Do electrons 'diffuse' through materials Consider a material $A$ having $5N_A$ electrons per $m^3$ and another material $B$ with $10N_A$ electrons per $m^3$. Now a meter cube of material B has a rather larger concentration of electrons when compared to A.
Now is it possible for electrons of material B diffuse into A the way gases do (i.e. from high Conc. to Low Conc.). If it is so does it this diffusion follow Graham's law. And how would this diffusion change with change in temperature of either of the blocks 
$N_A$ represents Avogadro's Number $(\approx 6.022 \cdot 10^{23})$
 A: What is $N_A$ , you should define your terms and give links if possible, a screen capture of where it is used may be helpful.
In general, electrons are quantum mechanical entities, basic building block of matter and are in the table of particles in the standard model of particle physics.
Electrons in materials are bound to nuclei, nucleus+electron make a neutral atom. There are no free running around electrons in general at normal temperatures and pressures, although there are various materials  called conductors, where some of the electrons are collectively bound to the lattice and not individual nuclei.
So no, electrons are not like a gas and there is no diffusion in the sense you mean. Both material A and material B will be neutral. There are special conditions where electrons may exchange at surfaces between materials, might be bound in a collective lattice, but not a diffusion. Some materials show a voltage drop under pressure, piezoelectric materials, but again that is not diffusion.
In  a gas at very high temperatures there exists the plasma phase, where some nuclei separate into a gas of ions and electrons, but this is different than the content of your questions.
A: Diffusion of electrons
It can’t happen in metals
For electrons to diffuse, they need to move freely, so this would happen only in metals or semiconductors where a handful of electrons per atom can roam throughout the material. However, if these electrons are completely free to move, the diffusion will actually generate an electric field which will oppose it: as the electrons start to diffuse from B to A, then A become negatively charged and B positively charged, since the materials were neutral to start with. Therefore the electrons are repelled from A and attracted by B. Thus the electrons will tend to come back to B. The only case where this will not happen is in semiconductors because the electrons which have diffused from B to A get trapped in A because of a mechanism I will explain below.
But first, why does it work out differently in atomic/molecular gases? The key reason is that atoms/molecules are neutral: at best they weakly interact through so-called van der Walls forces which decreases as $1/r^7$ (Lennard-Jones potential, for the geeks) — $r$ being the distance between the atoms/molecules. On the contrary, the plain interaction between charges that would occur in your scenario has forces scaling as $1/r^2$. I think you can see how much weaker are the forces in the case of a gas.
It can happen in semiconductors
Issues to address
We need to address two:


*

*How to trap in A the electrons which have diffused from B, so that the emerging electric field does not send them back to B

*How to get more freely moving electrons in B than in A in the first place


It turns out that those two issues can be addressed in a semiconductor. I am now going to explain how and this will be very lengthy and rather complex but this is the nature of it. If, for now, you are happy knowing it can be done, you can save the rest of this answer for later!  
How electron can effectively diffuse in semiconductors
I have to cover a lot of ground in atomic and solid physics, which means quantum physics at such small scales. I will cut corners to keep it simple but I will keep lies small. It will feels much more like a chemistry course than a physics one, note.
You have probably heard that the energy of each electron in an atom are quantized: this energy can only take discrete values. More importantly, an electron with that energy will have a high probability to be found in a characteristially shaped region of space, a region which is localised near the nucleus of the atom. Each such shape is called an orbital and it is denoted by a specific symbol. The simplest example is the lowest energy level. There is only one orbital, denoted $1s$, whose shape is a sphere. But let’s jump to the most relevant orbitals for our problem: those corresponding to the highest energy level. For a Silicon atom, those orbitals, denoted $sp_3$, are shaped like 4 lobes.

On this image, A denotes the position of the Silicon atom. The big lobes are the regions with the most probability to find an electron. It turns out that each of these 4 orbitals can be occupied by 2 electrons. But each atom of Silicon contribute only 1 to each of them, simply because it has only 14 electrons in total and the other 10 are accomodated by lower energy levels. However, the $sp_3$ orbitals of two neighbour Silicon atoms overlap and as a result the combination they form is fully occupied, by 1 electron from one atom and 1 electron from the other one. This overlapping and sharing is the best model we have for a chemical bond. Since there are 4 orbitals, each atom of Silicon is bonded to 4 neighbours, which are in turn bonded to 3 more Silicon atoms, etc. Because the geometry of the bonding is the same for all atoms, they are regularly arranged and solid Silicon is therefore a crystal. So we say that Silicon has a valence of 4. And by extension we call the electrons in those combined $sp_3$ orbitals valence electrons and the corresponding energy level, valence level. The important point to remember for the following is that those valence electrons are forced to stay between the pair of Silicon atoms they came from.
However in Silicon crystal, about $1\ \mathrm{eV}$ above the valence level (this is the energy of a photon of visible light), there is a band of energy levels such that the electrons occupying any of them are not localised anymore: it is called the conduction band and electron on there can be anywhere in the crystal. In a way, this band is shared by all the atoms.
In pure Silicon, if some energy is given to a valence electron by either a photon or thermal excitation, then it can jump to the conduction band. This is why Silicon is used for solar panel and also why the electrical conductivity of a semiconductor increases with temperature but I disgress. So if your material B was a semiconductor with more electrons in its conduction band than another semiconductor A, then these conduction electrons would diffuse from B to A. Then if the electrons moving over to A were to fall back onto the valence level (i.e. into one of the combined $sp_3$ orbital) and stay there, they would effectively be trapped and the electric field that has build up as I explained in the beginning of this answer would not be able to move them back to B (remember: the electrons on those orbitals are forced to stay around the nuclei of the two bonded atoms; naively the electrostatic interaction of the nuclei beats the electric field build up by the diffusion). As a result, the diffusion would be effective.
But how to achieve that? The two issues I raise earlier can now be rephrased more precisely.


*

*In pure Silicon, the valence level is completely full, except for those electrons which jumped to the conduction band, as explained above. Said differently, at the temperature of absolute zero, and in total darkness, the valence level of all atoms is full. And at room temperature, the valence level of most of the atoms is full. This is clearly a problem for the trapping scenario of my previous paragraph, as there is very little room for the electrons which have diffused to A to fall onto the valence level. 

*If A and B are both made of Silicon, there won’t be more conduction electron in B than in A. We could rely on temperature: if B is hotter than A, then more electrons will jump on the conduction band in B than in A. Or shine light on B only. But there is a more clever solution.


This solution is called doping and it solves issue (1) too. Doping consist in replacing a fraction of the Silicon atoms by another chemical element, between 1 every thousand to 1 every billion. We will dope A and B with different foreign elements.
In A, the foreign chemical element is chosen so that it can contribute only 3 electrons to its valence level, all other electrons being on lower energy levels. So it can contribute 1 electron to each of 3 $sp_3$ orbitals but there is one orbital to which it cannot contribute any. The neighbour Silicon atoms will still contribute 1 electron to each $sp_3$ orbital, resulting in 3 full orbitals (and therefore 3 bonds) and one orbital with a free slot. So now the electrons diffusing from B to A can fall onto that latter orbital, thus solving issue (1). Such a material is called p-type Silicon, “p” as positive because a missing electron is like a positive charge, called a hole. An practial example of such a dopant is Boron.
In B, the foreign chemical element is chosen so that it contributes 5 electrons to its valence level. The classic example is Phosphorus. So Phosphorus has 4 $sp_3$ orbitals and it contributes one electron to each of them whereas each Silicon neighbour shares one electron to each of them, resulting in 4 full combined orbitals (i.e. 4 bonds). But Phosphorus has still one more electron to account for. This electron will go on a spherical orbital denoted by the symbol $3s$. The beauty of this is that this electron can very easily jump onto the conduction band, much more easily than the Silicon electrons on the valence level. As a result, this so-called n-type Silicon will have more electrons on the conduction band than pure Silicon and also more than the p-type Silicon of the last paragraph. The letter n is as “negative” because more negative electrons.
And we have solved our two issues.
That’s all folks!
