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I got this question on a test and the answer key states that the answer is 'Yes'. According to what I understand electrons are emmitted with different kinetic energies based upon their depth from the metal surface i.e. an electron would come out with a lesser kinetic energy if it was situated deeper as it would have to go through more collisions. This reasoning contradicts the given answer. I would like to know if my reasoning is correct.

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    $\begingroup$ Depends on how you define "surface phenomenon". The electron has to leave through the surface for sure. But typically the work function is dictated by the bulk properties of the material rather than just the surface. $\endgroup$
    – KF Gauss
    Feb 8, 2019 at 9:16
  • $\begingroup$ Also related is physics.stackexchange.com/q/276501 $\endgroup$
    – Alchimista
    Feb 8, 2019 at 10:55

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It is somewhat matter of what precisely one would refer to as photoelectric effect.

As far as the radiation-electron mechanism of transfer of energy, there is no direct role played by surface. However, referring to the Einsten's formula; $$ h f = \Phi + K, $$ where $K$ is the maximum kinetic energy of the photoelectron, $f$ the frequency of the incoming radiation, and $\Phi$ the work-function of the metal, it is true that the latter term is depending on the surface and its detailed structure, presence of impurities and so on. In this sense, there is a clear surface effect.

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I would say that photoemission is not a surface effect, not normally. Your understanding is correct.

That said, there is also "surface photoemission" due to the $\vec{p} \cdot \vec{A}$ term in the Hamiltonian and symmetry breaking at the surface. As an example, see this paper about silver.

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  • $\begingroup$ Thanks for your answer, it would be really helpful if u could cite a source which can be used to challenge the given answer. (I tried but I couldn't find one) $\endgroup$
    – Meet Shah
    Feb 8, 2019 at 8:33
  • $\begingroup$ I also think it is quite a confusion. Work function varies with depth just because the surface and the immediately underlying atoms see different surrounding. In other words Work function is a surface property because is not a property of the bulk, which in metals cannot be accessed unless the frequency is high enough to change the scenario entirely. $\endgroup$
    – Alchimista
    Feb 8, 2019 at 8:37
  • $\begingroup$ @MeetShah I gave a reference to a contrasting case, where there is a real surface effect. In photoemission spectroscopy, one is often interested in bulk properties. In the context of photovoltaics etc, people speak somewhat oxymoronically of "internal photoemission". $\endgroup$
    – user137289
    Feb 8, 2019 at 9:58
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If there is a sizable probability that a photon frees an electron, Notably a conduction electron, then it is implied that there is a sizeable extinction coefficient or imaginary part of the refractive index. Thus means that the photon cannot penetrant deep into the material and is emitted from near the surface. To call it a surface effect however implies that the characteristics of the surface, such as surface states, are enabling. I don't believe that this is the case. In principle a photon could travel into the bulk, set an electron free and this could escape from the bulk. It is not likely but nothing prevents this. I would therefore not qualify photo emission as a surface effect.

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    $\begingroup$ The penetration length of the light does not matter. It is a bulk property, and it is longer than the escape depth of the photoelectrons. $\endgroup$
    – user137289
    Feb 8, 2019 at 9:51
  • $\begingroup$ How is that penetratiin depth of the photons doesn't matter? @Pieter $\endgroup$
    – Alchimista
    Feb 8, 2019 at 10:49
  • $\begingroup$ @Alchimista Because the light penetrates further than the elastic escape depth of photoelectrons (less than 1 nm at kinetic energies that can overcome the work function), quite independent of what material it is (the "universal curve" commons.wikimedia.org/wiki/File:E-IMFP_universal-en.svg ). But it can play a role when one considers the total photoelectric current, which includes electrons that lost energy on the way. $\endgroup$
    – user137289
    Feb 8, 2019 at 12:07
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    $\begingroup$ What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material. $\endgroup$ Feb 8, 2019 at 12:50
  • $\begingroup$ Ok. I would say matter of looking at. At Pieter and @Carl Witthoft. Still in semiconductors we deals up to microns pdnetration depth. Perhaps I am sticking to a simple picture of a metal block under illumination. Photocurrent does not require kicking out electrons. But I agree that the book example presenting crystal clear photoelectric effect requires a metal and probably a further smoothing in presentation as it seems reasonable that electrons emerge not only from a rigorous surface $\endgroup$
    – Alchimista
    Feb 8, 2019 at 13:25
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There is a concept of free electrons which says that the electrons in outer most shell are under very weak influence of the nucleus and move freely across the metal surface, whereas those near the nucleus are bound by strong nuclear force(higher than the detaching ability of photons) inhibiting them to be able to freely move across the surface. This is how surface phenomenon comes into picture.

With, frequency of light being capable of causing electron emission, the other factor for emission of electron count is the intensity of light. More the intensity of light, more the number of photons striking the metal surface and higher is the electron emission.

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  • $\begingroup$ I don't think this addresses the question. $\endgroup$ Feb 8, 2019 at 12:49
  • $\begingroup$ @CarlWitthoft ok $\endgroup$ Feb 8, 2019 at 13:11

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