Superb picture and really nice demonstration Max. My guess is the current and voltage are different here. Looking at the two images I would say the red image has a higher current and lower voltage across the lamp. Where as the voltage is higher with lower current in the blue image. This way you can select which energy transition occurs in the gasses.

Thanks! You're very close (both voltage and current are high in the bottom case), but you're missing something crucial here.

Ah, I'm intrigued. If the lower image has a high current and voltage then the electrodes will also be in thermionic emission and lower work-function. Not sure if that's getting closer or still missing the crucial point and going down a rabbit hole though.

You're not there yet. Something important is missing...

ok, I had been thinking it was a Penning mixture but paying a bit more attention see it’s mercury and neon. So, a couple of things now come to mind. The lamp temperature could be low enough there is no mercury present in a vapour state. Also, could you be using a DC supply? I think mercury can migrate but that would mean the other electrode would be blue.

Not there yet... there is no electrophoresis or cold temperature involved here. Mercury is present in the vapor phase everywhere in the tube in both cases. So, what's different?

You got me a bit stumped. I think I have gone full circle and back to the relationships for the transitional electron energies required for the neon and mercury. This being lower for mercury than neon. Hence controlling the likelihood of one or other discharge by voltage (eV). The only other relationship which I am not fully sure of is from observation with high frequency supplies appearing to influences particular transitions over other, but I haven't got a good explanation for that yet. Anything to do with kinetic energy in between the gas atoms, am I going to kick myself when you tell me?

You're getting there, but that's not it. Here's a hint: In the bottom case both the voltage and current are high. That goes against the natural propensity of the gas discharge's impedance to decrease at higher currents. So, how did I manage to run the bottom tube in a regime that is not affected by this natural characteristic?

Did wonder a bit what you meant by high V and I at the same time. If the magnitude is quite high, the only way I think this is achievable would be a pulse discharge similar to a xenon flash lamp. This fits with one observation I have in the lower image; if the current is high I'd expect a thermal glow from the electrode. As this isn't apparent either the current is too low or the discharge duration is very short. Does this make any sense?

Finally, we have the right answer here! Congratulation The bottom tube is fed with an unipolar pulsed current whereas the top tube is driven with a DC current. The reason why the voltage remains high under pulsed operation is because I used microsecond pulses, too short for the discharge to stabilize in any kind of steady-state ionization balance (the stabilization of the plasma's E field following a change in the applied voltage is never instantaneous, it depends mainly on the diffusion rate of charged particles). The high voltage is the result of a high electric field in the discharge, which causes the fast acceleration of electrons between the electrodes. The consequence is that the electron gas remains highly energetic there and neon atoms are the main species that are excited (the electrons are too energetic to excite mercury atoms in their lower energy states).

Phew! Bit annoyed that the back of my mind registered the lack of thermal emission but the niggle didn’t trigger thinking until your clue. Did enjoy the mental work out, that was quite a chewy one.

I do love these technical discussions.!

Glad you enjoyed it, and please don't hesitate to chime in! There are so many fascinating topics related to gas discharges and lamps in general that it's unlikely we'll run out of interesting discussions.