Is photoelectric effect a surface phenomenon?












12












$begingroup$


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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  • 1




    $begingroup$
    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
    20 hours ago










  • $begingroup$
    Also related is physics.stackexchange.com/q/276501
    $endgroup$
    – Alchimista
    19 hours ago


















12












$begingroup$


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.










share|cite|improve this question









New contributor




Meet Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    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
    20 hours ago










  • $begingroup$
    Also related is physics.stackexchange.com/q/276501
    $endgroup$
    – Alchimista
    19 hours ago
















12












12








12


1



$begingroup$


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.










share|cite|improve this question









New contributor




Meet Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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.







photoelectric-effect






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Meet Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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Meet Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 21 hours ago







Meet Shah













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asked 22 hours ago









Meet ShahMeet Shah

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Meet Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Meet Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    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
    20 hours ago










  • $begingroup$
    Also related is physics.stackexchange.com/q/276501
    $endgroup$
    – Alchimista
    19 hours ago
















  • 1




    $begingroup$
    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
    20 hours ago










  • $begingroup$
    Also related is physics.stackexchange.com/q/276501
    $endgroup$
    – Alchimista
    19 hours ago










1




1




$begingroup$
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
20 hours ago




$begingroup$
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
20 hours ago












$begingroup$
Also related is physics.stackexchange.com/q/276501
$endgroup$
– Alchimista
19 hours ago






$begingroup$
Also related is physics.stackexchange.com/q/276501
$endgroup$
– Alchimista
19 hours ago












4 Answers
4






active

oldest

votes


















11












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
      21 hours ago










    • $begingroup$
      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
      21 hours ago












    • $begingroup$
      @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".
      $endgroup$
      – Pieter
      20 hours ago





















    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      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.
      $endgroup$
      – Pieter
      20 hours ago












    • $begingroup$
      How is that penetratiin depth of the photons doesn't matter? @Pieter
      $endgroup$
      – Alchimista
      19 hours ago










    • $begingroup$
      @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.
      $endgroup$
      – Pieter
      17 hours ago








    • 1




      $begingroup$
      What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
      $endgroup$
      – Carl Witthoft
      17 hours ago










    • $begingroup$
      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
      16 hours ago



















    -2












    $begingroup$

    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.






    share|cite|improve this answer








    New contributor




    Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$













    • $begingroup$
      I don't think this addresses the question.
      $endgroup$
      – Carl Witthoft
      17 hours ago










    • $begingroup$
      @CarlWitthoft ok
      $endgroup$
      – Devanshu Kashyap
      16 hours ago











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    11












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      11












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        11












        11








        11





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 22 hours ago









        GiorgioPGiorgioP

        2,843321




        2,843321























            5












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
              21 hours ago










            • $begingroup$
              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
              21 hours ago












            • $begingroup$
              @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".
              $endgroup$
              – Pieter
              20 hours ago


















            5












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
              21 hours ago










            • $begingroup$
              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
              21 hours ago












            • $begingroup$
              @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".
              $endgroup$
              – Pieter
              20 hours ago
















            5












            5








            5





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 22 hours ago









            PieterPieter

            8,29131432




            8,29131432












            • $begingroup$
              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
              21 hours ago










            • $begingroup$
              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
              21 hours ago












            • $begingroup$
              @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".
              $endgroup$
              – Pieter
              20 hours ago




















            • $begingroup$
              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
              21 hours ago










            • $begingroup$
              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
              21 hours ago












            • $begingroup$
              @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".
              $endgroup$
              – Pieter
              20 hours ago


















            $begingroup$
            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
            21 hours ago




            $begingroup$
            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
            21 hours ago












            $begingroup$
            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
            21 hours ago






            $begingroup$
            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
            21 hours ago














            $begingroup$
            @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".
            $endgroup$
            – Pieter
            20 hours ago






            $begingroup$
            @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".
            $endgroup$
            – Pieter
            20 hours ago













            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              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.
              $endgroup$
              – Pieter
              20 hours ago












            • $begingroup$
              How is that penetratiin depth of the photons doesn't matter? @Pieter
              $endgroup$
              – Alchimista
              19 hours ago










            • $begingroup$
              @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.
              $endgroup$
              – Pieter
              17 hours ago








            • 1




              $begingroup$
              What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              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
              16 hours ago
















            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              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.
              $endgroup$
              – Pieter
              20 hours ago












            • $begingroup$
              How is that penetratiin depth of the photons doesn't matter? @Pieter
              $endgroup$
              – Alchimista
              19 hours ago










            • $begingroup$
              @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.
              $endgroup$
              – Pieter
              17 hours ago








            • 1




              $begingroup$
              What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              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
              16 hours ago














            1












            1








            1





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 21 hours ago









            my2ctsmy2cts

            5,2222618




            5,2222618








            • 1




              $begingroup$
              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.
              $endgroup$
              – Pieter
              20 hours ago












            • $begingroup$
              How is that penetratiin depth of the photons doesn't matter? @Pieter
              $endgroup$
              – Alchimista
              19 hours ago










            • $begingroup$
              @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.
              $endgroup$
              – Pieter
              17 hours ago








            • 1




              $begingroup$
              What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              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
              16 hours ago














            • 1




              $begingroup$
              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.
              $endgroup$
              – Pieter
              20 hours ago












            • $begingroup$
              How is that penetratiin depth of the photons doesn't matter? @Pieter
              $endgroup$
              – Alchimista
              19 hours ago










            • $begingroup$
              @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.
              $endgroup$
              – Pieter
              17 hours ago








            • 1




              $begingroup$
              What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              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
              16 hours ago








            1




            1




            $begingroup$
            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.
            $endgroup$
            – Pieter
            20 hours ago






            $begingroup$
            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.
            $endgroup$
            – Pieter
            20 hours ago














            $begingroup$
            How is that penetratiin depth of the photons doesn't matter? @Pieter
            $endgroup$
            – Alchimista
            19 hours ago




            $begingroup$
            How is that penetratiin depth of the photons doesn't matter? @Pieter
            $endgroup$
            – Alchimista
            19 hours ago












            $begingroup$
            @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.
            $endgroup$
            – Pieter
            17 hours ago






            $begingroup$
            @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.
            $endgroup$
            – Pieter
            17 hours ago






            1




            1




            $begingroup$
            What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
            $endgroup$
            – Carl Witthoft
            17 hours ago




            $begingroup$
            What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
            $endgroup$
            – Carl Witthoft
            17 hours ago












            $begingroup$
            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
            16 hours ago




            $begingroup$
            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
            16 hours ago











            -2












            $begingroup$

            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.






            share|cite|improve this answer








            New contributor




            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              I don't think this addresses the question.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              @CarlWitthoft ok
              $endgroup$
              – Devanshu Kashyap
              16 hours ago
















            -2












            $begingroup$

            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.






            share|cite|improve this answer








            New contributor




            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              I don't think this addresses the question.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              @CarlWitthoft ok
              $endgroup$
              – Devanshu Kashyap
              16 hours ago














            -2












            -2








            -2





            $begingroup$

            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.






            share|cite|improve this answer








            New contributor




            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            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.







            share|cite|improve this answer








            New contributor




            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer






            New contributor




            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 20 hours ago









            Devanshu KashyapDevanshu Kashyap

            11




            11




            New contributor




            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





            New contributor





            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            Devanshu Kashyap is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.












            • $begingroup$
              I don't think this addresses the question.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              @CarlWitthoft ok
              $endgroup$
              – Devanshu Kashyap
              16 hours ago


















            • $begingroup$
              I don't think this addresses the question.
              $endgroup$
              – Carl Witthoft
              17 hours ago










            • $begingroup$
              @CarlWitthoft ok
              $endgroup$
              – Devanshu Kashyap
              16 hours ago
















            $begingroup$
            I don't think this addresses the question.
            $endgroup$
            – Carl Witthoft
            17 hours ago




            $begingroup$
            I don't think this addresses the question.
            $endgroup$
            – Carl Witthoft
            17 hours ago












            $begingroup$
            @CarlWitthoft ok
            $endgroup$
            – Devanshu Kashyap
            16 hours ago




            $begingroup$
            @CarlWitthoft ok
            $endgroup$
            – Devanshu Kashyap
            16 hours ago










            Meet Shah is a new contributor. Be nice, and check out our Code of Conduct.










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            Meet Shah is a new contributor. Be nice, and check out our Code of Conduct.
















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