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Why is He I 6678.151 line used for investigating variations in Be stars? I mean, for instance, asymmetry and radial velocity. Many thanks

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Neutral Hydrogen exhibits a Balmer series which is a transition of the electron from a higher energy level down to the n=2 energy level. The n=3 to n=2 transition, otherwise known as Ba-α has a wavelength of 6562.79 Angstroms. Neutral Helium (a.k.a He I) can exhibit emissions similar to the Hydrogen Balmer series. And in fact, the He I 6678.151 Angstrom emission line is closest in energy to the H I Ba-α emission line, making it a good proxy for the H I Ba-α line.

So you may be asking why I'm talking about the Balmer series here. The reason I'm talking about it is because the Balmer series is a very easily observable series given how prevalent the lines are in spectra (which comes down to the prevalence of Hydrogen and the likelihood of this transition occurring) and how strong (i.e., how deep the emission line is and thus how easy it is to measure) the Balmer lines can be. Because they're so abundant in many (but not all!) stars, and because they're so easy to observe, they make a good set of wavelengths to watch out for and track the metrics of across a wide variety of stars.

And in fact, neutral helium can behave like hydrogen by having one electron in the 1s orbital (which acts to electrically screen the nucleus, making it appear to have on a single proton to the outer electron) and the second electron can make transitions like it might in Hydrogen, jumping from higher energy levels to lower ones. For this reason, B type stars can exhibit neutral helium lines quite strongly and the 6678.151 Angstrom line, being the closest thing to Ba-α, is likely to be one of the strongest lines one can observe. That makes it an obvious choice to observe in the type of stars that exhibit these lines.

In the above, I mostly discussed why this line itself is important, but there's also the dimension of the question which is why is it important for Be stars specifically. I can let others fill in the details there.

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  • $\begingroup$ Thank you very much. $\endgroup$ – Elena Greg Jun 4 at 2:59

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