## Nitrogen-14

Having an even number of neutrons and/or an even number of protons tends to make a nucleus more stable against radioactive decay:

• Wikipedia, Even and odd nuclei.

I just learned there are only 5 stable nuclei with an odd number of neutrons and an odd number of protons:

• deuterium (hydrogen-2), with 1 proton and 1 neutron.

• lithium-6, with 3 protons and 3 neutrons.

• boron-10, with 5 protons and 5 neutrons.

• nitrogen-14, with 7 protons and 7 neutrons.

• tantalum-180, with 73 protons and 107 neutrons.

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, though I was lying slightly when I said it was stable: theoretically it’s predicted to decay, though with such a long half-life—over 1016 years—that it’s never actually been seen to decay. It’s also weird because it’s the only nuclear isomer found naturally in nature: that is, a nucleus that’s in an excited state, not its ground state. To add to the weirdness, the ground state of tantalum-180 is less stable than the excited state: it decays into tungsten or hafnium with a half-life of 8 hours!

Anyway: 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 the most abundant isotope of an element despite it being doubly odd? Or more precisely, why is it the only isotope with this property?

### 13 Responses to Nitrogen-14

1. nuclei —-> neutrons

2. John Baez says:

Keith McClary writes:

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.”

Primary nitrogen is nitrogen formed before stars using the CNO cycle:

Nitrogen is mainly produced by the CN cycle of the CNO reactions which catalyze hydrogen burning in stars. In the CN cycle, the reaction 14N(p,γ)15O which destroys nitrogen has a very small cross section, which allows 14N to accumulate with time (e.g. Arnett 1996). Thus, 14N is usually the daughter element, hence a secondary element, of the CNO initially present in stars. If nitrogen is secondary, the increase in the abundance of 14N should be proportional to the initial CNO content and consequently the 14N-content will be proportional to the square of the CNO and metal content in a galaxy. If 14N comes from a primary production, i.e. is produced from the original hydrogen and helium, the 14N-abundance is proportional to that of the other primary heavy elements.

3. John Baez says:

PM 2Ring writes:

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.

• From the same Wikipedia article:

Of the nine primordial odd–odd nuclides (five stable and four radioactive with long half lives), only 14 7N is the most common isotope of a common element. This is the case because it is a part of the CNO cycle. The nuclides 6 3Li and 10 5B are minority isotopes of elements that are themselves rare compared to other light elements, while the other six isotopes make up only a tiny percentage of the natural abundance of their elements. For example, 180m73Ta is thought to be the rarest of the 252 stable nuclides.

• John Baez says:

Thanks. Somehow I missed that! Of course, the CNO cycle destroys as much nitrogen-14 as it creates!

But I guess if something creates either carbon-12 or oxygen-16, the CNO cycle will turn some of that into nitrogen-14.

This is vaguely connected to my next post:

which concerns how carbon-12 gets produced in the first place.

I guess carbon-12 gets produced from collisions between beryllium-8 and helium-4, and then some of it becomes nitrogen-14.

• I should know this but I don’t. If the cycle is complete, then, yes, nitrogen is essentially a catalyst and neither created nor destroyed. But perhaps it is not always complete?

I’m sure that someone knows the answer to this!

• John Baez says:

When a star gets old and starts emitting lots of gas (a “planetary nebula”) or blows up (a “nova” or “supernova”), presumably some nitrogen still hanging around gets out.

The CNO cycle must be sort of like a carousel in a fairground: when it stops going round and round, you get off wherever you are.

4. MMM says:

Could it do with the # of protons + neutrons being half of the magic number to fill a nuclear shell? E.g., “These numbers of protons or neutrons (2, 8, 20, 28, 50, 82, and 126) make complete shells in the nucleus”, and I’ll note that Boron is half of 20 and Nitrogen is half of 28, and those are your two exceptions.

Alternatively, is the N15 capture cross-section just larger than the N14 capture cross section, so by kinetics you end up with a lot of N14 and little N15. (but that raises the why? question, which takes me back to maybe half of magic numbers)

• John Baez says:

Half magic! Hmm, that’s an interesting idea. One problem is that nuclei are especially stable when either the number of protons or the number of neutrons — or both — is a magic number. The 7 protons and 7 neutrons in nitrogen-14 are each a quarter of the magic number 28. I don’t even know if something nice happens when the sum of the number of protons and neutrons is a magic number, much less half a magic number.

• SteveB says:

Deuterium is half helium-4, which in some ways deserves to be called an independent particle in its own right, in, for example, alpha decay.

5. Gerard Westendorp says:

A wild speculation, but perhaps fun:

Picture:

In the Fano plane, there are 7 “points”-> Protons, and 7 lines ->Neutrons. You can create a 3D version of the Fano plane in the shape of an octahedron: The points are the 6 vertices plus 1 center point. The 7 lines are 4 alternate faces of the octahedron plus 3 axis. (XYZ).

Each proton is then surrounded by 3 neutrons, and each neutron is surrounded by 3 protons.

The strong force (as I just learned from Wikipedia), attracts best between adjacent protons and neutrons with the same spin. But proton-proton and neutron-neutron with the same spin don’t like being adjacent because of Pauli. So when the numbers are even, it is easier to form pairs of the same spin, while keeping the total spin zero.

6. John Baez says:

Over on Physics StackExchange, ProfRob writes:

I think it is down to the production mechanism in the case of 14N vs other N isotopes.

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 that might process the nitrogen 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 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”) or in classical novae.

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