Why is an electric arc needed for spectroscopy of clear gases?

[Sophrosyne]

To obtain spectroscopy on a gas like hydrogen or helium, an electric arc is passed through a container of the material, and then the emitted light is viewed through a prism which breaks it up into its component wavelengths. There are sharp lines which are formed, caused by excitation of the electrons in the atoms to the respective higher level orbitals corresponding to the wavelength of absorbed light, emitting that light when they drop back down. Many of these occur in the visible range of the electromagnetic spectrum (eg, the Balmer series in the hydrogen atom).

My question is: why do you need an electric arc to get the electrons to do this? The wavelengths of photons in the Balmer series are present in plentiful quantities in regular ambient light. They should be enough to cause excitation of the electrons to their higher energy levels, shouldn't they? And then as the electrons drop back down they should be emitting those light wavelengths. That is how other colored objects in our world work. But hydrogen gas under ambient light conditions is colorless. Why? Is it because it absorbs and emits the photons instantaneously? Or is there something else going on?

Thanks.

 

[Charles Link]

There may be other reasons why this doesn't work, but for the Balmer series of hydrogen, it is a transition from a higher level down to n=2. In order to populate n=2, you need a photon energy of $E=13.6(\frac{1}{1^2}-\frac{1}{2^2})$ eV, which is about 10 eV and is in the lower end (shorter wavelength\higher energy region) of the UV. The n=2 simply wouldn't get populated by a white light source, and higher n's would be even less populated.

 

[Drakkith]

Hmm... Well, the Balmer series is the result of electrons falling from $n ≥ 3$ to $n=2$. My best guess is that most of the gas in a container of hydrogen at STP exists in their ground states, so you need more energy than visible light to get them up to a shell of 3 or more. That's just a guess though. Don't take my word for it.

 

[Sophrosyne]

Hmm... Well, the Balmer series is the result of electrons falling from $n ≥ 3$ to $n=2$. My best guess is that most of the gas in a container of hydrogen at STP exists in their ground states, so you need more energy than visible light to get them up to a shell of 3 or more. That's just a guess though. Don't take my word for it.

Oh I see. I will look into this some more, but this explanation does make a lot of sense.

 

[ZapperZ]

To obtain spectroscopy on a gas like hydrogen or helium, an electric arc is passed through a container of the material, and then the emitted light is viewed through a prism which breaks it up into its component wavelengths. There are sharp lines which are formed, caused by excitation of the electrons in the atoms to the respective higher level orbitals corresponding to the wavelength of absorbed light, emitting that light when they drop back down. Many of these occur in the visible range of the electromagnetic spectrum (eg, the Balmer series in the hydrogen atom).

My question is: why do you need an electric arc to get the electrons to do this? The wavelengths of photons in the Balmer series are present in plentiful quantities in regular ambient light. They should be enough to cause excitation of the electrons to their higher energy levels, shouldn't they? And then as the electrons drop back down they should be emitting those light wavelengths. That is how other colored objects in our world work. But hydrogen gas under ambient light conditions is colorless. Why? Is it because it absorbs and emits the photons instantaneously? Or is there something else going on?

Thanks.

The problem here is that (i) photon absorption mechanism is not as often as the usual gas excitation using either an electric arc or discharge tube and (ii) even if it happens, due to such low occurrences and the scattered emission, you are less likely to see this light.

Getting a photon of just the right wavelength to just hit and excite an atom, and then to get the emitted photon during decay to hit your eye so that you can observe the light, is not a very probable scenario. Even if there is an absorption, it doesn't mean that you get to see the light when it decays back down. This is why we have "absorption spectra" where there are "missing lines" in the spectra corresponding to the absorbed transition, and the light being absorbed was scattered in a different direction than the incoming one.

The fact that the gas is of low density, and is "transparent", already indicates that the probability of light having a significant absorption cross-section with the gas atom/molecules is already very, very low.

Zz.