Can any1 explain this process and why it supposedly only works on metals.
A paradox that the person that explained it is now asking how it works
Electrons in metals are more loosely bound. Therefore it takes less energy to cause them to be ejected. With high enough energy photons, the effect can occur in anything.
So if the photon energy is great, everything we look at is ejecting electrons?
Not exactly. It depends on the number of electrons in the outer shell as to wether or not it will gain or loose electrons easlily or at all. For example any stable element and/or compound will not be affected by it. As for semi-conductors.... Well, inorder to tell you that, I would have to be given the exact amount of enery the photon gives off in order to see if it could dislodge one of the electrons from its orbit. But Mainly it works well on conductive materials that have a low stability of 1 to 3 electrons in their outer shell where semi-conductors have 4 to 6 and insulators have 7 or 8(it is stable if it has 8). So it depends on how many electrons are in the outer shell and the amount of energy given off from the electron in order to determine wether or not you can dislodge the electrons to create electric flow.
Correct me if I am wrong about the amount of electrons in the outer shell for conductive,semi-condutive, and insulative materials. I'm a little rusty on the facts but I beleive that is right.
To clarify my point, if you hit anything with gamma rays, electrons will be knocked out. With ordinary light, it only works with loosely bound electrons.
The energy of photons increases in this order:
Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays.
There is a wide range of energies within each of these categories of photons; for example, red light is low energy visible photons, and blue light is high energy visible photons.
When a photon hits an electron it "jumps up" but the positively charged nucleus pulls it back down (well, that's a simplistic explanation, but it will do for now).
Pardon me if I hit your toes as I step backward here:
If a photon with enough energy hits the electron, the electron will jump sufficiently high to escape the pull of the nucleus. For the loosely bound electrons of metals (aka "free electrons") the required energy happens to coincide with the energy of visible photons.
This only applies to one or two of the electron in a metal. THe rest of the electrons are tightly bound.
For other things, a much higher energy photon is required to knock away an electron ; X-rays will do a fine job taking away electrons leaving positively charged atoms behind, hence the category: of "ionizing radiation."
HOwever even a high energy photoelectric effect would not work on non-conducting materials (or would it?) . Anyway, blasting stuff with x-rays would require more energy then you could possibly recover from the effect.
One example of the photoelectric effect -Lasers used to create plasma.
How this works, you take a high power laser (it can be IF, visible, or UV, it doesnt matter.) well actually you should use 2, when you cross their beams, enough energy is present to completely knock off all electrons, thus creating plasma. This is generally how plasma is made for experiments. If you dont use two and you use one very strong laser then you just get a nice little ball a few centimeters in front of the focusing lenses.
Something they forgot to tell you about the effect
The photoelectric effect is based entirely on the frequency(energy) of the photon(piece of light) hitting the metal target, and it does not matter how many photons hit the piece of metal, when determining IF an electron will be ejected from the metal's surface.
To make an easily understandable metaphor....If you were at a shooting range and your target was a 1 cm thick piece of metal and your objective was to knock out a piece of the back of the metal target. You could never achiever your goal by shooting a pellet gun at the target, even if you shot 1000(amplitude) BBs at the target, they would not eject any tiny piece of metal out the back of the target. BUT, if you fired a single shot from a high powered (high energy/high frequency) AK-47 at the target, you WOULD penetrate the target and pop out a small piece of the target at the bullet pushes through it.
Intuitively, if you shot 1000 rounds from the AK-47 at the target, you WOULD pop out 1000 tiny pieces of the metal target.
In a sentence: Its the power of the bullet, not the number of bullets, that determines whether you 'punch' out a piece of the metal target or not.
Light carries a certain amount of energy (proof: put you hand in sunlight and feel the heat). That energy "knocks-off" electrons.
You've said "supposedly" - why. Photoelecrtic effect id used in cinematography where sound is "written" like dark and transparent lines on film stripe - that's latter on transfered into variable electrical signal when light is passed/blocked on that stripe - that signal is brought (amlified) to speakers - and that's sound.
The most important thing u r all ignoring is that the element used 2 carry out photoelectic effect were metals, so the could lose their electrons (everybody knows that). But most of the things u see around urself are not free atoms( like the metals), they are complex compounds who have reacted forming molecular orbitals and have reached the maximum satability they can achieve. Also the aren't just simple compounds they are very complex and huge compounds.Eg:- Millions of photons strike our body in a day. If we were losing electrons we might have become +ve charged and might have touched any -ve ly charged object or just a small current will be able to attract us towards itself thus we should all have been dead but this doesn't happen. The only clear reason visible to my eyes is that our body is made up of cells which themseleves are made up of millions of atoms. When we strike the photon to any one of the atom then ( as we all know the electron is in molecular orbitals) this energy soon spreads as the other bonding atoms try to stabalize it and thus the photon is unable to excite and remove the electrons.
But in the case of radiations with higher energy like the UV rays strike the are not able to eject the electrons completely but the are able to distort the chemicals i.e change their structure and thus causing cancer.
If more stronger radiations fall like the gamma rays we would all die cause gamma rays ( except heat) is a reason for the death of living beings in case of explosion of an atom bomb.
Thus the bottom line is --- Common things around us are compounds having big structures and we should consider these molecules rather than individual atoms in our discussion.
A short but clear explanation
The photons can only eject electrons when the frequency is greater than the work function. we know that the energy of the photon would be equal to hf. For eg if hf=phi than f is known as the threshold frequency here phi means the workfunction,now if hf>phi than the remaining energy converts to kinetic energy of the electron.
Metals are preffered over other materials for the only reason that they have lower workfunctios.
I hope it helps.
Look, the phenomenon occurs everywhere, you simply need a metal plate so that you can apply a potential difference across it to guage the maximum kinetic energy of the electrons being ejected. That is not to say that there isn't any number of electrons being 'ejected', we only measure the maximum so that we can find the slope of the line which describes the constant h, plank's constant.
Dear einstein like you asked why this effect happens only in metals well this happens only
in metals because of the property of the metal to conduct electricity very good
In simple terms, the electrons in good conductors are loosely bound, it therefore takes less energy to liberate these electrons than those in other materials (hence a lower work function). In materials where the electrons are more tightly bound, the work function is so high that the incomming light simply does have sufficient energy to liberate the electrons.
First of all, this is a rather old question.
Secondly, the photoelectric or photoemission effect in general does not just happen in metals. It can happen in dielectric as well if you have a photon with high enough energy. There are many photoemission studies on metals and insulators. In fact, in RF photoinjectors, we are now using semiconductors as our photocathodes as high quantum efficiency electron sources.
So no, it does not just happen in metals.
i have to congratulate you, you know about this subject.
My name is Ricardo if you want to keep in touch i am richard14 in the forum.
What is the Photoelectric Effect?
Asked by: Kevin Mcgill
The photoelectric effect refers to the emission, or ejection, of electrons from the surface of, generally, a metal in response to incident light.
"GENERALLY in a metal!
Why? Because for within the typical range of visible light, metals have the LOWEST work function, and so, they are the ones most often used.
But the photoelectric effect, or more GENERALLY, photoemission, as defined by the emission of electrons from solids via light, can also be seen from semiconductors and insulators. The parent compound of high-Tc superconductors is a ceramic insulator and many photoemission studies have been done of them. Would you like me to give you exact citation? Here's one example:
I currently am doing photoemission work on the semiconductor cesium telluride as a high QE photocathode.
Thank you for the complement; but I don't know anywhere near as much about photoemission (or pretty much anything else) as ZapperZ, as he said he currently researches photoemission. Take note of what Zz is saying, photoemission can happen in non-metals but it is much more likely in metals for the reason both me and Zz have stated above.
you are right zapper absolutely like you said it can happen also in semiconductors
like in photodiodes etc...
Separate names with a comma.