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One of the things I've encountered in my travels is the mass-5 roadblock. Rod Nave writes about it on his excellent educational hyperphysics website:

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The helium-4 nucleus or alpha particle with a mass of 4 is particularly stable. But there are no is stable isotope with a mass of 5. Helium-5 isn't stable, nor is lithium-5. They decay almost immediately. Lithium-6 however is stable, but it has a lower binding energy than helium-4, which sounds relevant. Something else that sounds relevant is that lithium-7 is stable too but lithium-8 is not, and nor is beryllium-8, or boron-8. The $64,000 question is why?

Why are there no stable isotopes with an atomic mass of 5 or 8?

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  • $\begingroup$ Think about the magnetic dipole moments (commonly called intrinsic spin) of the involved protons and neutrons. Two as well as eight of theme are in perfect equilibration, see Symmetries in atomic orbitals, which holds for the magnetic dipole moments in nucleus too. $\endgroup$ – HolgerFiedler Feb 21 '17 at 19:07
  • $\begingroup$ Unwise to accept an answer if you don't actually think it has answered your question, since (rightly or wrongly) it discourages others from posting an answer. $\endgroup$ – Rob Jeffries Aug 3 '17 at 14:13
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The extreme stability of He-4.

Look at the decay modes of the the eights and they produce two alphas and if it is necessary convert a neutron/proton to a proton/neutron with an appropriate beta decay.
The production of two alphas is energetically favourable.

Li-6 and Li-7 lack nucleons to form two alphas.

Li-5 kicks out a proton and He-5 kicks out a neutron to form He-4.

More about 5 in LubošMotl's answer.

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  • $\begingroup$ I may be wrong, but I was guessing John might be wondering why He was so much more stable than isotopes of mass 5 and 8. $\endgroup$ – Rob Jeffries Feb 7 '17 at 7:19
  • $\begingroup$ @RobJeffries Does not the Shell model of the nucleus explain that? The 1s orbitals for neutrons and protons are full and to add another nucleon is energetically unfavourable? $\endgroup$ – Farcher Feb 7 '17 at 7:51
  • $\begingroup$ Then perhaps that is the answer. $\endgroup$ – Rob Jeffries Feb 7 '17 at 8:33
  • $\begingroup$ @Rob Jeffries : I think the shell model is certainly an attempt at an answer, but IMHO it isn't satisfactory. Electrons in the 1S atomic shell of the helium atom are there because the nucleus is there. They are bound to it. The 4 nucleons of the nucleus are bound together. Motl's answer is something of a non-answer because he doesn't tackle the reason why a 5th nucleon can't join the other 4. $\endgroup$ – John Duffield Feb 7 '17 at 13:46
  • $\begingroup$ Because of magic numbers and such-like - the details of the strong nuclear force. The protons and neutrons are more stable when paired up in a closed shell. The additional proton or neutron is not bound. You would be better revising your question to say that you know that an answer on one level is that the 5 and 8 isotopes are energetically unfavourable, but say that you really want to know why that is the case. @JohnDuffield $\endgroup$ – Rob Jeffries Feb 7 '17 at 16:16

protected by Qmechanic Aug 3 '17 at 11:22

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