8
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ 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? $\endgroup$
    – naturallyInconsistent
    Commented yesterday
  • 2
    $\begingroup$ "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
  • 4
    $\begingroup$ 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. $\endgroup$
    – Sebastian Riese
    Commented 4 hours ago

3 Answers 3

Reset to default
26
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ In other words, the answer is "But they do!" ;-) $\endgroup$
    – Peter - Reinstate Monica
    Commented 13 hours ago
  • 2
    $\begingroup$ 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? $\endgroup$
    – Peter - Reinstate Monica
    Commented 13 hours ago
  • 3
    $\begingroup$ @Peter-ReinstateMonica indeed - see my edit! $\endgroup$
    – ProfRob
    Commented 12 hours ago
8
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$
    – Aarunya Kumar
    Commented yesterday
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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? $\endgroup$
    – Aarunya Kumar
    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
0
$\begingroup$

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 $\sim 1800$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.