photon is not equal and thus is unknown.

It is pretty certain that there are errors in the computed binding energies because there is one nuclide that has a negative computed binding energy. That nuclide is the berylium isotope Be5 with four protons and one neutron. Its computed binding energy is −0.768 MeV. Thus ΔK must be at least 0.768 MeV.

The measured masses can be recovered by adding the published binding energy of a nuclide to the sum of the masses of its nucleons with one proton having a mass of 938.27231 MeV and one neutron a mass of 939.56563 MeV. Regression analysis may then be used to find the values of proton and neutron mass that best explain measured masses assuming no mass deficits.

The form of the regression equation is

#### m(p, n) = mPp + mNn + u

where u is a random error term. Initially it is assumed that u has a variance that is independent of the level of m(p, n).

The results of the regression analysis are:

#### m(p, n) = 948.80708p + 945.58357n                [7015.5]        [9964.1]

The coefficient of determination (R²) for this equation is 0.999999874 and the standard deviation of the random term is 50.5 MeV. The t-ratios for the coefficients, shown in the square brackets [ ], are so high that the probability that the correlation of m(p, n) with p and n is just due to chance is infinitesimally low.

However the problem is that the value for mN is significantly less than the value for mP. It is known that mN must be larger than mP because a neutron decays into a proton, an electron and a neutrino. The neutrino has negligible mass, but the electron has a mass of 0.511 MeV so the mass of a neutron must exceed that of a proton by at least 0.511 MeV.

There is a modification of the regression which is justified. The variance of the random variable u is likely proportional to m(p, n). Thus v=u/√m(p,n) would have a variance independent of the dependent variable in the regression equation The condition in which the variance of the random term is not independent of the level of the dependent variable is called heteroskedasticity. The adjustment is made to achieve homoskedasticity.

#### √m(p,n) = mPp/√m(p,n) + mNn/√m(p,n) + v

The results for this regression equation are:

#### √m(p,n) = 948.10438p/√m(p,n) + 946.23111n/√m(p,n)                [7772.7]        [10813.9]       R² = 0.999999835

So again the estimate of mN is less than that of mP.

The regression coeficient for n can be forced to be equal to that for p by regressing m(p,n) on (p+n). That gives a mass of a nucleon equal to 946.91191 MeV. The R² for this equation is 0.999999866; almost as high as for the equation using two variables. The standard error of the estimate is only slightly higher (52.1 MeV) than the 50.5 MeV found for the two variable equation.

An adjustment for heteroskedasticity gives an estimate of the mass of a nucleon of 953.3 MeV.

Since there is an empirically known value for the mass of a proton of 938.27231 MeV it is appropriate to compute (m(p,n)−938.27231p) and regress this quantity on n. That procedure gives an estimate of the mass of a neutron of 952.9 MeV, quite a bit higher than the conventional value of 939.56563 MeV. The coefficient of determination for the equation is 0.999998898. Here is what a graph of the data looks like. These results stem from the implicit assumption that there are no mass deficits for the nuclei.

## Conclusions

The exercise found some interesting results but nothing that would replace the conventional esimateof the mass of a neutron.