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Binding energy per nucleon is a measure of how stable the nucleaus is. is this true.? I looked at few answers on this site but there is no clear agreement on it.

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  • $\begingroup$ If every nucleon is happier, the whole nucleus is happier... $\endgroup$
    – Jon Custer
    Apr 27 at 15:59
  • $\begingroup$ No clear agreement? Please give some links so we can purge this site. $\endgroup$
    – my2cts
    Apr 27 at 16:53
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Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts

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The mass of an atomic nucleus is less than the sum of the individual masses of the free constituent protons and neutrons, according to Einstein's equation E=mc2. This 'missing mass' is known as the mass defect, and represents the energy that was released when the nucleus was formed.

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In the periodic table of elements, the series of light elements from hydrogen up to sodium is observed to exhibit generally increasing binding energy per nucleon as the atomic mass increases. This increase is generated by increasing forces per nucleon in the nucleus, as each additional nucleon is attracted by other nearby nucleons, and thus more tightly bound to the whole.

bindener

The region of increasing binding energy is followed by a region of relative stability (saturation) in the sequence from magnesium through xenon. In this region, the nucleus has become large enough that nuclear forces no longer completely extend efficiently across its width. Attractive nuclear forces in this region, as atomic mass increases, are nearly balanced by repellent electromagnetic forces between protons, as the atomic number increases.

Finally, in elements heavier than xenon, there is a decrease in binding energy per nucleon as atomic number increases. In this region of nuclear size, electromagnetic repulsive forces are beginning to overcome the strong nuclear force attraction.

You ask:

Binding energy per nucleon is a measure of how stable the nucleaus is. is this true.?

From the above quoted paragraphs it should be clear that it depends on the position of the nucleus in the periodic table, the number of nucleons and the number of charged to neutral. So the binding energy per nucleon is related to the stability but is not a measure of it.

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If the binding energy per nucleon is great, that means it would take a great amount of work to disrupt it, and the nucleus is resistant to things that want to perturb it. This means in practical terms that it is more stable.

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The binding energy of a nucleus is the amount of mass that was converted to energy (via $E=mc^2$) during formation of that nucleus from its constituent nucleons$\dagger$ (protons and neutrons) You can divide that number by the total number of nucleons, if you like.

The binding energy could be better named: 'nuclear energy of formation'.

The more negative this energy of formation is, the more stable is the nucleus. We can see that because to decompose the nucleus would require a large amount of energy: it is stable.

$\dagger$ Note that in reality this is not how nuclei are formed.

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By definition a nucleus is more stable if it has a higher binding energy per nucleus.

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Yes, It is because it is the measure of how much mass loss in form of energy takes place from the parent nucleus on decay. Though Gibbs energy is the real measure of spontaneity but Most of the exothermic reactions are stable Thus, justifying the Role of Ebn

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    $\begingroup$ It is because it is the measure of how much mass loss in form of energy takes place from the parent nucleus on decay. No, that's incorrect. $\endgroup$
    – Gert
    Apr 27 at 16:21
  • $\begingroup$ No, It is Correct $\endgroup$ Apr 28 at 4:03

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