As far as I know, number of protons is less that or equal to the number of neutrons in any atomic nucleus. But is there any possibility that there exists a nucleus where the number of protons exceeds number of neutrons (apart of course, from the trivial case of hydrogen)?

Actually I wanted this to be a discussion involving binding energy. It's my mistake: I did not correctly write my query. As many of you have pointed out, protium has in fact more protons than neutrons. But in protium there is only 1 proton, so there is no involvement of binding energy.

For an atomic nucleus to be stable, the repulsive force between protons must be less than the binding energy. But is there any atomic nuclei which is stable whose $n/p$ ratio is less than 1?

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    $\begingroup$ "As far as I know , number of protons is less that or equal to the number of neutrons in any atomic nuclei." - I don't understand how you came to this belief. The very first element in the periodic table forms atoms with one proton and no neutron... $\endgroup$
    – ACuriousMind
    Commented Oct 6, 2019 at 13:48
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    $\begingroup$ Did you check the periodic table? Some prior research is required. $\endgroup$
    – my2cts
    Commented Oct 6, 2019 at 13:51
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    $\begingroup$ Welcom to this forum, overkill - I hope you will enjoy what you learn here; I see you already have some good answers. Now, people, overkill is a new contributor, so try to rein in your irritation with what to you may be obvious. We all started out ignorant, and it was mostly by asking naive questions that we progressed, at least to start with. At least in my case, if I hadn't asked loads of "stupid" questions, I wouldn't have known a fraction of what I know today - and I still ask that kind of questions all the time. $\endgroup$
    – j4nd3r53n
    Commented Oct 7, 2019 at 7:34
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    $\begingroup$ @j4nd3r53n This has nothing to do with "irritation" or "what may be obvious" - prior research is expected and none has been executed here before asking. Remember, this is not a forum! Cheers. $\endgroup$ Commented Oct 7, 2019 at 12:02

4 Answers 4


What you are looking for is isotopes with neutron–proton ratio N/Z less than 1. You can find these isotopes, for example, in this list from Wikipedia. As you can see, you are looking for members of the table with N less than Z.

enter image description here

In these table you are looking for isotopes that are roughly above the gray zone (also known as band or belt of stability).

The colors indicate how stable the isotopes are, grey isotopes are stable, white isotopes have a half-life of less than a day, other colors are somewhere in-between. According to the table there are only three isotopes with less neutrons than protons and a half-life of more than a day. hydrogen-1 and helium-3 which are stable and beryllium-7 with a half-life of around 53 days.

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    $\begingroup$ An legend for the colors would be lovely. Also if you point out that the only ones are He3 and Be7 (which is unstable) then I think this would be the best answer. $\endgroup$
    – Eph
    Commented Oct 7, 2019 at 11:52

According to Wikipedia:

Other than protium (ordinary hydrogen), helium-3 is the only stable isotope of any element with more protons than neutrons.

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    $\begingroup$ The OP didn't ask about stable nuclei. $\endgroup$
    – user4552
    Commented Oct 7, 2019 at 0:41
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    $\begingroup$ @BenCrowell If you've got a better answer, by all means please post it. $\endgroup$
    – tparker
    Commented Oct 7, 2019 at 0:55
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    $\begingroup$ The "Common" (somewhat stable) unstable ones are C11 and N13, with half-lives of 20 and 9 minutes. They decay by positron emission. They are made in cyclotrons and used for Positron-Emission Tomography. He3 is the only stable one. $\endgroup$
    – nigel222
    Commented Oct 9, 2019 at 9:07

For an atomic nucleus to be stable, the repulsive force between protons must be less than the binding energy.

This doesn't really make any sense. You can't compare binding energy to repulsive force. This would be obvious if you tried to quantify the claim - it's a lot easier to say energy needs to be bigger than the force; it's a lot harder to put it in an equation :)

Rather, there's two binding energies involved - one from electromagnetic force, and another from strong nuclear force (we can ignore the other fundamental interactions). So, is it enough for the electromagnetic binding energy to be smaller than the strong binding energy?

Stability is not about absolute numbers, or about relative energies. It's about whether there's a lower energy state that you can occupy, reachable with the energy you have (e.g. no energy barrier, or a barrier small enough for quantum tunnelling, or a barrier small enough for the given temperature), and without breaking any of the conservation laws. A proton has a lot more energy than an electron; but it can't decay because there's nothing for it to decay into. A neutron has only a tiny bit more energy than a proton, but it readily decays into a proton (+ an electron and an electron anti-neutrino). Free neutrons aren't stable.

If you have a nucleus where the absolute value of electromagnetic binding energy is larger than the strong nuclear binding energy, it's not going to be stable. But that's not enough; if this were the case, you could keep adding neutrons to nuclei indefinitely. The binding energy of a He-8 is very large. But it still decays almost immediately, because there's other configurations that have lower total energy - in particular, it will very quickly release a positron, and transmute into Li-8.

But to answer your question directly, yes, there are stable nuclei with more protons than neutrons. The most obvious one is H-1, and the other He-3. He-2 is very close to being stable, but not quite (the strong force is higher than the electromagnetic repulsion, but there's other effects that make its binding energy positive).


Nuclear forces are short range forces are , if no. of proton are greater then number of neutron, then nucleii is not stable and it will decay into more stable daughter nuclei, energy related during this process due to mass defect called binding energy. Now coming to your question there is an exception helium-3 is the only stable isotope of any element with more protons than neutrons . with protium.


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