Iron has the highest binding energy per nucleon in the entirety of the known elements. But why Iron specifically? What makes it have the highest binding energy per nucleon?
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3$\begingroup$ The Bethe-Weizsäcker formula offers some explanation for nuclear binding energy. However, why the minimum is attained for iron depends on the parameters of the Bethe-Weizsäcker formula, which in turn depend on the specifics of the forces between the nucleons and asking why they are, how they are, is unphysical. $\endgroup$– Sebastian RieseCommented Jan 31, 2016 at 13:25
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$\begingroup$ Related: physics.stackexchange.com/q/182525/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jan 31, 2016 at 13:57
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$\begingroup$ read about the stability and nucleus levels in Wikipedia : Nuclear shell model $\endgroup$– user46925Commented Jan 31, 2016 at 17:01
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$\begingroup$ see this:-chemistry.stackexchange.com/questions/40407/… $\endgroup$– Nilay GhoshCommented Feb 1, 2016 at 8:36
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2 Answers
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The existence of nuclei is dependent on a number of quantum mechanical boundary conditions. They appear as solutions to a problem where there is a balance of: a) the attractive spill over color force that binds the quarks into a proton or a neutron, b) the repulsive electromagnetic force between protons, c) the Pauli exclusion principle, d) the instability of not strongly bound neutrons to a weak decay. There are additional factors entering once electrons get trapped around a nucleus, but that is another story.
To answer "why" the element with 26 protons and 30 neutrons is stable (or the one with 26 protons and 32 neutrons) and has close to the maximum binding energy, one needs a specific quantum mechanical model for the collective potential of the above factors. Shell models are fairly successful in classifying the periodic table.
The real answer about iron though would be phenomenological, that is what we observe and fit phenomenologically with the Weizsaecker formula, which is based on a liquid drop model. The way the effective potential works, the inclusion of more and more nucleons in the potential well after iron stops creating a deeper effective potential well, due to the increase of the effect of repulsive forces described above.
Please note that it is Ni62 that is more tightly bound in the binding energy curve.
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$\begingroup$ Looking at that link, Ni58 seems incorrect as it is not even on the chart. Do you mean Fe58, Ni62, or Ni60? Presumably Ni62, but in that case a comment on why Fe56 is so much more abundant might be in order. $\endgroup$ Commented Jan 31, 2016 at 17:39
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$\begingroup$ @PieterGeerkens you are correct, I will edit. As for abundance, it would be a different question $\endgroup$– anna vCommented Jan 31, 2016 at 17:55
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$\begingroup$ @annav Would you have a look on my elaboration About the distribution of electrons magnetic dipole moments in atoms, please? $\endgroup$ Commented Jan 31, 2016 at 19:19
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$\begingroup$ Available in German (it's the original) academia.edu/18391105/… $\endgroup$ Commented Jan 31, 2016 at 20:48
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$\begingroup$ Sbould be binding energy per nucleon. The binding energy of heavier nuclei is larger than that of iron. $\endgroup$– ProfRobCommented 4 hours ago
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What determines the most stable element (Fe) is the trade off between the nuclear binding (attractive) and the coulomb repulsion between protons. Nucleons feel binding forces that can be described as bulk and surface forces. The bulk forces are those associated with the saturation of nuclear forces (nuclear density in the interior of heavy atoms is relatively constant). Surface terms are associated with the reduced binding at the nuclear surface due to the reduced density. Coulomb repulsion builds up as proton number is increased. It just happens that a transition occurs near Fe (the coulomb repulsion begins to dominate over the enhanced binding due to reducing the surface to volume ratio).
answered Jan 31, 2016 at 13:54
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$\begingroup$ That doesn't explain why a neutron ball wouldn't be the most stble of all! $\endgroup$– JDługoszCommented Feb 1, 2016 at 3:30
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1$\begingroup$ If you are talking about neutron stars, gravity plays a role. $\endgroup$ Commented Feb 1, 2016 at 13:22