I am just confusing myself with binding energy and the binding energy curve. I want to know whether I have interpreted the graph right. So when both nuclei have a nucleon number over 56 and are fused together, the product nucleus must have a binding energy which is less than the binding energy of the two lighter nuclei.

This means that because the binding energy has decreased, the mass defect must have also decreased which therefore means that mass has been gained.

Now assuming this, does this mean that no energy is released from the nuclear fusion (because no mass has been lost) or does it mean that because a small amount of mass has been gained, very little energy is released? I think its the latter...

Oh also! I just figured out why I am not understanding! So the basic laws of conservation of energy state that energy cannot be destroyed or created... and so where does this energy come from in order for the mass to be gained?

Thanks for any help!

  • $\begingroup$ Hello there! Please follow a couple helpful rules when posting: provide sources and, in this case, post the graph you're looking at (and ideally the equations behind the graph) so we know better what you are asking about. $\endgroup$ – Carl Witthoft Sep 18 '20 at 12:14
  • $\begingroup$ In any case, it's well known that once you're past fusing to get Fe, fusion is endothermic, not exothermic. I think that's what you're asking: energy has to be subsumed in the fusion process and thus the total mass increases. $\endgroup$ – Carl Witthoft Sep 18 '20 at 12:15
  • 1
    $\begingroup$ One should be careful with broad statements. There are a number of fusion reactions between nuclei with A > 56 that result in increased binding energy. However, you are not starting with stable nuclei. (For example, Ca-57 + Ga-57 to make Sb-114 is exothermic.) $\endgroup$ – Jon Custer Sep 18 '20 at 14:13
  • $\begingroup$ But how is that possible? I thought the binding energy per nucleon decreases as the nuclei have a nucleon number larger than 56? Surely, Sb-114 has a smaller binding energy than the two 57 nucleon nuclei? $\endgroup$ – Phoooebe Sep 18 '20 at 16:06
  • $\begingroup$ Consider that, while Sb-114 is indeed unstable, Ca-57 and Ga-57 are even more unstable... You might google for the Atomic Mass Evaluation pdf and look at all the data there. $\endgroup$ – Jon Custer Sep 21 '20 at 22:00


This is the binding energy curve:


So when both nuclei have a nucleon number over 56 and are fused together, the product nucleus must have a binding energy which is less than the binding energy of the two lighter nuclei.

you should add per/nucleon . OK, lets take 56+57. That makes 113 nuclei, Nihonium an unstable elemen, breaks up to nucleons with higher binding energy per nucleon, i.e. energy is released.

Are you confusing energy per nucleon to total binding energy?

  • $\begingroup$ Thanks for the reply but I am still confused because if you follow the curve, the nucleus of Nihonium will still have a TOTAL binding energy less than the total binding energy of 56 and 57 (not per nucleon). This means that mass has been gained so does this mean that less energy is released? $\endgroup$ – Phoooebe Sep 19 '20 at 6:29
  • 1
    $\begingroup$ Fusion cannot happen over 56 .fusion means that two low binding energy per nucleon merge to give one with high binding energy, releasing the excess energy in the process. 55 and 56 have high binding energy whereaw 113 low. The energy balance numbers have negative sign. $\endgroup$ – anna v Sep 19 '20 at 8:20
  • $\begingroup$ Ok I understand now that the energy balance will be negative, but what exactly does that mean? That external energy is needed or that it will take energy rather than releasing it? $\endgroup$ – Phoooebe Sep 19 '20 at 10:21
  • 1
    $\begingroup$ negative energy in this case means that it cannot be done. $\endgroup$ – anna v Sep 19 '20 at 10:26
  • $\begingroup$ Ohh okay thanks! I understand now $\endgroup$ – Phoooebe Sep 19 '20 at 10:49

The usual "binding energy curve" that you see is (negative) binding energy per nucleon for the most stable nucleus at that mass number. The net binding energy of larger stable nuclei continues to increase, but the binding energy per nucleon goes down.

If you fuse two nuclei with mass numbers greater than 56, then the average binding energy per nucleon of the product is smaller. Since the number of nucleons is conserved, this means the net amount of binding energy of the product is smaller than the sum of the binding energies of the two reactants.

The mass of the nucleus is the mass of the nucleons minus the binding energy and thus mass enes up being created in this process.

This is not exothermic, it will not release energy. It requires energy in order to get it to happen and create that extra mass.

Another way of thinking about this is to imagine you have a bag of nucleons and you allow them to form whatever nuclei they like. The way this would happen is so that the total energy density of the material in the bag is minimised. This will happen when the (negative) binding energy per nucleon is maximised. i.e. The nucleons will tend to form nuclei at the peak of the binding energy per nucleon curve. To split these nuclei up, or to fuse them together to make heavier nuclei will cost energy.

Note though that there is a further problem that is not obvious from the binding energy curve. As nuclei get heavier, the most stable n/p ratio increases (this is because of competition between the strong force acting between adjacent nucleons vs Coulomb repulsion affecting all the protons in the nucleus. But fusion of nuclei will conserve both the number of protons and neutrons separately. This means the nucleus that is produced by fusion of two nuclei with a mass number of about 56 will not be the most stable nucleus at a mass number of 112, it will have far too low a n/p ratio and a much lower binding energy per nucleon as a result. Thus the effects discussed in the first part of my answer are exacerbated by this and make it even harder and more energetically unfavourable to fuse two heavy nuclei together.


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.