Light waves from a normal house hold globe effecting a dc circuit. Why?

uart

Hi Simon. The photon energy of visible light only extends up to about 3eV. ([itex]E= h c/\lambda[/itex]). This gives about 3 eV for blue light of lambda = 400 nm.

The work function of common metals is around 3 to 5 electron volts while the first ionization energy of neon is around 21eV.

Of course the photon doesn't need to provide all of the energy, as the applied electric field is already close to ripping the electrons from the surface anyway. Given the energies involved however, I suggest it is far more likely the photoelectric effect at the neon bulb cathode is the main culprit here.

Re my previous reply on this topic.

calvinator

Excellent. Thank you. Would you know of any text/ebooks that can give me detailed accounts of his study. Or just google? I have so much to learn.

Simon Bridge

You can just google "photoelectric effect"... it's very well documented. (It's what got Einstein his Nobel Prize.)

I'm not going to argue the toss over where the photoelectric effect is occurring :) but I've certainly added to my understanding here.
Something of a "doh!" experience.

uart

Yeah the wikipedia article is pretty good.

Ok. :) But realize that a 21eV photon is extremely hard UV radiation (almost getting up to xray territory).

As the tube voltage is increased, it is the surface of the cathode that is the "weak link", not the gas molecules themselves. This can easily be seen by the fact that gas tubes will begin their discharge at a lower voltage as the pressure of the gas is reduced. This is something that confuses many students. They reason that if the gas is somehow allowing the discharge current to flow then more gas should allow easier conduction, not less.

The actual reason for this is that as the voltage is increased, the first thing "to give" is the electrons at the cathode surface, and not the gas atoms ionizing by themselves. When an electron gets ripped from the cathode it begins accelerating in the electric field, but if the mean free path is too low (gas density too high) it collides with a neon molecule before it has gained the required energy (>21eV) to cause ionization, and release further electrons to sustain the process.

BTW. This is the same reason that we heat the cathode of electron tube devices. The physical mechanism is quite different, but the effect much the same, the electrons are more easily ripped away from the cathode by the electric field.

Simon Bridge

Oh sure but since the applied field is so high you'd get both wouldn't you? All things being equal, you'd expect photoelectric to be at least an order of magnitude stronger than photoionization. I didn't get a good look at the tube package - sometimes the cathode is in a blacked-out bit at the bottom so light may not reach it so much... not arguing not arguing! Arrgh!

This is pretty neat though ... OP could just apply a voltage to the bulb and crank it down until it stops glowing (in the dark), then switch the light on and presto.

I suppose to be proper scientists we should propose an experiment .... maybe first check the cathode is exposed to light then try painting out more of the bottom of the bulb and see what effect it has. A directional light source would help too.

Hmmm ... on student CRT rigs, some are blacked out and others left clear - I don't remember seeing any difference in the intensity of the beam when there was light falling on the heated cathodes but I'm still not arguing... really...