The (physical) temperature is defined as the energy per degree of freedom (times k.) A nucleus is usually at its ground state, save for radioactive ones, its temperature is then 0 K. The energies of the lowest states of a nucleus are very far appart, that's a quantum effect, so that the first exited state is already above "room temperature" (~ 20°C.) In nuclear reactions near the Fermi energy (~ 50 MeV) per nucleon, hot nuclei can be produced with a temperature around 5 MeV. It is measured by the energy distribution of the evaporated light particles, considered in thermal equilibrium with the nucleus.

As the proton have an electric charge, there is an electric field in the nucleus that repels the protons from one another. But the electrostatic energy can't be lower.

Dear colleague!

To estimate the core temperature you can use the following model:

To consider the nucleus as a sphere of radius R, which is a gas of quarks, the methods of Monte Carlo or molecular dynamics to calculate the average kinetic energy and equate

Ekin =3/2RT

Similar work: . A. A. Sobko, S. A. Garelina, Calculation of the neutron-proton mass difference by the Monte Carlo method, Sciences of Europe, vol 3. № 5(5), 2016, Physics and mathematics.

If You're interested I can send print.

Sincerely A. A. Sobko.

Dear J.P.

The first thing to put in perspective is that the macroscopic "temperatures" that we measure are due at the atomic level to excitation of the electronic orbitals in excess of the electrons least action energy levels.

This excess excitation is caused by transfer of energy via convection (radiation) or conduction (in bodies where atoms are in immediate contact). When the levels of energy (excitation) exceeds the allowed energy level sufficiently, the electron is ejected from its rest orbital to move to a metastable orbital further away from the nucleus, which it will soon leave to return to its rest orbital, point at which it will release an electromagnetic photon that carries away the excess energy.

If the "temperature" is lowered to zero K, all excess energy is removed, but the electron remains stabilized with only the default energy that maintains it in its least action resonance state. In the case of the hydrogen atom, this default amount is 27.2 eV.

There is no possibility for the electron in the ground state of the hydrogen atom to be induced with less than this amount of energy by the Coulomb force.

The same has to apply to the charged components making up the nucleons in the nuclei of atoms since they are involved by means of the Coulomb force with the accompanying electrons.

It is a very good issue.

But I am no expert in that field

Addition to Your question:

All scientists and engineers find the subject Energy of biggest importance.

But what do we really know about the basic nature of energy .

It could be a subject to discuss .

Kurt Wraae

An alternative view: Temperature of a body indicates its 3D matter-content level with respect to 3D matter-content level of a reference body at specific 3D matter-content level.

Entire space, outside basic 3D matter-particles, is filled with an all-encompassing universal medium, structured by quanta of matter. Photons are the most basic 3D matter-particles. 3D matter-content levels of all 3D matter-bodies, including nuclei, depend on external pressure on them from surrounding universal medium.

A single photon, in universal medium away from all other 3D matter-particles, has minimum external pressure on it and hence has highest 3D matter-content level. This photon may be regarded to exist at absolute zero temperature. Even those photons in the vicinity of other 3D matter-bodies may have lower 3D matter-content levels (higher temperature).

All superior 3D matter-particles/bodies are formed by various combinations of photons. In order to organize photons into different types of 3D matter-particles/bodies, additional distortions in universal medium are employed. Additional distortions increase external pressure on these superior 3D matter-particles/bodies. Increase in external pressure is bound to lower 3D matter-content level of combined bodies. Therefore, no superior 3D matter-particle/body can achieve absolute zero temperature.

Temperature of an atomic nucleus depends on external pressure on it. This includes external pressure on whole of atom and whole of 3D matter-body, of which the atom is a constituent. Higher the external pressure, lower will be its 3D matter-content level and higher the temperature. See: ‘MATTER (Re-examined)’, www.matterdoc.info

Talking about the core temperature does not make sense. If the nucleus had a temperature according to the Stefan-Boltzmann law - would radiate electromagnetic radiation is proportional to T4, and this radiation is not observed. Protons and neutrons or quarks are in their levels and emit only during transitions.

Talking about the core temperature does not make sense. If the nucleus had a temperature according to the Stefan-Boltzmann law - would radiate electromagnetic radiation is proportional to T4, and this radiation is not observed. Protons and neutrons or quarks are in their levels and emit only during transitions.

Temperature is a measure of the average total energy of the smallest objects resp. particles within a certain amount of substance. The particles itself have a great velocity distribution resp. distribution of energy. Therefore we can not measure the temeperature of a single particle. As André said: Temperature is a macroscopic property. Furthermore: The temperature is measured as the average energy of the whole particles and not as energy of parts of it. Therefore we can not define a separate temperature of the atomic nuclei.

But the atomic nuclei (and all the other known particles) have also very different energies. It is impossible that all the nuclei of a mole oxygen or radon have all the same internal energy. This leads to Heisenbergs uncertainty principle and it is the real cause of the β-spectre and unepected decays of atomic nuclei which are termed as effects of neutrinos.

The decay and half-life of unstable nuclei is also depending on the various internal energy of the nuclei and at least also on the macroscopic temperature.

Remark: I'm missing the correction option. It should be corrected as: ... unexpected decays of atomic nuclei ...

Hans-G> I'm missing the correction option.

On Q&A, if you hoover the cursor over the time marker in the upper right corner of your post, a menu including an EDIT option appears.

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Dear all

I want to thank all your valuable comments, which have helped me to understand a little better a possible answer to my question.

First, is it reasonable to talk about the Temperature of atomic nucleuses?

I boldly asked if we could measure the Temperature (T) of an atomic nucleus, and I realize we cannot.

However, the circumstance of being unable to measure it (and to measure it with acceptable accuracy) seems not to jeopardize the circumstance that each atomic nucleus has a property, associated with its energy, which may be described by T. This is coherent to Heisenberg's uncertainty principle, for which I cannot accurately measure the exact position and speed of a particle (although it is reasonable to assume that each particle is, at time t, at an unknown precise position, and moving with an unknown exact speed).

Similarly, I am unable to measure the Temperature of a single electron; nevertheless, let us consider a large population of electrons, that we find in a piece of matter. Although each electron i exhibits a Temperature, Ti, the large population has a statistically meaningful Temperature distribution with an average value which we call the Room Temperature, To. I really do not care to disturb the privacy of each single electron, as long as I may define an average Temperature of the population, To.

An analogous reasoning can be made about an electric current in a conductor, or semiconductor. Each electron flowing in the electric current exhibits a given speed, and I only care that, for a given drift electric field, the moving population of electrons flows at an average speed, e.g., associated with a 1 mA current (associated with a mC amount of charge per second).

Hence, as we have a large population of atomic nucleus in a solid piece of matter (in the order of 10 to 23 atoms per cubic cm), I really do not care (and cannot measure) the Temperature of each nucleus. However, it is reasonable to assume that an average value of T of the atomic nucleuses in the target population exists.

Really, as André Michaud stated, Temperature is a macroscopic property. But it seems that the effect it describes goes down to the microscopic level.

If so, allow me to comment what Aleksandr A. Sobko wrote. He wrote: "Talking about the core temperature does not make sense. If the nucleus had a temperature according to the Stefan-Boltzmann law - would radiate electromagnetic radiation is proportional to T4, and this radiation is not observed". However, following this reasoning, could it not be that radiation is not observed due to the fact that T is almost zero?

Thierry De Mees also stated "Yes!" to my comment: "Is it possible that a cloud of electrons, either orbiting the atoms nucleuses or moving as free electrons, play a shielding effect?". If we are understanding it right, then a large population exhibits an average To Temperature, shielding atomic nucleuses, which may exhibit an average T almost 0ºK, thus inhibiting the nucleuses from radiating.

Now, again I ask: Can we assume that the impact of EM fields ultimately vanishes as T approaches 0ºK? Could this help explaining why protons in an atomic nucleus stay together, and are not violently scattered away from each other?

Although it appears that elements are at room tem. however in nuclear physics, the nuclear temperature is defined as energy =kT, so BE/nucleon =8 MeV ; consider 10 MeV; so kT=10MeV; T=11609*10^7K; So Nuclear Temp=1.17*10^11 K.

This is the corresponding nuclear tem. as per the given BE

Dear Munish Kumar

Do you mean that the Temperature of atomic nucleuses is in the order of 10^11 (100000 million) degrees Kelvin? Is it reasonable for you? Each nucleus at a Temperature higher than the Temperature of the Sun?

Aleksandr, you are speaking about the case where there in no quantum effect. The T4 law no longer apply even for ordinary (atomic) matter at low temperature. An excited nucleus does emit radiations, though not only electromagnetic but mainly light particles, because strong and weak interactions are important too. But as I explained, the first excited level of a nucleus is way above "room temperature." The quantum effects for a nucleus are some orders of magnitude greater than for ordinary matter.

A nucleus is not a single particle but a system of particles. A temperature can then be defined, although it has not much meaning for a light nucleus. This temperature can, and is measured by the energy spectrum of the emitted light particles, in a similar way as with the black body radiation. The energy entering the definition of temperature is only kinetic energy, and not binding energy or other. Heat is disordered energy, and only kinetic energy of many particles can be disordered. Too much false information in the answers of this question. Just do a search with "hot nuclei."

According to my prediction the temperature of atomic nucleus is near to 10^12 k or greather than it. Because nucleus is not a stabe in microscopic view. Within nuclius proton are no longer proton they are in the form of quark and gluons ( i predict gluons are gravitons with mass near to 10^-70kg)

I predict that the structure of nucleus has two parts, one is core which is behaves like a micro black hole and temperature of this core is 10^12 k or more than it. Please view the pdf

I don´t believe it possible, that the nucleus temperature is in the order of 10^12K.

Electromagnetic radiation is a direct function of the temperature. At 10^12K the strength of the radiation of each individual nucleus would be, in my understanding, overwhelming, which does not seem to be in harmony with the observed reality. I wonder, at what temperature would be the nucleus of radiative matter…

Claude-Pierre Massé, , following your idea that the temperature of an object can be inferred from its emitted radiation, , and since nuclei emit only gamma rays, , I would say that hey have a rather high temperature (especially if you assume them to be black bodies, and employ Wien's law).

However, I don´t think this is correct, because temperature is defined dfor systems with a very large number of particles---- of the order of Avogadro´s number---.

Mass and electric charge are not inseparable, there are massive neutral particles. Temperature is only definable if you have a lot of particles, here nuclei. Just a few wont define

temperature.

A look at the Bose factor will tell you how the level occupancy changes with temperature for photons or phonons. (ie. Black body radiation), but here it is stat. Mech. for many bodies.

The odd or even spin of nuclei will tell you the kind of statistics you need.

If you exite nuclei to a higher energy state, they will emit gamma rays, if you wack them with something, but this does not have to do with temperature, just energy input.

The botom line, if you do not talk about ensembles, do not use temperature.

Temperature is the measure of 3D matter-content level of an object. 3D matter-content level of a body depends on the external pressure on it. An atom in free space has the least external pressure on it and hence it is in its coolest state (highest 3D matter-content level). As external pressure on it increases by having neighboring atoms, external pressure on it increases and its 3D matter-content lowers. This increases its temperature.

Nainan

It is pointless to talk about the temperature of the nucleus, as it is pointless to talk about the temperature of electronic shells in the atom. The nucleus is a quantum gas or liquid of quarks that are at their energy levels. When quarks move from level to level, beta and gamma radiation (beta and gamma kernel decay) is observed.

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Raul Simon

The temperature is the average energy of motion, more precisely of thermalised motion. A nucleus with ~ 100 nucleon has an energy standard deviation about a tenth of the energy average, which then is good enough defined. A huge number like the Avogadro number is an overkill. Such a nucleus with a temperature of 5 MeV, which is roughly the highest temperature it can sustain, emits principally neutrons and protons, and even alpha particles and heavier nuclei. They are perfectly admissible radiations, since they as well expel energy. Experimentally it is seen that they have an energy distribution with an exponential fall off, just as the blackbody radiation. There is a whole science about hot nuclei, you'd better study it.

I guess IV should. ery interesting answer.

Dear J.P. Teixeira,

You have made a good question (containing a beautiful answer in it…) My co-author, Eng. Robert Therriault, of the article:

https://www.researchgate.net/publication/327201504_Gravity_a_paradym_shift_in_reasoning

It has a similar concept like you: atom's nucleus temperature is close to 0ºK.

In 2018 autumn in time of discovering intuitionally one principle of earthquakes, had an intuitional moment connecting to the structure of the atom (I have written it somewhere) and the same it the base of temperature close to 0ºK… Actually this seems the best explanation….

Please read the proposed article and tell your opinion about it, because indirectly can build a simple atomic model onto it…

Regards,

Horvath L.A.

Dear J.P. Teixeira,

My co-author, Eng. Robert Therriault, of the article 'Gravity a paradym shift in reasoning ' wrote to me yesterday to your question:

'the model I have for atoms is based on least stacking lattice structures . . .  it is only based on EM the attractive forces between electrons and protons in the nucleus . . . with Coulomb's law to be the basis of the understanding.'

Regards,

Laszlo

The concept of temperature is a statistical concept that makes sense for a large number of particles. More precisely, the temperature is the maximum of the Maxwell-Boltzmann distribution for particles in terms of velocities. This is for classical particles, which the nuclei of atoms are not. And for quantum particles, depending on their spin, Fermi-Dirac and Bose-Einstein distributions are used. This concept does not apply to 1, 2, 3, etc.particles, but only to a pile of particles. So there is some doubt that the question is correct and meaningful (making sense).

Dear colleagues

I have been quite busy; this is why I haven’t been answering the valuable comments you are making to my question: “What is the Temperature of atomic nucleuses?”

First, I wish to thank you all for your significant contributions to uncover the correct answer to my question, which I really consider relevant.

The concept of Temperature is a key issue to understand our physical world. It started as a macroscopic concept, associated with matter, and vague – some object is cold, warm, hot… Then, quantification has been introduced, and e.g., zero and 100 degrees centigrade were determined by the Temperature at which waters freezes, or evaporate. The concept of absolute Temperature and the limit associated with zero degrees Kelvin (0ºK) have been a milestone to characterize the physical world.

I agree that “The concept of temperature is a statistical concept that makes sense for a large number of particles”. Associated with “statistical” is the concept of “average”, as we move from macroscopic into microscopic level.

Let me give you an example, to clarify my reasoning.

In a semiconductor, let’s say at room Temperature, T=300ºK, if an Electric field is applied, the (macroscopic) result is the flow of an electric current, whose value depends on the field intensity, E, and of the material resistance. If the field intensity is constant, the electric current value, I, is constant.

Microscopically, what is responsible for this current flow? Inside the semiconductor, we have a set of charge carriers, namely electrons and “holes” (electrons moving in the valence band, from one atom to a neighboring atom). To simplify, let’s assume that the doping density of impurities is high enough, so the majority of carriers are free electrons in the conduction band.

Dependent on the semiconductor’s Temperature, electrons experience a random motion associated with their thermal energy. As two electrons approximate each other, the electric field associated with their electric charges produce an inelastic collision, and both are scattered in opposite directions. This random movement occurs even in the absence of an external electric field, E.

When E is applied, each individual electron is accelerated by a force, f, due to E, in the time interval between collisions (f=qE, where q is the electron electric charge, q=1.6 x10-19 C). In the case of a constant E, uniform acceleration occurs. In the following subsequent inelastic collision, the electrons loose energy, and then again are accelerated, and so forth. For low values of E, there is a proportion between the average velocity of charge carriers, v, (due to the external electric field), and the field intensity. The concept of drift mobility (µ) is defined, as the constant of such proportionality: v = µE.

This mobility concept is macroscopic, and statistical in nature. In average, the set of free electrons move coherently as the result of the external field, E, so we may measure the resulting electric current, I, according to the semiconductor’s resistance. The drift mobility and the resistance are dependent also on the Temperature.

Why I give this example?

The movement of each free electron depends on its thermal energy, and on the external field, E. Its thermal energy depends on its Temperature. We talk of the “Thermal voltage”, defined as kT/q, where k is the Boltzmann constant. At room Temperature, T=300ºK, kT/q = 0.0259 V. At each time instant, t, each electron has a movement, has a space position, a velocity, and so on. We know it is not possible to measure these attributes for each electron, but really we don’t care. We are concerned with the behavior of a set of electrons, driven by the external field. What each humble electron is doing, or where it is, at instant t, is, for us, irrelevant.

Hence, although the “concept of temperature is a statistical concept that makes sense for a large number of particles”, it seems that each particle should have its own energy, motion and Temperature, even if we can’t measure it.

Applying this reasoning to atom’s nucleuses, although each one is microscopic and build with relatively few “particles”, the real question seems not to be if the nucleus has a Temperature, or not (which we cannot measure).

The real important question, in my opinion, is: the physical behavior of the atoms nucleuses is consistent with a nucleus’s almost at 0ºK, or (exclusive or) at an almost infinite Temperature?

It is reasonable to assume that only one of these answers is true, that is, consistent with reality. Hence, the identification of which hypothesis is more trustworthy is not irrelevant. It is, in my opinion, very relevant.

I am looking forward to read your comments. Have a nice day!

Nuclei are made up of neutrons and proton (nucleons)in about equal numbers for light nuclei, excess neutrons for the heavy.

As far as I know temperature is not a frequent concept here, the number of nucleons is not high.

RAther the comon discussions related are fission , fusion,

radioactivity, energy levels in, poorly understood nuclear forces, or quarks and gluons on a more fundamental level.

Without electrons and their frequencies, the nuclei are considered as fixed points. So a bold but logical statement would be 0 Kelvin, as long as they don't vibrate. If they vibrate, they are by simplicity, attributed the same temperature as the rest of the molecule.