66
$\begingroup$

In this answer to Is lithium considered a metal in astronomy? this image from Wikimedia of the abundances of elements in the universe was shared:

A log graph of atomic abundances in the universe
Credit: Wikimedia Commons user 28bytes, under C.C.-by-S.A.-3.0.

I thought it was fascinating that there is a near-constant zig-zag of abundancies.

Are these values accurate, and if so, what would cause this zig-zag?

Improve this question
$\endgroup$
5
  • 11
    $\begingroup$ Pretty sure this is because of nuclear stability - note the "peaks" are all atoms with even atomic numbers, which are inherently more stable than those with odd atomic numbers. $\endgroup$
    – Allure
    Oct 28, 2020 at 1:41
  • $\begingroup$ I'd guessed something like that, or alternatively that it results in their emissions being stronger somehow. $\endgroup$
    – curiousdannii
    Oct 28, 2020 at 2:04
  • 3
    $\begingroup$ I suspect that the most common synthesis events involve a helium-4 nucleus and something heavier, so stepping up by two atomic numbers is common. Beryllium is rare because Be-8 (which would come about from the fusion of two He-4 nuclei) is unstable, and only the Be-8 that gets turned into C-12 survives. Someone more learned than I will have to fill in the gaps. $\endgroup$
    – EvilSnack
    Oct 28, 2020 at 3:43
  • 4
    $\begingroup$ There should actually be constant zig-zags, but some of the elements are missing from the chart (43 Tc, 61 Pm, lots after Bi). The only exception seems to be the first four elements. $\endgroup$
    – BlueRaja - Danny Pflughoeft
    Oct 28, 2020 at 21:06
  • 1
    $\begingroup$ THAT'S AMAZING. Taking a step back, I'm in awe. Something fundamental we know about how atoms work, and it just naturally pops out of data like this. Sorry for low-quality comment, but that plot is just great. $\endgroup$
    – Joel Reid
    Oct 29, 2020 at 12:31

5 Answers 5

Reset to default
46
$\begingroup$

Nuclei are more stable if they have an even number of protons Z, and are also more stable if they have an even number of neutrons N. This is because the particles form pairs. (Almost all nuclei with both N and Z odd are unstable with respect to beta decay.)

If a nucleus is going to have significant cosmic abundance, it must be either absolutely stable or at least have an extremely long half-life. This is more likely to happen when Z is even (and also when N is even).

Stability also effects probabilities of different types of decay. E.g., if a nucleus can undergo both beta+ and beta- decay, then it is usually more likely to do the decay that results in the more stable daughter.

Improve this answer
$\endgroup$
3
  • 4
    $\begingroup$ A notable exception is nitrogen-14, though it is less abundant than C or O. $\endgroup$
    – ProfRob
    Oct 28, 2020 at 8:11
  • 1
    $\begingroup$ What happened to Beryllium? $\endgroup$
    – Jack Aidley
    Oct 29, 2020 at 15:14
  • 6
    $\begingroup$ @JackAidley The most natural channel for creating Beryllium would be the fusion of two alpha particles. However, this turns out to be energetically unfavorable, so the newly created Beryllium-8 nucleus immediately splits back into the two alpha particles that created it. That means all Beryllium is Beryllium-9, and had to form through other, slower methods. $\endgroup$
    – eyeballfrog
    Oct 29, 2020 at 16:04
25
$\begingroup$

TL;DR: Nature favors packing nucleons in pairs of anti-parallel spins. (electrons as well!)

The zig-zag is a nucleosynthesis artifact.

H, He and Li are "primordial" (made out of the abundant neutrons in the first hours after the Big Bang.)

Everything else is made in the stars.

There are a few main processes that make heavier elements (oversimplified!):

  1. Hydrogen burning (makes helium as we already don't have great deal of it)
  2. Helium burning (makes C and O)
  3. Helium (even element itself) + any even element (makes even elements up to Fe)
  4. Slowly adding neutrons to everything and waiting for it to beta-decay (called s-process, makes odd elements out of even elements and elements heavier than Fe, up to some point)
  5. Rapidly adding neutrons without waiting much anything to beta-decay before adding the next neutron (r-process, makes all elements up to uranium out of existing elements from previous processes)

Every one of these processes gets its material from the previous ones and has less yield. That's why we have heavier elements less than lighter elements, even elements more than odd elements and r-process elements less than s-process elements.

Edit: Both s- and r-processes, as well as other ones not listed here, leave ultimately more even elements as they are in general more stable against both beta-decay and capturing the next neutron. Why they are more stable? Because both protons and neutrons fill their very own shell-like orbitals inside the nucleus and nature favors packing nucleons in pairs of anti-parallel spins. (electrons as well!) See Nuclear shell model

Improve this answer
$\endgroup$
10
  • 1
    $\begingroup$ Nice answer, but Big Nucleosynthesis lasted for about 17 minutes (from t=3 m to t=20 m). It produced some beryllium, mostly Be-7, but that decays with a half-life of 56 days to Li-7. Most beryllium in the modern cosmos is Be-9, which was mostly produced long after the BB, via cosmic ray spallation. $\endgroup$
    – PM 2Ring
    Oct 28, 2020 at 11:51
  • 1
    $\begingroup$ True, but they aren't doing very much, apart from decaying. ;) I guess a few of them would have enough energy to convert Be-7 to Be-8 (which immediately decays to He-4) and He-3 to He-4. Nothing else in existence at that stage has a significant cross section for thermal neutrons. $\endgroup$
    – PM 2Ring
    Oct 28, 2020 at 12:14
  • 2
    $\begingroup$ @Sean It's complicated. There are alpha process ladder reactions as well as symmetric fusion. Certainly, in carbon burning, symmetric fusion is dominant. But neon burning is predominantly alpha process, and at that temperature photodisintegration becomes significant, so some neon nuclei decompose to oxygen + alpha. Oxygen burning is primarily symmetric fusion. $\endgroup$
    – PM 2Ring
    Oct 29, 2020 at 10:40
  • 2
    $\begingroup$ Hmm, no mention of the CNO cycle? I would've thought that accounted for the unexpectedly high abundance of nitrogen-14, despite it being an odd-odd isotope. $\endgroup$
    – Ilmari Karonen
    Oct 29, 2020 at 14:50
  • 3
    $\begingroup$ @IlmariKaronen the question is about the zigzag. A lot of other things simplified away. $\endgroup$
    – fraxinus
    Oct 29, 2020 at 14:55
9
$\begingroup$

A very informative, short, You Tube video discusses this - see the 4:20 mark.

Apart from hydrogen, helium, lithium and beryllium, all the elements were formed as the products of fusion in stars.

If you look at the scale for the vertical axis, it's logarithmic. Also, the elements on top of the zig-zag have even atomic numbers and those on the bottom have odd atomic numbers. The significance of this is the elements with even atomic numbers are about ten times more abundant than those with odd atomic numbers.

The reason for the zig-zag pattern is the processes of fusion. Fusion involving helium (which has two protons and two neutrons) is more likely than fusion involving hydrogen (which has one proton). When helium fuses with an even numbered atomic nucleus another even numbered atomic nucleus is produced, but when hydrogen fuses with an atomic nucleus it produces the element next in the periodic table of elements.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ This doesn't answer the question. You are just repeating that the production of even-numbered nuclei is more likely, without explaining why. $\endgroup$
    – ProfRob
    Oct 28, 2020 at 8:04
  • $\begingroup$ Comment only: Iron is "star ash" (nicely stable) - with an abundance about 2 magnitudes higher than might otherwise be expected from a somewhat linear zig-zag. Yes? $\endgroup$
    – Russell McMahon
    Oct 31, 2020 at 1:27
  • 1
    $\begingroup$ In addition, how would this apply beyond iron, since these elements aren't formed in fusion reactions... $\endgroup$
    – ProfRob
    Oct 31, 2020 at 16:52
2
$\begingroup$

The zig-zig pattern might be related to Oddo-Harkins rule. It states:

An element with an even atomic number is more abundant than both elements with the adjacently larger and smaller odd atomic numbers.

The basis of this abundance pattern might have something to with nucleon stability. The rule argues that elements with even atomic numbers have their protons paired, with each member of the pair balancing the spin of the other thus enhances nucleon stability while elements with odd atomic numbers have one unpaired proton, thus not stable.

But there are many big exception to this rule. One of the exception is hydrogen-helium-lithium-beryllium.

Improve this answer
$\endgroup$
2
  • 4
    $\begingroup$ The nuclear physics is a little garbled here. "Balancing the spin" isn't what enhances stability. It's just that when the angular momenta are opposite, the two orbitals have a lot of spatial overlap, so the attractive interaction is fairly strong. $\endgroup$
    – user15381
    Oct 28, 2020 at 3:43
  • 2
    $\begingroup$ " more likely to capture another" This doesn't really make sense. Capture another proton? Protons in nuclei don't capture other protons. There is electron capture, but that reduces the atomic number. $\endgroup$
    – user15381
    Oct 28, 2020 at 3:44
2
$\begingroup$

This is complementary to other answers and not intended as a full answer in its own right.

I saved the diagram below to disk in March 2009 - source not (yet) known.
Update: Wikipedia :-) - with an excellent discussion of the reasons for the variations in the terrestrial versus universe wide distributions here

This shows elemental abundance in the earth's elemental crust rather than in the universe.
It will be seen that the clear "zig-zag" in the original diagram is not as clear cut here and that some relative ratios are substantially different.
The reasons given for the zig zag shape remain the same but having a partial sample of conditions at the bottom of a tectonically stirred, atmosphere and ocean covered gravity well probably go a long way toward explaining most of the differences.

The thin shell of biosphere probably help account for some of the differences in eg O & H ratio and abundance.

Pb : Bi: Th : U are much the same, but (sadly) the rarest metals are 2 to 3 orders of magnitude down in abundance - but this suggests that asteroid mining may yet prove 'interesting'.

enter image description here

Versus original from above:

enter image description here

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .