Electron vs Photon: The Photoelectric Effect Explained

[Dannyinjapan]

Forgive me if this question is too simple or in the wrong place, but my texts have confused me and I need some clarification.

The photoelectric effect says that photons strike a metal surface and are absorbed by outer electrons, giving them enough energy to break free and escape.
So, what do we have left?
Does that mean that, if we left a light shining on a piece of metal long enough, that eventually there would be a hole burned in the metal?
Are these electrons replaced?

What happenes to the atom? It has absorbed a photon and lost an electron.
Is the total energey of the system lower or higher?

 

[El Hombre Invisible]

The number of atoms in the metal is not reduced - just the number of electrons. The metal becomes more positively charged due to the loss of electrons so, no, they are not replaced. The energy of the photon is lost as the kinetic energy of the emitted electron. This is how they discovered that the intensity of the light is due to the number of photons - brighter light did not result in faster electrons, just more of them. It was the frequency/wavelength of the light that caused electrons to have higher energies - i.e. the wavelength/frequency corresponded to the energy of the photon.

 

[ZapperZ]

Forgive me if this question is too simple or in the wrong place, but my texts have confused me and I need some clarification.

The photoelectric effect says that photons strike a metal surface and are absorbed by outer electrons, giving them enough energy to break free and escape.
So, what do we have left?
Does that mean that, if we left a light shining on a piece of metal long enough, that eventually there would be a hole burned in the metal?
Are these electrons replaced?

Yes. The metal is always grounded in a typical photoelectric effect experiment. So the emitted electrons are almost instantaneously replenished. If not, you have what is known as the "charging effect", and eventually, no more electrons will be emitted since it will have to not only overcome the work function, but also the additional net coulombic attraction.

What happenes to the atom? It has absorbed a photon and lost an electron.
Is the total energey of the system lower or higher?

It has nothing to do with an "atom". Remember that you are doing this on a METAL. When a glob of atoms form a solid like a metal, most often they lose their individuality and the valence electrons now form their own energy state, often a continuous band. This is what makes the study of materials very different than the study of atomic-molecular physics. In a metal, you now have to deal with electrons in a conduction band that really are not localized, or don't belong to a single atom. Rather, it belows to the whole bulk material. So the energy state of the atom is really not affected. And since the electrons are continuously replenished, no "excited" state of the metals occur in this case.

Zz.

 

[Dannyinjapan]

Replenished from where?
(thank you, by the way)

 

[ZapperZ]

Replenished from where?
(thank you, by the way)

Do you know what it means when a conductor is "grounded"? This knowledge will be something you would need when you do electromagnetic theory and electrical circuits.

Zz.

 

[DaveC426913]

Replenished from where?
(thank you, by the way)

"The metal is always grounded in a typical photoelectric effect experiment."

It's replenished from ground. i.e. Earth is a giant undepletable source of electrons.

You are creating a current flow.

When you electrocute yourself, it's usually because you've created a path from a current source "to ground" (such as a puddle of water). This means the circuit has access to an undepletable source of electrons (or undepletable "sink" for electrons to fill).

 

[DaveC426913]

Doh! Beat by ZZ

 

[Dannyinjapan]

I knew about grounding, but it didnt click. Can you tell me how these negatively charged particles of matter somehow go from the ground to the atoms that are missing electrons?

 

[masudr]

The "ground" is connected by a low-resistance path to the metal plate. So electrons can come (and go) to the Earth through this path from the metal.

 

[Dannyinjapan]

You see, I guess this is where I get confused.
People say "electrons" and the book says electrons are particles.
Matter.
How can matter from the Earth just jump up and replenish the missing matter from a piece of metal?
Are we saying that electrons are both matter and energy?

What is the true basic unit of energy?
The quanta?
The photon?
The electron?

I guess I am going to go back to my books for now and maybe I can come back to the forum in a while, I am just having a little trouble digesting all of this...
Definitely interesting though...

 

[ZapperZ]

You see, I guess this is where I get confused.
People say "electrons" and the book says electrons are particles.
Matter.
How can matter from the Earth just jump up and replenish the missing matter from a piece of metal?
Are we saying that electrons are both matter and energy?

What is the true basic unit of energy?
The quanta?
The photon?
The electron?

I guess I am going to go back to my books for now and maybe I can come back to the forum in a while, I am just having a little trouble digesting all of this...
Definitely interesting though...

You will then have trouble not only with the photoelectric effect/quantum physics, but also electrical circuit, and classical electromagnetism. In BOTH areas of study, there is such a thing as a "ground" connection. There's nothing special or unique about it. It is simply a "charge reservoir". And if you have problems with a concept of a "reservoir", then you will also have problems with classical thermodynamics that has a "heat" and "cold reservoir".

My point here is that this is not a unique concept that is only applied to quantum physics or the photoelectric effect. I do not understand why you are finding it so difficult to comprehend. A charge reservoir simply replenish whatever deficiencies an object has so that it has the same "potential" as ground, the very same way a heat reservoir will supply an object with the necessary heat to get to its "heat potential".

If you are having issues with this already, then you will face with more daunting stuff to understand later that are even more complex.

Zz.

 

[SpaceTiger]

How can matter from the Earth just jump up and replenish the missing matter from a piece of metal?

When an object is "grounded", it means it's in contact with a much larger body that has many electrons that can be easily carried from one point to another. How do they get from one point to the other? By the electric force, of course.

Imagine that I have a much smaller conductor, like a little metal sphere, that isn't grounded, so there are no electrons available to it from outside. If I shine light of the appropriate energy on it, some of its electrons will be removed and it the sphere will now have more positive charges than negative charges. Now the interesting thing about a conductor is that this net positive charge will be now be distributed "uniformly" on its surface, such that the surface is all at the same potential. Why? Well, if it's not distributed this way, then the charges on one part of the sphere will be pulling on the charges on another part of the sphere. Since charge is free to move, this force will result in motion of charge. The charge will keep moving around until it has reached some sort of balance where the net force on any charge in the sphere is zero. Not surprisingly, this balance occurs when the charge is uniformly distributed on the sphere (this just follows from symmetry).

What if this conductor is much larger, like the earth? The effects are basically the same; that is, the charge will still distribute to make an equipotential, but this time the excess charge will be much more spread out. In the case we're discussing, you're effectively removing a charge of a few electrons from a conductor the size of the Earth (assuming the experiment is grounded). If this charge were to distribute uniformly on the surface of the earth, the net overdensity of charge (just from the electrons you removed) in one square meter would be

$$Q=\frac{Nq_e}{4\pi R_e^2}$$

where N is the number of electrons removed, q_e is the charge of the electron, and R_e is the radius of the earth. For 100 electrons, this equates to ~10^-36 Coulombs! You won't get a strong force from that.

In practice, the Earth isn't quite as ideal a conductor as this, but the idea is the same. Removing such a small amount of charge from a conductor as large as that will result in the charge being effectively "replenished" at the point where the charge was removed. Charge is still conserved, however, and if you did enough photoelectric experiments (billions of billions!), the charge difference on the Earth might become noticable, but such things make no practical difference.

It's also worth noting that the Earth does carry a noticable amount of charge on its surface and the potential difference between the surface and upper atmosphere is quite large. This is because lightning storms (tens of thousands per day) will periodically deposit large numbers of electrons on the surface. So, another way to look at why your experiment makes no difference to the Earth's charge is to consider how it compares to tens of thousands of bolts of lightning.

 

[vanesch]

The "ground" is connected by a low-resistance path to the metal plate.

Ha, that would mean that a photo-electric cell doesn't work in a satellite :-)
It is true that the photo-cathode is "grounded", but that doesn't necessarily mean that it is connected to the planet Earth. It just means it is connected to, eh, the reference conductor (usually the metal box in which things happen). Now, the liberated electrons from the photocathode usually go somewhere, usually a dynode or something else... and that one is then ALSO connected to the reference conductor (through resistors and a power supply or anything of the kind). So you finally have some closed circuit in which the electrons leave the photocathode, go somewhere, and through some circuitry get back to the reference electrode, and back to the photocathode (which is connected to it). There can of course be some "temporary borrowing" of electrons, but in a steady state functioning, there is always a closed circuit. And the "ground" is just one of the electrodes in that closed circuit.

cheers,
Patrick.

 

[ZapperZ]

Ha, that would mean that a photo-electric cell doesn't work in a satellite :-)
It is true that the photo-cathode is "grounded", but that doesn't necessarily mean that it is connected to the planet Earth. It just means it is connected to, eh, the reference conductor (usually the metal box in which things happen). Now, the liberated electrons from the photocathode usually go somewhere, usually a dynode or something else... and that one is then ALSO connected to the reference conductor (through resistors and a power supply or anything of the kind). So you finally have some closed circuit in which the electrons leave the photocathode, go somewhere, and through some circuitry get back to the reference electrode, and back to the photocathode (which is connected to it). There can of course be some "temporary borrowing" of electrons, but in a steady state functioning, there is always a closed circuit. And the "ground" is just one of the electrodes in that closed circuit.

cheers,
Patrick.

But this isn't the same thing, is it? The photocells used in here are not exactly a "photoelectric effect" in actions. You are not liberating electrons into vacuum, but merely promoting electrons into the conduction band to create a potential difference - the electrons stay within the material. Furthermore, the photocells use p-n junction diodes made up of semiconductors.

I don't think you can make a fair comparison between the two.

Zz.

 

[vanesch]

You are not liberating electrons into vacuum, but merely promoting electrons into the conduction band to create a potential difference - the electrons stay within the material. Furthermore, the photocells use p-n junction diodes made up of semiconductors.

Well, the point (which is very basic) I wanted to make was more accute for "oldfashioned" PM than for photodiodes, but basicly comes down to the same thing. If some process "liberates" electrons (whether it is physically removing them from a metal such as in a PM, or just changing them from one band to another), there are 2 possibilities: an electrical circuit CLOSES the circuit and the electrons (well, not the same ones maybe :-) which left a place finally end up going around the circuit and fill up the holes again ; or the circuit is NOT closed, in which case an E-field will build up (as you said), finally stopping any emission (or band transition, due to the band deformation because of the growing potential difference).
A working measurement system always has a CLOSED circuit in one way or another ; usually ONE conductor in that closed circuit is labeled "ground" and usually it is connected to the metal box in which you do the things. I just wanted to point out that there is no need to invoke the EARTH.
You can, if you want to, connect the conductor, that is called "ground" also to Earth (except if you're in a satellite). But that has more to do with problems of security than anything else, and certainly isn't an essential part of the detection circuitry.

Now back in the OLD days (19th century), the Earth WAS important, and somehow this is still kept in traditions and (mis) understandings: old telegraphy wires were SINGLE wires, and the Earth was used as the return conductor. Earth is still used also in power distribution systems, because you don't want a power device floating at 300000 V, because YOU can then close the circuit with Earth as an unwanted return conductor. That's why electrical appliances need to be somehow connected to earth: if you touch them, that you don't, by accident, make a new closed circuit through which current can flow (and kill you). But normally, Earth doesn't have anything to do with the correct functioning of any device or circuit.

The reason to label a conductor in a working, closed circuit, GROUND, and to connect it to the metal casing is something else. The reason is essentially that this is the best conductor that can serve as a potential reference, because it is all pervasive, present anywhere, and of very low impedance. Being the casing, it is also the conductor that is most exposed to potential fluctuations from external influences. So it might seem, at first sight, paradoxial to take the most perturbed conductor as reference ! But because it is taken as a reference, and EVERYTHING in the circuit works with it as a reference, these potential fluctuations are done away with. If we would have taken another conductor as a reference, it would then see this casing change in potential, and so would be all the other conductors. Capacitive coupling from the casing to any other conductor would then induce fluctuating potentials and currents everywhere and the circuit would be perturbed.

cheers,
Patrick.

 

[Dannyinjapan]

I think I understand all that you guys have just said, but somehow my fundamental question is still here.
Let me try it again.

The physics book says that an electron is a particle of matter. It has mass and charge. The way you talk about it though, you make it sound as though the electron was just energy.
Is it matter or is it energy?
If it is energy, then how can there be a positive electron (positron) ?
If it is matter, how can it just "flow" across a copper wire?

 

[Gza]

The physics book says that an electron is a particle of matter. It has mass and charge. The way you talk about it though, you make it sound as though the electron was just energy.
Is it matter or is it energy?
If it is energy, then how can there be a positive electron (positron) ?
If it is matter, how can it just "flow" across a copper wire?

From what I've read so far, it doesn't seem that anyone is trying to describe electrons as energy, since that would not really make much sense in this case. Anything with mass has a rest energy associated with it (with the average layman jumping up to blurt out E=mc^2), but don't think about the problem that way. Electrons are just little particles, "flowing" across a copper wire (yes the electrons flow across the wire, and bump into the wire's atoms on the way; don't believe me? touch a wire that's been connected across a battery or any potential difference and you will feel the heat as a result of the collisions of the electrons with the internals of the wire.)

 

[Dannyinjapan]

Ok, now we're getting somewhere. Thanks
So, when these electrons move across a wire and through a circuit to do some kind of work, in say, a light bulb, what do the electrons do after they have passed through the light bulb? Are they converted to photons or are they changed and part of their energy used to create potons?
Im sorry if this seems really basic, but like I said, I am new to this stuff..

 

[Omegatron]

Ha, that would mean that a photo-electric cell doesn't work in a satellite :-)

See http://www.eas.asu.edu/~holbert/eee460/spc-chrg.html for info about spacecraft charging from the photoelectric effect. :-)

In BOTH areas of study, there is such a thing as a "ground" connection. There's nothing special or unique about it. It is simply a "charge reservoir". And if you have problems with a concept of a "reservoir", then you will also have problems with classical thermodynamics that has a "heat" and "cold reservoir".

In electronics, "ground" is just the point in the circuit defined as 0 volts. You can define any point as ground and measure other voltages relative to that point (though there are conventions for deciding which point to label). It isn't a reservoir in this instance. This shows the water analogy to electronics and includes a big ground reservoir. But if you think about it, no water will ever move from the reservoir into the rest of the circuit, since the circuit is already full of water. I don't understand the point of the reservoir in their drawing, either. They don't use it on this version , though.

Anything with mass has a rest energy associated with it

Nooooo! That will just confuse people! :-)

Ok, now we're getting somewhere. Thanks
So, when these electrons move across a wire and through a circuit to do some kind of work, in say, a light bulb, what do the electrons do after they have passed through the light bulb? Are they converted to photons or are they changed and part of their energy used to create potons?
Im sorry if this seems really basic, but like I said, I am new to this stuff..

Nothing happens to the electrons! They just keep moving through. The same number of electrons go into the light bulb as come out of the light bulb. They are not any slower or any faster on either side. Imagine a water in pipe analogy like above. You have a continuous loop of pipe that is completely filled with water. At one end you have a pump. At the other, you have a waterwheel to be pushed against. The same water continuously moves through the entire circuit, at the same speed at all points, yet energy is transferred from the pump to the load. All they do in the lightbulb is they are confined to a very thin wire, so the passage of lots of electrons random movement through all the atoms of the wire heats up the wire through friction, and it starts to glow from the heat. This is called blackbody radiation and comes from all objects, but not in the range of visible light until things get hot (where the phrases "red hot" "white hot" etc come from).

Electrons are not energy! Electrons are just little things stuffed in a pipe that can be pushed down the pipe.

 

[ZapperZ]

In electronics, "ground" is just the point in the circuit defined as 0 volts. You can define any point as ground and measure other voltages relative to that point (though there are conventions for deciding which point to label). It isn't a reservoir in this instance. This shows the water analogy to electronics and includes a big ground reservoir. But if you think about it, no water will ever move from the reservoir into the rest of the circuit, since the circuit is already full of water. I don't understand the point of the reservoir in their drawing, either. They don't use it on this version , though.

The reason why I used the analogy of a "reservoir" was because this specific question was on where the "charges" are coming from. So in this instance, it is appropriate to consider such a thing, especially when the OP is having some unspecified problem in understanding how the emitted charges are replenished. Considering such difficulties, I certainly didn't want to bring out further complications by pointing to the "zero potential" of ground.

Note that the photoelectric effect can also be run in a different way, by biasing the "cathode" negatively and "grounding" the anode. Many photocathode during fabrications are monitored this way.

Nothing happens to the electrons! They just keep moving through. The same number of electrons go into the light bulb as come out of the light bulb. They are not any slower or any faster on either side. Imagine a water in pipe analogy like above. You have a continuous loop of pipe that is completely filled with water. At one end you have a pump. At the other, you have a waterwheel to be pushed against. The same water continuously moves through the entire circuit, at the same speed at all points, yet energy is transferred from the pump to the load. All they do in the lightbulb is they are confined to a very thin wire, so the passage of lots of electrons random movement through all the atoms of the wire heats up the wire through friction, and it starts to glow from the heat. This is called blackbody radiation and comes from all objects, but not in the range of visible light until things get hot (where the phrases "red hot" "white hot" etc come from).

Electrons are not energy! Electrons are just little things stuffed in a pipe that can be pushed down the pipe.

Keep in mind that this answer is prefectly acceptable if we consider the level that we have to explain to. This is precisely the case why I had to resort to the "reservoir" analogy. However, once we dive into this in greater detail, such a picture that you have described no longer apply (i.e. you can't tell which electron is which in a fermionic distribution), nor can you say that the wire heats up through "friction", because this implies the same type of "rubbing" between two surfaces. Scattering off lattice ions are not normally called "friction".

Moral of the story: if you can bring up a more complicated answer/scenario, so can I! :)

Zz.

 

[Omegatron]

The reason why I used the analogy of a "reservoir" was because this specific question was on where the "charges" are coming from.

Yeah. I was just clarifying, which I guess means I am guilty of making things confusing, too. :-)

nor can you say that the wire heats up through "friction", because this implies the same type of "rubbing" between two surfaces. Scattering off lattice ions are not normally called "friction".

oops. yeah.

 

[novjo]

The physics book says that an electron is a particle of matter. It has mass and charge. The way you talk about it though, you make it sound as though the electron was just energy.
Is it matter or is it energy?
If it is energy, then how can there be a positive electron (positron) ?
If it is matter, how can it just "flow" across a copper wire?

Correct, an electron has both mass and charge. When talking about energy, a conversion is 1eV x (1.6E-19 Joules/ 1eV), where eV stands for electron Volt. This is equal to the amount of kinetic energy gained by the electron when it accelerates through a one volt potential difference. Now, the mass of an electron is approximately 9.11E-31 kg. This is very very small! So when a potential difference is applied in a circuit, the voltage "pressure" easily moves the electrons through a copper wire. To answer your question, an electron has mass which requires energy (i.e. a related force) to move the electron. To say it has energy is different from saying it is energy. There are many forms of energy (i.e. electromagnetic, potential, thermal, etc.) which explain the relationship between a particular force and its symmetric particle. Electric charge is simply a value which helps explain a charged particle's interaction in an electromagnetic field.

 

[ZapperZ]

Just so you know that you are responding to a thread that had its last activity in 2005. The ship has left the harbor.

Zz.

 

[v2kkim]

The physics book says that an electron is a particle of matter. It has mass and charge. The way you talk about it though, you make it sound as though the electron was just energy.
Is it matter or is it energy?
If it is energy, then how can there be a positive electron (positron) ?
If it is matter, how can it just "flow" across a copper wire?

- Electron can have kinetic and also potential energy in a solid metal or semiconductor.
- There exist positron which is used in medical area to detect abnormal tissue.
- If you think electron as a hard object, then its flow in metal may be difficult to imagine,
but remember that electron behaves mostly as a quantum mechanical wave. For example, in semiconductor electrons move in silicon lattice being scattered by silicon lattice. In this electron movement the electron is treated as a wave and quantum mechanical calculation shows good agreement with experiment. If you treat electron as a simple hard object then it is almost impossible to get a good prediction on electron in semiconductor.

 

[cragar]

Do photons create a gravitational field.

 

[novjo]

Do photons create a gravitational field.

Photons are stated as having no mass. Therefore, they cannot create a gravitational field. They may be able to create a gravitational "wave" perpendicular to their path, perhaps.

 

[jtbell]

In general relativity, the stress-energy tensor (whose components depend on both energy and momentum) is the source of gravitation, not mass alone (or rather its associated energy). Someone once posted in these forums the equations for the attraction or repulsion of two parallel light beams, but I can't find it right now.

 

[cragar]

ya photons do create gravity there are two ways we can show this .
Newtons 3rd law.
if light can be bent in a gravitational field then the light has an equal but opposite reaction. and another way we can show that light creates gravity is that if we had
matter and antimatter in a sealed can with mirrors when the matter and anti matter collided they would produce photons and the gravitational field would still be there it wouldn't dissappear.
and in general relativity mass or energy creates a disstortion in space-time.

 

[cragar]

can some tell me where photons come from , like i know that an electron gets kicked up
to a higher energy level then whn it goes back down it releases a photon , but does the energy in making the electron move up to the next energy level is that the energy the transalated into the photon.

 

[v2kkim]

Photons come from many sources including atoms, solid material like semiconductor LED, matter/antimatter combination, charged particle oscillation, atomic bomb reaction.. etc

 

[cragar]

right but like are the photons inside the matter and then they are released or what am i
missing.

 

[v2kkim]

The photon is created from something like a vacuum using energy ? I wish I can answer. Maybe the vacuum is something to be ready to release photons. The vacuum may have a lot of photons and anti-photons.

 

[cragar]

there are no anti-photons, the photon is its own anti particle.