Incandescent bulbs emit light by heating a filament using electricity, this would lead to a continuous spectrum according to Kirchoff's first law. However, the glass casing around the filament is filled with Argon (or other inert gases), its presence should make the spectrum discontinuous due to the absorption of wavelengths corresponding to Argon's emission spectrum.
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3 Answers
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Seeing thin absorption lines is difficult. You need pretty good equipment to see them over an extended body. If you're just looking at it with a prism, it will overlap enough that such lines are obscured and the spectrum appears continuous. We describe sunlight as a continuous spectrum even though there are absorption lines from elements in places above the photosphere.
But absorption is a numbers game. Even the strongest peaks in an absorption spectrum are not perfect absorbers. In this case, a few centimeters of atmospheric pressure argon is too thin to matter. Yes, the molecules will occasionally interact and remove a few photons, but most will go right through.
A full pot of coffee is fairly opaque, but a thin film of it at the bottom of your mug is nearly transparent. In the same way, the thin film of argon in the lightbulb doesn't materially affect the spectrum.
answered Feb 17, 2021 at 7:53
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$\begingroup$ please refer to this youtube.com/watch?v=7u3rRy97m9Y, such a small amount of sodium vapor was able to create very apparent absorption lines. This makes it seem like smaller amounts should still create a difference. $\endgroup$– ApporvCommented Feb 17, 2021 at 11:24
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$\begingroup$ @Apporv sodium lines are extremely strong. This is more obvious in some emission experiments, where trace quantities dominate attempts to see anything else, but is also true in absorption. The vapour pressure may also have an effect - I don;t know what it was for Na $\endgroup$– Chris HCommented Feb 17, 2021 at 15:52
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3$\begingroup$ If the light travels 2 centimeters inside the bulb and then several more meters outside of the bulb, it will interact with more argon atoms outside of the bulb than inside. Both of those effects will be negligible, though, as this answer makes very clear. $\endgroup$ Commented Feb 17, 2021 at 16:55
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4$\begingroup$ @Apporv That was not a "small" amount of sodium vapor in that video. They threw about a tablespoon of sodium bicarbonate into a flame and got a column density that was briefly high enough to create a noticeable absorption line. A lightbulb has nowhere near that much material in it. $\endgroup$– NobodyCommented Feb 17, 2021 at 17:41
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2$\begingroup$ @Apporv, When doing emission, even a mild peak can be distinguished against a zero-output background. 0 -> 0.1 is very detectable. But a thin absorber is much tougher. Recognizing 100 -> 99.9 is harder. $\endgroup$ Commented Feb 17, 2021 at 19:20
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Sure, argon absorbs some light at its favorite wavelengths. It probably does create some tiny dips in the lamp spectrum - and off my head I am sure there is no that much precise measurement technique that will detect them. The light simply passes too little argon and argon is not absorbing much in the first place.
Why stop at argon in the bulb? The air around is much more and contains ~1% argon, as well as much more colorful gases (oxygen comes to mind). If you look at something that is tens of kilometers away in clear air (like, mountain tops looked from another mountain top), you will see that the more distant is the object, the more blue it is (it has dips in the red parts of the spectrum).
If you look for a gas with visible color (i.e. deep absorption lines) in small amounts, see chlorine. It is visibly green when it is just ~5cm tick (at normal pressure). Well, it is not really good for filling lightbulbs.
And, there is no such thing as a true discontinuous spectrum. The well-known banded picture of the Sun spectrum is in fact continuous, the dark lines having much less (but still measurable) intensity. Our lovely Sun achieves this with like ~4000km layer of gas that absorbs lines in the more or less incandescent spectrum coming from below.
Compare these 4000km with ~2cm of gas in the bulb, even if the gas in the bulb is somewhat denser.
p.s. modern incandescent light bulbs contain krypton and/or xenon instead of nitrogen and argon used in older ones. These gases carry away less heat from the filament and make the bulb more efficient. And some of them (they are called halogen bulbs) have some iodine as well. None of these gases affects measurably the spectrum of the bulb by absorption of the light.
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3$\begingroup$ "the more distant is the object, the more blue it is" — this is more due to air perspective (inscattered light) than absorption by oxygen. $\endgroup$– RuslanCommented Feb 17, 2021 at 8:39
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$\begingroup$ please refer to this youtube.com/watch?v=7u3rRy97m9Y, such a small amount of sodium vapor was able to create very apparent absorption lines. This makes it seem like smaller amounts should still create a difference. $\endgroup$– ApporvCommented Feb 17, 2021 at 9:34
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1$\begingroup$ @apporv leaving aside the extremely strong absorption pattern that sodium is well known for, that is not even remotely a "small amount" of sodium vapor, it's an enormous amount. The amount of argon in a light bulb is miniscule in comparison with the amount of sodium being used in the video. $\endgroup$– barbecueCommented Feb 17, 2021 at 16:26
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For the argon inside a light bulb to affect the passage of light through it in a measurable way, it needs to be hot enough to be ionized. Then it will begin absorbing and emitting its characteristic line spectrum. Furthermore, there has to be a lot of it present and ionized to be detectable in this manner.
Inside a tungsten-filament light bulb there is only a small amount of argon and the only part of it that is really hot is right next to the filament- which is at that moment screaming out its blackbody spectrum, which overwhelms everything else.
The only way you can get a tungsten bulb to emit a line spectrum is to accompany the filament with a pair of electrodes in parallel with it, very close by, and put lots of gas (like xenon) in the bulb. Then, when you turn on the bulb, the tungsten filament heats the xenon at the same time the filament voltage is present across the electrode pair. Eventually the gas between the electrodes gets hot enough to ionize and it begins conduction electricity.
As the quantity of ionized gas in the electrode gap grows, its resistance drops and more and more of the electricity flows through the gas- and less and less through the filament.
When the bulb is fully "on", most of the current is flowing through the gas which keeps it hot enough to be ionized and to emit its characteristic line spectrum, with a blackbody continuum (from the hot gas) superimposed on it.
This is how those brilliant blue-white xenon headlights for cars work.
answered Feb 17, 2021 at 17:08
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$\begingroup$ For the argon inside a light bulb to affect the passage of light through it in a measurable way, it needs to be hot enough to be ionized. Could you explain this? 1. Why do you consider ionization only, and not transition between electron shells? 2. Why does the gas need to be hot? $\endgroup$ Commented Feb 17, 2021 at 23:36
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$\begingroup$ Heat promotes ionization. Ionization renders the gas electrically conductive. Once conductive, it self-heats via the flow of electricity through it. The easiest ionization is always that involving the outermost (valence) shell, so even if the next inner shell loses electrons (which takes more energy) the process always starts with the outermost one. $\endgroup$ Commented Feb 18, 2021 at 7:58