I am watching a Science Channel program on the collapse of massive stars and it got me wondering... What is the distribution of heavier than iron elements in the universe. (It is my understanding that some of these elements are formed by the shock wave compressing the outer layers of the star as it explodes)... This lead me to the Wiki about the iron peak and the chart there with the different amounts of elements in the universe:

Element abundance chart

At the end of the chart we see there is more mercury and significantly more lead in the universe than gold. Why is this?


The high abundance of lead (and to a lesser extent mercury) compared to gold comes about from their origins in different nucleosynthetic process in different types of stars. There are three main reasons:

The first is that lead is the final stable element that is built up (mainly) by the s-process. The s-process builds up the abundances of certain heavy elements in stellar interiors through a series of slow (hence "s") neutron captures onto existing elements around the iron-peak, followed by radioactive decays.

The process stops at lead (and bismuth) because there is no way to reach any of the heavier stable elements (U, Th) because the intermediate products (e.g. Po, At, Rn, Fr) are unstable and decay too rapidly to allow any build up of a significant abundance of heavir elements. In fact the terminus of the s-process is dominated by a circular lead-bismuth-polonium cycle, which at equilibrium ends up producing mostly the 206 and 208 isotopes of lead. So you can think of this as a sort-of pile-up process a bit like passengers on a train, where many get off at intermediate stops, but anyone left on the train has to get off at the last station.

Mercury is also an element that is produced in some abundance by the s-process, but in contrast, gold is not one of the stations on the s-process train. Most of the gold in the universe is produced by the rapid r-process of neutron capture. Here, in sites with a much higher neutron flux, many neutron captures happen consecutively followed by decays. This is a way to build many heavy elements that would be unreachable by the s-process because the intermediate nuclei are highly unstable. The r-process gets around that by simple fact of having such a high neutron capture rate that the unstable nuclei don't get chance to decay before they encounter another neutron. Gold is certainly one of the more (but not the most) abundant r-process nuclei, but in general the r-process elements are rarer for a reason I explain next.

Contributions of the r- and s-process to elemental abundances in the solar system. (from Sneden & Cowan 2003), note the logarithmic scale).

Contributions of the r- and s-process to elements in the solar system

The second reason lead is common is that the s-process operates in relatively abundant low-mass star. By low-mass, I mean stars between 1 and a few solar masses, that have had time to be born, live and die in the lifetime of the universe, but will not explode as supernovae. The lead is produced by s-process neutron capture in their interiors during their giant phases and is then expelled through mixing and strong stellar winds into the interstellar medium.

In contrast the r-process either acts in the interiors of core-collapse supernovae, the final stage of life for more massive stars, or in the ejecta from colliding neutron stars, which themselves are the products of core-collapse supernovae in massive stars. Ultimately gold is produced from massive stars or the remnants of massive stars.

The preponderance of lead over gold is therefore also due to their different stellar origins. Low-mass stars are much more common than high-mass stars. The "stellar birth mass function" goes as something like $n(m) \propto m^{-2.3}$. Therefore we would expect a priori that elements produced in the interiors of low-mass stars would be more common than those in high-mass stars, although of course the production efficiency matters too.

The third reason is that the abundance of lead is continually increased from the other direction by the radioactive decay of heavier elements. Most radioactive elements, including the relatively abundant U and Th have decay routes that lead them to stable lead nuclei. This is not as important as the other two reasons, since the production rate and abundance of these heavier elements is low compared with the abundance of lead produce in the s-process.

It is fair to say that although this basic picture is understood, there is a phenomenal amount of progress to be made in understanding exactly what fraction of each chemical elements was made by what process in what type of stars. This is a highly active research topic and I highly recommend the review by Frebel (2018) as further reading.

| cite | improve this answer | |
  • $\begingroup$ Since the S-Process stops at lead, it should have stable steps along the way but the author writes "but in contrast, gold is not one of the stations on the s-process train." Why is gold skipped entirely by the s-process, given that the atomic number for lead is 82, and gold is 79? What are the stops along the way to lead? $\endgroup$ – Sheldon Aug 24 at 1:22
  • 1
    $\begingroup$ @Sheldon as you can see from the plot, all elements are produced to some extent by the s-process. I may revise my analogy. Note that odd nuclei are less stable than even. This plays a role too. $\endgroup$ – Rob Jeffries Aug 24 at 6:58

Heavy elements come from Supernovas and from Neutron star explosions. They may make heavy elements in different proportions, and there are more of one than the other.

Many of the elements made in a supernova are unstable and decay into lighter elements such as lead in particular. Over billions of years the amount of lead has steadily increased.

Here is a NASA article


| cite | improve this answer | |
  • 2
    $\begingroup$ About half the abundance of heavy elements beyond iron are not made in supernovae or merging neutron stars. They are made by the s-process in relatively low-mass stars that do not explode. This includes lead, which is mainly made by the s-process. Unfortunately the NASA "article" is quite wrong. In particular, the contribution from radiactivve decay of heavier elements is probably small. $\endgroup$ – Rob Jeffries Aug 17 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.