 # Is the neutron actually less massive than the proton is?!

• Here is a conundrum I cannot solve myself, or at least I wish to expose myself and share this and find out if I simply made an error in my thinking or math, or that there is perhaps reason to redefine the mass belonging the the "neutron" and or perhaps there is a completely different conclusion to be drawn?

Looking at the Periodic Table of the Elements (PTE) and assuming the protons and electrons of the elements and isotopes are always the same value, we see that as element increase in size, the mass-defect increases steadily. The atom is made up of protons, electrons and neutrons allegedly (in principle, even if you assume they are in turn made up from smaller stuff).
The proton and the electron have been experimentally measured pretty accurately as far as I understand and the only thing that was at first unclear was the mass belonging to the neutron. So many attempts were made to figure out the mass of the neutron as it could not be measured directly...

* Conversion MeV to mass - 1 u = 931.49410242 MeV

If one uses the neutron -> proton + electron + 0.78 Mev (released as gamma/photon) (yes also an anti-neutrino allegedly, but does not carry any real energy or mass, so we ignore it here)

1.00866491588 u -> 1.007276466621 u + 0.000548579909 u + mass 0.00084 u

As far as I could figure out, this is how the value of the neutron was determined and accepted. We know all values through measurements etc and the neutron was thus calculated to the 1.00866491588 u.

Looks simple and true, right?!

However....

Lets look a bit more at this.

Assembling and disassembling a deuteron is also used to figure out BE for example.

Hydrogen 1 + Hydrogen 1 ↔ Deuterium + Photon (reaction goes both ways!)
is the basic reaction used for that purpose. It of course means that when we fuse two H1 together we see a mass-defect occurring as a result. Problem is that in effect we have transformed one proton and electron into a neutron and as a result the newly created deuterium is now a bit less massive, so the neutron must be less massive than the proton! So how come?

When the Deuteron is hit by a 2.225 MeV gamma it disintegrates again, but usually by creating a neutron and a H1. obviously the neutron then in 15 mins or so disintegrates into a H1 (proton plus electron) again.

Deuterium + photon ↔ Neutron + Hydrogen 1 (n -> p + e + 0.78 MeV) -> H1 + H1

rewriting the formula gives me

Deuterium ↔ Neutron + Hydrogen 1 - photon

and

Neutron ↔ Deuterium - Hydrogen 1 + Photon

gives us (In MeV)
939.565854484 ↔ 1,876.12392797 - 938.783073486 + 2.225

939.565854484 ↔ 937.340854484 + 2.225 (MeV) → 1.00866538 u (mass)

seems correct but the thing is.... we attributed the gamma mass value to the neutron this way!

So lets do the numbers again.

* The mass-defect for Deuterium is 1.442 MeV
* A neutron is proton + electron + 0.78 MeV = H1 (in mass) + 0.00084 u
* photon is 2.225MeV / 0.0023886 u

Hydrogen 1 + Neutron ↔ Deuterium + photon

gives

Neutron (+0.78) – Photon (2.225) ↔ Deuterium - Hydrogen 1

or

neutron <-> Deuterium - Hydrogen 1 + photon - 0.78MeV (note that 2.225 - 0,78 MeV= 1.442 MeV equates the actual mass-defect for the deuterium)

--> neutron (+ 0.78 MeV) - 0.0023886 <-> 1,876.12392797 - 1.00782503190
--> neutron = 2.0141017781 - 1.00782503190 + 0.0023886 - 0.00084 = 1.0078253462 u which of course leaves us with the value of the proton plus electron since we already took the 0.78 MeV into account.
But in reality we have created a neutron by fusing two Hydrogen atom together. The mass-defect is the gamma ray that is released in the process and if we reduce the neutron value by that number we end up with
(neutron) = (H1 + 0.78) - 2.225 MeV -> mass-defect -> 1.0078247462 - 0.0023886 + 0.00084 = 1.0062761462 mass for the neutron !

Finally,

H1 + H1 -> D + 1.442 (mass defect) Where D is the total of proton + electron + neutron

---> H1 + H1 -> (neutron + proton + electron) + 1.442 (MeV)

In Mass (u)

1.00782503190 + 1.00782503190 <-> 1.0062761462 + 1.007276466621 + 0.000548579909 = 2.01410119273 Which is the correct mass value for Deuterium, BUT! The equation is not correct anymore, meaning 1.00782503190 + 1.00782503190 = 2.0156500638 u and the outcomes is 2.01410119273 with an obvious difference of 0.001548 which is the mass-defect again.

What does this mean?

I think that the two proton of the H1 systems fusing together loose energy, but in effect we would have transformed a proton and electron into a neutron. The only conclusion in my mind is that there is no neutron as a fundamental particle. According the the Structured Atom Model we have only protons and nuclear electrons in the nucleus. That would imply here that the protons loose mass when they become a proton that is connected to another proton forming a nucleus. The neutron is not the current value, although it all appears to be correct, but that is because the emitted gamma ray of 2.225 MeV is conveniently forgotten and taken into account with free-neutron decay.

The protons in the H1 system are an entity in the universe on their own, only by releasing their (potential ?) energy in relation to each other can they be fused together and now act as one collectively as an energy system. That means they are as far as the rest of the universe is concerned one, or a construct that is bonded together.

Coming back to the PTE, adding more H1 to an existing nucleus creates more neutrons and energy release which represents the mass-defect. All other components are known fixed values, meaning the proton and electron. So the only way for that system, if one accepts neutrons to be real, is that the neutron releases that energy somehow, meaning the neutron in the nucleus is a less massive entity than the proton is and the difference is radiated away.... Same problem therefore and the neutron cannot be more massive than the proton in this logic! By isolating the free neutron and forgetting where that free neutron came from, namely a nuclear reaction such as fission of deuterium, can we see the whole picture and that shows us, in my mind, that the story is far from complete.

Hydrogen1 + Hydrogen1 or p+e + p+e <-> Deuterium + Energy and if deuterium is indeed made up of electrons and protons then the protons in the nucleus must be according to the SAM logic have shed some of the mass.... or energy, or charge? So the protons in the nucleus in SAM are "down-graded" protons! with the difference radiated away as gamma rays.
The difference between the 2.225 MeV gamma ray and the mass-defect of 1.442 MeV is the inner electron that has a "self-binding" energy of 0.78 MeV. Now all this detail is completely obscured by attributing the difference to the neutron and not looking at the whole picture...

So please please help me out here and comment on this topic with hopefully a healthy skeptical look but also an open mind, because this is not so much about being right or wrong for me, but to really understand what is going on at this most fundamental level.

• I show the exact neutron,Deuterium,4-He etc. structure in detail in my paper about SO(4) physics. There are no neutrons inside 4-He. Same for Deuterium the neutron gets formed when you add 2.25MeV to Deuterium what obviously is more than the neutrons excess energy.

All charges in the nucleus are based on the electron internal bond energy that is 1183.x eV for one rotation, strong Nuclear bonds always are double connected or 3/5 fold connected for electro weak bonds.

• I show the exact neutron,Deuterium,4-He etc. structure in detail in my paper about SO(4) physics. There are no neutrons inside 4-He. Same for Deuterium the neutron gets formed when you add 2.25MeV to Deuterium what obviously is more than the neutrons excess energy.

All charges in the nucleus are based on the electron internal bond energy that is 1183.x eV for one rotation, strong Nuclear bonds always are double connected or 3/5 fold connected for electro weak bonds.

That would mean you (more or less) agree that there is an issue with the neutron (mass) as I tried to describe? And if so, am I correct in understanding you already implemented that understanding in your own model?

I do not understand the second remark yet and would humbly ask for a link to your paper so I can try to read that, although I fear my lack of (complex) math is a problem..

Also, Am I correct in understanding that you see the nuclear electron as truth also? If so, perhaps we are not that far away from each other in understanding atoms then. I know that we may not agree on everything and probably both think we have the correct model, but perhaps the truth does lie in between, since I started with the structural arrangement and am working my way back to math and you based it on the math? Perhaps we are simply seeing "the truth" in a different manner. The electron in my mind is in the first place a connection between protons, or making that connection possible.

The 0.78 MeV released with free neutron decay is what energy is needed to have the nuclear electron and that can only hold true in the case of at least two protons, hence the single free neutron, which is like a delayed reaction from another nuclear reaction, decays since it cannot maintain the state it was in.

Taking into account the nuclear electron and the 0.78 MeV combined with the binding energy of 2.225 MeV per proton-proton connection structurally in the SAM model leaves me with pretty accurate BE values and mass numbers for all elements and isotopes. It is not perfect, but very close and neatly correlates with the know values. The only exception is that for He we have structurally only one tetrahedron and the (2) nuclear electrons hence overlap all p-p connections so we must double the calculated value leaving us with again almost precisely the He4 BE value.

If anyone is interested in this follow this link to the presentation I gave on the topic

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(BE stuff starts at around 20 mins in)

or this link to our website with the 3D atom-viewer and all the data put into it such as calculated BE only based on the 2.225 MeV per proton-proton connection.
https://structuredatom.org/atomizer/atom-viewer

• this line

" H1 + H1 = (neutron + proton + electron) + 1.442 (MeV)"

Are you saying that if two hydrogen atoms were to form a deuterium

the net energy released should be 1.442 Mev?

which is a combination of the mass defect from the effect of n-p-e bondingin deuterium(2.225Mev

and the excess energy of the neutron (0.78?) ...)0.783 ? relative to p-e?

this sounds reasonable...

but to say that deuterium forms from ( p+e) +(p+e)

and that the mass defect should be the same as the net energy released by comparing against 2p+2e

goes against convention

the convention sort of says that deuterium is formed by n+(p+e)

and the mass defect is by comparing against (n+p+e)..

"The deuterium would have been formed in the first 1,000 seconds after the big bang, when the rapidly expanding universe had cooled off just enough to allow protons and neutrons to combine..

But conventions can be wrong

The only conclusion in my mind is that there is no neutron as a fundamental particle.

The neutron is unstable as shown by its excess energy.. it does not exist as a free neutron very long outside the nucleus and probably does not exist inside the nucleus as circles depicted in simplistic 2D pictures..it is a 4D particle,,,bonding in 3D/4D,.. welcome to 4D/6D headaches,...

• The idea that the Neutron is not a particle is a rather popular one. Santilli proposed the Neutron not being a particle but the sum of a proton and an electron. He calls his model the Rutherford-Santilli neutron and also calls it pseudoneutron.

I certainly Hope to see LENR helping humans to blossom, and I'm here to help it happen.

• I do not understand the second remark yet and would humbly ask for a link to your paper so I can try to read that, although I fear my lack of (complex) math is a problem..

The only exception is that for He we have structurally only one tetrahedron and the (2) nuclear electrons hence overlap all p-p connections so we must double the calculated value leaving us with again almost precisely the He4 BE value.

4-He the first time has a fully parallel flux (4D 2x2) of the protons EM flux. This leads to an increase of the nuclear binding force. This is the main reason why nobody ever could measure the nuclear force as it varies...

Whenever the flux doubles one more time the force increase. There is an other issues with the second torus norm (7/4), But first you should try to understand the simple things.

• https://www.researchgate.net/p…ances_and_fusion_products

4-He the first time has a fully parallel flux (4D 2x2) of the protons EM flux. This leads to an increase of the nuclear binding force. This is the main reason why nobody ever could measure the nuclear force as it varies...

Whenever the flux doubles one more time the force increase. There is an other issues with the second torus norm (7/4), But first you should try to understand the simple things.

How do you deal with structure in your model? Or how is the nucleus organized in the SO(4) model?

In SAM we see normally each time a nucleon (proton) is added that another tetrahedron form is created which means that each proton actually touches 3 other protons. There is straight up a very nice correlation with the more classical idea that there are three quarks per nucleon. When I talked to another scientist who is more versed in QM and mentioned that unfortunately in SAM we sometimes see 4 or 5 connections between the protons whenever the icosahedron form (Carbon is the first element that has 5 connections per proton) so this is probably not a true correlation then. He answered to me that in fact there is such a thing in QM up to the penta-quark. This is another strong correlation I found and was ready to give up on, but in fact was strengthened.
In effect is means that the structure is an important, crucial thing even, to understand and work with the elements when we talk about Binding Energy and mass-defect.

• The formula comes from mainstream papers that deal with this all throughout the last 100 years or so.

In short, by measuring the gamma ray accurately and the electron and proton mass as well as the deuterium mass, we are left in conventional thinking with the mass of the neutron as an unknown but we can now calculate it. Those papers use the formula D + gamma -> H1 + n (which decays). When the gamma ray is stronger the deuterium tends to split directly into two H1 (whether or not the electron is bound to the proton or not) so in effect D + gamma -> H1 + H1. Also since we are talking fusion on this forum, H1 and H1 should be able to fuse in some way, although that seems to be a rather difficult thing to do or explain. Point is, this is a formula and to calculate the neutron mass. that is why I think we can rewrite the formula according to mathematical rules which are so simple even I can do that....

This paper is one the papers I used

## Files

• He answered to me that in fact there is such a thing in QM up to the penta-quark. This is another strong correlation I found and was ready to give up on, but in fact was strengthened.

You cannot use QM (Dirac equation) for modelling a nucleus. Further quarks are 100% nonsense and have only be postulated to cheat politics to get more money for accelerators. So just forget the standard model and try to be much better.

5 rotations are standard in SO(4) physics strong force modelling. So so we really can have 5 connections (waves). E.g. the neutron has 5 rotation connection. You also can see it internally in Deuterium.

But physics has rules and does not follow simplistic thinking in brick structures. So step one try to understand mechanics in 6D.

• You cannot use QM (Dirac equation) for modelling a nucleus. Further quarks are 100% nonsense and have only be postulated to cheat politics to get more money for accelerators. So just forget the standard model and try to be much better.

5 rotations are standard in SO(4) physics strong force modelling. So so we really can have 5 connections (waves). E.g. the neutron has 5 rotation connection. You also can see it internally in Deuterium.

But physics has rules and does not follow simplistic thinking in brick structures. So step one try to understand mechanics in 6D.

Please believe me when I state that I have broken with the standard model, I was however pointing out how structure is important for nuclear physics. Perhaps I can entice you to take a look at the structure in SAM and how all that works otherwise you may get the idea that I support things like colliding neutron stars forming the heavy elements... (Which I do not) • Pharis Williams created a theory of everything which has a single force. The various expectations of gravity, electromagnetic, strong and weak forces are combined by using a non-singular potential. It seems to have thoughts like in this thread.

For example, this is a quote from Fusion for Earth and Space. "For example, the first proton-proton scattering produced high energy results that deviated from scattering predicted by the singular coulomb potential and it is now argued that this deviation is due to a strong nuclear force that does not appear in the Maxwell equations for electromagnetism. The non-singular potential predicts that a neutron is a proton trapped in a nuclear orbit around an electron (Williams, 1983). This proton orbit is the result of a positive energy well from which it is predicted to tunnel with a half-life matching that experimentally measured (See Figure 1). All disintegrations of neutrons are accompanied by the emergence of a proton and an electron together with motion of the center of mass. ... weak nuclear forces are predictable as the force between un-like particles with non-singular electrostatic potentials (Williams, 1983, 2001). (6) Fusion for Earth and Space (researchgate.net)

So like SAM, Williams predicts electrons and protons but no neutrons in the nucleus.

Nucleons within the nucleus are indistinguishable. But if there is a standard mass for a proton and an electron, then something makes up the balance of mass/energy to explain mass differences between elements and their isotopes.

My model proposes states that an electron can form a string with an electron. This new kind of electron (R-electron) can form a hydrogen atom. When these string states of the hydrogen atom decay, the energy must correspond to this equation; E = n2 (13. 58378 eV). The value 13.5837844145 and a mass for the neutrino of 0.108670274 are derived by fitting special relativity to the change in mass (energy), decay time and radius of string state of the hydrogen atom and of a neutron. These would be pre-nuclear states. R-electron could exist without being part of an atom.

R-electron states make R-electrons attractive at small separation and repulsive at a large separation. R-electrons can form EVOs. An EVO in this model is a planetoid held together by electro-gravity. So, an electron in the planetoid must have energy sufficient to get beyond the escape horizon to get out. By balancing at the escape horizon, the repulsive coulomb potential to electro-gravity, one finds that coupling constant for electro-gravity is 42 orders of magnitude stronger than universal gravity.

So, a "nuclear electron" could a string state of a neutrino and an electron. Matsumoto propose that within the nucleus the string could be an electron, a neutrino and a positron. So maybe strings with or without a positron. He calls them itons. Itons cause a degrading of nuclear structures within an atom but also between atoms since the string states are not restricted to the nucleus.

The usual structured model of a neutron does result in gravitation attraction so strong that a "neutron" cluster can collapse under the extreme gravity to produce something like black-hole or even a wormhole. Matsumoto has images produced by radiation from cold fusion reaction which he explains as results of electronuclear collapse.

• The neutron is not a fundamental particle but the electron is. As such, it must have a size (which I have not being able to nail down) and a structure. In order to take a free electron or an electron in a hydrogen atom and squeeze it down to partake in the nucleus takes work (energy). Ignoring that energy may be the cause of your problem.

When Randell Mills forms a hydrino he squeezes the electron somewhat. So, are his emitted photons missing some energy?

• So like SAM, Williams predicts electrons and protons but no neutrons in the nucleus.

Electrons and protons are the only two mass like stable form factors for EM energy. There is no classic potential force on nuclear level as the shape must be toroidal. Charge is a topological effect of nested EM-flux. This is the key for understanding dense mass/particles!

All EM mass can pair but then the electron must reorganize and this has nothing to do with 13.58 potential to a proton.

• The neutron is not a fundamental particle but the electron is. As such, it must have a size (which I have not being able to nail down) and a structure. In order to take a free electron or an electron in a hydrogen atom and squeeze it down to partake in the nucleus takes work (energy). Ignoring that energy may be the cause of your problem.

When Randell Mills forms a hydrino he squeezes the electron somewhat. So, are his emitted photons missing some energy?

Mills mistakes the photons from E = n2 (13. 58378 eV) for emitted photons. Compare this equation to his.

Conversion to a smaller diameter hydrogen requires energy per this equation. However, this equation also controls what emission occurs after fusion as well. Each n,(energy level) shares energy by equal energy partitioning. Partition happens because each energy level is a function of the same base energy (13.58378 eV). Energy levels are a boson condensate of lower n energy levels. Fusion occurs near the escape horizon of a planetoid (that is at where MeV level energies are found) but most of energy of reaction is partitioned back to the various energy levels of the planetoid. Most of the nuclear energy doesn't escape with the transmutation products. Neither does the mass of planetoid allow the formation of much in the way on gamma rays etc.

Edited once, last by Drgenek ().

• Conversion to a smaller diameter hydrogen requires energy per this equation.

No this is only Mills illusion. 13.59 eV is an equivalence relation of the total forces at action between e/p. Mills cannot give the force for lower orbits hence he cannot explain a potential that does not exist at all. The Bohr level is a saddle point and there is no way down.

H*-H* the only Di-Hydrino is a weak nuclear bond and the claimed potential energy (495.x eV) Mills measured can be exactly calculated by SO(4) physics and is not a classic potential!

• The neutron is not a fundamental particle but the electron is. As such, it must have a size (which I have not being able to nail down) and a structure. In order to take a free electron or an electron in a hydrogen atom and squeeze it down to partake in the nucleus takes work (energy). Ignoring that energy may be the cause of your problem.

When Randell Mills forms a hydrino he squeezes the electron somewhat. So, are his emitted photons missing some energy?

That is the 0.78 MeV, which is needed to keep the electron in the nucleus apparently.

• Sometimes mainstream science is more exciting than speculation...

Neither neutrons nor protons are fundamental particles:

They have measurable non-zero diameter

The constituents can be seen

Deep Compton Virtual Scattering has shown structure inside neutrons A deep look inside the neutron
The internal structure of the neutron is closer to being understood thanks to novel scattering experiments.
www.natureasia.com

Now, a large international team led by Meriem Benali and Malek Mazouz from Faculté des Sciences de Monastir in Tunisia, and Carlos Munoz Camacho from Institut de Physique Nucléaire d’Orsay in France, has observed, for the first time, rare events of a process called deeply virtual Compton scattering (DVCS) off the neutron. DVCS involves firing very high-energy electrons at nucleons in order to probe the quarks, which in turn emit high-energy photons. These photons carry information on the quark dynamics encoded by GPD.

“The neutron and the proton are actually two manifestations of the same object: the nucleon,” explains Benali. “Experimentally, the study of the neutron is very challenging, but essential. Otherwise it would be like studying humans by focusing only on men!”

Benali and co-workers used data from DVCS experiments performed at Thomas Jefferson National Accelerator Facility in the USA, where the targets were liquid hydrogen (with single-proton nuclei) and liquid deuterium (with nuclei of one proton and one neutron). By comparing the data from both targets, the team deduced the DVCS reactions occurring on quasi-free neutrons. Thanks to the different quark flavour content of protons and neutrons, and prior knowledge of proton structure, the team were able to deduce the likely contributions of the two main quark flavours to the scattering from neutrons.

The study demonstrates the strong potential of these novel techniques to impose new constraints on GPD theory.

Proper paper: Deeply virtual Compton scattering off the neutron - Nature Physics
The internal structure of the neutron has now been probed by highly energetic photons scattering off it. Combined with previous results for protons, these…
www.nature.com

Working out the mass of nuclei is complex because a lot of the mass comes from the binding energy of the constituent quarks (as GRMattson says above) - and that as you can imagine is difficult to calculate given the "hot soup" of quarks and the strong interactions between quarks.

In fact, the quark binding energy (they move at very near the speed of light) forms 99% of the energy of a neutron. Wikipedia summary but you can go to the source papers if you can bear it:

Most of the mass of hadrons is actually QCD binding energy, through mass–energy equivalence. This phenomenon is related to chiral symmetry breaking. In the case of nucleonsprotons and neutrons – QCD binding energy forms about 99% of the nucleon's mass. That is if assuming that the kinetic energy of the hadron's constituents, moving at near the speed of light, which contributes greatly to the hadron mass, is part of QCD binding energy. For protons, the sum of the rest masses of the three valence quarks (two up quarks and one down quark) is approximately 9.4 MeV/c2, while the proton's total mass is about 938.3 MeV/c2. For neutrons, the sum of the rest masses of the three valence quarks (two down quarks and one up quark) is approximately 11.9 MeV/c2, while the neutron's total mass is about 939.6 MeV/c2. Considering that nearly all of the atom's mass is concentrated in the nucleons, this means that about 99% of the mass of everyday matter (baryonic matter) is, in fact, chromodynamic binding energy.

We have come a long way in understanding what we are made of from the time of neutrons, protons, and electrons!

• Also, Am I correct in understanding that you see the nuclear electron as truth also? If so, perhaps we are not that far away from each other in understanding atoms then. I know that we may not agree on everything and probably both think we have the correct model, but perhaps the truth does lie in between, since I started with the structural arrangement and am working my way back to math and you based it on the math? Perhaps we are simply seeing "the truth" in a different manner. The electron in my mind is in the first place a connection between protons, or making that connection possible.

Indeed - electron wave functions can overlap the nucleus - (must do so).

However electrons are just much too big (ironic, since they are a point particle, but due to particle-wave duality also true) to fit into the nucleus. More precisely, an electron wave packet squeezed into the diameter of the nucleus would have momentum much too high to stay there (and this would also be observed). Can an electron exist in the nucleus?
Answer (1 of 26): Electron (e-) =fuel cell ; it generates the electricity. They are the shells, the atom bond. Nitrogen(Ne)carbon dioxide =Air fuel. Carbon…
www.quora.com

Harris and McClennen gave answers that are technically correct, but I think miss the point. Sure, the wave function gives a non-zero probability of the electron being found in the nucleus, but it cannot stay there.

Why? Because as far back as the 30s, physicists realized there were several reasons there can be no electrons abiding in the nucleus.

The first was: because then the spin of the nucleus, if there were only protons and electrons in there, would be all wrong, they would not match experiment for many elements, such as Nitrogen.

But there is another more fundamental reason: although orbitals are not orbits, e.g., they do not form trajectories, there are still some principles derived from orbits that still do apply, such a conservation of energy and of angular momentum: the electron has way too much energy and angular momentum to stay in the nucleus. That is, just as planets around the Sun all occupy orbits with definite energies, and it takes a huge expenditure of energy to get Jupiter to crash into the Sun, electrons have too much energy to crash into the nucleus; something would have to expend energy to reduce the electron’s energy, and it would be a large expenditure. Where would it all come from?

But there is a yet more fundamental reason: Heisenberg’s Uncertainty Principle: if you try to confine an electron with its high momentum and low mass into a small area, the small variation allowed in distance times the variation in momentum must be greater than half of h bar, so this means the variation in momentum shoots up to ridiculously high values, values which should be easy to observe — yet have never been observed. Such a high momentum should result in escape from the nucleus anyway.

(could give more scholarly references if anyone was interested).

• Harris and McClennen gave answers that are technically correct, but I think miss the point. Sure, the wave function gives a non-zero probability of the electron being found in the nucleus, but it cannot stay there.

Just posting nonsense. Protons cannot be described by the Dirac equation. And a neutron is completely out of reach... Learn SO(4) physics!