Is photoelectric effect a surface phenomenon?
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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.
photoelectric-effect
New contributor
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add a comment |
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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.
photoelectric-effect
New contributor
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1
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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.
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– KF Gauss
20 hours ago
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Also related is physics.stackexchange.com/q/276501
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– Alchimista
19 hours ago
add a comment |
$begingroup$
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.
photoelectric-effect
New contributor
$endgroup$
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
photoelectric-effect
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New contributor
edited 21 hours ago
Meet Shah
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asked 22 hours ago
Meet ShahMeet Shah
644
644
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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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
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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)
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– Meet Shah
21 hours ago
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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
add a comment |
$begingroup$
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.
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1
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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.
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– Pieter
20 hours ago
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How is that penetratiin depth of the photons doesn't matter? @Pieter
$endgroup$
– Alchimista
19 hours ago
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@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.
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– Pieter
17 hours ago
1
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What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
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– Carl Witthoft
17 hours ago
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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
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– Alchimista
16 hours ago
add a comment |
$begingroup$
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.
New contributor
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I don't think this addresses the question.
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– Carl Witthoft
17 hours ago
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@CarlWitthoft ok
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– Devanshu Kashyap
16 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 22 hours ago
GiorgioPGiorgioP
2,843321
2,843321
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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.
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– Pieter
17 hours ago
1
1
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What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
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– Carl Witthoft
17 hours ago
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What @Pieter said. CCDs most certainly produced photoelectrons in the bulk material.
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– Carl Witthoft
17 hours ago
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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
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– Alchimista
16 hours ago
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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
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– Alchimista
16 hours ago
add a comment |
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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.
New contributor
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I don't think this addresses the question.
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– Carl Witthoft
17 hours ago
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@CarlWitthoft ok
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– Devanshu Kashyap
16 hours ago
add a comment |
$begingroup$
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.
New contributor
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I don't think this addresses the question.
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– Carl Witthoft
17 hours ago
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@CarlWitthoft ok
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– Devanshu Kashyap
16 hours ago
add a comment |
$begingroup$
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.
New contributor
$endgroup$
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.
New contributor
New contributor
answered 20 hours ago
Devanshu KashyapDevanshu Kashyap
11
11
New contributor
New contributor
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I don't think this addresses the question.
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– Carl Witthoft
17 hours ago
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@CarlWitthoft ok
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– Devanshu Kashyap
16 hours ago
add a comment |
$begingroup$
I don't think this addresses the question.
$endgroup$
– Carl Witthoft
17 hours ago
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@CarlWitthoft ok
$endgroup$
– Devanshu Kashyap
16 hours ago
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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
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@CarlWitthoft ok
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– Devanshu Kashyap
16 hours ago
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@CarlWitthoft ok
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– Devanshu Kashyap
16 hours ago
add a comment |
Meet Shah is a new contributor. Be nice, and check out our Code of Conduct.
Meet Shah is a new contributor. Be nice, and check out our Code of Conduct.
Meet Shah is a new contributor. Be nice, and check out our Code of Conduct.
Meet Shah is a new contributor. Be nice, and check out our Code of Conduct.
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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.
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– KF Gauss
20 hours ago
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Also related is physics.stackexchange.com/q/276501
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– Alchimista
19 hours ago