12. HELIUM BURNING AND THE C-N CHAIN

The p-p chain has produced various hydrogen and helium nuclei and would seem natural that the process of build up by one proton or neutron at a time. However there is a gap that must be crossed. Extending the p-p chain reactions only result in products that decay back down to He-4. So here we are in a first-generation star with a lot of He-4 around and running out of H-1.

What happens next is that the He-4 core contracts and gets hotter and the H-1 shell expands and gets cooler. The star becomes a red-giant. At some point the He-4 starts to interact in a process called helium burning. In effect the star has a new coal for the fire. Helium Burning starts with a series of reactions that are called triple-alpha process where alpha is another name for the He-4 nucleus. The process can continue with the addition of He-4 producing in turn O-16, Ne-20, Mg-24, Si-28, etc. However, the process is limited by reaction rate above O-16. Then after most of the helium is gone, a similar process continues at higher temperature at least though Ca-40 by using helium from the breakup of neon.

It appears that ten of the elements up though at least Ca-40, and may be beyond, could be considered to be multiples of He-4. This may well be the case which means that beryllium-8 which is not stable is an exception among the low mass elements. It could also mean, and will be assumed here, that a nucleus which is stable with regard to alpha decay can have only one spin 0 helium-4 in it, for whatever reason, representing a sort of EPSM Helium-4 exclusion principle.

The triple-alpha process has been proposed to either occur in a single step by the reaction of three He-4 at once or in a two step process with Be-4 as a intermediate. It can be readily seen in Figure 2-13 that three He-4 EPSMs could form a Pi Layer (PiL) as the Carbon-12 EPSM. This is the same ring structure shown in Figure 2-11. It appears that this structure should be very stable. It is a beautiful EPSM and would make a beautiful carbon-12 EPSM; however, it does not have a free negative dimensionalities which will be needed in the Carbon-Nitrogen (C-N) cycle which will be looked at later.

Figure 2-13: Shows the Pi Layer can be formed from 3 Helium-4 EPSMs. A possibility for carbon-12 EPSM with Spin 0 and even parity.

A second possibility for a carbon-12 EPSM is shown below in Figure 2-14. In this case there are two layers; a Si Layer (or SiL) and a He-4 layer. The SiL is formed from the two He-4 EPSMs shown in Figure 2-7. This carbon-12 has room to attach a proton in the Carbon-Nitrogen (C-N) cycle.

Figure 2-14: Another possible Carbon-12 EPSM formed by a SiL and a He-4.

THE CARBON-NITROGEN CHAIN

After the hydrogen and helium burning processes in the stars have produced carbon-12 another process called the carbon-nitrogen or C-N cycle can also produce helium-4. In addition, the C-N cycle is suppose to account for carbon-13, nitrogen-14 and nitrogen-15 production. The cycle starts with carbon-12 and ends and with carbon-12. The use of an ESU in positron decay has been discussed several times. Recalling that the two positrons will annihilate two electrons, it is seen that each C-N cycle turns four H-1 and two electrons into one He-4 just as the proton-proton cycle does. Leakage from the cycle is said to account for the other stable carbon nuclei and the other stable nitrogen nuclei.

₆C¹² + ₁H¹ > ₇N¹³ + g

₇N¹³ + ESU > ₆C¹² + b⁺ + n

₆C¹³ + ₁H¹ > ₇N¹⁴ + g

₇N¹⁴ + ₁H¹ > ₈O¹⁵ + g

₈O¹⁵ + ESU > ₇N¹⁵ + b⁺ + n

₇N¹⁵ + ₁H¹ > ₂He⁴ + ₆C¹²

Based upon the carbon-12 EPSM of C-12 (1He,1SiL) there is only one place for the first proton in the C-N cycle to go and that is on the negative free dimensionality of the helium-4 to form nitrogen-13 as shown in Figure 2-15. The proton cannot go onto the Si Layer because a Si Layer cannot accept a proton.

Figure 2-15: Nitrogen-13 EPSM. As seen in Figure 2-14, the proton has only one place to attach to on the carbon-12 EPSM.

Next through beta + decay an "electron" is added to the nucleus to obtain carbon-13. It would seem logical to add the electron at a point where two positive dimensionalities are close together as shown in Figure 2-16. The C-N cycle then requires another proton to be added to obtain nitrogen-14 as shown in Figure 2-17.

Figure 2-16: Carbon-13 EPSM consisting of a SiL ESPM and a He-5 EPSM.

Figure 2-17, below, is a great looking nitrogen-14 in that it looks extremely stable. However, it is a dead-end for the C-N cycle because a proton cannot be added as required by the next step.

Figure 2-17 : Nitrogen-14 EPSM.

There are several alternatives EPSM schemes that can be examined with regard to the C-N cycle all of which may be possible at the 10 million degree Celsius at which the reactions are occurring in stars. One of these EPSM alternative C-N cycles which tracks the nuclei parities and spins will be examined next. Figure 2-18 shows a carbon-12 EPSM formed from three He-4 with spin 0 but without the formation of a Si Layer. Figures 2-19 and 2-20 show nitrogen-13 EPSM and carbon-13 EPSM respectively formed by the first proton addition and the first beta(+) decay.

Figure 2-18: Carbon-12 EPSM formed from three He-4 with spin 0. The assumed starting point for the C-N cycle as described in the text.

Figure 2-19: Nitrogen-13 EPSM. A proton has been added to the carbon-12 EPSM shown in Figure 2-18.

Figure 2-20: Carbon-13 EPSM. An electron by beta(+) decay has been added to the nitrogen-13 EPSM shown in Figure 2-19.

The addition of the second proton results in an activated spin 0 nitrogen-14 EPSM as seen in Figure 2-21. Next, the nitrogen-14 EPSM in Figure 21 splits into a nitrogen-14 EPSM with spin 1 (sum of spins of He-4, H-2, and 1SiL EPSMs) and an even parity as shown in Figure 2-22.

Figure 2-21: Activated nitrogen-14 EPSM. It is assumed that the addition of the proton develops conditions that result in a hydrogen-2 splitting off.

Figure 2-22: Nitrogen-14 EPSM with spin 1 and even parity after the hydrogen-2 EPSM broke off.

Figures 2-24 and 2-25 show the addition of the third proton and the second beta(+) decay resulting in oxygen-15 and nitrogen-15.

Figure 2-23: Oxygen-15 EPSM after a proton was added to the nitrogen-14 EPSM shown in Figure 2-22.

Figure 2-24: Nitrogen-15 EPSM after a beta(+) decay of the oxygen-15 EPSM shown in Figure 2-23.

Finally, the addition of the last proton as shown in Figure 2-25 results in two spin 0 helium-4 being in the same nucleus which results in an alpha decay by the EPSM helium-4 exclusion principle. The alpha decay of the oxygen-16 in Figure 2-25 results in the carbon-12 EPSM shown in Figure 2-14.

Figure 2-25: Unstable oxygen-16 EPSM. One of the spin 0 He-4 layers will leave the nucleus by alpha decay resulting in a He-4 EPSM and a carbon-12 EPSM.

BACK TO HELIUM BURNING

The Helium Burning process can continue beyond C-12 with the addition of He-4 producing in turn O-16, Ne-20, Mg-24, Si-28, S-32, Ar-36, Ca-40, Ti-48, Cr-52, and Fe-52. Each of these nuclei can enter into other reactions to form other nuclei. Also, Helium burning can add a He-4 after another reaction had occurred.

Oxygen-16, O-16, 8O16 (99.76%) (spin 0, parity +1)

The oxygen-16 EPSM is O-16 (2SiL). The products, nitrogen-14 (Figure 2-22) and helium-4, of the reaction that occurs when oxygen-16 is bombarded with H-2 add support to this configuration.

Neon-20, Ne-20, 10Ne20 (90.48%) (spin 0 , parity +1)

The straight forward addition of a helium-4 to the oxygen-16 EPSM (2 SiL) would lead to Ne-20 EPSM (1He,2SiL).

Magnesium-24, Mg-24, 12Mg24 (78.99%) (spin 0 , parity +1)

Magnesium-24 is the result of the helium burning of neon-20 or carbon burning in stars. It is reasonable that a helium-4 should be added to the neon-20 EPSM to obtain the magnesium-24 EPSM Mg-24 (3SiL). The alternative configuration of Mg-24 (1He,1SiL,1PiL) is also feasible.

Silicon-28, Si-28, 14Si28 (92.23%) (spin 0 , parity +1)

The silicon-28 EPSM could be Si-28 (1He,3SiL) or Si-28 (2Si,1PiL).

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