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In astrophysics elements heavier than helium are termed “metals” collectively. These elements are more opaque and radiate more efficiently than hydrogen and helium. As a result, metallicity has a profound effect on stellar evolution. For example, high metallicity enables the protostars to contract faster and ignite the nuclear fusion sooner, which prevents the formation of extremely massive stars. High metallicity stars also more easily lose the outer envelope at the late phase of stellar evolution.

But I don’t understand why heavier elements are more opaque. It’s not hard to understand that helium is very transparent because helium has the highest 1st ionization energy. However, the 1st ionization energy of hydrogen is lower than nitrogen, oxygen and neon, which are among the most abundant metals. Does that mean these elements should not be regarded “metals”?

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    $\begingroup$ Some light transmission occurs through scattering, and a lighter nucleus obviously has less inertia and would be more susceptible to this -- transmission isn't necessarily contingent on emission spectra. $\endgroup$
    – meltyness
    Nov 1, 2022 at 20:53

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Atoms bigger than H and He have more complex electron orbital structures with more available transitions in them- by the time you get up to silicon and iron and molecules containing them, there are lots that are active and available to interact with infrared. This makes an intergalactic cloud with iron, silicon, and oxygen compounds (including carbon monoxide) in it a lot more effective at radiating away heat generated by gravitational compression than a gas cloud with nothing in it but (H)2 and He.

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    $\begingroup$ Carbon monoxide gas is a major "metallic" radiator in molecular clouds. Molecular H_2 (a symmetric molecule) is not. $\endgroup$
    – ProfRob
    Nov 2, 2022 at 16:32
  • $\begingroup$ @ProfRob, thanks, will edit. -NN $\endgroup$ Nov 2, 2022 at 18:08
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Niels Nielsen's answer applies to absorption between bound states, when the atoms are neutral or only partially ionised. But metals actually still cause more opacity per unit mass even when they become ionised. This is more relevant in the interiors of stars.

The reason is that the cross-sections of free-free and bound-free processes between photons, electrons and nuclei, like bremsstrahlung, photoelectric absorption and pair production, all increase strongly (as $Z^2$ or steeper) with the charge $Ze$ on the atomic nucleus.

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The idea of a metal has to do with atomic bonds. Some substances form discrete molecules, like H$_2$ or H$_2$O.

Some form extended crystal lattices like SiO$_2$. A simple explanation is that electrons are tied up in bonds that connect two atoms. Or NaCl, where electrons from Na stick to the Cl. The Na become positive, and Cl become negative. Eletrostatic attraction holds the crystal together.

Some elements form lattices where the electrons spread out through the lattice. They are free to move around. These are metals. Some characteristics of metals are

  • High electric and thermal conductivity because moving electrons can carry change and energy around.
  • Reflective surfaces when smooth. Light is a vibrating electromagnetic field. It vibrates electrons. Vibrating electrons radiate light. This is the reflection.

Na and Fe and are metals. NaCl and Fe$_2$O$_3$ (rust) are not because electrons are tied up in bonds.

N, O, and Ne are not metals.

H is usually not a metal. But under high enough pressure (around the conditions at the center of Jupiter) it becomes metallic.

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    $\begingroup$ The term "metal" in astronomy has nothing to do with metallic solids. To astronomers, N, O, and Ne are metals. $\endgroup$
    – John Doty
    Nov 1, 2022 at 22:32
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    $\begingroup$ @JohnDoty -- your statement is probably true. None-the-less, it's really confusing to anyone who is not an astronomer, and a different term should have been thought of. I do realize that once something has become a convention, it's hard to escape from it. But sulfur isn't a metal, sorry, it just isn't! $\endgroup$ Nov 2, 2022 at 14:18
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    $\begingroup$ @JosephDoggie A little insight on why it's named the way it is: physics.stackexchange.com/questions/221873/… $\endgroup$ Nov 2, 2022 at 15:13
  • $\begingroup$ @NathanOliver -- thanks. I have heard of Fraunhofer before, in a diffraction context for optics. $\endgroup$ Nov 2, 2022 at 16:13
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    $\begingroup$ @JosephDoggie I am not responsible for terminology that became standard before I was born. $\endgroup$
    – John Doty
    Nov 2, 2022 at 16:13

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