<p>Both free-free and bound-free absorption are strongly dependent on the atomic number <span class="math-container">$Z$</span> of nuclei in the gas.</p>
<p>For free-free absorption it is simply that the <em>emissivity</em> per unit volume of electrons accelerating in the field of an ion scales as <span class="math-container">$Z^2$</span> (there are <span class="math-container">$Ze$</span> electrons accelerating near a charge of <span class="math-container">$+Ze$</span>). Since by Kirchoff's Law, absorptivity is directly related to emissivity then the absorption coefficient and opacity are also proportional to <span class="math-container">$Z^2$</span>.</p>
<p>There is a similar argument for bound-free absorption, where since electrons are highly accelerated in the field of a strongly positively charged ion, there is a <span class="math-container">$Z^4$</span> dependence of the absorption cross-section.</p>
<p>These very strong dependences translate to weaker dependences on the mass-fraction of heavy elements because the total absorption coefficients must be averaged over all the constituents of the gas - most of which are hydrogen and helium with <span class="math-container">$Z=1$</span> and <span class="math-container">$Z=2$</span> respectively. Nevertheless there still ends up being a fairly strong, positive correlation between metallicity and opacity.</p>
<p>Even in the outer parts of stars where the opacity is dominated by H<span class="math-container">$^{-}$</span> ions, it is still metals that supply almost all the free electrons to make those ions and the opacity ends up being proportional to the metallicity.</p>

CC BY-SA 4.0
2021-09-01T20:47:52.473Z
Why do metals provide so many more free electrons than hydrogen? The most common metals don't have significantly lower first ionization energies than hydrogen. Is it just because metals have more electrons to give?
zucculent
CC BY-SA 4.0
2021-09-01T21:23:24.273Z
@zucculent actually hydrogen has a relatively high ionization energy. Most atoms have lower ionization energy than hydrogen. Judging from [Wikipedia: Ionization energy](https://en.wikipedia.org/wiki/Ionization_energy) only He, N, O, F, Ne, Ar and Kr have higher ionization energy that hydrogen. Li and Na have almost three times less ionization energy.
Prallax
CC BY-SA 4.0
2021-09-01T21:53:48.430Z
@Prallax that's true, but of the elements you mentioned, N, O, and Ne are some of the most abundant metals. Li and Na are much less so.
zucculent
CC BY-SA 4.0
2021-09-01T22:56:22.303Z
Nevertheless, in the photosphere it is Na that supplies the free electrons.
ProfRob
CC BY-SA 4.0
2021-09-02T03:02:41.810Z
Oh so hydrogen's ionization energy is too high to contribute significantly to the free electrons?
zucculent
CC BY-SA 4.0
2024-05-09T19:23:53.647Z
@zucculent Both H and Na contribute some free electrons in the solar *photosphere*. More from Na than H because its ionisation is much lower, which compensates for its much lower abundance. I think actually that Mg may make an even larger contribution.
ProfRob
CC BY-SA 4.0
2021-09-01T18:01:37.777Z
2021-09-01T18:01:37.777Z
ProfRob
<p>A vague answer I've seen is that it has something do with a greater number of possible excited states, but I don't know what this means for a partially ionized plasma - much less a fully ionized plasma - nor how this would allow the plasma inside a star to absorb more radiation.</p>

<star><metallicity><opacity>
Why does metallicity increase the opacity inside stars?
zucculent
# Why does metallicity increase the opacity inside stars?

(Asked by: username_0 on 2021-09-01)

A vague answer I've seen is that it has something do with a greater number of possible excited states, but I don't know what this means for a partially ionized plasma - much less a fully ionized plasma - nor how this would allow the plasma inside a star to absorb more radiation.

**Tags:** star, metallicity, opacity

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## 1 answers

### Answer by username_1 on 2021-09-01. Score: 10

Both free-free and bound-free absorption are strongly dependent on the atomic number $Z$ of nuclei in the gas.

For free-free absorption it is simply that the *emissivity* per unit volume of electrons accelerating in the field of an ion scales as $Z^2$ (there are $Ze$ electrons accelerating near a charge of $+Ze$). Since by Kirchoff's Law, absorptivity is directly related to emissivity then the absorption coefficient and opacity are also proportional to $Z^2$.

There is a similar argument for bound-free absorption, where since electrons are highly accelerated in the field of a strongly positively charged ion, there is a $Z^4$ dependence of the absorption cross-section.

These very strong dependences translate to weaker dependences on the mass-fraction of heavy elements because the total absorption coefficients must be averaged over all the constituents of the gas - most of which are hydrogen and helium with $Z=1$ and $Z=2$ respectively. Nevertheless there still ends up being a fairly strong, positive correlation between metallicity and opacity.

Even in the outer parts of stars where the opacity is dominated by H$^{-}$ ions, it is still metals that supply almost all the free electrons to make those ions and the opacity ends up being proportional to the metallicity.

> **Comments:**
> **username_0**: Why do metals provide so many more free electrons than hydrogen? The most common metals don't have significantly lower first ionization energies than hydrogen. Is it just because metals have more electrons to give?  
> **username_2**: @zucculent actually hydrogen has a relatively high ionization energy. Most atoms have lower ionization energy than hydrogen. Judging from [Wikipedia: Ionization energy](https://en.wikipedia.org/wiki/Ionization_energy) only He, N, O, F, Ne, Ar and Kr have higher ionization energy that hydrogen. Li and Na have almost three times less ionization energy.  
> **username_0**: @Prallax that's true, but of the elements you mentioned, N, O, and Ne are some of the most abundant metals. Li and Na are much less so.  
> **username_1**: Nevertheless, in the photosphere it is Na that supplies the free electrons.  
> **username_0**: Oh so hydrogen's ionization energy is too high to contribute significantly to the free electrons?  
> **username_1**: @zucculent Both H and Na contribute some free electrons in the solar *photosphere*. More from Na than H because its ionisation is much lower, which compensates for its much lower abundance. I think actually that Mg may make an even larger contribution.  

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