It is quite common to list metallicity for stars. For example for Teegarden's star, notable because one of its two planets (b) has an Earth similarity index of 0.95, the highest listed on the habitable-exoplanets-catalog, has a metallicity listed as -0.19±0.16 [Fe/H]. It would appear that the Fe/H estimator of metallicity is the most commonly used measurement, which makes sense in that most of the primordial matter was hydrogen (H) and the end product of a simple fusion cascade stops at iron (Fe) as iron has the [Sic, stable] isotope with the (Edit) least practicable binding energy available to fusion per nucleon of any element. But note comment by @johndoty and Once the chain of reactions reaches Fe, there is no reason to examine heavier or more stable nuclei, because the conditions are such that they are barely produced.

But what is -0.19 [Fe/H]? Perhaps the logarithm of the ratio of concentrations? And if so, where is sol in that regard and what range of [Fe/H] do we expect in what context? At the risk of seeming to ask multiple questions, all I am really asking for here is some simple explanation of metallicity so that when I see it cited I have some concept of what it means.

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    $\begingroup$ Does this answer your questions? en.wikipedia.org/wiki/Metallicity#Chemical_abundance_ratios $\endgroup$ Commented Jan 1, 2022 at 7:42
  • $\begingroup$ You may find other questions related to metallicity: astronomy.stackexchange.com/search?q=metallicity $\endgroup$ Commented Jan 1, 2022 at 9:18
  • $\begingroup$ @DaddyKropotkin Partly, although the distributions in ProfRob's answer are more satisfying. $\endgroup$
    – Carl
    Commented Jan 1, 2022 at 17:12
  • $\begingroup$ @NilayGhosh In point of fact I looked at all of the prior Q/A searching for a simple answer, and only when I didn't find one did I posit the question above. $\endgroup$
    – Carl
    Commented Jan 1, 2022 at 17:17

1 Answer 1


[Fe/H] is the logarithmic ratio of the number of iron nuclei per proton in the star to that in the Sun, $${\rm [Fe/H]} = \log_{10} \left(\frac{n({\rm Fe}_*)/n({\rm H})_*}{n({\rm Fe}_\odot) / n({\rm H}_\odot)}\right) .$$

On this scale, [Fe/H]$_\odot$ is exactly zero.

If [Fe/H]$=-0.19$, this means that the concentration of iron nuclei in the star (or at least, measured in its photosphere) is $10^{-0.19}=0.65$ times that of the Sun (though note the large error bar that contains a confidence interval of 68% of the possibility).

In the solar neighborhood of the Galactic disc, the average [Fe/H] is about -0.1, and has a standard deviation of about 0.1. The distribution is asymmetric with basically no stars with [Fe/H]$>0.4$ but a tail of low metallicity stars down to [Fe/H]$<-1$. The distribution in the solar neighborhood is shown as the pink histogram in the plot below (taken from Adibekyan 2019, who also discuss metallicity estimates). Note that the histograms have been normalised so that the total area is 1. It is a well-known fact that the metallicity distribution of stars hosting giant exoplanets, especially "hot Jupiters," is shifted towards higher metallicity (shown in blue on the plot).

Metallicity distributions

Using [Fe/H] as a proxy for metallicity is often done, though "metals" includes anything heavier than helium and iron is not the most common metal nucleus. However, iron is the easiest metal abundance to measure in stellar spectra (usually lots of absorption lines, both neutral and singly ionised), except in very hot stars. The assumption is then made that all the other "metal" elements scale with iron. e.g. that the ratio of say oxygen to iron is the same in all stars as it is in the Sun. This is a reasonable assumption for stars near the Sun, although old stars, which tend to have low [Fe/H], have relatively more "alpha elements" (oxygen, magnesium, silicon).

A couple of other comments on your question: The nucleus with the highest binding energy per nucleon is an isotope of nickel. The end point of the chain of nuclear fusion reactions is $^{56}$Ni, which then decays to iron.

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    $\begingroup$ See physics.stackexchange.com/questions/182525/… for more on the last point. $\endgroup$
    – John Doty
    Commented Jan 1, 2022 at 16:45
  • $\begingroup$ Thanks. Least available binding energy to fusion is what I was trying to say, so I edited it. As $^{56}$Ni has a half-life of approximately 6.075 days, and undergoes an exothermic beta plus decay, and subsequently a 77 day decay to $^{56}$Fe, then $^{56}$Fe has a lesser mass per nucleon than $^{56}$Ni and yet I get your point that the fusion cascade proper ends at $^{56}$Ni, but since $^{56}$Fe production proceeds causally it is not improper to consider it as an end point of the fusion process. A fine point to be sure. $\endgroup$
    – Carl
    Commented Jan 1, 2022 at 17:02

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