Nuclear fusion does not occur with nuclei (with a nucleon number higher than 56) because the binding energy of the product nucleus is lower than the binding energy of the nuclei fusing together. Due to the product nucleus having a lower binding energy (and thus lower mass defect), this must mean that mass has been gained through the reaction which is impossible because energy is released?

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2022-08-28

Nuclear fusion does not occur with nuclei (with a nucleon number higher than 56) because the binding energy of the product nucleus is lower than the binding energy of the nuclei fusing together. Due to the product nucleus having a lower binding energy (and thus lower mass defect), this must mean that mass has been gained through the reaction which is impossible because energy is released?

Answer & Explanation

Kaleigh Patel

Kaleigh Patel

Beginner2022-08-29Added 5 answers

It is correct. For instance the binding energies of an alpha particle (28.30 MeV) and a 56Ni (483.99 MeV) nucleus sum to 512.29 MeV. They fuse to give 60Zn, which has a binding energy of 514.99 MeV. i.e. the reaction is still exothermic.
There are two factors that stop this happening to a great degree in stars.
First, the alpha particle has to come from somewhere. If it has to be ripped out of another 56Ni nucleus then that does require additional energy.
Second, to overcome the increasing Coulomb barrier for fusion onto nuclei with higher proton numbers requires increasingly high temperatures. At these temperatures, the photons in the gas are capable of photodisintegrating the heavier nuclei. Thus they are not present in any great concentration.
Heavier elements are predominantly made by neutron captures and decay.

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