If you have more neutrons than protons, then there will be more strong force present to counteract the repulsive forces between protons. Why is it that above bismuth, no nucleus is stable, regardless of its N:Z ratio?

  • $\begingroup$ see the chosen answer here physics.stackexchange.com/q/9098 $\endgroup$
    – anna v
    Jul 13 '15 at 11:15
  • $\begingroup$ FYI: all isotopes of bismuth, including Bi-209 are unstable in terms alpha-decay. The decay of Bi-209, the least unstable, has been observed directly. $\endgroup$
    – DJohnM
    Jul 13 '15 at 15:53

The answers by Alex and userTLK are correct but incomplete.

It is true that whilst the strong force essentially only acts between nearest neighbours, whilst coulomb repulsion acts between all protons, it is actually the weak force that prevents the building of extremely large nuclei.

For example, one must explain why you can't build nuclei with more and more neutrons, simply increasing the neutron to proton ratio.

The answer is that neutrons decay (via a weak interaction) into protons (and electrons) providing there is a spare quantum state for the proton to drop into. If this is not the case then the beta decay is "blocked" by the Pauli exclusion principle. Thus highly neutron rich nuclei will be unstable to beta decay.

Interestingly, in a neutron star crust, beta decay can also be blocked by surrounding degenerate electrons (if their Fermi energy is high enough). There, you can build enormous (atomic mass more than 300) neutron-rich nuclei.

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    $\begingroup$ To be totally accurate, beta decay requires not only a"spare" quantum state for the proton but also that the daughter nucleus is more bound than the parent nucleus so that the decay is energetically favorable. Plenty of nuclei with more neutrons than protons are stable against beta decay, eg calcium-48 or lead-208. $\endgroup$
    – ragnar
    Jul 13 '15 at 14:34

The reason is because the strong force isn't cumulative but the electromagnetic force is. Now, the strong force is a bit more complicated as it does change based on the number of protons and neutrons, but it doesn't build continuously as more protons or neutrons are bound to the nucleus, but the electromagnetic force does.

Say you have a helium atom, 2 protons, 2 neutrons, each is tightly bound by the strong force and the 2 protons are only repelled only by each other. So, it's 1 strong force of attraction, and 1 electromagnetic force of repulsion and the strong force wins. The strong force is 137 times stronger

Now, take Uranium, 92 protons. Each Proton and Neutron is bound to the nucleus by the strong force, but it's only 1 strong force attraction but each proton is now repulsed by 91 other positively charge protons. Hence you have 91 little forces pushing it away. This is much less stable.

Quantum instability always happens at 83 or more protons (1-82 are mostly stable, except for Technetium and Promethium, with 43 and 61 protons respectively), which in and of itself is rather curious. The strong force binds tighter with specific combinations, and as a general rule, even numbers of protons are more often stable than odd numbers. I'm not sure why that is but it seems consistently true.

More on that here: https://en.wikipedia.org/wiki/Even_and_odd_atomic_nuclei and, loosely related, here: http://io9.com/the-oddo-harkins-rule-shows-the-universe-hates-the-odd-1446581327

Also, please see Rob Jeffries answer, as he mentions the weak force - I think his answer is more correct than mine.

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    $\begingroup$ Nice answer. The even numbers are more stable because of the effect of nuclear pairing, which is essentially the nuclear equivalent of super conductivity. $\endgroup$
    – ragnar
    Jul 13 '15 at 6:50
  • $\begingroup$ @ragnar sorry but i could not relate the analogy given to superconductivity. I would be grateful if you could give reference to this as it is interesting that nuclea physics and condensed matter physics are related $\endgroup$ Jul 13 '15 at 8:37
  • $\begingroup$ I wasn't familiar with that analogy till I read ragnar's post, but I found this on it, which gives a very brief explanation: en.wikipedia.org/wiki/… $\endgroup$
    – userLTK
    Jul 13 '15 at 9:55

Although there are more neutrons than protons to counteract the electrostatic repulsion, there still is proton repulsion. This repulsion grows with larger and larger atoms. By emitting alpha radiation or helium nuclei, an atom can transition from a high energy state to a lower energy state. This is why it is favored as a decay. The more protons the more repulsion and higher energy state.

It's similar to electrons which go from higher energy states to lower ones by emitting photons. The more energetic the electron the more likely it is to emit a photon and transition down to a lower energy. Nuclei are similar but under the influence of different forces.


As the mass number increases, the repulsive effect becomes more significant, because the strong nuclear force (that binds all the nucleons) is very short range, while Coulomb force is not: that is why energy tied up in repulsive Coulomb force bound between protons increases more rapidly as the mass numbers increases than energy tied up in attractive strong nuclear force bound between any two nucleons.

That's why heavy nuclei become unstable.

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    $\begingroup$ Fails to explain why you can't simply add neutrons. $\endgroup$
    – ProfRob
    Oct 5 '17 at 6:52

large nuclei contains more number of protons.As a result the coloumb's force of repulsion between protons increases and becomes dominant over the short range attractive nuclear force between nucleons.As a result the unstability of large nuclei increases.

  • $\begingroup$ So why not add even more neutrons, say in a 3:1 ratio to protons? You didn't address the question. $\endgroup$
    – Bill N
    Jul 1 at 19:14

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