electron physics question - Page 1
Forum section: electronics / projects

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: February 28, 2016, 02:09:48 am
content: It occurred to me: what is the electrical difference between the photoelectric effect popping electrons out of a metal, and thermionic emission popping electrons out of a metal?In other words, the cathode of a (directly heated) vacuum tube can offer x amps per square meter through thermionic emission, how much light would you need to shine on the same cathode to get as much density? Is it even possible?Is the photoelectric effect a surface thing, and thermionic is obviously a bulk thing, but in either case, the surface has to pop electrons out, no? Does the photoelectric effect "target" only specific electrons in the shell, but the thermionic effect makes more electrons in the material "available"?What I'm trying to ask, is what prevents the use of a strongly illuminated surface as a cathode? What's the physics? What can I read up on?

author: Andy Watson
author_rank: Super Contributor
timestamp: February 28, 2016, 03:39:44 am
content: It's getting a bit late in the day here and my brain is fading... so briefly:Quote from: Alex Eisenhut on February 28, 2016, 02:09:48 amBasically, you need to supply enough energy to allow the electrons to escape from the surface - they need to be able to overcome the work-function of the metal. Light or heat will do.QuoteYes it is possible - though perhaps not practical. It is the principle used in photo-multiplier tubes. You can calculate the number of electrons emitted if you know the work-function of the metal and the frequency of the light.QuoteThe light would not penetrate much into the metal, and, almost by definition, an electron needs to leave the surface - otherwise it is still trapped in the body of the metal - which is usually considered to be "a sea of electrons".QuoteThe frequency of the light needs to exceed the work-function energy of the metal. What do you mean by "strong" - note that the photon energy is not the same thing as the intensity of the light.The photo-electric effect was one of the anomalies that lead to thinking about quantum mechanics - most QM texts begin with a consideration of this phenomenon.

author: IanB
author_rank: Super Contributor
timestamp: February 28, 2016, 03:41:23 am
content: Quote from: Alex Eisenhut on February 28, 2016, 02:09:48 amNothing? For example:https://en.wikipedia.org/wiki/Video_camera_tube

author: CatalinaWOW
author_rank: Super Contributor
timestamp: February 28, 2016, 04:31:57 am
content: While it works, it seems much harder than the thermal path. QE of photoemission ranges from a little above .2 (for special materials at the best wavelength) down. Down by several orders of magnitude. So you generate a large amount of light. Collect as much as you can and point it at your cathode, as best as you can. Then only a small fraction of your photons generate electrons.Assume you use LEDs as the light source. Maybe as many as 20% of your electrons generate photons (very generous estimate). You then manage to get half of those to your cathode. Again generous. And then get 10% of those to generate photoelectrons if you can get a good match between cathode material and your LED. 1% efficiency. And some fancy LEDs if you need very much current out of your cathode.It does give you a potentially much higher speed way of controlling the available electrons than is possible with thermoemission.

author: vk6zgo
author_rank: Super Contributor
timestamp: February 28, 2016, 04:32:34 am
content: Yes,the Image Orthicon was just such a device,as is the "garden variety" photo-multiplier tube.https://en.wikipedia.org/wiki/PhotomultiplierThe I.O. had the interesting distinction of being replaced by a device with lower performance.The later generation Plumbicons were "nearly as good",but vidicons were pretty dire.Using bright light to cause emission in normal amplifier tubes would use far more power than thermionics,as well as being mechanically messy to implement.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 28, 2016, 08:58:16 am
content: As far as the equivalence, I don't know that optimal thermionic surfaces are also optimal photoemissive surfaces. A standard oxide cathode is white as snow, so obviously, it isn't absorbing much in the visible range (say 1-3 eV), at least at room temperature (but I don't think the color changes with heating, either). Meanwhile, when heated to a fairly modest temperature (red heat, ~1eV), it gives off bountiful emission. (The energy level of the emission is comparable, as one should expect; the voltage offset is something like 0.5V, with an exponential tail. This is visible in the curves of sensitive tubes (virtual cathode effect), and low level measurements, likethis plot of 6AL5 at low voltage.)Mechanisms: obviously, physical surfaces are "sharper" than optical waves. Atoms in a crystal are fractional nm apart, while visible photons are hundreds of nm: thousands of atomic layers. Metal layers of similar thickness are required to shield EM radiation in this range, or generally, to interact with it at all. (Metals exhibit resistance, so they have a predominant lossy e_r characteristic. The relatively large |e_r| reduces the speed of light, makes the skin depth shallower, and the loss causes attenuation and phase shift through the thickness of the layer. So actually, only hundreds of atomic layers are necessary.)So, it stands to reason that the photoelectric effect occurs within the bulk of a substance, and only extremely rarely on the (atomic) surface as such. It also follows that some material thickness is necessary, though it can be quite a thin film (~10s nm?). The excited electron has to exit the material and enter space, which is itself probably a low-yield phenomenon. Consider photon emission from an LED: it's limited by index of refraction (and thus internal reflection). As far as I know, electrons rather "like" to be inside crystal lattices, and in the same way that a photon's velocity changes in a crystal, the electron's effective mass also changes, which I expect makes it prone to internal reflection.For what it's worth, the electron "sizes" we're talking about here are on the order of 10-100 atomic layers (1.7nm de Broglie wavelength at 0.5eV, in space), so there should be meaningful propagation/reflection within a crystal. Which of course would lead to a low quantum efficiency.So, we also must be very careful what we mean by surface. Because an atomic surface, relevant to phenomena like gas adsorption, shallow-angle x-ray reflection/scattering, various kinds of microscopes (STEM, AFM) and so on, is around a tenth of a nanometer "thick" (basically, the reach of atomic valence orbitals in the crystal), while the optical depth "surface" is tens of nanometers (perhaps no more than 200 nm?).Apparently, oxide cathodes utilize a thin layer of barium as the active substance, which is achieved during manufacture (something to do with decomposition of the original material, and preferential evaporation). The layer is too thin to see (i.e., to interact with light to a significant amount), but works just fine for electrons heated internally by the bulk.It's also interesting that the oxide is porous. At high current densities, the space charge can be stripped away, exposing surface grains to electric field; but still more current is available (it doesn't suddenly go into electron saturation), because of space charge tucked away in the pores of the material. There's also absorption and re-emission between grains, leading to resistive losses (the effective cathode voltage rises at high currents). I wonder how much of the surface area is important to the emission capacity: is it desirable to use a fractal surface? How does a fractal (or at least rough/featured) surface compare in terms of efficiency (emission/watt radiated)?Tim

author: kikinou
author_rank: Contributor
timestamp: February 28, 2016, 09:21:28 am
content: Above replies probably answered your question.Quote from: Alex Eisenhut on February 28, 2016, 02:09:48 amCohen Tannoudji, chapter 13, complement DAhcroft/Mermin, chapter 18

author: RogerRowland
author_rank: Regular Contributor
timestamp: February 28, 2016, 09:24:12 am
content: Quote from: Alex Eisenhut on February 28, 2016, 02:09:48 amIf I've understood the point of your question correctly, you might also like to read about field emission (FE), which is not exactly the photoelectric effect but does have some useful background of the general physics:https://en.wikipedia.org/wiki/Field_electron_emissionWe use both FE and thermionic emission to produce electron beams in scanning electron microscopes (SEM). In some cases, we use a combination of both.

author: wolf32d
author_rank: Contributor
timestamp: March 18, 2016, 05:53:17 pm
content: Quote from: Alex Eisenhut on February 28, 2016, 02:09:48 amThere are two basic ways to produce electrons:- by thermoionic emission.- by photoelectric effect, i.e. shining a laser (it's photon energy must exceed the metal's work function) on a cold cathode.The former method is very easy to implement: you take a crappy tungsten filament and you wire it to a power supply. The downside is that the emitted electrons have a random spin distribution.The latter method is in general more costly and costly but allows to produce a lot of spin-polarized electrons (their angular momentum doesn't have a random distribution)

author: TimFox
author_rank: Super Contributor
timestamp: March 18, 2016, 08:55:43 pm
content: Normally, photoemission is one electron emitted per photon interaction, where the photon must have sufficient energy to overcome the work function (barrier that keeps electrons inside the metal).Thermionic emission involves electrons at the high-energy "tail" of the energy distribution for electrons in a high-temperature metal, that can overcome the work function due to their high energy.Field emission results when the applied field gives a net potential (combination of potential from the field and the work function of the material) through which electrons can tunnel. Quantum mechanics allows a small probability of tunneling through such a "wall" when the energy outside the wall is lower than the energy inside the wall (over-simplifying a bit). Classical mechanics would not allow any tunneling in this case.

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: March 18, 2016, 11:39:15 pm
content: Thanks. It's clear I've forgotten a lot since college; I am now baffled by energy level diagrams and valence bands and what not. I know I did very well in physics back then, I threw out all my school notes last fall!But I've been going through the basics again at leisure. I'm still not sure what the "eta" (the Greek letter) is between the two effects; is thermionic emission more efficient somehow? I get that it's "easier", if you skip over all the technical details of getting a tungsten coil to do what you want.Basically I want to know why we aren't shining bright lights at cathodes?

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 19, 2016, 12:12:17 am
content: So put numbers on it. Generating one amp of photocurrent is generating ~6E18 electons/sec. QE is quantum efficiency, or number of electrons generated for each incident photon. From datasheets on phototubes, which have presumably been optimized for generation of electrons the QE peaks at about 0.2. So about 3E19 photons/sec are needed to generate an amp of photocurrent. The phototubes work in the visible so photon energy will be about 3E-16 Joules. So 9E3 Joules in one second, which is an optical power of roughly 10,000 watts. Both tough to generate, and far more than needed to generate a comparable thermo current.It would be impossible to directly compare these. A sizable fraction of the photons which don't generate photoelectrons will be converted to heat. So there will be huge thermoemission just from the waste heat of the optical path.Only at very low power levels, where the waste heat is insufficient to generate thermoelectrons will the photocurrent dominate. Which is why the only practical application of this effect has been in detectors operating at very low currents.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 19, 2016, 12:48:32 am
content: I think you slipped a decimal in there. The power required for a 100% efficient photoemission process, is simply the photon energy in eV, times the resulting electric current, divided by e, to get eV * I / e = V*I = P in watts. For visible photons, that's 1-3W, or for a quantum efficiency in the 20% range, 5-15W.(A flux of 10kW upon even a rather large cathode would also heat it more than enough to achieve thermal emission, so the quantum efficiency wouldn't be nearly so bad in that case either...)Tim

author: John Heath
author_rank: Frequent Contributor
timestamp: March 19, 2016, 11:39:23 pm
content: Quote from: T3sl4co1l on February 28, 2016, 08:58:16 amNice post. In the case of CRTs with older color TVs the surface area of cathode electrons availability would degrade over time. Not sure why . We used a brute force method of discharging a condenser from cathode to grid. This would resurface the cathode to improve emission. Some of the old farts such as myself will remember doing this. CRT rejuvenation was the buzz word. Back to the original question. Blue photons have an energy of 2 to 3 eV electron volts so it should have enough force to kick a few electrons off a cathode. This can be measured directly with the forward bias voltage a blue LED. It is not often that an atomic event can be measured with a simple voltage meter. Interesting note is the electron mass energy and photon energy are described in the same units of blue photon energy 2 to 3 eV and electron mass energy .511 MeV .

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 20, 2016, 02:42:30 am
content: Quote from: T3sl4co1l on March 19, 2016, 12:48:32 amYou are right. Long day and away from home and I skipped a power of 1000 in the photon energy.You and I agree about thermal emissions in a high current photocathode. Not as bad as with my botched numbers, but in any real case trying to operate at one amp type current levels it will be a real issue. The waste heat from the photons is of the same order as the power used to heat normal thermionic cathodes.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 20, 2016, 08:48:40 pm
content: Using just a tungsten wire for thermal emission to get a current in the 1 A range also needs quite some power. I remember a setup where it took something like 50 W to get 0.5 A out. The special oxides are much more efficient though.So something like 15 W of optical power in the blue light is not that bad - 405 nm laser diodes are quite efficient, though usually lower power. For a tube current in the more usual 1-10 mA range this would not be so bad.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 20, 2016, 10:07:52 pm
content: Still not really a competitive approach. 15W optical power, but you have to either put the diodes inside the vacuum envelope, or have reflection losses. You need a clear optical path through the same general region that you want to put grids and plates. And due to LED efficiency, electrical power is comparable at best.Added complexity for no real benefit.

author: d-smes
author_rank: Regular Contributor
timestamp: March 23, 2016, 04:46:36 am
content: Quote from: TimFox on March 18, 2016, 08:55:43 pmSay you managed to generate huge electron emission. Can one design an anode to generate photons due to electron bombardment? If so, and this created a regenerative process such that more and more emission occurred, wouldn't anode voltage collapse and quench the process? Or would it stabilize at some saturation voltage? I ask because processes like field emission and my questioned anodic photon emission depend on a high anode voltage. And usually the objective is as low an on-state voltage as possible when a tube is saturated with electron flow.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 23, 2016, 09:42:11 am
content: Magnetrons behave roughly that way, I suppose. Below threshold, electrons spiral back to the cathode and current flow is small; above threshold, electrons are pulled with sufficient force to spiral around and eventually impinge on the anode, delivering current. All the while, the tangential motion couples to the resonant EM field, driving energy into it.The anode voltage isn't an obvious quantum figure (~1meV photons produced from a ~1kV voltage drop), because energy is delivered to the field many times during the flight of a given electron. It's given by crossed-fields electron physics instead. (Even neglecting noise, it's doubtful the voltage-to-frequency ratio should be visibly quantized anyway, because space charge can act to 'cushion' particle effects, namely shot noise.)I don't think there are many materials which exhibit, uh, electrophotic effect? That is, giving out light of characteristic (optical range) wavelengths due to electron bombardment above a threshold energy. The most common case seems to be x-ray production (bremsstrahlung), but that's a high energy phenomenon; I've never heard that electron bombardment results in visible light, except with phosphors, which are known to emit light from most any energetic stimulus (including more light!).Of course you could always strap a GaAlAs or whatever junction, onto an anode in a vacuum, and bombard that and get light... but that's no different than hooking wires to it as an LED.(Which isn't as silly as it sounds. Semiconductors have been used for image intensification and storage (some Vidicon camera tubes), or in hybrids with a photodiode (avalanche or otherwise) beneath a photoemissive layer, or a multiplier chain or MCP. I think Hamamatsu still makes such HPDs.)Tim