Hi. Stable isotope of helium called helium-3 and a rare and radioactive isotope of hydrogen called tritium. Helium-3 contains two protons and one neutron, while tritium contains one proton and two neutrons.

“…Another contradiction appears in the following case: Hydrogen 3 (H3 or Tritium) has 2 neutrons and 1 proton while Helium 3 has 1 neutron and 2 protons. We would tend to think that since the nuclear force (which is independent of the charge) is the same in both cases and since Tritium has only one proton, the Coulomb force will further destabilize Helium 3 ( Coulomb interaction is zero for Tritium)….”

- yeah, that “since the nuclear force (which is independent of the charge) is the same in both cases” [and so in SM there exists such quantum number as “isospin”] is some SM postulate, however this quantum number is rather artificial characteristics of hadrons, unlike, say, real spin of particles; and so its application really isn’t universal. Including that

“….But all the same there is a paradox:In effect, the system in this case evolves towards more stable but less bound nucleus!!….”

- is an example; though more evident can be that deuteron is a stable nucleus, though nucleons’ binding energy is equal to 2.224 MeV.

The QFT of the fundamental Nature force “Strong Force” is now based on too many practically purely ad hoc assumptions, parameters, “spontaneous breaking” of the parameters, etc., to be a really an effective theory.

Cheers

Mr Preston,

When you say that the neutron is unstable, it should be noted that the free neutron (outside the nucleus) is unstable, it decays by beta (-) radioactivity and becomes a proton after about 15 minutes. Inside a nucleus the neutron is in most cases perfectly stable.

I do not compare the neutron inside and outside the nucleus. I compare two configurations of the same basic system (3 nucleons)

In my question, the energy balance sheet (binding energy, total energy) in the comparison between the 2 systems is perfectly clear: in the end you have a less bound but more stable nucleus which is paradoxical because the more a system is bound, the less it has chance to decay.

Where I agree with you is that a complete and coherent theory of nuclear decay remains to be done.

There is no paradox, and you gave the answer. The binding energy is only a part of the energy balance of the system, and the total mass-energy of H3 is higher than that of He3+e+anti neutrino.

It is meaningless to consider separately the energy balance of the nucleus. It is just like considering the water pumped into a water tower without considering the electricity used by the pump.

Here the transformation of n into p is the fuel that gives the energy necessary to push the nucleus into a state with less binding energy.

Mr. Tondeur,

The text will be long, but this is necessary for clarity.

Of course a radioactive decay is nothing more than a nuclear reaction and the total energy is always conserved before and after the nuclear reaction. I never said (or implied) otherwise.

I think you misunderstood my point. The paradox is not in the energy balance but in the fact that the decay goes in the direction of a more bound system (tritium) towards a less bound system (Helium 3). In very clear terms: tritium which has a higher binding energy decays into Helium 3 which has a lower binding energy. On the other hand, in nuclear physics, we often are told that the higher the binding energy per nucleon, the more stable the nucleus. In the case that I cited, the one that has the highest binding energy per nucleon is tritium. In theory it is therefore more stable than Helium 3, it should not decay into Helium 3. In reality it is the opposite. This is where the paradox lies. If stability is not the binding energy per nucleon, what would be the "absolute indicator" of the stability of a nucleus? By "prudence", we should then specify that the binding energy per nucleon is not always an indicator of stability.

Apart from that, the total mass of tritium (and thus its total energy) is larger than that of Helium 3 and the decay and the system evolves from the higher energy to (another) lower energy system (Helium 3). This is quite normal. In this regard I quote a paragraph from the URL:

https://en.wikipedia.org/wiki/Nuclear_binding_energy#The_binding_energy_maximum_and_ways_to_approach_it_by_decay

"Radioactive decay where a neutron or proton in the radioactive nucleus decays spontaneously by emitting either particles, electromagnetic radiation (gamma rays), or both. Note that for radioactive decay, it is not strictly necessary for the binding energy to increase. What is strictly necessary is that the mass decrease. If a neutron turns into a proton and the energy of the decay is less than 0.782343 MeV, the difference between the masses of the neutron and proton multiplied by the speed of light squared, (such as rubidium-87 decaying to strontium-87), the average binding energy per nucleon will actually decrease." (end of quote).

If the text of the URL given just above insisted on specifying this "situation", it is because many people (not just me) found this aspect at least odd (or even paradoxical). By the way, the quoted necessary condition is not a sufficient condition. Finally, there is no "absolute" theoretical or empirical rule to quantify the stability of a nucleus. The only (approximate) rule is the one that has been quoted (binding energy per nucleon).

Best regards

I just wanted to thank Pedro L. Contreras E. for suggesting this question. Okay, the first big input is for B. Mohammed-Azizi you and François Tondeur seem to agree that a complete and coherent theory of nuclear decay remains to be done.

Therefore, I would like us to debate my hypothesis of a two-particle universe, where unfortunately the strong force does not exist. I have made two quick citable presentations

Presentation The Rutherford Conjecture, Why the Taijitu symbol better rep...

Presentation How Muons Further Illustrate the Rutherford Conjecture

but the paper is under construction. Basically, Rutherford suggested in 1919 or so that there were electrons in the nucleus but eventually abandoned that idea over time. Today we know that neutrons decay into an electron and a proton and are created in lightning strikes: an electron + energy becomes massive enough, (unlike a muon, which just shrinks the atom's diameter) to fuse with a proton to become a neutron. Look: if electrons are inside the nucleus then there is no such thing or at least no raison d'etre for the existence of gluons, hence no strong or weak force. There then is a binding "quantum dance" between protons and electrons in the nucleus. Knowing that electrons over quantum distances can occupy more than one place at once thus in deuterium two protons "share" one electron in the nucleus. Evidently, the three electrons and four protons in Tritium have a 50 % probability of misstepping once in 15 minutes. Once the "link" is broken, a neutron decays with the resulting atomic transmutation occurring. Seeming the diameter of the nucleus plays a role in missteps. The larger the diameter the more likely an entangled or embroiled electron will be too far away to "connect". I don't think it is an accident that the diameter of the tritium nucleus is 4.2 fermi four times larger than helium whose distance alone possibly

"Inside a nucleus, the neutron is in most cases perfectly stable." then, as I mentioned to Ara Giaquinto is the opposite of radioactive decay, and the weak force is not a force at all but simply a manifestation of the lowest energy state which is present throughout our universe.

Dear Mr Azizi

I think that the people who told you "that the higher the binding energy per nucleon, the more stable the nucleus" were not good professors. You are right in pointing that this formulation is only approximate. The words "more stable" themselves have no precise meaning. If "less unstable" can mean to have a longer half-life, what about "more stable"?

Referring to other fields of physics, "more stable" could mean that more energy is needed to transform the nucleus by an external action. But then we are out of the field of radioactivity, which only considers spontaneous transformations.

Let us avoid these approximate formulations.

There is nevertheless an absolute rule.

A nucleus is stable if there is no mechanism allowing it to spontaneously transform into a state of lower total mass-energy.

Dear Mr. Tondeur,

You say:

" I think that the people who told you "that the higher the binding energy per nucleon, the more stable the nucleus" were not good professors."

For you the one who "connects" the nuclear binding energy to the stability of the nucleus is "not a good professor".

And yet, such assertions are quite frequent, but, you don't have to go too far to find "not good professor"." (according to your own definition). Here are some examples (and there are many more).

Go to the following url:

===========================================

URL N°1

https://en.wikipedia.org/wiki/Island_of_stability#Nuclide_stability

you will read exactly the following sentence:

"The stability of a nucleus is determined by its binding energy, higher binding energy conferring greater stability…"

===============================================

We can read in the same url as the one above ,the following sentence:

"In nuclear physics, the island of stability is a predicted set of isotopes of super heavy elements that may have considerably longer half-lives than known isotopes of these elements. It is predicted to appear as an "island" in the chart of nuclides, separated from known stable and long-lived primordial radionuclide. Its theoretical existence is attributed to stabilizing effects of predicted "magic numbers" of protons and neutrons in the super heavy mass region."

But it is well known that magic nuclear numbers and shell effects influence quantitatively the binding energy.

=================================================

URL N°2

Such statements can also be found in another url:

https://www.thoughtco.com/island-stability-discovering-new-superheavy-elements-4018746

states that:

" Glenn T. Seaborg coined the phrase "island of stability" in the late 1960s. Using the nuclear shell model, he proposed filling the energy levels of a given shell with the optimal number of protons and neutrons would maximize binding energy per nucleon, permitting that particular isotope to have a longer half-life than other isotopes, …"

================================================

URL N°3

https://link.springer.com/referenceworkentry/10.1007%2F978-3-642-11274-4_1076

"Those nuclei with the highest binding energy per nucleon are the most stable with respect to destructive nuclear reactions".

===============================================================================

URL N°4

https://courses.lumenlearning.com/introchem/chapter/nuclear-binding-energy-and-mass-defect/

"Nuclear binding energy curve:

This graph shows the nuclear binding energy (in MeV) per nucleon as a function of the number of nucleons in the nucleus. Notice that iron-56 has the most binding energy per nucleon, making it the most stable nucleus".

===================================================

I don't want to go back over what has been said before and I think that basically we agree. What I mean is that the idea that binding energy per nucleon provides good stability is a common concept and in this case there would be a lot of "not good professors".

Mr Tondeur when you define instability by the spontaneous possibility that the nucleus has to evolve towards "less total mass" you repeat in another form the fact that the system evolves in the "good sense" (expression that I have used in my first post), otherwise it would have been illogical for the system to evolve towards more mass. Now in the definition you give you speak of "mechanism". This mechanism (you avoid citing it) is precisely the radioactivity (or possibly fission). In other words you are saying in a roundabout way that a nucleus is stable if it is not radioactive! You specify in the rule that the evolution must be done towards less mass. This condition is necessary but obviously not sufficient, otherwise any nucleus would disintegrate into a smaller nucleus.

The case I cited (tritium toward helium3) is moving towards less binding energy. It's still astonishing (if you don't want to be surprised, it's your right!). Indeed let us suppose that a nucleus "X" disintegrates towards a nucleus "Y" with less binding energy and that the nucleus "Y" disintegrates towards a nucleus "Z" with once again less binding energy, if we repeat this process (decay chain) we move away from the valley of stability. This is illogical, so the fact that the binding energy decreases each time cannot be physically possible and must not be repeated too times.

Here is for example the most probable decay chain of Th-228 with the corresponding nucleon binding energy per nucleon:

Th228 (7.645MeV) ->Ra224 (7.677)-> Rn220 (7.717) ->Po216 (7.758) ->Pb212 (7.804) ->Bi212 (7.800)->Po212 (7.810) ->Pb208 (7.867MeV)

It can be seen that the binding energy per nucleon always increases except once when it decreases very slightly. So the remark made about the binding energy in the decay (tritium->He3) is quite normal.

sorry I've come late but has anyone discussed muon helium which acts chemically as if it were a hydrogen isotope? https://www.newscientist.com/article/dn20049-atomic-disguise-makes-helium-look-like-hydrogen/

Mr Azizi

Why don't you include in the discussion of the decay chain of 228Th the binding energy of the nucleons that are emitted as alpha particles, i.e. 7.074 MeV per nucleon? For them the binding energy does not increase but decreases. Another paradox?

Not at all, the simple rule is the decrease of the total mass-energy. There is no need of a rule for the binding energy.

Mr Tondeur,

Let examine the following case:

mass Pb208 =207.976652071 u

mass He4 = 4.002603254 u

mass Hg204=203.9734939 u

Let us assume (only assume) that:

Pb208-->Hg204+He4 (spontaneous alpha decay)

Thus, sum masses of Hg204 and He4 = 207.976097154 u

Applying your rule:

Then, mass Pb208 > sum masses of Hg204 and He4

Thus in this case the total mass decreases as the condition imposed by "your rule".

But, how useful is your rule for this case about stability of Pb208? your rule says only that this decay is possible and therefore Pb208 could be unstable (which is obviously wrong).

Conclusion:

As far as I'm concerned I'm not saying that your rule is wrong (spontaneously the total mass must obviously decrease). I am only saying that your rule does not allow to say that a nucleus is stable or not. If a rule does not allow to conclude if a nucleus is stable or not, it cannot define the stability of a nucleus.

Once again there is no rule (even "universal" as such as yours) which allows to conclude about the stability of a nucleus.

The case of 208Pb is indeed more subtle than 3He which was your first concern.

3He is stable because there is no state wit lower mass for the three nucleons, even if the transformation mechanism exists.

208Pb is expected to be unstable, because the nucleons can be rearranged in states with less mass, and the mechanisms exist, but the probability of the spontaneous transformation is expected to be so low that the chance to observe it is negligible.

There were similar cases in the past , e.g.. for 209Bi, of nuclides initially considered to be stable, but that were found to be radioactive with a very long half life. It cannot be excluded that in the future, with new experimental techniques, other cases will be observed, but probably 208Pb is not the easiest candidate for that research.

Thus it could be useful to distinguish two types of nuclear stability:

- nuclei that are absolutely stable according to the known laws of physics, like 3He.

- nuclei that are presumably unstable but their decay cannot be observed with our present experimental capacity, and are considered to be stable in practice.

The rule of minimal mass combined with our knowledge of decay mechanisms is sufficient to determine if a nucleus belongs to the first category. With this rule you can also determine which nuclei are beta-stable.

The question wether a beta-stable nucleus is "stable in practice" or not is a question of practice, i.e. observation, rather than the application of a rule.

Dear Mr. Tondeur,

First of all I notice that your message this time is much longer. There is a striking contrast between the 2 lines of your "universal rule" stated with confidence and the number of lines and arguments that you put forward this time to resolve the contradictions that appear.

My approach is simple, you state a rule and I challenge it with a counter example. You say that this case (that of Pb208) is more "subtle" than the previous one. This word "subtle" tends to make one think that there is a trap in the counter-example and therefore the argument would be fallacious. There is no malicious intent in the example given (Pb208). It is an example that has the merit of being very clear and leads to contradicting (paradoxically) your rule.

Let's come now to the notion of stability, you say that there are 2 types of stability:

1) an "absolute" stability

2) an "empirical" stability because in theory the nucleus (Pb208 in our case) should be unstable but the present experimental means do not allow to prove it because the half-life of Pb208 is too long.

In fact this classification seems to me to be based on an unspoken and implicit argument. Indeed, when the experiment goes in the direction of the theory, it is good, when it is the opposite, it is bad.

You will tell me that in the case of Pb208, it is still theoretically unstable, but that this will be difficult to prove. We can turn the argument around and tell you that as long as you have not proved its instability, it can be considered stable as and that it is the theory that needs to be revised.

I do not see any reason to revise the theory. 208Pb is expected to be unstable because a state with lower mass exists and a mechanism to reach it exists. This mechanism is some kind of splitting of 208Pb into two lighter fragments, e.g. alpha-decay or fission. The so-called Coulomb barrier would classically prohibit this mechanism, but in quantum physics this is turned into a very low probability of tunnel effect. This probability can be calculated, at least approximately, and the result will explain why the decay of 208Pb is not observed. Theory is thus in full agreement with experiment.

I think it will be a very hard job for you if you decide to revise it. Good luck !

Mr. Tondeur,

You are satisfied with the theory, that's your right. For my part, I think that physics is ultimately based on experimentation. Therefore, as long as there is no experimental proof of the instability of Pb208, I will not adopt your point of view. However, if one day, by chance, you discover an experiment that proves that Pb208 is unstable, although I doubt it, do not hesitate, let me know anyway.

Good evening

Thank you for this nice discussion. I am impressed to see that the Strutinski method, to which I contributed some 40 to 50 years ago, is still alive thanks to you.

F. Tondeur

Dear Mr Tondeur,

it was a pleasure for me to chat with you even though at times there was a little tension. The main thing in the discussion being to scientifically argue the scientific questions.

Between 1977 and 1982 I did my magister thesis (just before the doctorate), the theme of my thesis was the equilibrium shape of nuclei (ground state) by the macroscopic-microscopic method (Strutinsky method). I used this method again in my doctoral thesis (the seven microscopic functions of the BOHR collective Hamiltonian in which the deformation energy is calculed by Strutinsky"). So I went into the question well. Concerning the period 1977-1982, I actually read some of your articles on this method (it seems to me that you were the sole author of your articles).

Once again thank you for the exchange of ideas.

Aref Wazwaz I see your point.

Than you so much dear Jose Risomar Sousa .

Happy New Year 2022 to all RG colleagues.