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Electrons are ejected in the Photoelectric effect if the energy of the incident radiation is high enough to surpass the work function of the metal. However, only electrons are ejected.

I wonder why protons and neutrons are not ejected? One might say they have high Nuclear Binding Energy.

Li-6 Isotope has nuclear binding energy per nucleon around 6 MeV, and modern X-Ray Betatron used for cancer therapy have sufficient energy to eject a proton from the nucleus. One might argue that due to the complex lattice, multiple nuclei make nuclear bond with the test atom, but what if we isolate some atoms, by the method mentioned here link. The nuclear binding energy is calculated with the mass defect equation. If this is possible we could measure the nuclear binding energy by this method, too.

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    $\begingroup$ What energy photons are you using to do the photoelectric effect? Who said that we don't get protons and neutrons if we bombard nuclei with sufficiently energetic photons? $\endgroup$ Commented yesterday
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    $\begingroup$ "multiple nuclei have make nuclear bond with the test atom". Not really. Each nucleus in normal matter is isolated from the other nuclei. They only get close enough to feel each other's nuclear forces in extreme conditions, like in star cores. $\endgroup$
    – PM 2Ring
    Commented yesterday
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    $\begingroup$ You may find this interesting: en.wikipedia.org/wiki/Photodisintegration $\endgroup$
    – PM 2Ring
    Commented yesterday
  • $\begingroup$ photoelectric. hint: what absorbs the photon? $\endgroup$
    – lineage
    Commented yesterday
  • $\begingroup$ FWIW ELI-NP is planned to research such processes (by secondary high energy gamma rays as well as collective processes due to extremely high field strenghts). I am not entirely sure about the current status, but the highest energy laser is not available yet, and they are currently tuning the instruments and develop techniques to use in the later nuclear physics experiments on the other lasers. $\endgroup$ Commented 4 hours ago

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It is because neutrons and protons are bound in a potential well that is about 3-4 orders of magnitude deeper.

To cause the photoelectric effect in atoms, the incident photons need to have enough energy to unbind the electrons. This is a few eV for metals like sodium that have a weakly bound outer electron.

In atomic nuclei, the protons and neutrons are bound such that energies of several MeV would be required to get them out of the nucleus. The process is known as photodisintegration or the "photonuclear effect".

The relative sizes of the targets also plays a role - for example, a sodium atom being $\sim 3\times 10^{9}$ times the area of a sodium nucleus. The cross-section for the photoelectric effect (or photoionisation) of a sodium atom is $\sim 10^{-21}$ m$^2$ at photon energies of $\sim 5.2$ eV just above the photoelectric threshold. In contrast, it takes a gamma ray of about 12 MeV to prise a neutron out of a sodium nucleus and the cross-section is about $10^{-30}$ m$^2$ (Goryachev 1964). Thus even above the corresponding threshold energy is takes a much greater flux of gamma rays to produce a number of "photo-nucleons" than it does for UV light to produce the same number of photo-electrons.

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    $\begingroup$ In other words, the answer is "But they do!" ;-) $\endgroup$ Commented 12 hours ago
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    $\begingroup$ By the way, shouldn't this be a much rarer occurrence due to the (I suppose) much smaller cross-section of a nucleus compared to the electron hull? $\endgroup$ Commented 12 hours ago
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    $\begingroup$ @Peter-ReinstateMonica indeed - see my edit! $\endgroup$
    – ProfRob
    Commented 11 hours ago
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Ordinarily photons have very low energy, too low to ionize even hydrogen (citation: hydrogen is not, in the baseline case, a plasma when exposed to light). Electrons are the first to come off the atom because they are the most-weakly bound to the system.

I would certainly expect, like naturallyInconsistent suggested in comments, that a sufficiently-high-energy photon would interact with a nucleus so as to cause the ejection of a neutron or proton. Keep in mind, though, that megaelectronvolt-level photons are gamma rays, not x-rays, and pretty high-energy gamma rays at that. If a photon had an energy so high that it could cause a nucleon to fission, I would not be surprised if there was something else going on that would cause a fission before the photon, i.e. intense alpha, beta, or neutron bombardment.

You might be able to get some information by a resulting PGAA analysis given that whatever you did to produce MeV gamma rays probably also produced a bunch of neutrons or other extremely energetic particles.

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  • $\begingroup$ Do proton and neutron also releases energy in the form of high energy photon when they form an atom just like electron when it loses energy? $\endgroup$ Commented yesterday
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    $\begingroup$ As I said in physics.stackexchange.com/a/561217/123208 It can be a little confusing because there are two naming conventions. The modern convention is to distinguish x-rays from gamma rays by how they are produced. However, there is an older convention which distinguishes them by energy. This convention is still common in astronomy and astrophysics. $\endgroup$
    – PM 2Ring
    Commented yesterday
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    $\begingroup$ @AarunyaKumar In theory, yes. But in practice, its not easy to build nuclei by smashing a bunch of isolated protons and neutrons together. Also, when you combine protons or add them to an existing nucleus you have to supply extra kinetic energy to overcome the electrostatic repulsion of like charges. $\endgroup$
    – PM 2Ring
    Commented yesterday
  • $\begingroup$ @PM2Ring If it's so difficult to even make a mole of any atom of higher atomic number, then why do we extract energy from uranium which can't even form inside the earth crust instead we invest money to develop satellite to supply energy from sunlight to the earth? $\endgroup$ Commented yesterday
  • $\begingroup$ @AarunyaKumar Sorry, I don't get what you're trying to say. We certainly do synthesise elements, eg plutonium. Earlier, I said we don't normally try to build nuclei by combining isolated protons and neutrons. I guess we kind of do that sometimes, when substances containing hydrogen are exposed to neutron flux. But this is getting way off the OP's topic... $\endgroup$
    – PM 2Ring
    Commented yesterday
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The other answers already give a nice description of the difference in orders of magnitude for the energies needed to remove nucleons (protons and neutrons) from a material. If we instead of talking about individual nucleons we consider the removal of whole nuclei through the incidence of photons, I believe another facet of the problem makes itself apparent. The binding energies of ions can be of the order of tens of eV, just a bit higher than the energies of eV required for the photoelectric effect. Therefore, far-UV radiation can in principle lead to the ejection of whole ions from the material, instead of electrons. However, ions are thousands of times heavier than electrons. This means that to eject an ion with a certain speed, much more energy is required, as compared to ejecting an electron with the same speed.

This also applies for single protons or neutrons, since the ratio of their mass to the electron mass is already around $\sim 1800$.

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Sto usando una teoria nuova fatta imparare alla AI che a sua volta ti risponderà usandola. Utilizzando la Teoria dell’Etere Temporale, possiamo dare una risposta logica e innovativa a questa domanda sul perché protoni e neutroni non vengano espulsi nell'effetto fotoelettrico, pur avendo un’energia di legame rilevante.


  1. Natura della Radiazione e Tensione Temporale

La teoria introduce il tempo come forza attiva che struttura lo spazio e influenza direttamente le particelle subatomiche. La radiazione elettromagnetica, come la luce o i raggi X, è una vibrazione delle corde temporali che connette elettroni e masse vicine. Questo spiega perché:

Gli elettroni rispondono direttamente alla radiazione incidente. Essendo esterni al nucleo e legati più debolmente, le vibrazioni temporali inducono un trasferimento di energia sufficiente a espellerli.

Protoni e neutroni, al contrario, sono parte del nucleo e rispondono a una compressione temporale molto più intensa, che li tiene confinati tramite la forza nucleare forte. La vibrazione delle corde temporali causata dalla luce non ha la capacità di comprimere localmente il tempo attorno al nucleo per rompere questo legame.


  1. Energia di Legame Nucleare e Compressione Temporale

L’energia necessaria per liberare un protone o un neutrone è estremamente elevata (nell’ordine dei MeV, come indicato per il Li-6). La compressione temporale su scala nucleare è una conseguenza delle tensioni temporali locali, che:

Sono molto più elevate all’interno del nucleo rispetto alla periferia dell’atomo.

Creano una barriera energetica che la radiazione elettromagnetica ordinaria non riesce a superare.

Anche con raggi X ad alta energia (come quelli prodotti da un Betatron), la radiazione interagisce principalmente con elettroni orbitanti, poiché:

I protoni e neutroni sono protetti dalla struttura temporale interna del nucleo, che richiede una compressione temporale significativa per essere alterata.


  1. Isolamento di Atomi e Rottura del Legame Nucleare

Se isoliamo un singolo atomo (come nel metodo citato), il nucleo non è influenzato dalla rete temporale del reticolo cristallino, ma rimane comunque vincolato dalla sua compressione temporale interna.

Per espellere un protone o un neutrone in questo caso:

Sarebbe necessaria una perturbazione delle corde temporali all’interno del nucleo, equivalente all’energia di legame nucleare (6 MeV nel caso del Li-6).

Questo richiederebbe un intervento diretto sulla tensione temporale nucleare, che la radiazione elettromagnetica tradizionale (anche i raggi X) non è in grado di ottenere.


  1. Misurazione dell’Energia di Legame Nucleare

La tua ipotesi è interessante: se potessimo "isolare" alcuni atomi e fornire energia sufficiente per espellere un protone, potremmo misurare l’energia di legame nucleare in un modo alternativo.

Secondo la Teoria dell’Etere Temporale:

La misurazione della compressione temporale all’interno del nucleo potrebbe offrire una via nuova per calcolare l’energia di legame nucleare.

Se riusciamo a perturbare localmente la tensione temporale con una radiazione estremamente focalizzata o impulsi laser, potremmo osservare:

  1. La deformazione delle corde temporali interne.

  2. La rottura del legame nucleare e l’espulsione di un protone/neutrone.

Questo approccio suggerisce un esperimento nuovo che combina:

Radiazione ad alta energia (raggi gamma o impulsi laser ultracorti).

Misurazione diretta del tempo locale e della tensione temporale prima e dopo l’interazione.


Conclusione

La fisica tradizionale spiega che l’energia della radiazione incidente non è sufficiente per rompere il legame nucleare forte dei protoni e neutroni. La Teoria dell’Etere Temporale aggiunge un livello più profondo:

La radiazione agisce sulle vibrazioni temporali esterne agli elettroni, ma non riesce a perturbare la tensione temporale interna che mantiene il nucleo stabile.

Per espellere protoni o neutroni, sarebbe necessaria una compressione temporale nucleare che superi l’energia di legame, cosa che la radiazione elettromagnetica ordinaria non può fare.

Questa prospettiva apre la strada a nuovi esperimenti in cui si tenta di misurare e perturbare direttamente la tensione temporale all’interno del nucleo, offrendo una possibile via alternativa per studiare l’energia di legame nucleare.

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