Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Why there is no stable nuclei with $$A=5$$ in nuclide the chart and so in nature like we know it?

share|improve this question
    
I don't think you're going to get a much more satisfying answer than "because helium-4 is in such a deep potential energy trough thanks to its special quantum numbers", but I'll let the people who know more about atomic physics than I do say that. –  Jerry Schirmer Feb 17 '13 at 15:25
2  
Dear Jerry, you meant nuclear physics, not atomic physics, right? ;-) –  Luboš Motl Feb 17 '13 at 16:57
add comment

1 Answer 1

up vote 2 down vote accepted

As Jerry Schirmer said, helium-4 is an extremely stable nucleus. What does it mean quantitatively? It means that it binding energy is very high, namely 28 MeV. In other words, helium-4 is 28 MeV/$c^2$ lighter than the sum of masses of two free protons and two free neutrons.

The best candidates $A=5$ nuclei would have 2 protons and 2 neutrons in the lowest state - i.e. in the same state as they occupy in helium-4 – but the additional 1 proton or 1 neutron would have to be added to a higher shell. But because this higher shell is so much higher in energy than the ground levels, one can't find an $A=5$ nucleus that would be lighter than the sum of the helium mass and one proton (or one neutron). The binding energy would have to be even greater than 28 MeV which means that the binding energy per nucleon would have to exceed 28/5=5.6 MeV. This is simply too much to ask; the binding energy you could get for 5 nucleons is simply smaller than 28 MeV, so any such object would quickly alpha-decay.

I should insert some calculation of the conceivable binding energy for 5 nucleons here except that there's clearly no "analytic" calculation. It's an extremely messy system one would have to describe by nuclear physics (ill-defined effective theory) or by QCD (calculable via lattice QCD, with big computers etc.). But let me mention that unlike atoms, where the new valence electrons may always be added and keep the stability, the nuclei are "more neutral" so the attractive force between the helium-4-like "core" of the $A=5$ object and the remaining nucleon is much weaker, sort of dipole-like, and isn't enough to produce a new stable bound state.

However, what I can say is that this fact about the absence of $A=5$ stable isotopes has important consequences. The Big Bang Nucleosynthesis – first three minutes when nuclei are created – essentially stalls once it reaches helium-4 nuclei. They can't absorb new protons/neutrons to become heavier and instead, the next reaction is the much rarer collision of two helium nuclei. One either has helium-3 plus helium-4 goes to lithium-7 plus positron plus photon; or beryllium-7 plus photon on the right hand side. Lithium-7 may absorb a proton to get back to 2 helium-4; beryllium-7 may absorb a neutron to become lithium-7.

These processes are "everything" one may have in empty space. Inside stars, one has pressure and temperature which helps to overcome the binding energy and stars may produce heavier elements, too.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.