curiousatlarg said:
Photons striking elecrons in metal will add energy and excite the electron. If the energy of the photon is greater than the threshold energy for that metal, the electron will be ejected. However, if the photon is not, the electron will return to it's origional orbit, but in the process, shouldn't it release a photon? The reason I mentioned IR is because it is lower energy, and it is comon as thermal radiation under these conditions. Also, since IR is readily emitted by the tungsten, shoudn't it be readiy absorbed? The main thing I am not clear on is the converson of energy from photons to metal. If we assume all of the energy striking the filament is from photons, then, what are the principle ways that the energy is conducted in the filament? As for the generation of current, I was referring to Herts's experiment. As for the conversion to thermal energy, wouldn't IR be one of the principle forms? Other forms relate to the motion of the atoms. Sorry that I am not more concise.
curiousatlarg said:
OK, that makes sense. Next, what happens to the EM radiation? In some cases, electrons are ejected (for wavelengths below 293 nm), other cases, photons are emitted. In others a current is produced (if the metal object is the right length. For example, would a nano antenna generate electricity from light? Heat is also produced. In other cases, the EM passes around the metal or is reflected.
Strangely enough, your question here isn't really about "light traveling in wires", but rather the details of a photoemission process! You are asking what happened to the electrons that never made it out, and to be able to answer that, you need to understand what exactly is going on in the photoemission process at every step of the way.
We can do this by following Spicer's 3-Step model for the photoemission process.
Step 1: A photon is absorbed by the metal, and a conduction electron is excited to the vacuum state (i.e. above the work function of the metal).
Step 2: via random diffusion/scattering process, the "lucky" electron migrates to the surface.
Step 3: the electrons that make it to the surface then escapes the metal.
Now, what happens to the ones that did not escape. There are several scenarios:
1. In Step 2, since this is a purely random process, it means that statistically, HALF of the electrons will end up moving in the wrong direction, i.e. they will have components of their velocity moving away from the surface. Those will not make it out. This is why in a simple, standard metal, you'll never get a quantum efficiency (QE - number of electrons out per photon) greater than 50%. In fact, for metals, the QE is often 0.01% - 0.0001%! These "lost" electrons will eventually lose their energy and decay back into the conduction band.
2. Also in Step 2, these excited electrons can bump into other electrons (both excited and not excited). Remember, these are metals, and there are a lot of conduction electrons. Electron-electron scattering tends to remove energy from the energetic ones. So in just one single collision with another electron, that excited electron can lose its energy and drops back into the conduction band, below the work function. Here, you can already see that the primary mechanism for energy loss isn't the emission light, but a transfer of energy to the rest of the conduction electron.
Zz.