Why is Helium an anomaly on the binding energy curve? So I'm failing to understand why helium is anomalous on the binding energy curve. I know it includes concepts such as pairing, shell correction, and the liquid drop model. However, these were simply stated by my professor. He didn't explain any of them. 
Just requesting if someone could explain these concepts and their relation to helium on the binding energy curve. 
Also, are there other concepts involved explaining helium's position? 
 A: 
I'm failing to understand why helium is anomalous on the binding energy curve. I know it includes concepts such as pairing, shell correction, and the liquid drop model. However, these were simply stated by my professor. He didn't explain any of them.

From hyperphysics

The mass of a nucleus is always less than the sum of the individual masses of the nucleons(n,p..) which constitute it. 
  The difference is a measure of the nuclear binding energy which holds the nucleus together  and can be calculated from the Einstein relationship:
Nuclear binding energy = Δmc^2
For the alpha particle Δm= 0.0304 u  =equivalent to   28.3 MeV.

If you look up the BE curve Iron-56 is abundant and with a binding energy per nucleon of 8.8 MeV, it is the third most tightly bound of the nuclides.
Iron-56 average binding energy per nucleon is exceeded only by 58Fe and 62Ni, the later  being the most tightly bound of the nuclides.
Therefore Alpha particle is not the most tightly bound nuclei.
The BE curve gives an impression of "Anomolous" BE of alpha particle as the curve rises from deuteron , a nuclei which is a purely "loosely" bound state having about 1.1 MeV binding per nucleon.
One can visualize Alpha -4 as composed of two deuterons and Carbon-12 as a cluster of  three Alpha particles.
And the BE per nucleon rises substantially .
As if the liquid drop had floating blobs of bound groups/clusters and these  may  get out by tunneling the nuclear potential wall  (Alpha decay).
From applet-magic.com:

The binding energy of tritium, H 3, is 8.481 Mev, but that of He 3, which also contains three nucleons but two protons instead of two neutrons, is 7.718 Mev. 
Thus the electrostatic repulsion of the two protons reduces the binding energy of He 3 by 0.763 Mev or 9 percent.
There would be a similar reduction for the alpha particle. And such a correction would likely bring the calculated binding energy for the alpha particle fairly close to the actual value.

From liquisearch.com:

High-energy electron-scattering experiments show He-4 charge to decrease exponentially from a maximum at a central point, exactly as does the charge density of helium's own electron cloud. 
This symmetry reflects [...] that the pair of neutrons and the pair of protons in helium's nucleus obey the same features as do helium's pair of electrons (although they  are subject to a different nuclear potential), [...] so that all these fermions fully occupy 1s1s orbitals in pairs, zero  orbital angular momentum, and each canceling the other's intrinsic spin.
Adding another particle would require angular momentum, and would release substantially less energy (in fact, no nucleus with five nucleons is stable). 
  This arrangement is thus energetically extremely stable for all these 
  particles, and
   this stability accounts for many crucial facts regarding helium in nature.
For example, the stability and low energy of the electron cloud of helium > causes helium's chemical inertness , and also the lack of interaction of helium atoms with each other (producing the lowest melting and boiling points of all the elements).
Some stable He-3 is produced in fusion reactions from hydrogen, but it is a very small fraction, compared with the highly energetically favorable production of Helium-4. [...] It is also partly responsible for the fact that the alpha particle is by far the most common type of baryonic particle to be ejected from an atomic nucleus - in other words, alpha decay is far more common than is cluster decay.

For more details see also https://en.wikipedia.org/wiki/Helium-4
