electron VS. photon


by Dannyinjapan
Tags: electron, photon
Dannyinjapan
Dannyinjapan is offline
#1
May20-05, 08:15 AM
P: 8
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?
Phys.Org News Partner Physics news on Phys.org
Sensitive detection method may help impede illicit nuclear trafficking
CERN: World-record current in a superconductor
Beam on target: CEBAF accelerator achieves 12 GeV commissioning milestone
El Hombre Invisible
El Hombre Invisible is offline
#2
May20-05, 08:26 AM
P: 1,017
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
ZapperZ is offline
#3
May20-05, 08:30 AM
Mentor
ZapperZ's Avatar
P: 28,783
Quote Quote by 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?
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
Dannyinjapan is offline
#4
May20-05, 09:12 AM
P: 8

electron VS. photon


Replenished from where?
(thank you, by the way)
ZapperZ
ZapperZ is offline
#5
May20-05, 09:18 AM
Mentor
ZapperZ's Avatar
P: 28,783
Quote Quote by Dannyinjapan
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
DaveC426913 is offline
#6
May20-05, 09:20 AM
DaveC426913's Avatar
P: 15,325
Quote Quote by Dannyinjapan
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
DaveC426913 is offline
#7
May20-05, 09:21 AM
DaveC426913's Avatar
P: 15,325
Doh! Beat by ZZ
Dannyinjapan
Dannyinjapan is offline
#8
May20-05, 10:31 AM
P: 8
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
masudr is offline
#9
May20-05, 11:36 AM
P: 932
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
Dannyinjapan is offline
#10
May21-05, 01:39 AM
P: 8
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 Im going to go back to my books for now and maybe I can come back to the forum in a while, Im just having a little trouble digesting all of this....
Definitely interesting though....
ZapperZ
ZapperZ is offline
#11
May21-05, 09:23 AM
Mentor
ZapperZ's Avatar
P: 28,783
Quote Quote by 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 Im going to go back to my books for now and maybe I can come back to the forum in a while, Im 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
SpaceTiger is offline
#12
May21-05, 10:08 AM
Emeritus
Sci Advisor
PF Gold
SpaceTiger's Avatar
P: 2,977
Quote Quote by Dannyinjapan
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

[tex]Q=\frac{Nq_e}{4\pi R_e^2}[/tex]

where N is the number of electrons removed, qe is the charge of the electron, and Re 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
vanesch is offline
#13
May21-05, 12:18 PM
Emeritus
Sci Advisor
PF Gold
P: 6,238
Quote Quote by masudr
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
ZapperZ is offline
#14
May21-05, 01:41 PM
Mentor
ZapperZ's Avatar
P: 28,783
Quote Quote by vanesch
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
vanesch is offline
#15
May22-05, 02:25 AM
Emeritus
Sci Advisor
PF Gold
P: 6,238
Quote Quote by ZapperZ
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
Dannyinjapan is offline
#16
May23-05, 07:50 PM
P: 8
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
Gza is offline
#17
May23-05, 10:15 PM
Gza's Avatar
P: 525
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 thats 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
Dannyinjapan is offline
#18
May24-05, 07:33 AM
P: 8
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, Im new to this stuff..


Register to reply

Related Discussions
Any Known Similarities between the electron and the photon? General Physics 5
Photon-Electron Collision Introductory Physics Homework 1
Photon-Electron Collision Quantum Physics 1
re:photon, not the electron... General Physics 1
photon, not the electron... General Physics 29