Why the binding energy per nucleon has specific pattern?

The binding energy per nucleon for each element has a specific pattern. It increases from hydrogen to iron and then decreases again. What is the fundamental reason why the binding energy per nucleon has this pattern?

According to my investigation, the range of action of the strong force is narrower than that of the Coulomb force. Therefore, if the atomic nucleus is smaller than iron, the strong influence is greater, so the larger the atomic nucleus, the more stable it is. However, if the atomic nucleus is larger than iron, the influence of the Coulomb force is greater, so the smaller the atomic nucleus, the more stable it is.

Is this right? If this is correct, I would like to ask for a more detailed explanation. In particular, why does helium have a greater binding energy per nucleon than lithium?

Very qualitatively yes, but there are other relevant effects too. In particular the nucleon pairing effects which essentially make even-even nuclei more stable.

hongseok said:

The binding energy per nucleon for each element

An element does not have a specific binding energy per nucleon, or a specific binding energy, or a specific number of nucleons.
Observe that isotopes do.
Of first 35 nucleon counts, 2 have no stable isobars and the other 33 each have a single stable isobar. For their average binding energies, see
https://en.wikipedia.org/wiki/List_of_nuclides
It is defined relative to free neutrons, so protium has a "binding energy".
What would make sense is keep track of not just average binding energy but marginal binding energy. It is, in practical sense, the separation energy of a nucleon (I´m picking the one whose separation gives a stable nucleus, not unstable); in fundamental sense, it is the binding energy of the nucleon on the outermost orbit.

1. 1 - H - average 0,782 MeV (the decay energy of neutron)
2. 2 - also H - average 1,503 MeV - total 3,006 - n - marginal 2,228
3. 3 - He - 3,094 - 9,283 - p - total marginal 6,277, without proton binding energy the practical number is 5,495
4. 4 - He - 7,465 - 29,860 - n - 20,577
5. 5 - nothing bound
6. 6 - Li - 5,724 - 34,341 - since the only thing to separate would be α+d, that would be 1,475 MeV
7. 7 - Li - 5,942 - 41,591 - n - 7,250
8. 8-Be is unbound, 8-Li and 8-B bound but shortlived
9. 9 - Be - 6,810 - 61,294 - the separation would be n+2α, at 1,574 MeV
10. 10 - B - 6,866 - 68,663 - p - 7,369/6,587 MeV
11. 11 - B - 7,283 - 80,117 - n - 11,454
12. 12 - C - 8,071 - 96,856 - p - 16,739
13. 13 - C - 7,831 - 101,802 - n - 4,946

Can you see a pattern?
The energy of adding a nucleon starts off at 2,2 MeV at d, then grows a lot at He-3 to 5,5 MeV, then jumps to 20,6 MeV... when you reach He-4. Which is exactly pairs of proton and neutron each... precisely filling the 1s orbitals with opposite spin pairs.
A fifth nucleon is not bound to be alone in an outer shell, at all. 2 nucleons in outer shell would be bound by around 3,7 MeV between both of them (but note that they could also leave together as a d). A third nucleon is better bound... but when you add a fourth nucleon, the problem is not so much that it is weakly bound relative to the previous ones, but that it is best bound in one of two α particles, not in 8-Be. From 9-Be on, the binding energy climbs with each added nucleon, passes 11 MeV at 11-B... and jumps past 16 MeV at 12-C. And drops back below 5 MeV at 13-C. The pattern is that another orbital fills with C-12. The peak of last nucleon filling a shell is there, though not as great as for α. So is the dropoff for a lone nucleon in outer shell, though not as drastic as the case of 5, which is always unbound.