# Some questions about photoelectric effect

by Fuego
Tags: effect, photoelectric
 P: 43 Hey people, I'm doing some last minute revision for my physics A-level (not literally last minute, i have a few days till the exam) and I need something clarifying. The photoelectric effect is supposed to demonstrate particle behaviour of light. From what I have read, the energy of the photoelectrons is independent of the intensity of the light, but dependent on its frequency. Supposedly that means that light is a particle, but I can't see why. I think it has something to do with the equation $$E = hf - \phi$$. My text book says that the photoelectric effect is explained by saying the "the energy exchanged with the wave is always $$nhf$$ where n is an integer". By why does n have to be integer? Why can't n be any number? Thanks in advance.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,552 Experimental evidence. There is no "a priori" reason why the energy is always an integer multiple of hf just as there is no "a priori" reason for light to be a particle. It was, among other things, the experimental result that the energy always IS an integer multiple of hf that showed that, in this situation, light has to be treated as a particle (photon) and, eventually, lead to quantum physics.
 P: 51 n has to be an integer because it either exites an electron or it does not. If it does, it gets "absorbed" and as such transfer all of its energy to the interacting electron. The energy of the photon is E=hf as you said, so ...
 Sci Advisor HW Helper P: 1,772 Some questions about photoelectric effect ...so if it took (theoretically) 1.5 eV of energy to bounce a photoelectron out of a metal, a single photon of 1.0 eV would do nothing. Two photons of 1.0 eV each would also do nothing since they can't add up their energies. A single photon of 3.0 eV could NOT bounce out two photo electrons because it can't divide up its energy. The extra energy becomes the KE of the electron after it has escaped the metal. Many times people will bring up the argument such as: "well consider a soccerball that someone kicks up into the air. THis kick makes the ball go 3/4 of the way over the fence. What if, at just the right time another player jumps up and kicks the ball while it is still in the air and the ball then goes over the fence. Why can't this happen to electrons?" It can't happen because electrons can't exist in an equivalent state of "up in the air." Electrons don't exist "between shells." If the photon is not quite enough energy to get the electron to escape, the electron doesn't "almost escape," the electron simply "doesn't."
Emeritus
PF Gold
P: 29,238
 Quote by Chi Meson ...so if it took (theoretically) 1.5 eV of energy to bounce a photoelectron out of a metal, a single photon of 1.0 eV would do nothing. Two photons of 1.0 eV each would also do nothing since they can't add up their energies. A single photon of 3.0 eV could NOT bounce out two photo electrons because it can't divide up its energy. The extra energy becomes the KE of the electron after it has escaped the metal. Many times people will bring up the argument such as: "well consider a soccerball that someone kicks up into the air. THis kick makes the ball go 3/4 of the way over the fence. What if, at just the right time another player jumps up and kicks the ball while it is still in the air and the ball then goes over the fence. Why can't this happen to electrons?" It can't happen because electrons can't exist in an equivalent state of "up in the air." Electrons don't exist "between shells." If the photon is not quite enough energy to get the electron to escape, the electron doesn't "almost escape," the electron simply "doesn't."
Er.... hum. I think I'm in a bit of a difficult position here. Do I just let this go, or should I try to correct this.....?

A while back, I described to someone who was looking for a project to write on, an experiment that seemingly "violate" the conventional photoelectric effect. This is called the multiphoton photoemission. What happens here is that you CAN create a photoelectron using photons with energy LOWER than the material's work function! However, this requires the use of high-powered light source that has quite a large intensity, or density of photons per unit area hitting the material's surface. Let me give you an example...

Let's say you have a metal with work function of 3.6 eV. You are using photons of energy 3.3 eV. Single-photon photoemission (which is the conventional photoelectric effect) will never produce photoelectrons. However, since the conduction band in metals has a CONTINUOUS energy state, a single photon can excite an electron at the top of the occupied band up to a state 0.3 eV below the vacuum level, so it is still confined within the metal. If we have a high intensity light source, there is a substantial probability of a second, subsequent absorption by this excited electron that would give it enough energy to escape the metal. This second step has to occur very quickly since the excited electron has a liftime in that state of the order of picoseconds. The higher the intensity of the light source, the more probable this effect can be observed. Most of the light sources used in a typical photoelectric effect experiment (such as from a dischaged lamp) does not come close to having this type of intensity, which is why this effect is not seen in those experiments.

Multiphoton photoemission is in fact very well-known. It is used to investigate the image states of various metals. So yes, in some cases, the photon energies from 2 or more photons can "add up" to cause a photoemission.

Zz.
Emeritus