Nuclear Binding energy

The nuclear binding energy, is the energy that is needed to seperate the nucleons in a nucleus. And binding energy is also defined as the energy given out when a nucleus forms from nucleons.

Also the larger the nucleus is, the more energy is required to break it apart, so why doesn't that mean that larger nuclei are more stable? I mean Uranium has a lower binding energy per nucleon than Iron, but there are many many more nucleons in Uranium that Iron so the total binding energy is going to be much greater.

Basically I don't understand why whether an element gives out energy by fusion or fission (why the lighter element provide energy by fusion not fission and vice versa for heavy elements) depends on binding energy per nucleon and not "total" binding energy

Answer

Your basic nuclear reaction conserves the number of nucleons present.¹

That is important, because at a bit less than 1 GeV each the mass of the nucleons dominates the total energy of all these states.

So the only place available to get or lose energy in a reaction is by

1. 
   

   Changing the flavor of nucleons. Every neutron converted to a proton gets you a neutrino and some gammas (once the positron has captured and annihilated).

   
   
   


2. 
   

   By changing the total binding energy. Notice that getting one nucleon more bound doesn't help if another one gets less bound by a large amount. The decision to express this in terms of the average binding energy is completely arbitrary because for $N$ total nucleons (which doesn't change, remember?) $E_\text{total} = E_\text{AVG} * N$.

   
   

Another questions addresses what is so special about iron that it has the highest binding energy per nucleon.

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¹ This is more or less required by Baryon number conservation in the Standard Model, and we will ignore the need for Baryon non-conservation in most beyond the Standard Model candidate theories.