Reason for peaks in graph of binding energy per nucleon A similar question was asked before, but it asked for a different thing. My question here is: What is the reason for spikes in this graph? The graph initially has spikes and then shows a constant decrease. Is it related to something called magic numbers as it is seen in multiple of 4?

 A: There are two effects that lead to the presence of the small jagged peaks and valleys in the binding energy per nucleon.  (The main shape of the curve is given by the semi-empirical mass formula, derived from the liquid drop model of the nucleus.  The model has a positive binding energy proportional to the number of adjacent nucleon-nucleon pairs, a Coulomb repulsion term related to the square of of the number of protons, and a term related to the proton-neutron imbalance.) On top of this, producing the zigzags, there are pairing effects and magic number effects.
The pairing effects come from the fact that the bound nucleons have slightly lower energies when they are correlated in in proton-proton or neutron-neutron pairs.  That tends to make the binding energy per nucleon for an odd-odd nucleus like $^{10}$B less than the odd-even $^{11}$B.  Or $^{17}$O is less tightly bound than $^{16}$O and $^{18}$O on either side of it, since $^{16}$O and $^{18}$O are both even-even.
The other effect is due to the presence of magic numbers, which are related to filled nuclear orbitals.  Just like atomic electrons are most stable when they form a filled outer shell, nuclei are most stable when the protons and/or neutrons fill certain nuclear shells.  For example, $^{4}$He is much more tightly bound than $^{3}$H or $^{3}$He, since the $^{4}$He has two protons and two neutrons, with each pair filling a $1$S shell.  Another (double) magic nucleus is $^{16}$O, with the eight protons and neutrons each filling the $1$S and $1$P shells.  (The notation for shells differs a bit here from that used with electrons.  Nuclear shells are denoted by $n$S, $n$P, etc., where $n$ starts separately at one for each value of the angular momentum.)  The $^{18}$O nucleus has to have its two extra neutrons shunted into the higher-energy $2$S shell, lowering the binding energy per nucleon.  Another magic number occurs at 10, which is why $^{20}$Ca is especially stable; the ten protons and ten neutrons fill up the $1$S, $1$P, and $2$S shells.  (The pattern of magic numbers gets a bit more complicated than this, because of the strong spin-orbit coupling in the nucleus, but this is a reasonable picture of the general behavior.)
A: It has to be with the pairing term. Nature likes even-even pairs of nucleons. I mean, an even number of protons and an even number of protons. The reason is ultimately related to spin couplings.
So, odd-even pairs are more or less over the curve. Even-even isotopes, like $C^6$, or $O^18$, are especially stable. On the other hand, odd-odd pairs are especially unstable, but there are only 4 stable nuclei which are odd-odd.

Edit:
So, odd-even pairs are more or less over the curve could you elaborate this point?
Okay, I'll ellaborate.
Let's take the liquid drop model, which is empirical, but explains quite good what is happening. It has 5 parameters, though.
Let $B=B(Z,A)$ be the binding energy of the nucleus. The more energy, the more stable. Because that's the energy you have to overcome if you want to separate the nucleus.
The liquid drop model stablishes
$$B(Z,A)=a\cdot A  -b\ A^{2/3} - s \frac{(A-2Z)^2}{A} \ -d \frac{Z^2}{A^{1/3}} - \delta\frac{Z^2}{A^{2/3}} $$
That's the function that fits the curve you are showing, with
$a=15,835 MeV; \quad  b=18,33 MeV;  \quad   s=23,20MeV;  \quad  d=0,714 MeV$
The first term (a) is due to volume. It is responsible that the curve saturates at a certain value. $B/a=cosnt$ for large $A$.
The second term is due to the surface. Sicne small $A$ have much more surface, they are more unbounded. HEnce the stong decay at the beginning.
The 3rd term (s) is due to symmetry. Note that the fraction contains $N-Z$. If there is a big unbalance of nucleons, the nucleus will be unstable.
And the next one (d) is the Coulomb's repulsion. Check that $R\propto A^{1/3}$.
And what about $\delta$? Well,
$$\delta=\begin{cases} +11,2 MeV & if\ even-even \\ 0 & if\ odd-even \\ -11,2 MeV & if\ odd-odd \end{cases}$$
So, the curve that is usually plotted is the function without this delta-term. That's why I say that that "odd-even nuclei are on the curve". However, even-even will be above and odd-odd will be below.
In your curve, they have joined thereal nuclei, which is good.
But since this function $B(Z,A)$ is defined by parts, it is easier to represent it without the delta term. The curve is like this much more smooth. IT's like the "mean curve". Then, if we add the delta, we find more peaks. That's what I meant.
