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We have some protons and neutrons and want to create a nucleus.

We must do work against the Coloumb force to bring these protons and neutrons together to form the nucleus, this increases the energy of the system. Hence the energy (or mass) of the nucleus should be greater than the energy (or mass) of the constituent parts. this is not correct though and I’m not sure why

Likewise, when we want to disassemble the nucleus, this energy should be released.

However, this is not the case, because energy is required to split a nucleus.

So why is my logic incorrect?

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  • $\begingroup$ Are you sure that this energy is not released? How do nuclear power plants and bombs work then? $\endgroup$ – infinitezero May 11 at 22:33
  • $\begingroup$ @infinitezero oh yes that is true, but then why does my other logic not work? $\endgroup$ – PhysicsMathsLove May 11 at 22:34
  • $\begingroup$ It's wrong because you're not taking into account the nuclear force which overcomes the Coulomb force. $\endgroup$ – Gert May 11 at 22:39
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"We must do work against the Coloumb force to bring these protons and neutrons together to form the nucleus."

I think you are assuming that the only force between nucleons is the Coulomb (electrostatic) repulsion. But that can't be the case or the newly assembled nucleus would fly apart again. There is also a much shorter range attractive force between nucleons, that prevents the flying apart. Work has to be done against this force to pull the nucleons apart, more work than is contributed by the repulsive Coulomb force trying to separate the nucleons.

Because a net amount of work has to be put in from an external source (e.g. a subatomic 'missile' hitting the nucleus) in order to separate the nucleons, the nucleus must have less energy than the separated nucleons.

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  • $\begingroup$ Ok yes I see thank you :) $\endgroup$ – PhysicsMathsLove May 11 at 22:45
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You have to do work against the Coulomb force to bring the nucleons close to one another, but once you pass a certain distance, the appropriately-named strong force pulls them together, which releases much more energy than was required to push them close in the first place.

Flip it around: in order to pull a nucleus apart into its constituent nucleons, you need to do an enormous quantity of work against the strong force. You only get a small fraction of this energy back from the repulsion of the Coulomb force.

Note that you can still get energy from nuclear fission, splitting a nucleus into multiple smaller nuclei. One way to reconcile this is to imagine taking the larger nucleus apart into its fragments, which always requires $E_{\rm binding}$ of work, but then reassembling it into two smaller nuclei, which releases $E_{\rm 1}+E_{\rm 2}$ energy. Whether this fission requires or releases energy overall depends on the relative binding energies of the nuclei.

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  • $\begingroup$ Ok yes I see thank you ;) $\endgroup$ – PhysicsMathsLove May 11 at 22:45
  • $\begingroup$ @PhysicsMathsLove No problem. So you know, though, comments just saying "thank you" are discouraged here. See this help page. $\endgroup$ – Chris May 11 at 22:47

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