UV mercury lamps: Changing the 185nm vs 253.7nm ratio

[neanderthalphysics]

According to Wikipedia,

From 85% to 90% of the UV produced by these lamps is at 253.7 nm, whereas only 5–10% is at 185 nm.

(citation needed)

https://en.wikipedia.org/wiki/Ultraviolet

What would alter this proportion of emitted radiation to favor more 185nm radiation?

Preferably not with a filter that filters out 253.7nm (mathematically would also increase the ratio of radiance of 185nm:253.7nm) but an option that causes an increase in intensity of emitted 185nm radiation.

Would heating the discharge tube alter the ratio of emitted radiation, all other things being equal?

 

[kuruman]

Would heating the discharge tube alter the ratio of emitted radiation, all other things being equal?

Probably not. As I understand it, the emission lines are produced by electric discharge that vaporizes and ionizes the mercury. Providing external heat won't do much as the vaporized mercury is already "hot". If your goal is to actually increase the 185 nm component at the expense of the 253.7 nm, i.e. in a zero sum situation, I believe it cannot be done without changing the quantum transition probabilities which, as far as I know, are not externally controllable.

 

[Vanadium 50]

Would heating the discharge tube alter the ratio of emitted radiation, all other things being equal?

A tiny bit. Both transitions are to the same daughter state (6s ¹S₀), so it's a question of populating the (6p ¹P₁) parent state (relatively) more than the lower energy state (6s ³P₁). Increasing T will do this, but T for 185 nm is about 80000 K. And you're not going to increase the heat much compared to 80000 K.

 

[etotheipi]

A tiny bit. Both transitions are to the same daughter state (6s ¹S₀), so it's a question of populating the (6p ¹P₁) parent state more than the lower energy state (6s ³P₁). Increasing T will do this, but T for 185 ns is about 80000 K. And you're not going to increase the heat much compared to 80000 K.

I have a question, what does the ^aS_b and ^aP_b notation mean?

 

[Vanadium 50]

https://en.wikipedia.org/wiki/Term_symbol

 

[etotheipi]

https://en.wikipedia.org/wiki/Term_symbol

danke schön! ☺

 

[kuruman]

A tiny bit. Both transitions are to the same daughter state (6s ¹S₀), so it's a question of populating the (6p ¹P₁) parent state (relatively) more than the lower energy state (6s ³P₁). Increasing T will do this, but T for 185 nm is about 80000 K. And you're not going to increase the heat much compared to 80000 K.

Doesn't this assume that the mercury atoms are in thermal equilibrium with a reservoir at some temperature T? I was under the impression that the electric discharge creates a non-thermal equilibrium situation where ionized atoms in various excited states continuously recombine with electrons, are re-ionized and so on. Why would $\frac{N_i}{N}=e^{\text{-}\frac{E_i}{kT}}$ be meaningful in this case?

 

[neanderthalphysics]

Thanks for the input, everyone.

According to the blackbody calculator linked at the end of this post, the blackbody radiation peak for 185nm is found at 15650K. Therefore, in theory, would the maximum 185nm:253.7nm emission ratio, all other things being equal, be found at 15650K? How did you calculate the 80000K figure, Vanadium?

Kuruman, I think it changes the ratio a bit because heating shifts the black body peak.

What about if you lower the pressure in the discharge tube and increase the AC voltage? Will this cause more higher energy transitions to occur?

What about if you drove the discharge tube like a flash tube but with high voltage pulses? What would the spectral distribution look like in a high voltage flash, compared to a lower voltage AC mode of operation?

https://www.spectralcalc.com/blackbody_calculator/blackbody.php

 

[kuruman]

The spectrum of a mercury lamp is not the result of blackbody radiation but of atomic transitions as @Vanadium 50 indicated. If you try to convert a mercury lamp into a blackbody radiator at 15,650 K, you will melt vaporize it long before it reaches the desired temperature. I don't see the relevance of a blackbody radiator which gives you a continuous spectrum at any temperature. Shifting the peak closer to 185 nm will give you more of the 185 nm that you want but will also give you much more of the 253.7 nm that you don't want.

What are you trying to do anyway? What is your ultimate goal?

 

[Vanadium 50]

Doesn't this assume

It would if it were an actual calculation. It just shows that the order of magnitude is such that changes on the scale of tens of kelvins - or even many tens of kelvins - are tiny compared to the scale of the problem.

It's like someone planning on jumping over a skyscraper. We can discuss the properties of various kinds of shoes, but they really don't make a difference.

How did you calculate the 80000K figure, Vanadium?

6.7 eV / (300K = 1/40 eV).

By far the easiest thing to do is filter the low frequency light and if it's not bright enough, buy more bulbs.

 

[neanderthalphysics]

@kuruman, yes you're right about the standard mode of operation of a UV mercury lamp, but because I was asking about modifications you could do to the lamp to make it emit more 185nm UV, I included heating and therefore blackbody radiation.

Ultimately the spectra you get from the UV lamp is a superposition of the atomic transitions + blackbody radiation, but as we have seen, the blackbody radiation effect is small at "reasonable" temperatures.

Just asking questions re: heating of the lamps or using high voltage pulses through the mercury discharge tubes to expand my understanding.