To Mr H. Yan

We can only compare the behavior of matter if the conditions in which it "evolves" are comparable.

Indeed, one cannot compare a nucleus at "normal" temperatures with a neutron star where extreme conditions prevail.

I quote from a documentation:

“Inside neutron stars matter is ultra-dense (10 to 20 times denser than atomic nuclei) and ultra-hot (100 billion Kelvin). At such extraordinary densities, atoms themselves are 'collapse and atomic nuclei are so tightly packed together that new states of matter are formed, including exotic condensates of elementary particles and the possible formation of a plasma made up of the most basic building blocks. fundamentals of matter, quarks ".

In 2012, Spyrou et al. reported the observation of the emission of dineutron (n2) during the decay of Be-16. The dineutron splits into two neutrons at the moment of emission. However, they found that the two neutrons flew in almost the same direction. If the two neutrons had been emitted separately, the direction would have been more inconsistent. However, It does not exclude the possibility that, if they were clustered in the nucleus, they could still be correlated even though they were emitted separately.

Article First Observation of Ground State Dineutron Decay: Be 16

To Kengo Shibuya

Thank you for your remark but the Dineutron is not a bound particle.

From wikipedia: https://en.wikipedia.org/wiki/Neutronium

"Dineutron: The dineutron, containing two neutrons, was unambiguously observed in 2012 in the decay of beryllium-16. It is not a bound particle, but had been proposed as an extremely short-lived resonance state produced by nuclear reactions involving tritium".

We will try to give a simplified answer (so that everyone can understand).

The Pauli exclusion principle states that 2 identical particles (which are fermions, i.e. with half odd integer spin) cannot be in the same quantum state.

Take, for example, oxygen 16. This nucleus has 8 neutrons (fermions) and 8 protons (fermions). Each nucleon is in a quantum state with at a defined energy level.

(a) As a first approximation the lowest energy of oxygen 16 is the sum of 16 lowest energy levels (each nucleon on each level).

(b) However, due to the fact that neutrons and protons are not identical particles, the Pauli exclusion principle makes it possible to put a proton and a neutron on the 8 lowest levels (i.e. a pair neutron-proton on the same energy level) and the energy will be 2X 8 lowest levels values.

It is then not difficult to deduce that solution (b) leads to a lower energy than solution (a). As a physical system is always heading towards a lower (internal) energy state it is obvious that the "8 neutrons + 8 protons" system will always be more stable than the 16 neutron or 16 proton system (for extreme cases).

So the Pauli exclusion principle and the principle of minimum energy (internal energy of nucleus) favors the formation of nuclei with N = Z. Indeed for the lightest nuclei, the stable nuclei follow the N = Z "rule." The fact that heavier stable nuclei have N greater than Z is explained by the fact that the Coulomb interaction tends to "beat" the "pure" nuclear interaction, because the latter is short range. For large nuclei, nucleons far from each other no longer "feel the nuclear interaction while they" feel "the Coulomb interaction. For heavier and heavier nuclei, it is therefore necessary to add more and more neutrons (more nuclear force and less Coulomb force) to "stabilize" the system (nucleus). It is for this reason that the beta stability valley (region of stable nuclei in thee (N,Z) chart with respect to beta radioactivity) deviates more and more from the N = Z rule as the nuclei get larger.

Note: This explanation is not valid for neutron stars because we have here a medium brought to temparatures of billions of degrees Kelvin, the excitations of the nuclei are of another order of magnitude.

As the simplest example of the explanation given just above there is the deuteron (stable n-p system) while the di-proton (p-p system) or di-neutrons (n-n system) do not exist, i.e. do not form bound states.

I just found in ResearchGate a discussion on this simple example with similar arguments but "more mathematical" and "less pedagogical" (not for the layman)

https://www.researchgate.net/post/As-we-know-there-is-no-stable-bound-state-for-ppproton-proton-and-nnneutron-neutron-okay-Why

It depends on your definition of a nucleus. You may apply tremendous pressures on matter to cause electrons to clash with protons, resulting in just neutrons, but you no longer have a nucleus.

B. Mohammed-Azizi

I agree with Mr Khaled Ali and to summarize, I quote a piece of article read in the internet:

"At the heart of the atom, the nucleus. In the nucleus an excessively dense matter, nuclear matter. At normal temperature and density, it is known to be made up of particles, protons and neutrons, themselves made up of quarks.

However, its structure and phases are still poorly understood. To explore the thermal and mechanical properties of this nuclear material, researchers are using collisions between nuclei caused by particle accelerators. At very high temperature or density, the particles release their constituents and the matter becomes a plasma of quarks and gluon. The discovery of this new state of matter was announced by CERN in Geneva at the beginning of 2000." (end of quote)

It is easy to see that outside of certain limits, i.e. "normales limits", nuclear matter no longer has the same behavior.

I think it is equal to the law of the same charges but here are the same particles

A similar topic close to the present question:

https://www.researchgate.net/post/Why_a_free_neutron_is_unstable

there's no physical reason for the non-existence of dineutrons or trineutrons or...A diproton may not exist due to the huge Coulonb repulsion, and their fusion in star core occurs through weak interaction indeed, which change the quark flavour inside one proton, turning it into a neutron and producing a deuteron, a positron and a neutrino, but with a very low probability

To S. Bartalucci,

I don't understand your reasoning. You say that there is nothing to prevent that nuclei made only of neutrons can exist. However, the scientific approach is the opposite of what you advocate. First we observe, then we try to explain what we observe. In the Isotope tables there are no nuclei made of only neutrons or only protons (simple observation). From there I asked the question concerning nuclei with only neutrons because the case is simpler than that of protons since apparently the nuclear force must bind the neutrons without the coulomb force counteracts such effect.

As for your explanation that the di-proton does not exist because the coulomb interaction prevents two protons from binding. In fact for a two di-proton system one must first consider the nuclear force which is much more important than the coulomb force. To accept your argument, one must assume that at short range, the coulomb force exceeds the nuclear force. This is not true.

Indeed, the di-proton does not exist but this is not due to your argument.

Kindly yours

you should inform yourself much better: dineutrons have been observed, but they don't give rise to stable particles

About diproton or He-2: Pauli's principle forces the two protons to stay in an antisymmetric state, so the normal fusion in s-wave may occur only if the two spins are antiparallel but in this case nuclear force becomes slightly repulsive (see Blatt-Weisskopf) and the nucleus becomes unstable

If we think of the atom, and not just separate nucleus, that is, nucleus and the electron, could answer be because there is a balance of electrical charge?

As we all know, positive proton and negative electron both have a positive electric charge equal to 1.6 x 10-19 C.

Therefore, it may be that this balance of charges makes nuclei need a positive charge.

To Filipe Wiltgen

Dear Colleague,

The answer you give is simple. By following your logic, the "necessity" of the global neutrality of the atom implies that it suffices to put as many electrons as protons (In this case, the number of neutron does not matter) to obtain a "good nucleus" which exists and which is stable. However, it is the nuclear forces which decide the existence and the stability of the nucleus. The number of electrons is the consequence of the one of the protons not the inverse.

If you reason only in terms of the global electrical neutrality of the atom, I propose an atom with a nucleus composed only of 2 protons (charge ++) and 2 electrons (charge - -) which realize the electrical neutrality (atom of charge 0). This atom is perfectly neutral. But this element (di-proton) does not exist. In other words it is not the 2 electrons in orbit that will "force" the 2 protons to stay together. A nucleus is composed of nucleons held together by the nuclear force, the electron does not influence the nucleons inside the nucleus

Besides the atom can perfectly lose its electrons (losing the balance of charges) without any consequence for the nucleus. Monoatomic ions are an example.

https://en.wikipedia.org/wiki/Monatomic_ion

In addition, the nucleus can even capture an electron from its atom (electron capture radioactivity for nuclei deficient in neutrons) to change into another nucleus and therefore into another atom. In this process an electron is "stolen" by the nucleus. The other nucleons have no influence in this process.

In fact, the dineutron has no bound state and this conclusion seems to be unanimous in the literature. Other aspects of "this particle" are more or less controversial. In any case the topic of the dineutron itself has already been discussed in a ResearchGate question, see the URL below:

https://www.researchgate.net/post/As-we-know-there-is-no-stable-bound-state-for-ppproton-proton-and-nnneutron-neutron-okay-Why

In the same order of ideas and arguments you can follow the discussion ( titled: Nuclear Physics: Hydrogen3 (Tritium) vs Helium3, a paradox?) at the url:

https://www.researchgate.net/post/Nuclear_Physics_Hydrogen3_Tritium_vs_Helium3_a_paradox

To Mr Mahfuz Siraz,

Thank you for your kind comment. In the same frame of mind, I will add that:

Neutrons need companionship, they can't live alone. A free neutron (outside of nucleus) has a half life of about 10.2 min. In the nuclei they do not like to be in the minority compared to the protons. In neutron stars they are billions. There, they are happy.

Dear Mohammed-Azizi

It is no possible that a nucleus formed only by neutrons exists, since nuclear force needs electrical charge to interact. This is the reason why a neutron without being bounded to a proton decays (beta decay) and in a nucleus with protons is stable. Moreover, a nucleus with only protons does exist: 2He, 3Li. Then: to have a nucleus with only neutrons is no possible because of the nuclear force interaction.

To Mr Osvaldo Moreira

I do not understand your answer which states that the nuclear interaction "needs" a charge in order to act. In your opinion, the nuclear interaction between 2 neutrons cannot exist.

Can you clarify your answer?

Dear Mohammed-Azizi

In a nucleus protons and neutrons are no more identifiable, that means, there is an interchange between them. This is how nuclear interaction acts. For example, in 3H, each nucleon is 1/3 proton and 2/3 neutron; it is similar to the linear combination of the neutral pion structure, for example. Nuclear interaction interchange charged particles with no charged particles. Since charge is a component of the nuclear interaction (but not the only one), you have nucleus with only protons, although they have a very short life, since there is no neutrons to interchange. In the case of two neutrons for example, since you do not have charge, it is no possible to exist, even with a short life: you do not have protons to interchange and also you do not have charge.

To Mr Osvaldo Moreira

You say that the charge is a component of the nuclear interaction. In which book did you read that?

The charge independence of nuclear force is well known and is a widely accepted property. This mean that the nuclear strong force "do not distinguish" between the two types of nucleon. You contradict this assertion.

Thus, from wikipedia: https://en.wikipedia.org/wiki/Nuclear_force

The nuclear force is nearly independent of whether the nucleons are neutrons or protons. This property is called charge independence.

The fact that n-n interaction and p-p interaction do not form a bound state is only due to Pauli exclusion principle not to the charge of the nucleons.

For those who say neutron stars are "giant nuclei". They must know that the cohesion inside the neutron star is ensured by the gravitational forces and not the nuclear forces as in an atomic nucleus. These two systems, although both made up of nucleons, are dissimilar.

B. Mohammed-Azizi I will second your opinions here. Interesting question for sure.

I disagree with professor M.M. Mahfuz Siraz

Dear Colleague,

I think Mr Mahfuz Siraz was obviously joking.

The repulsion between like charges is too strong, that's why.

Mr Jose Risomar Sousa, I don't know if you read the title correctly.

Assuming a nucleus composed only of neutrons, Coulomb repulsive forces do not exist.

Dear B. Mohammed-Azizi,

- firstly a lot of thanks for the papers recommendations, and to

“Nuclear Physics: Why are there no nuclei formed only of neutrons ?”

- the answer is on one hand rather simple, on another is too hard. The simple answer is: that is so, because of in this case Matter, i.e. particles, substances, bodies, cosmological objects, etc., is as it is,

- i.e. to compose of essentially neutral atoms, that are the main constitutions of substances, etc., it is necessary to compose a system of ± electric charges, and if that would be enough, then atoms would consist of protons and electrons only; thus – since if nuclei would be formed only of neutrons, then atoms wouldn’t exists at all, and the thread question really would be – why neutrons exist at all?

However really, because of the electric repulsion, whereas the fundamental Natural Electric force has action range much larger than Strong Force, even di-proton stable system doesn’t exist, and even for He nucleus it is necessary to have at least one neutron to enhance and to spread Strong Force action distance between nucleons so, that He nuclei, and He atoms are stable.

The hard question is “let it be so, but for what reason Matter is as it is?” – the answer on this question is fundamentally outside physics; more see, say, section “Cosmology” in https://www.researchgate.net/publication/355361749_The_informational_physical_model_and_fundamental_problems_in_physics , https://dx.doi.org/10.20944/preprints202110.0453.v1

Cheers

Dear colleague shevchenko,

I could have asked a similar question, why an atomic nucleus would not be formed only of protons. The argument that the atom would not have been neutral no longer holds, since in this case it is possible to ensure the neutrality of the atom.

In fact, having a nucleus made only of protons is a more difficult hypothesis because in this case the coulomb repulsion forces would make things more difficult compared to the case of a set of neutrons.

According to the liquid drop model with the quantum asymmetry term, the minimum energy of a nucleus would be obtained for N=Z. The fact that N is greater than Z would come from the fact that the coulombic force (which destabilizes the nucleus) is longer than the nuclear force and that more neutrons are needed to restabilize the system by introducing more binding nuclear interaction.

Dear B. Mohammed-Azizi . Neutrons are required for the stability of nuclei, with the exception of the single-proton hydrogen nucleus. Neutrons are produced copiously in nuclear fission and fusion. They are a primary contributor to the nucleosynthesis of chemical elements within stars through fission, fusion, and neutron capture processes. See the link: https://en.m.wikipedia.org/wiki/Neutron

Dear Mr Wazwaz ,

1) Since you say that "neutrons are required for the stability of nuclei", so why does a nucleus composed only of neutrons not exist?

2) The nuclear force being attractive and the Coulomb force being repulsive (protons), I do not see any inconvenient in that the nuclei made only of neutrons do not exist.

Obviously, we know that such nuclei do not exist. But the question is why?

3) This is not a simplistic question as your answer suggests (I myself am a theoretical nuclear physicist).

Dear colleague B. Mohammed-Azizi,

“…I could have asked a similar question, why an atomic nucleus would not be formed only of protons. The argument that the atom would not have been neutral no longer holds, since in this case it is possible to ensure the neutrality of the atom.…..”

- the neutrality of atoms isn’t some atoms’ arbitrary purely “own” property, that is necessary, again, using rather strong Electric Force to compose wide diversity of material objects – molecules, bodies, etc., including large cosmological objects, till extremely weak Gravity Force becomes to dominate, and so Matter is as it is. However the problem – why Matter is as it is, i.e. why in logical system Matter just four Forces exist, and have just such strengths and Forces mediators as Forces have, is outside physics.

Including Matter has became just this Matter in the nucleosysntesis, which mostly was determined by Strong Force, however very essentially by Weak Force also, which causes neutron’s decay with energy 0.782 ± 0.013 MeV.

At that really physics hasn’t clear and rational enough theories of Forces; besides, in certain extent, of Electric and Gravity Forces [Newton gravity law, not, of course the GR], however even in the case of rather easily study-able Electric Force, official physics theories are based on postulates that Forces interactions are mediated by some “virtual” mediators, i.e. here “virtual photons”.

Strong Force is mediated by “virtual gluons” inside most of particles, and “virtual π-mesons” at interactions of protons and neutrons in nuclei; though π-mesons have masses well more 100 MeV, whereas the binding energy of the baryons in nuclei is ~7 MeV.

That looks as rather strange – it seems for any normal human quite clear, that in Matter real interactions are mediated not by virtual but real mediators, and the mediation must not so essentially violate the energy conservation law; etc.

However the theories work rather adequately in applications in practice, and the problem of at least dineutron is evaluated on the level, which is consistent with experiments, which show, that free di- and more-, neutron systems decay in Strong Force times, i.e. ~10-20s, and possibly exist only in some exotic nuclei baryons configurations.

Including in existent theory, say, the dineutron binding energy drastically differs from, say the n+p system 2.24 MeV, and is estimated as Qnn = 0−14 KeV [ https://arxiv.org/pdf/0904.1807.pdf ], what is well lesser than the neutron decay energy above.

More fundamentally reliable theory of Forces, including Strong Force, which will be based on real mediators and real interactions, are, however, possible, but for that is necessary to understand firstly – what is Matter, and everything in Matter, including “particles”, “Forces”, etc.,

- what is, in turn, possible only in framework of the Shevchenko-Tokarevsky’s informational physical model, https://www.researchgate.net/publication/354418793_The_Informational_Conception_and_the_Base_of_Physics ; https://arxiv.org/abs/0707.4657 , where, though, now only initial models of Gravity and Electric Forces are proposed,

- however what is for sure is that the theories will be based on that Matter’s spacetime is the fundamentally absolute, fundamentally flat, and fundamentally Euclidian, [5]4D spacetime with metrics (cτ,X,Y,Z,ct), where the Matter’s utmost fundamental base – the [5]4D dense lattice of [5]4D fundamental logical elements [FLE] is placed; and practically everything in Matter, including particles and Forces, is determined by the logical construction and properties of FLEs.

But now about what are the particles theories/approaches in official physics see SS comments to series of corresponding theories/approaches in https://www.researchgate.net/project/Creative-Particles-of-Higgs-or-CPH-Theory/update/6192445cf5675b211b1005c6and in links in this link.

Cheers

It seems that the simplest answer to this question is that the configuration which gives the minimum energy (and therefore the most stable) for a nucleus is that given by the Pauli exclusion principle which leads to a "neutron mixture. -protons "for which N = Z (N> Z for heavy nuclei is easily explained due to the long range of the Coulomb force compared to the nuclear force)

In addition to all the interesting answers Prof. B. Mohammed-Azizi, I know from the Landau and Lifshitz statistical physics course, that there are neutron starts, right? (they can be magnetic or not magnetic); So there is another delicate question as you predicted in the thread: there are neutron stars but not neutron nuclei?

Feliz Navidad, Best Regards.

Article The Equation of State of Neutron-Star Matter in Strong Magnetic Fields

How do we know there are not clumps of neutrons? The question seems to suggest "nuclei" as in of atoms around which electrons circulate. The STOE suggests the ruling force is magnetic. This force decreases by inverse cube of distance. this is the rate of magnetic force decrease perpendicular to the center of N-S poles. This force is much stronger than the coulomb force. The neutron in combination with a proton provides a unique magnetic field which hold the atom together. But combinations of neutrons are possible but the unique magnetic field is not present for an atom to be formed. Simplistically, consider a neutron is a proton with an electron attached/included. But the N-S pole structures are neutralized very rapidly so can be only a few (2-4). The lack of the bonded electron between 2 protons in atomic nuclei causes a more stable, unique magnetic field - thus atoms.

A nucleon-nucleon pair in a singlet spin state has a negative scattering length, so it cannot give rise to a bound state (see e.g. Segrè, Nuclei and particles, p.458). Pauli's exclusion principle forbids n-n and p-p 3S1 states (total wave function would be symmetric), while the n-p state is allowed, of course.

"spin state" = model of the effect of a particle's magnetic field which is demonstrated in the Stern-Gerlach experiment.

Happy New Year 2022 to all RG colleagues.

There have been no experimental outcomes found dineutrons because weak nuclear force would form at the existence of dineutrons. Given that the dineutron isn't bound, the diproton is guaranteed not to be bound.

Shuraik Kader

We know that nuclear forces are strong and attractive at short range (although repulsive at very short range).

B. Mohammed-Azizi Yes but that "short range" is need to be defined and compared relevant to "required range" for the existence of strong nucleus

Shuraik Kader

"Yes but that "short range" is need to be defined"

1) In nuclear physics the unit of length (typical of nuclear interaction) is the fermi 1fm=10-15m. Two nucleons feel the nuclear interaction if they are a few fermi apart. It is a basic knowledge in nuclear physics.

"and compared relevant to "required range" for the existence of strong nucleus"

2) I have 45 years of seniority in nuclear physics and the tone in which you speak to me seems to indicate that you know more about nuclear physics than I do, but you seem very young and nuclear physics does not seem to be your specialty.

What gives you so much confidence?

B. Mohammed-Azizi Thanks for your information

The free neutrons are instable and decay in 15 minutes. Two neutrons will decay almost immediately as well (see refernces below). Protons are lighter and stable. In stable nucleus occurs the dynamic equilibrium between continuously converting heavy particles. This explanation was given in the 70th in university lectures "General Physics". May be this explanation in outdated now.

The actual state of the experimental research of nuclei with low number is summarized at: Probing Few-Body Nuclear Dynamics via 3H and 3He (e,e′p)pn Cross-Section Measurements

R. Cruz-Torres et al. (Jefferson Lab Hall A Tritium Collaboration)

Phys. Rev. Lett. 124, 212501 – Published 26 May 2020

The actual state of the understanding of hypothetical short-leaving dineutron particles was reported in the above sources.

[astro-ph/0312388] The Effect of Bound Dineutrons upon BBN (arxiv.org)

A Method for Unambiguous Detection of a Hypothetical Bound Two-Neutron System | Few-Body Systems (springer.com)

The other hypothetical particles are tetraneutrons and octoneutrons:

Phys. Rev. 137, B345 (1965) - Further Evidence for the Nonexistence of Particle-Stable Tetraneutrons (aps.org)

Gerald Rosen, Are dark matter WIMP's of mass 8.6 GeV octoneutrons? (m-hikari.com)

Eventually the tetraneutrons have somewhat considerable existence time before decay.

“…The free neutrons are instable and decay in 15 minutes. Two neutrons will decay almost immediately as well (see refernces below). …. In stable nucleus occurs the dynamic equilibrium between continuously converting heavy particles. …". May be this explanation in outdated now.…..”

- this explanation isn’t in outdated now, however really that isn’t explanation since doesn’t explain – why and how neutrons in nuclei are stable because of “the dynamic equilibrium between continuously converting heavy particles” – so that a nucleus at the “converting” remains be always the same, i.e. the system that contains the same numbers of protons and neutrons;

– and so doesn’t explain why there cannot be some . “dynamic equilibrium” between neutrons in some system of neutrons.

However that for some really unknown reasons neutrons really don’t compose stable systems is seems well established experimental fact, while, say, in

- “[astro-ph/0312388] The Effect of Bound Dineutrons upon BBN” – the “Dineutrons upon BBN” look as rather strange base to think that dineutrons really can exist;

- “A Method for Unambiguous Detection of a Hypothetical Bound Two-Neutron System” – this method didn’t dectected any real Bound Two-Neutron System;

- “Phys. Rev. 137, B345 (1965) - Further Evidence for the Nonexistence of Particle-Stable Tetraneutrons” - that looks as rational publication. A number of “detections” of N-neutrons systems, when “ neutron clusters ” are observed at decays of neutrons-rich nuclei, that really can be some cases when some shell a few neutrons clusters in a nucleus indeed exist, and at a decay this cluster is radiated, but that doesn’t mean convincingly enough hat outside the nucleus the neutrons remain be bounded by Nuclear force bounded “neutron nucleus”;

“Gerald Rosen, Are dark matter WIMP's of mass 8.6 GeV octoneutrons?” – this question looks as too strange at all.

Real answers to the questions about what exists and happens in nuclei can be obtained rather convincingly only if nuclei and nucleons are considered on the Planck scale;

- and so at development of the Shevchenko-Tokarevsky’s 2023 Planck scale initial model of Nuclear Force; more see

https://www.researchgate.net/publication/369357747_The_informational_model_-Nuclear_Force

Cheers

The report on the first observation of dineutron emission in the decay of 16Be:

Phys. Rev. Lett. 108, 102501 (2012) - First Observation of Ground State Dineutron Decay: 16Be (aps.org)

It was also discussed about extremely short-leaving resonant states of dineutron ([1512.00314] Experimental Evidence of a Bonded Dineutron Existence (arxiv.org)

Though in SS post on page 5 it is pointed that though really mainstream physics doesn’t know what Nuclear Force is, and real understanding can be, as that is shown in SS&VT informational physical model, only at consideration of observed this Force interactions on Planck scale,

- and, first of all, at experiments where observed parameters of interactions – particles, spins, momentums, angular momentums, probabilities of reactions/decays channels, etc., are maximally certain and measured with maximal precision,

- in the thread a couple of really too questionable experiments examples again appeared, where the authors claim that they observed “Dineutron nuclei”. At the experiments, again – see the SS post on page 5 – decays of some too exotically neutrons enriched nuclei, about shells states in which physics knows practically nothing, though it looks as would not be surprising if in the nuclei’ shells some 2n clusters really would exist in some essential time intervals, and so probability of radiation of 2n would be non-zero.

However simultaneous detection of two neutrons even with a small angle between their motion directions doesn’t mean that the neutrons composed in some time moment some “2n nucleus”, as in

“……The report on the first observation of dineutron emission in the decay of 16Be:

Phys. Rev. Lett. 108, 102501 (2012) - First Observation of Ground State Dineutron Decay: 16Be (aps.org)…..”

- at the experiment besides detection of the neutrons pair nothing more is measured – no “2n nucleus” spin, no its half-life value, etc., while, again, in this case the 16Be nucleons shells are practically uncertain, correspondingly the experimental results can have other interpretations, where “2n nucleus” doesn’t exist; an example see in https://arxiv.org/abs/1204.5946, though, again, both versions really are too vague.

That

“…..It was also discussed about extremely short-leaving resonant states of dineutron ([1512.00314] Experimental Evidence of a Bonded Dineutron Existence (arxiv.org) ….”

https://arxiv.org/ftp/arxiv/papers/1512/1512.00314.pdf

- where “Experimental observation of 159Tb(n, 2n) reaction product was performed” – really as observation of “2n decay” of 158mTb, which was made when “Several instrumental [gamma] spectra of Tb specimen were measured with HPGe spectrometer in 1.5 years after last irradiation”, i.e. at that no neutrons were detected at all, /since “the reasonable

assumption that the transition 2n → d occurs via β –decay”;

- and so neutrons [and β] detectors weren’t used at experiment; what looks as rather, if too, strange, and so in the result

“…With these binding energy limits set one can make an estimate for the half-life of the dineutron (t1/2 ) based on the reasonable assumption that the transition 2n → d occurs via β−-decay [14]… we obtain 0.0074 s