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Say you have a vacuum tube, such as the kind used in old amplifiers, wherein electrons are accelerated from the cathode to the anode through an electric field. Presuming, for the sake of argument, that electrons coming off of the cathode have a zero energy, and the anode, made of a material with a work function of 4eV, is at a potential of +24V relative to the cathode, then wouldn't a photon with an energy of 28eV be emitted upon impact of the electron on the anode? Using E=h*f and c=lambda*f, then wouldn't the wavelength of this emitted photon be ~44.28nm? That's beyond the highest reaches of ultraviolet light and into X-ray territory, so why are vacuum tubes safe? Is there something wrong with my understanding of how all this electron/photon business works?

Second side question if you care to answer it:

I turn off the filament in my vacuum tube and let the cathode get cold, but I keep it plugged into the circuit, with the anode still at +24V relative to cathode, then I shine a light at 635nm on the cathode. Supposing my cathode has a work function of 4eV as well, do electrons hop off of the cathode due to photoelectric effect or not? Without the anode there, I would say no, because h*f-4eV is negative (-2.05eV), meaning the electron doesn't have enough energy to escape the metal plate. With the anode however, there's an electric field pulling the electron as well, so can it make the jump? I calculate it would reach the anode with 25.95eV of energy if it did.

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3 Answers 3

When the electron strikes the anode there are many ways it can dissipate it's 24eV worth of energy. It could eject an electron from some atom in the anode, and that excited atom could emit a photon as it recombines with an electron, however it is vastly more likely that the energy will end up as lattice vibration i.e. heat. The amount of radiation emitted from the anode is very small, but it does happen and is used as the basis of soft x-ray emission spectroscopy.

In any case 24eV would be an exceedingly soft X-ray, if it could be called an X-ray at all. At that energy any radiation would be rapidly absorbed by pretty much everything including the glass of your valve.

Re your second question, in the absence of the anode the photoemission would charge the cathode and increase the work function. So you'd get some photemission as first, but it would stop after a while. With the anode I guess the electrons would indeed arrive with more than 24ev energy.

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Due to the work function of the anode, isn't the energy dissipated 28eV? In your last paragraph, do you mean charge the cathode? Also, I thought that, since the energy gained by an electron would be less than the 4eV work function of the metal due to 635nm light, that no electrons would jump off of the cathode unless the 24V potential to anode was there to reduce the work function. I could be mistaken, but can you clarify? –  Big Endian Jan 16 at 19:09

"..Say you have a vacuum tube, such as the kind used in old amplifiers, .."

So let's say we do. Nobody would build a vacuum tube with a plate made out of a 4 Volt work function metal. The electron colliding with the plate, is absorbed, and its KE simply heats the anode. So the anodes of vacuum tubes are made from alloys that have very low thermionic emission; so secondary electron emission is suppressed.

Also, good luck on finding a vacuum tube (as used in old amplifiers) that would do much of anything, with just 24 Volts on the anode. 100 to 500 Volts might be a more typical range; but maybe as low as 50-60, in some "starvation" amplifiers. 75-225 would cover most of the designs, I'm familiar with.

So the effect you postulate does not happen.

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@ george e. smith. Actually vacuum tubes have existed that were able to operate at low voltages. They were used at times where radio's were battery powered and high voltage batteries were expensive. These tubes contained an extra grid right after the cathode at some small positive voltage that collected space charge, thus allowing the tube to operate. This can also be achieved with modern tubes. See pi4raz.nl/… for a receiver working at 6V. The article is in Dutch but the schematics will be clear. –  Urgje Jan 16 at 10:26
    
Well, I believe the OP said vacuum tubes "such as the kind used in old amplifiers." That would explicitly exclude the more modern types you describe. I never designed with any tubes after Nuvistors; and none of them contained a positive first grid. But the exception defines the rule. –  george e. smith Jan 17 at 0:06
    
I looked at your Dutch radio schematic. The pentodes look conventional, but the circuit definitely is not. The grid is at cathode Voltage and the input signal is fed to the suppressor grid, so it operates at a quite low gm value. But considering that crystal set radios operate with no active signal gain at all, I can believe this radio works. The early battery operated radios, I have built, used the Eveready 467 battery for plate Voltage supply, which was a 67 1/2 Volt battery Voltaic pile battery that was quite economical. A 45 Volt version was also available. –  george e. smith Jan 17 at 23:55
    
In the text it is explained that the first grid is simply not used. The second grid is used to remove the space charge (note the potentiometer for adjusting its voltage) and the third one is now the active one. I was surprised that this setup actually worked. The gain of the tube will be miserable. This was also true for the original tubes used way back and it explains why the concept never became popular. –  Urgje Jan 18 at 11:31
    
A small search showed that space charge tubes are not a relic from the 1920th but had a revival in the 1950th in connection with car radios. See junkbox.com/electronics/lowvoltagetubes.shtml –  Urgje Jan 19 at 12:24

Well it won't let me comment, so a new answer.

Urgie, I couldn't read the Dutch, but I pretty much figured it out.

Actually a quite cunning circuit.

By grounding (to the cathode), the grid, the tube is operating at zero bias, instead of what would normally be perhaps -2 Volts.

So at zero bias, the current would normally be quite high; but that lets them run the tube at the low plate Voltage. The first grid, is on the border of having a grid current, which normally would be bad, but doesn't matter in this case since the only grid to cathode Voltage, would be a work function difference, and any grid current isn't going to hurt anything.

So now the screen grid acts normally to extract electrons from the cathode region, and send them on to the plate through the third "suppressor" grid. Because the current is lower than normal, and the suppressor grid is a long way from the cathode, the gm for the grid three signal input, is dismal, but not zero, so you do get some useful gain.

The tube behaves more like a triode than a pentode, and now the suppressor grid to plate capacitance is much bigger than the grid 1 to plate capacitance, so the Miller effect of the g3-p capacitance is large, so the circuit can't be very fast, but probably ok for a radio.

Yes quite a clever use of the pentode. A lot of pentodes have the suppressor grid grounded to the cathode, so you can't connect to it signalwise. It is mainly to suppress secondary emission from the plate in normal operation.

As a practical matter, the anode is virtually always completely enclosing everything else, so in the event that any EM radiation was emitted from collision with the anode, very little could escape to the outside anyway; so OP's concern is unfounded.

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