San José State University

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 The Structure and Binding Energy of the Alpha Particle, the Helium 4 Nucleus

One of the major problems of the physics of nuclei is the explanation of the high binding energy of Helium 4 nucleus, the alpha particle. A deuteron (proton-neutron pair) has a binding energy of 2.225 million electron volts (Mev) but an alpha particle has a binding energy of 28.3 Mev. (The higher the binding energy of a particle the more energy is required to break it up into its constituent parts.)

If an alpha particle is considered to be the combination of two deuterons then there is a gain in energy of 23.85 Mev from putting them together. From one perspective the creation of an alpha particle is the same as the creation of a deuteron but with double weight constituent particles. For the analysis here the difference between the masses of the proton and the neutron will be ignored; i.e., the mass of the neutron will be taken to be the same as the mass of the proton.

## The Orbit Radius of Deuterons in the Alpha Particle

The basis for the following analysis is given elsewhere for the deuteron. The orbit radius r1 is found through solving the following transcendental equation

#### ez = ε + ε²μ²z² or, equivalently z = ln(ε) + ln(1+εμ²z²)

where z=2r1/r0. The parameter r0 is the range parameter for the nuclear force and is equal to about 1.5×10−15 m. The parameter ε is (γ/l)² where γ=H/(hc) and l is an integer. The angular momentum of the constituent particles is hc)l. The parameter γ is in the nature of a fine structure constant for the nuclear force is equal to 1.59, in contrast to the fine structure constant for the electrostatic force of about 1/137.The parameter μ equal to mc²r0/(2hc) where m is the mass of one of the constituent particles. If k is the ratio of the mass of one of the constituent particles to the mass of the proton then μ=1.11656k.

The value of ε is 2.53566/l². The allowable values for l are given by the condition

#### l ≤ k(mpc²r0/(2hc)) which evaluates tol ≤ 1.778k

Thus for the alpha particle where k=2, l≤3.556 which means the allowable values for l are 1, 2 and 3. Any subset of these values could be excluded on the basis of other criteria.

The parameters for the transcendental equation and its solution are presented below.

 Angular MomentumNumber l ε μ εμ² Solutionz r1/r0 1 2.53566 2.2331 12.6449 7.50 3.75 2 0.63392 2.2331 3.1612 2.78 1.39 3 0.28174 2.2331 1.4050 negativesolution

The value of the potential energy and hence the binding energy is given by

#### V(r1) = (H/r0)(∫ν∞(e−s/s²)ds)

where ν=r1/r0

For l=2, ν=1.39 for this value r1=1.0425 fermi and separation of the deuterons in the alpha particle is 2.085 fermis. The binding energy of the deuterons V(r1) evaluates to

#### V(r1) = [5.03776×10−26/(7.5×10−16)](0.061195) V(r1) = 4.11×10−12 joules = 25.7 Mev

This gives a total binding energy for the alpha particle as 25.7+2(2.225)=30.15 Mev. Since the actual value is 28.3 Mev this computed figure is in error by 6.5 percent. But the above analysis does not take into account the effect of the electrostatic repulsion of protons.

The binding energy of tritium, H 3, is 8.481 Mev, but that of He 3, which also contains three nucleons but two protons instead of two neutrons, is 7.718 Mev. Thus the electrostatic repulsion of the two protons reduces the binding energy of He 3 by 0.763 Mev or 9 percent. There would be a similar reduction for the alpha particle. A reduction of the calculated binding energy of the alpha particle by 9 percent would reduce the figure from 25.7 Mev to 23.4 Mev and give a total binding energy for the particle of 23.4+2(2.225)=27.8 Mev which differs from the actutal value by only 1.6 percent. A reduction of the calculated value by 0.763 Mev would reduce the figure from 25.7 Mev to 24.9 Mev which would be a total binding energy for the particle of 29.4 Mev which is 4 percent too high. The proper correction for the proton repulsion in the alpha particle would not be the same as for the He 3 nucleus but the figures given above indicate that such a correction would likely bring the calculated bining energy for the alpha particle fairly close to the actual value.

There is a definite possibility that the conventional estimate of the mass of the neutron is too low because it is based upon the presumption that the mass deficit of the deuteron is exactly equal to the energy of the gamma ray involved in the association or disassociation of the proton and neutron constituting a deuteron. When the change in kinetic energy is taken into account and the estimated mass of the neutron is adjusted the mass deficit of the alpha particle turns out to be 30 MeV. This makes the figure cited above of 29.4 MeV only 2 percent low. For more on the adjusted estimate of the mass of the neutron see Two Particle Spectrum.

Now back to the line of analysis presented above. The transcendental equation apparently does not have a solution for the i=3 case. The solution for i=1 corresponds to an energy of only 0.48 Mev and is obviously not in the ball-park. A reduction of the binding energy by 0.763 to allow for the electrostatic repulsion of the two protons means that the case of l=1 is likely unstable and would not be the one state observed in nature.

The analysis indicates that the extraordinarily large binding energy of the alpha particle can be readily accounted for by the model of the nuclear force presented here.

(To be continued.)