The graph of binding energy per nucleon (Hyperphysics) seems to peak at Iron-56. However, Wikipedia says that Nickel-62 has the highest binding energy per nucleon of any known nuclide (8.7945 MeV). Isn't this a direct contradiction? Why does the binding energy per nucleon peak at Iron-56, but then Wikipedia says Nickel-62 has the highest binding energy per nucleon?

  • 2
    $\begingroup$ Is this the graph from Hyperphysics? If so it explicitly says on the graph that Ni-62 is the most bound. If it is not can you quote and give a link to what you are referring to? The most relevant link I can on hyperphysics is find is hyperphysics.phy-astr.gsu.edu/hbase/NucEne/nucbin2.html#c1 $\endgroup$
    – Ian Bush
    Commented Jul 8, 2018 at 15:59
  • $\begingroup$ Ni-62 is not the most common nickel isotope so they have likely plotted the most common isotopes. Of those, Fe-56 is the most stable while Ni-62 is the most stable overall. I suspect there's some very interesting nuclear physics associated with how additional neutrons can stabilize a system and yet make it more unlikely to be created. $\endgroup$
    – jheindel
    Commented Jul 8, 2018 at 17:47
  • $\begingroup$ chemistry.stackexchange.com/a/32123/9961 covers this $\endgroup$
    – Mithoron
    Commented Jul 8, 2018 at 18:45
  • $\begingroup$ @IanBush, yes this was from hyperphysics but doesn’t the statement “nickel 62 has the most tightly bound nucleus” directly contradict the graph, which peaks at iron? $\endgroup$
    – coder
    Commented Jul 9, 2018 at 11:58

2 Answers 2


Can't be expertly comment on the plot without knowing exactly how the plot was constructed. But I can make the following observations...

  • User Mithoron makes a good point that it depends on how the Y-axis of the plot is taken. You can plot average mass per nucleon in which case iron-56 is the heaviest atom per nucleon. However if you plot binding energy per nucleon, then nickel-62 has the highest binding energy per nucleon.

  • For elements with stable isotopes I'd guess only the stable ones are shown. For elements with unstable isotopes than some selection process is necessary so as not to clutter up the plot too much.

  • I'll point out that the "peak" is marked as "Iron Group." It isn't entirely clear what isotopes that means. Wikipedia notes that in astrophysics The iron group in astrophysics is the group of elements from chromium to nickel which are substantially more abundant in the universe than those that come after them – or immediately before them – in order of atomic number.

  • In terms of stellar production much more iron than nickel is produced. The iron produced in stars is about about 92 % iron-56. Nickel is only about 3.6% Nickel-62.

  • I'd guess that the plot is a "best fit line." Here is a different plot showing more detail. You can see that not all isotopes fall on "the line."

enter image description here

  • Here is another plot at link pointed out by user IanBush in a comment. The "best fit" curve doesn't pass through all the data points.

enter image description here


Of some relevance is the course of the silicon-burning process, which represents the final stage of nuclear fusion in massive stars. Basically, the helium-4 nuclei formed by the primary hydrogen-fusion process "polymerize" to give a transient Be-8 then stable C-12, O-16, and so on. In the silicon burning phase the relatively stable Si-28 can react with helium nuclei photolytically generated from other nuclei to continue this chain, but products survive only as long as the additional fusion is sufficiently exothermic (stabilizing) for the product to hold up in the thermal and photonic environment of the stellar core.

That endpoint happens to be Ni-56, so helium/alpha-particle fueled stellar fusion processes can get only as high as that mass number -- and thus, after positron beta-particle decays, to iron-56. Hence the common occurrence of iron-56 (and where there is intelligent life, steel made therefrom) even though the more difficult to reach Ni-62 would be more stable in terms of binding energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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