San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Enigma of the Mass Deficits of Nuclei

When protons and neutrons form a nucleus the mass of the nucleus is less than the sum of the masses of the particles which make it up. Gamma rays are given off in the process of formation and the nucleus is broken up only if the nucleus absorbs a gamma ray of the same or greater energy than the one given off in its formation. The phenomenon of mass deficits has been known and accepted for many decades, without much comment on what an enigma it is.

The case of the deuteron, the combination of a proton and a neutron, is easy to visualize. It was first noticed in the 1930's that if gamma rays of at least 2.22457 MeV irradiated deuterium then deuterons would dissociate into free neutrons and protons. Later it was noticed that if slow neutrons were brought in contact with protons deuterons would form, accompanied by the emission of gamma rays of 2.22457 MeV energy.

When particles come together they lose potential energy which is transformed into an increase in kinetic energy. In atoms when electrons move into a lower energy state they lose more potential energy than they gain in kinetic energy and the difference goes into the emission of photons. In nuclear processes it is a puzzle as to why the system must lose mass as well as potential energy.

One process that involves the loss of mass is when a neutron disintergrates. Free neutrons are unstable and their half-life is about 15 minutes. A neutron decays into a proton, an electron and radiation. The mass of the neutron is estimated to be very close of 2.5 electron masses greater than that of the proton and thus 1.5 electron masses greater than the mass of a proton and electron together. The rest-mass energy of an electron is 0.511 million electron volts (MeV). Thus the disintergration of a neutron involves a loss of mass of 0.7823 MeV.

It is dogma that neutrons exist as particles in nuclei rather than breaking up into protons and electrons, but consider what such a breakup would entail for a deuteron. The electron would be positioned midway between the two protons. The protons would revolve around the electron at their center of mass as indicated in the following diagram.

There would be potential energies due to the nuclear force and the electrostatic force. Consider just the electrostatic force which has the form αq1q2/z, where q1 and q2 are the charges of two bodies and z is their separation distance. As indicated in the diagram the effect of the intermediation of the electron is to create a negative potenial energy equal to three times the positive potential energy created by the repulsion of the two protons. This possible configuration could be described as an electron bonded proton pair. In effect, the electron creates a bond between the two protons in the same way that orbital electrons create a bond between the positive nuclei in diatomic atoms such as H2, O2 and N2. The diatomic molecules are known to be stable and that indicates that the deuteron model would also be stable.

There is a bit of evidence for nuclei not containing proton/neutron pairs as subparticle. Alpha particles are sometimes ejected for nuclei and also electrons (beta rays) but no evidence of deuterons being ejected.

Some evidence of the magnitude of the potential energy involved in the repulsion of protons is available from the properties of tritium (3H1) nucleus and the (3He2) nucleus. The binding energy of a tritium nucleus is 8.482 million electron volts (MeV) but that of the Helium 3 nucleus is only 7.718 MeV. The 0.764 MeV difference is due to the mutual repulsion of the two protons in the He 3 nucleus.

If the 0.764 MeV figure is taken as the magnitude of the potential energy due to the mutual repulsion of the protons in the deuteron model, then the loss of potential energy due to the formation of a deuteron would be 3(0.764)=2.292 MeV. This is not far off from the 2.22457 MeV energy of the gamma ray involved in the formation or dissolution of a deuteron. The difference is only 3 percent.

However the loss of potential energy from the creation of an electron-bonded proton pair could not occur without also the loss of 0.7823 MeV of mass due to the disintergration of the neutron. But changes in potential energy might go in part to changes in kinetic energy as well as into the energy of emitted gamma rays. In the case of orbital electrons in an atom the change in potential energy for a transition of an electron to a lower energy level goes half into the change in kinetic energy and half into the energy of an emitted photon. It is not known what the division of potential energy change would be in a nucleus between the change in kinetic energy and the energy of an emitted photon. If all of the energy from the disintergration of the neutron and five eights of the loss in potential energy from the formation of the electron bond between the two protons went into the energy of the photon, the energy of the photon would be 0.7823+(5/8)(2.292)=2.215 MeV, a value less than 0.5 of 1 percent lower than the actual 2.22457 MeV value.

At the simplest level the binding energy might be conjectured to be proportional to the numbers of protons and neutrons. The results of the regression of binding energy on the number of protons, P, and the number of neutrons, N, with no constant term based on the data set for 2932 isotopes is:

B(P,N) = 10.53476P + 6.01794N
         (0.135)           (0.095)
         [145.5]            [63.3]
R² = 0.99

The regression coefficient of 10.53476 for P indicates that for every additional proton added to the nucleus, on average the binding energy increases by 10.53476 MeV. For an additional neutron there is the lesser amount of 6.01794 MeV. The numbers in parentheses below the regression coefficents are the standard deviation of the regression coefficient estimate. The numbers in the square brackets below the standard deviations are the ratios of the coefficients to their standard deviations, the so-called t-ratios. For a coefficient to be statistically significantly different from zero at the 95 percent level of confidence the t-ratio needs to be of a magnitude on the order of 2 or higher. The t-ratios for the regression coefficents indicate that they are statistically highly significantly different from zero. The coefficient of determination, R², indicates that 99 percent of the variation in binding energies of the nuclei is explained by the variations in their proton and neutron numbers.

A more refined hypothesis is that the mass deficit (binding energy) of a nucleus is proportional to the number of proton-neutron pairs formed with somewhat less of a contribution from a singleton proton or neutron. Let D=min(P,N), the number of proton-neutron (deuteron) pairs. Then the existence and nature of a singleton nucleon is given by values of #P=P-D and #N=N-D. The indicated regression results are:

B(P,N) = 16.654D + 2.398#P + 5.797#N
             (0.0439)      (1.0729)    (0.0984)
             [379.3]      [2.235]    [58.952]
R² = 0.991

While the coefficient for a singleton proton, #P, is significant at the 95 percent level of confidence it is just barely so. On the other hand the t-ratio for the coefficient for the number of proton-neutron pairs is astronomical, strongly suggesting that the mass deficit has its origin in a process involving these proton-neutron pairs. Even a singleton neutron has a significant effect. The division of the coefficient for D by 2 gives a mass deficit per nucleon of 8.327 MeV, a value notably higher than the value for a singleton neutron.

A still more refined hypothesis is that the mass deficits are connected with the formation of alpha particles within the nucleus. The alpha particle, the Helium 4 nucleus, has an unusually high mass deficit, indicating some energy efficient structure. Let α be the number of possible alpha particles. A singleton deuteron is indicated by the value of #D=D-2α. The existence of a singleton proton or neutron is then given by #P=P-2α-#D and #N=N-2α-#D. The indicated regression results are:

B(P,N) = 32.94963α + 36.57635#D + 0.64663#P + 5.88038#N
             (0.0908)       (1.6611)    (1.0578)    (0.0963)
             [362.8]     [22.02]      [0.61]    [61.8]
R² = 0.991

In this case the effect of a singleton proton on the mass deficit is not statistically significant at the 95 percent level of confidence. It is possible that this is the result of a positive effect being offset by the negative effect of the repulsion involved in the addition of a proton to a nucleus.

The t-ratio for the number of alpha particles is an astronomical 362.8, an order of magnitude higher than the t-ratio for the effect of a singleton proton-neutron pair. A singleton neutron does have a significant effect. The effect per nucleon for the effect of an alpha particle is 8.119 MeV, about the same level as for the previous regression.

The above regression equation gives a good fit for the heavier nuclei but a notably poor fit for the lighter nuclei. The statistical fit can be improved by including a quadratic term. The only variable that is different from 0 or 1 is α. The results for a regression equation involving α² are

B(P,N) = 38.45542α - 0.17207α² + 8.15586#D - 3.68489#P + 7.60537#N
             (0.0436)       (0.00104)       (0.5436)    (0.3295)    (0.03166)
             [882.3]     [-165.6]     [15.00]      [-11.2]    [240.3]
R² = 0.9991

A notable aspect of this regression equation is that it explains 99.9 percent of the variation in binding energy. In this case a singleton proton not only does not add to the binding energy it detracts from the binding energy by 3.68 MeV. This negative value is highly statistically significant. A singleton deuteron adds only 8.15 MeV and a singleton neutron 7.605 MeV.

The effect of an additional singleton proton or neutron could vary with the size of the nucleus. Such an effect can be captured by including variables which are the products of the alpha value and #P and #N. The results for such a regression are:

B(P,N) =
38.28163α - 0.16349α² + 8.10163#D - 1.71832#P - 0.1895α#P + 7.89910#N - 0.00957α#N
  (0.0978)       (0.0030)       (0.5430)    (0.3016)    (0.08515)     (0.1054)    (0.0311)
     [390.6]     [-53.9]     [14.9]      [-2.2]    [-2.2]     [77.5]    [-3.07]
R² = 0.9991

The inclusion of nuclear size adjustments for the effects of an additional singleton proton or neutron does not improve the overall R² value significantly, but improved the fit for the light nuclei. Again the singleton proton was the least significant variable.

The conclusion to be drawn from the results is that the mass deficit (binding energy) for a nucleus comes from the neutrons it contains. The effect is enhanced by the neutrons' combinations with protons but protons by themselves do not contribute significantly to the binding energy.

The dissociation of a neutron would account for only 0.7823 MeV. The potential energy loss in the formation of a proton-neutron pair accounts for about 2.3 MeV. The big factor in the mass deficit is the potential energy loss in the formation of an alpha particle, about 28 MeV, or about 7 MeV per nucleon. The internal structure of the alpha particle is not known and so this relatively high binding energy for such as light nuclei is another enigma of the physics of nuclei.

In addition to the mass deficit accounted for by the alpha particles themselves there is apparently some configuration of these alpha particles with in the nucleus that accounts for an additional increment in the mass deficit. For more on this topic see Nuclear Structure.

The Measurement of the Mass of the Neutron

The mass deficits are based upon the estimated mass of the neutron. The mass of the neutron is inferred from the measured masses of the deuteron, proton and electron and the mass equivalent of the energy of the gamma ray involved in the dissociation of deuterons. That is to say, the mass of the neutron is calculated as the mass of the deuteron less the mass of the proton and electron and less the mass equivalent of the energy of the gamma ray. This presumes that the energy of the gamma comes exclusively from the change in mass in the formation or dissolution of a neutron. A different allocation of the energy changes would produce a different estimate of the mass of the neutron and a different set of mass deficits for nuclei.

The published mass deficit for the Berylium 5 nucleus which contains 4 protons and 1 neutron is -0.750 MeV. This suggests that the mass of the neutron has been underestimated by at least 0.750 MeV. The figure of 0.750 MeV is about 1½ electron masses. This would mean that the neutron has a mass 4 electron masses greater than the proton instead of 2½ electron masses. Thus in the formation or dissolution of a deuteron the energy change would be (2.22457+1.5(0.511))=2.991 MeV, of which 3/4 goes into the energy of the gamma ray and the other 1/4 goes into a net change in potential and kinetic energy. Thus, an error in the neutron mass would be accounted for by the unverified assumption that the energy of the gamma ray represents 100% of the change in energy for the formation or dissolution of a deuteron. It is notable that the fraction going into the gamma ray energy is a simple fraction, 3/4. In atomic transitions the energy of the photon for an electron transition is 1/2 of the total energy change.

A statistical analysis of the binding energies recalculated on the basis of a neutron mass 1½ electron mass greater would result in the coefficients for those variables involving neutrons being larger by an appropriate multiple of the error in the neutron mass. This would make the difference between the effect of neutrons and the effect of protons larger by the amount of error in the neutron mass.

This is a matter that bears further investigation. Meanwhile however it may safely be concluded that the enigma of the mass deficits of nuclei has part of its explanation in the following sources:

When the binding energies of nuclides which could contain an integral number of alpha particles are compared with the binding energy which the alpha particle would contain there is an excess. This excess increases systematically with the number of alpha particles as shown below.

For the change for the range 0 to 2 alpha particles the increase in excess binding energy is essentially zero. From two alpha particles to fourteen the increase averages 7.3 MeV per alpha particle and from fourteen to twentyfive the increase averages 2.7 MeV per alpha particle.

(To be continued.)

For an alternate explanation of the mass deficits of nuclei see the Nature of Mass. For an implementation of the theory see Spectrum of Two-Particle Systems.

HOME PAGE OF applet-magic
HOME PAGE OF Thayer Watkins