As we know, anything to the left of Iron-56 in Binding Energy curve can be created by nuclear fusion reaction. The same with the fission products to the right side of Iron-56.

But in case of this nuclear binding energy curve, http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/nucbin.html, the fusion products are given between 2Mev to 7Mev and fission products between 7.5Mev(approx.) to 8.2Mev(approx.)

Can anyone please explain why the range of the fusion products is between 2 to 7 Mev instead of Deutrium to Iron?

What would be the range of the end products of fusion and fission reaction in the Binding Energy curve?

  • 1
    $\begingroup$ First off, elements with a higher number of nucleons than iron can still be created by fusion, it's just that the fusion would consume energy instead of releasing it. This is actually why stars collapse at the end of their life (iron core) and how neutron stars are created. Also note that binding energies are negative, so going from 2 MeV to 8 MeV means 6 MeV are released. $\endgroup$ – Wihtedeka Mar 12 at 11:42
  • $\begingroup$ Are you forgeting the untis "binding energy per nucleon" ? The high z nuclei have much more energy available in tota, $\endgroup$ – anna v Mar 12 at 14:20
  • $\begingroup$ I don't understand your question: " why the range of the fusion products is between 2 to 7 Mev instead of Deutrium to Iron?" What does "instead of" refer to? You can't compare energy to elements. And when you mention "range", to what quantity or objects are you referring? $\endgroup$ – Bill N Mar 12 at 16:12
  • $\begingroup$ Please correct me if my concept is wrong. Different elements have different binding energy and from this sense I told that (referring to your query about comparision between energy and element). As we know fusion products can be anything between Hydrogen-2 and Iron-56 (this is the range I am talking about), I asked "instead of this range, why they put a range between Hydrogen-2 and Helieum-4." $\endgroup$ – Nawaz Sharif Mar 12 at 16:21

Yes. This is binding Energy per nucleon vs Atomic Number curve of all Elements (has very little to do with Fusion Products strictly speaking).

As you have stated Iron-56 has the highest binding energy per nucleon and is there for the stablest. Nuclear reaction would prefer to approach Iron-56.

The reason why Fusion (Exothermic Fusion) occurs to the left of Fe is for exactly the same reason. They have low BE per nucleon hence fusion will take them up and closer to Fe.

Fission (Exothermic Fission) happens to the right of Iron because these elements have low BE and by fragmenting they can come close to Fe and hence be more stable.

The particular "yield of fusion" tagg in the graph is misleading, or equivalently the graph is pulled out of place without citation. The fusion products can be anything before Fe, for the sole reason that BE per nucleon energy of product is moving closer to Fe which has the most stable BE.

  • $\begingroup$ Actually my question is- does the fission reaction yield products within the range of 2 to 7 Mev as said in the graph? What is the reason behind this? $\endgroup$ – Nawaz Sharif Mar 12 at 12:00
  • $\begingroup$ ohh.. yes until you reach Fe all the products are on the left of Fe meaning they can be fused again. To be precise the fusion products can have binding energy from around $~2\,MeV$ to $~8.8\,MeV$ (value of Fe). The reason is they are not as stable as Fe and so can be fused again since they still have $Z<56$. $\endgroup$ – TheImperfectCrazy Mar 12 at 12:05
  • $\begingroup$ Thank you for the answer and please accept my apology for my english. In the graph, they gave a range between 2 to 7Mev and indicated that region as 'yield from nuclear fusion.' Why they did not give the range between 2 to 8.8 Mev for the products of nuclear fusion? $\endgroup$ – Nawaz Sharif Mar 12 at 12:25
  • $\begingroup$ That's absolutely fine. ohh no no... that particular graph is very out of place. That marking simply gives the yield of fusion of two Hydrogen-2 into a Helium-4. In general the by products of fusion can be any element before Fe, experimental confirmed in Stellar evolution. $\endgroup$ – TheImperfectCrazy Mar 12 at 12:46
  • $\begingroup$ Thanks a lot. :) $\endgroup$ – Nawaz Sharif Mar 12 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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