Why do stars only produce up to iron and nickel

  • #1
niko_.97
18
4
I know it's a common question but I've found no answers online so far. My professor has made a point out of saying that fusion reactions after iron and nickel do release energy but just not enough to keep the star from imploding. This didn't make sense to me. How would fission release energy if that were the case? Every where I've looked online says that it does in fact take energy for fusion after iron and nickel. I even did a little numerical calculation for for nickel-62 fusing with an alpha particle to make zinc-66 and found that it's preferable for it to stay as its constituents.
 
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  • #2
Actual picture is very complicated, therefore both you and your professor are partially correct.
From binding energy perspective, iron (N=26) to zirconium (N=40) nuclei are very similar and theoretically stable against fission once formed, therefore preference to Iron-56 in modern universe is mostly due kinetic, not thermodynamic effects.

Binding energy per nucleon (referenced to carbon-12)
Fe-56: 1.082 MeV
Ni-62: 1.077 MeV
Sr-88: 0.991 MeV
Zr-92: 0.962 MeV

Heavy elements are actually produced in r-process, because of non-equilibrium fluxes of protons and neutrons. If you have for example iron nuclei immersed in proton&neutron flux, the conversion of iron to lead still gains additional energy. Of course, as long as non-equilibrum flux of light particles vanishes, the heavy nuclei tends to disintegrate back to Fe-Zr range, albeit inefficiently in case of stellar explosion.
 
Last edited:
  • #3
trurle said:
Actual picture is very complicated, therefore both you and your professor are partially correct.
From binding energy perspective, iron (N=26) to zirconium (N=40) nuclei are very similar and theoretically stable against fission once formed, therefore preference to Iron-56 in modern universe is mostly due kinetic, not thermodynamic effects.

Binding energy per nucleon (referenced to carbon-12)
Fe-56: 1.082 MeV
Ni-62: 1.077 MeV
Sr-88: 0.991 MeV
Zr-92: 0.962 MeV

Heavy elements are actually produced in r-process, because of non-equilibrium fluxes of protons and neutrons. If you have for example iron nuclei immersed in proton&neutron flux, the conversion of iron to lead still gains additional energy. Of course, as long as non-equilibrum flux of light particles vanishes, the heavy nuclei tends to disintegrate back to Fe-Zr range, albeit inefficiently in case of stellar explosion.

Thanks for the reply. What do you mean by an r-process?
So, I emailed my professor and I see where I may have made a mistake in my calculation. If you calculate the energy of the nuclei using E=mc^2 then there is more energy in the alpha and Nickel nucleus than in the Zinc.

So, if what we actually care about is the difference in mass energy (which makes sense) why do we look at B/A and so often everywhere online talks about the difference in B/A as the energy given off in a fusion/fission reaction?
 
  • #4
niko_.97 said:
Thanks for the reply. What do you mean by an r-process?
So, I emailed my professor and I see where I may have made a mistake in my calculation. If you calculate the energy of the nuclei using E=mc^2 then there is more energy in the alpha and Nickel nucleus than in the Zinc.

So, if what we actually care about is the difference in mass energy (which makes sense) why do we look at B/A and so often everywhere online talks about the difference in B/A as the energy given off in a fusion/fission reaction?
https://en.wikipedia.org/wiki/R-process
 

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