 So in the last video we saw that the unit cell of any crystal can only have one of these seven different shapes. Now on top of that these unit cells could either be simple or body centered or face centered or ends centered. However, we also figured that not all the systems can have all the four types. Some of these unit cells were in fact identical to each other. So we were ultimately left with these 14 different types. Now studying about all these different unit cells, the kind of crystals that they can make and their properties is in itself a fascinating branch of science called crystallography. Now because it won't be possible for us to go into the details of all these different unit cells, so in this series of videos we are going to focus only on the cubic system. Now the main motivation behind choosing cubic and not the others is because it turns out that most metals have unit cells in the form of cubes, especially the face centered cubic. So why are most metals FCC? To answer this question we have to remember that metals are atomic solids. The constituent particles of metals are atoms, copper rods are simply made of copper atoms and gold bars of gold atoms. Unlike say in the case of molecular solids like ice, which are made of H2O molecules. Now molecules have definite three dimensional shapes and molecules of different compounds can have different shapes. Atoms of most metals can simply be thought of as spheres, right? So while crystals of ice are made of complex 3D molecules, crystals of metals can simply be thought of as a bunch of spheres that are packed together. Now whenever we try to pack spheres in a given volume, we will always be left behind with these gaps, right? However, so we try we will always have gaps that are going to be left. But depending on how we pack these spheres, we might either be left with bigger gaps or we will be left with smaller gaps. So this leads to what we call different packing efficiencies in different arrangements. And it turns out that arranging spheres in the form of FCC actually leaves one of the smallest gaps. So FCC has one of the highest packing efficiency amongst all unit cells. So arranging spheres in the form of a cube with a sphere at every corner of the cube as well as at each of the face centers. Such an arrangement turns out to be the most efficient way to pack spheres in a given volume over any other arrangement like say the BCC, which has a sphere at the center besides the ones at the corners or the simple cubic which can simply be thought of as spheres placed at the corners of a cube. Now to really see that FCC indeed has the highest packing efficiency, we should be able to calculate the packing efficiency of these different unit cells and compare them to each other, right? Now to calculate this packing efficiency, we need to first think about something called the effective number of atoms, effective of the unit cell. So in this video, we will try and understand what effective number of atoms means and calculate this A effective for different unit cells. And in the next video, we will see how we can use this A effective to calculate the packing efficiency. So what exactly is A effective? To understand, let us take the help of a simple cubic unit cell. Now if I replace these lattice points with the motif, which in case of metals is nothing but a single atom, then I'll get a unit crystal. Do remember that lattice points represents the center of an atom via convention. Now crystals do not contain a single unit cell but are instead made up of many unit cells. So now if you look closely at the atoms at the corners, we will realize that these atoms do not belong exclusively to a single unit cell but are instead shared between different unit cells. So these spheres do not lie a hundred percent inside the unit cell but only parts of these spheres does. A effective tries to capture this information and is defined as the total number of atoms or spheres that can be thought of as being a hundred percent inside the unit cell. Now that we have defined A effective, let us try our hand at calculating the A effective of this unit cell. Let us start by thinking about the sphere that is present at this corner. Now this sphere with its center at this corner penetrates not only into these two unit cells but also into these ones at the bottom, right? So this sphere having its center at this particular corner penetrates into four unit cells at the bottom. And similarly it also penetrates into these four unit cells at the top. So this sphere is shared between four unit cells at the top and four at the bottom. So ultimately only one eighth of this sphere lies inside the unit cell. Now because a crystal is made of infinite unit cells in all directions, so each of these corner atoms are also shared with eight other unit cells. So each of these also contribute one eighth each. So now if you think about it one eighth of each sphere at the corner lies inside the unit cell and there are eight such spheres because there are eight corners. So effectively we can say that there is one whole sphere that can be thought of as being a hundred percent inside the unit cell. So atoms at the corners are one eighth inside. What about atoms at some other position? Say at the face center. So what fraction of a face centered atom do you think lies inside the unit cell? Pause the video and think about this for a moment. Well a face centered atom is shared only between two unit cells, right? And because the lattice point always represents the center of a sphere, so this sphere is equally shared between these two unit cells. So the fraction of this atom that lies inside the unit cell is half, right? So corner atoms are one eighth inside while face centers are fifty percent inside. What about atoms at some other position? Maybe at this edge center. What fraction of an atom that lies at the center of any edge is within the unit cell? You can pause the video again and try and come up with an answer. Now any edge of the cube will always be shared with four other unit cells. So the atom at the eighth center will also be equally shared between these four unit cells and so an eighth centered atom will only be twenty five percent inside or one fourth inside. Let us take one final scenario in which we have an atom exactly at the center of the unit cell. So a body centered atom will lie a hundred percent inside the unit cell and we can say that the fraction that is inside is equal to one, right? So now that we know the contributions to the unit cell of atoms at different positions we are now ready to calculate A effective of any unit cell. Let us say we have a cubic unit cell with atoms only at these positions. There are four atoms at the corners and one at a face center. So what do you think will be the A effective of this unit cell? Well, we know that only one eighth of the atom at the corners are within the unit cell, but there are four such corners. So one by eight multiplied by four. And this atom at the face center is fifty percent inside, so plus half. So if we do the math, the A effective will come out to be equal to one. So this unit cell can be thought of as effectively having one whole sphere inside it. Let us take another example. What do you think will be the A effective of this particular unit cell? There are two atoms at these corners, one at the body center and two of these at the edge centers. You can pause the video and try to come up with your answer. Well this time we have two atoms at the corners and we know that corners contribute one eighth to the A effective. There is an atom at the body center and a body center is a hundred percent inside. So its contribution will be one. While these edge centers contribute one fourth each and there are two such edge centers. And solving this we will have an A effective of seven by four, which is 1.75 atoms per unit cell. So in this way we can calculate the A effective of any unit cell be it FCC, BCC or anything else. For example if we want to calculate the A effective of FCC, we just need to know the positions of the different atoms. So in FCC there are eight atoms at the corners and each corner contributes one by eight each. And additionally there are these face centered atoms at each of the six faces. So there are six faces and the face centered atoms contribute half each. So if we do a math, FCC can be thought of as having effectively four atoms inside the unit cell. You can now pause the video and think about the A effective of the BCC as well as the simple cubic. So BCC has eight atoms at the corners, so eight into one by eight. And there is also this extra body centered atom completely inside it. So the A effective of BCC comes out to be equal to two. Now the simple cubic only has atoms at the corners, so the A effective of a simple cubic is going to be equal to one, right? So in this video we learned how to calculate the A effective of different unit cells and we don't need to memorize these values for FCC or BCC, we can just calculate them. In the next video we will see how we can use this A effective to calculate the packing efficiencies.