Having an even number of neutrons and/or an even number of protons tends to make a nucleus more stable against radioactive decay. There are only 5 stable nuclei with both an odd number of neutrons and an odd number of protons:

  • hydrogen-2 (deuterium)

  • lithium-6

  • boron-10

  • nitrogen-14

  • tantalum-180

Deuterium is rare compared to hydrogen and helium-4. Lithium-6 is rare compared to lithium-7. Tantalum-180 is rare compared to tantalum-181 (and apparently not even completely stable, though its half-life is so long nobody has actually seen it decay). So in these cases one gets the feeling that odd-odd nuclei are hard for nature to produce. But 20% of boron is boron-10 (the rest being boron-11), and 99.6% of nitrogen is nitrogen-14 (the rest being nitrogen-15).

What's up with nitrogen-14? Why is it so much more abundant than any other odd-odd isotope?

  • $\begingroup$ Well, neither C-14 or O-14 are particularly appetizing, so N-14 it is... $\endgroup$
    – Jon Custer
    Commented Feb 1, 2021 at 0:04
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    $\begingroup$ From a 2001 paper: "For more than 20 years, the nucleosynthetic origin of primary nitrogen, which is one of the most abundant and important elements in the Universe, for life in particular, has remained a deep mystery in astrophysics" $\endgroup$ Commented Feb 1, 2021 at 5:21
  • 2
    $\begingroup$ Note that 14 nucleons have 42 (valence) quarks. ;) $\endgroup$
    – PM 2Ring
    Commented Feb 1, 2021 at 8:47
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    $\begingroup$ I suppose the high abundance of N-14 relative to N-15 is because it's a bit of a bottleneck in the CNO cycle "The limiting (slowest) reaction in the CNO-I cycle is the proton capture on N-14". But of course that doesn't explain why N-14 happens to be stable when C-14 has a half-life of 5730 years, and O-14 70.62 seconds. $\endgroup$
    – PM 2Ring
    Commented Feb 1, 2021 at 8:54
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    $\begingroup$ By the way, it takes work to explain the long half-life of C-14: researchgate.net/publication/… $\endgroup$
    – John Baez
    Commented Feb 5, 2021 at 23:28

2 Answers 2


It is down to the cosmic production mechanism in the case of 14N vs other N isotopes - all of which are produced chiefly inside stars.

Nitrogen-14 is the dominant catalysed by-product of the CNO hydrogen-burning cycle, which powers stars with mass $>1.5 M_{\odot}$. There has been plenty of time in the universe for such stars to have completed their lives and returned their nucleosynthetic products to the interstellar medium. Further, because the birth mass distribution for stars strongly favours lower masses ($ N(M)\propto M^{-2.3}$), as opposed to higher mass stars ($>8 M_{\odot}$) that might process the nitrogen much further, there are lots of "production sites" in a typical galaxy.

The reasons that 14N ends up having the highest equilibrium concentration of the C, N, O isotopes involved in the CNO cycle is discussed in https://physics.stackexchange.com/a/587699/43351 . In summary, the proton capture reaction that converts 14N to 15O is a relatively slow "radiative capture" involving the electromagnetic interaction and emission of a gamma ray. The other proton captures in the CNO cycle go via the strong interaction or are faster because they involve no Coulomb barrier (beta decays). In particular, 15N is very rapidly turned back into 12C by a strong proton capture followed by alpha decay. As a result, when the cycle reaches equilibrium, the 14N proton capture is the slowest reaction and the 14N content of the star builds up.

In intermediate mass stars ($1.5-8 M_\odot$) a significant part of the CNO cycle burning takes place in shells around the core. Towards the ends of the lives of these stars, there are instabilities that cause mixing of the nuclear processed material into the envelope, which can then be ejected into the interstellar medium via strong winds.

As for the other isotopes in your question, Deuterium, 6Li and 10B are all very fragile and are easily burned in stellar interiors at temperatures below that required for hydrogen burning. The same is true of 7Li and 11B (the dominant isotopes), but the temperature requirements are higher and there are also production mechanisms for these isotopes inside intermediate mass stars (see the "Cameron-Fowler mechanism", Cameron & Fowler 1971) or in classical novae.


One way to answer this question is to look at all of the elemental abundances

Elements abundance-bars.svg
By Swift - Own work, CC0, Link

and information about the elements’ origins,

Nucleosynthesis periodic table.svg
By Cmglee - Own work, CC BY-SA 3.0, Link

Tantalum is way out in the tail of the other heaviest elements. It’s rare for the same reason that all of the heavies are rare: they have to be produced from iron by the s-process and the r-process.

Deuterium, lithium, and boron are mostly not produced in stellar fusion at all. Deuterium is produced by from protium by neutron capture, but has a binding energy of only 2.2 MeV. An environment with enough neutrons to produce deuterium probably also has lots of thermal photons with enough energy to dissociate the deuterium. And there is not a low-temperature fusion pathway to produce lithium or boron. The earliest “zero-metallicity” stars produced carbon in the triple-alpha process, which enabled CNO-catalyzed fusion at lower temperatures in their higher-metallicity descendants. Some lithium was produced in the big bang, but no boron. The figure suggests that essentially all of the universe’s boron was produced as cosmic-ray spallation fragments.

In this picture, nitrogen outnumbers the other odd-odd isotopes just because it’s easier to make nitrogen in main-sequence low-mass stars than it is the other elements.

Another way to ask why nitrogen is the most abundant of the low-mass odd-odd nuclei is to ask what happens to each species in the kind of high-radiation environment where s-process element production might be taking place:

  • Deuterium will form from hydrogen, but be subject to photodissociation.
  • There is no s-process path from helium to lithium, because helium-5 isn’t stable.
  • Lithium-6 will absorb neutrons and fragment, $\require{mhchem}\ce{^6Li(n,\alpha){}^3He}$. Lithium-6 has a much higher neutron cross section than lithium-7 (though Li-7 is also destroyed by neutron capture), so a high-radiation environment will deplete the lithium-6.
  • Boron-10 likewise has a much higher neutron absorption/fragmentation cross section than boron-11, with significant contributions from $\ce{^{10}B(n,\gamma){}^{11}B}$ and $\ce{^{10}B(n,\alpha)^{7}Li}$.
  • Nitrogen under neutron irradiation, however, undergoes $\ce{^{14}N(n,p)^{14}C}$. After a few thousand years, the new carbon nucleus turns back into nitrogen by beta decay. (This is why there’s carbon-14 in Earth’s atmosphere; the neutrons come from spallation by cosmic rays.)

I don’t know if there’s a good, intuitive explanation for why nitrogen-14 is a hundred times more likely to undergo (n,p) than (n,$\gamma$). I suspect that, if the cross sections for those two processes were reversed, then the s-process would also deplete N-14 relative to N-15. But whether the s-process is actually relevant here, or whether it’s just a useful shorthand for “a place where element production happens and there are lots of neutrons” which happens to support a just-so story, isn’t something I’m terribly clear on.


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