Alpha cluster and alpha separation energy

The alpha separation energy is the energy to add int a nucleus, so that it will break up into an alpha-particle and the rest of the nucleus. If we define the mass of a nucleus with mass number A and charge number Z as , the alpha-separation energy is,

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The following plot the nuclides chart for alpha-separation energy,

The chart can be divided in 2 regions. In the upper region, nuclei have negative alpha-separation energies, i.e. nucleus gives out energy when emitting an alpha particle, thus, they are alpha-emitter. In the lower region, the alpha-separation energies are positive. And we can see that there are some local minimum for the .

According the alpha-decay theory, alpha-particle should be formed inside a nucleus before it tunnels through the nuclear potential and get out. the formation of the alpha-particle is described as the preformation factor.

It seems that, the alpha-separation energy, somehow, relates to the preformation factor.

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The alpha cluster is studied for the many light nuclei, particularly on the 8Be, 12C, 16O, 20Ne, 24Mg, 28Si, 32S, 36Ar, 40Ca, and 44Ti. [Ref??]

From the above plot, it seems that the alpha-separation energies have no correlation with the magic number, and also no correlation with the Z=even. In a naive imagination, alpha-cluster could appear at all Z=even, A = 2Z nuclei. A trivial example is the the 8Be, its alpha-separation energy is -0.09 MeV and it will decay or split into 2 alpha particles. In 12C, the Hoyle state at 7.7 MeV is a triple-alpha state, note that MeV.

When we use the shell model to look at 8Be and 12C, we will found that the alpha-cluster is “not” making any sense. The protons and neutrons occupy the 0s1/2 and 0p3/2 orbitals. Since 9Be is stable and ground state spin is 3/2. A 8Be nucleus can be obtained by removing a 0p3/2 neutron. We may guess that, the 4 nucleons at the 0s1/2 orbital, which is an alpha-particle, somehow, escape from the 8Be, leaving the 4 0p3/2 nucleons behind. And the 4 0p3/2 nucleons “de-excite” back to the 0s1/2 shell and becomes another alpha-particle. But it seems that it does not make sense. Also, How to use shell model to describe the triplet-alpha cluster?

Above is the the alpha-separation energy for light nuclei. We can see that, there are few local minimum around 8Be, 20Ne, 40Ca, and 72Kr. Near 20Ne, the 18F, 19F, and 19Ne are having small alpha-separation energies, relative to the near by nuclei. For 20Ne, it could be understand why the alpha-separation energy is relatively smaller, as the 20Ne can be considered as 16O core with 4 nucleons in sd-shell. And 20Ne is well deformed that, there is a chance all 4 nucleons are in the 1s1/2 and form a quai-alpha particle. But the situation is a bit strange for 19Ne, 19F, and 18F. In order to form an alpha-cluster, or a quasi-alpha-particle, one or two nucleon from the p-shell has to excited to sd-shell. But the situation may be even more complicated. For some heavy alpha emitter, the s-orbital nucleons are tiny fraction compare to the rest of the nuclei, so, how the alpha-particle is formed before the decay is still unknown. This suggests that the involvement of s-orbital is not needed in alpha formation.

, and there are many states around 4 MeV in 19F, I am wondering, one of these state is alpha cluster? If so, 19F(d,d’) reaction could excite those states and we will observed 15N + alpha. [need to check the data]. Similar experiment could be done on 18F, 19Ne, and 20Ne. If it is the case, that could provides some information on the alpha formation.

[need to check the present theory of alpha formation]

Analyzing power for proton elastic scattering from the neutron-rich 6He nucleus

DOI : 10.1103/PhysRevC.82.021602

This paper is based on the solid proton polarized target, and make improvement on the experiment. The proton polarization was monitored by NMR method and the absolute polarization is known. Therefore, the Analyzing power ( or the spin asymmetry) can be determined. The experimental result is different form the prediction of the t-matrix folding model.

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The 1st and 2nd paragraphs explain why the spin-asymmetry is important in nuclear research. a simple reason is, by manipulating the spin, we can reveal the spin-dependence of the nuclear-structure. For example, the orbital-spin coupling, which is the major factor in the nuclear potential.

the 3rd, 4th and 5th paragraphs talk about the solid proton polarization target and the advantage of operating under weak magnetic field. (already discussed in Here )

The 6th paragraph report the result from the 2003 experiment. and say that the unsatisfiable result on the spin-asymmetry.

The 7th to 11th paragraphs talk about the experiment procedures and conditions.

the 12th paragraph explains the angular distribution in center of mass frame of the differential cross section. the data of 6He is similar to the 6Li for angle less then 50 degrees, which is the forward angle in the lab frame. at the backward angle or angle more then 50 degree in C.M. frame, there are some different and it may be due to the halo structure of 6He.

the next paragraph turns the focus on the analyzing power, which can be determined at this time. the data is very different from 6Li data. the t-matrix folding method predicted that the analyzing power should be positive but the experiment result is different. this indicated that there is other mechanism to explain the nuclear structure of 6He.

The t-matrix stands for transition matrix.

the other mechanism may be the g-matrix folding model…..i am sorry, i don’t understand the following….

alpha decay

different decay cause by deferent mechanism, we first start on alpha decay.

i assume we know what is alpha decay, which is a process that bring excited nucleus to lower energy state by emitting an alpha particle.

The force govern this process is the strong force, due to the force is so strong, the interaction time is very short, base on the uncertainty principle that large change in energy leads to short time interval. however, the observed alpha decay constant is about 1.3 × 10¹⁰ year, which is about the age of our universe. That’s why we still able to find it at the beginning of nuclear physics : discovery of radioactive matter.

The reason for such a long decay time is due to the Coulomb barrier of the nuclear potential. since the proton carry positive charge, thus. it creates a positive potential wall in the nucleus. that potential not only repulse proton from outside but also the proton from inside which try to get out. thus, the inside protons are bounded back and forth inside the nucleus. due to the momentum carried by the protons, it has frequency 6  × 10²¹ per sec.

Due to the Quantum tunneling effect, the probability of tunneling is 4 × 10^-40. which is a very small chance. But , don’t forget there are  6  × 10²¹ trails per sec. Thus, the chance per sec is 2.4 × 10^-18 . and the mean life time is inverse of the probability, thus it is approx 1.3 × 10¹⁰ year.