 So, by now you have seen that to calculate the pecking fraction of any unit cell, you need to know its A effective, the radius of the atoms that make up the unit cell and its edge length A. Now, if you slightly rearrange this relation and write r cube by A cube as r by A cube, then you'll realize that even if you don't know the exact value of r and A, but if you simply know the ratio of r by A, then we'll still be able to calculate its pecking fraction. Sounds interesting, right? However, the really interesting bit, and the real reason why I'm writing it as r by A, is because it turns out that this ratio is constant for each of these unit cells. For example, the simple cubic unit cell has a r by A ratio of 1 by 2. And what it really means is that it doesn't matter what size of atoms you take. As long as you arrange them in the form of simple cubic, the ratio of the radius of the atom to the edge length will always remain the same. Cool, right? The same goes for BCC as well as FCC. As long as you arrange atoms in the form of BCC, the ratio of the radius of the atom to its edge length will always be root 3 by 4. And similarly for FCC, it will always be root 2 by 4. So in this video, we will try and figure out how these ratios came about, why they turn out to be constant, and ultimately use them to calculate the pecking fraction of each of these unit cells. So let's dive right into the video. Let us for a moment consider metal atoms to be circles rather than spheres. Now in a metal crystal, these metal atoms are not free, but are instead bonded to each other by some metallic bonds, right? Now let us assume that these atoms are bonded in such a way so that they form a square unit cell. Now in such a scenario, the side of the square or the edge of the square will always be equal to 2 times the radius of these circles which make up the square, right? So the side A will always be equal to twice R. So even if you don't know the exact values of A or R, we'll always be able to say that for a square unit cell, the R by A ratio will always be equal to 1 by 2, right? It doesn't really matter if these circles were bigger or smaller. If they form a square unit cell, the atoms at the corners will touch each other. So A will always be equal to twice R and R by A will always be equal to 1 by 2. Now if we had a scenario in which the circles did not touch each other, then the side A will not be equal to twice R. So we won't be able to figure out the R by A ratio just by looking at this configuration. In this scenario, we need to know the radius as well as the side to find out R by A. However, having said that, we should remember that such a scenario is not possible when it comes to a metal crystal. These atoms, they cannot lie just hanging in space. They have to be bonded at least to one other atom. Let us now take a look at a different configuration of atoms. This is a squared centered unit cell, right? This can also result in a valid metal crystal as each of these atoms are bonded to at least one other atom. We don't have any atom that is hanging around in space. So can we figure out the R by A ratio just by looking at the configuration? Well, let us see. Well, out here the atoms at the corners, they do not touch each other. So we can't find any relationship between A and R from here. But if we look across this diagonal, across this line, we will see that all these atoms, they touch each other. So the length of this diagonal, length of diagonal can be written as the radius of this circle R plus the diameter of this one, which is going to be twice R, plus the radius of this circle R. So the length of the diagonal in terms of the radius can be written as 4R. Now, if you look at this triangle, let me name this as ABC triangle ABC. We will realize that this is a right triangle where AB is equal to BC is equal to side of the square A. So now if we use Pythagoras theorem, we can say that AC square will be equal to AB square plus BC square. So AC square will be equal to A square plus A square, which is 2A square. So this implies that AC will be equal to root 2 times of A, right? So the diagonal AC can always be written in terms of the edges A as root 2 of A. So therefore, if we equate these two relations, then root 2 of A will be equal to 4R. So R by A will always be equal to root 2 by 4, right? In other words, it doesn't matter whether these circles are smaller or bigger because the atoms will always touch each other across the diagonal. So the length of the diagonal will always be 4R. And because the diagonal of a square is always root 2 times the edge A, so R by A will always be equal to root 2 by 4. We are now finally ready to calculate the packing fraction of different unit cells. So for a simple cubic unit cell, the A effective comes out to be equal to 1. What about R by A? Pause the video, look at this figure and try to come up with an answer. Well, R by A is going to be equal to 1 by 2, right? Because the atoms along the edges, they are touching each other, so A is going to be clearly equal to 2R. So the packing fraction will be 1 into 4 by 3 pi into 1 by 2 cube, which if you do the math will come out to be 0.524. So in terms of percentages, the packing efficiency of this cube comes out to be equal to 52.4%. Now I'd like to add out here that a crystal is not made of a single unit cell, but instead it contains infinite unit cells. Now all these unit cells are identical to each other in all respects, so the packing efficiency of every single one of them will be 52.4%, right? So we can now go ahead and say that the overall crystal also has a packing efficiency of 52.4%. So arranging atoms in the form of a simple cubic occupies only around 50% of the available volume and leaves the remaining 50% as gaps. Not a very efficient system if you ask me. Let us see what happens for the FCC. What would be the R by A ratio for FCC? Pause this video again and think about the answer. Well in FCC the atoms at these corners, they do not touch each other, right? So we won't be able to figure out a relationship between A and R from out here. So what do we do now? Well if you look at these atoms, the one across this face will call this the face diagonal. All the atoms across this face diagonal touch each other, right? So we might be able to figure out the relationship between R and A from out here. Now this line is the diagonal of this square. So we have seen that the diagonal of a square will come out to be equal to root 2A and this will also be equal to 4R as these spheres are in contact with each other. So this is going to be equal to 4R. So the R by A ratio will come out to be equal to root 2 by 4. Now it doesn't matter if these atoms are big or small. As long as they form FCC the atoms along the face diagonal will always touch each other. So this distance will always be equal to 4R and because the diagonal of a square can be written in terms of A as root 2A so R by A will always be equal to root 2 by 4. So if we now plug in the values for the packing fraction. So R by A will be root 2 by 4 whole thing cube while A effective for FCC is going to be equal to 4. So 4 into 4 by 3 pi into root 2 by 4 whole thing cube. So now if we do this calculation we'll find out that the packing fraction is comes out to be equal to 0.74. So arranging atoms in the form of FCC leads to 74% packing efficiency. So almost three fourths of the available volume is occupied and only one fourth is left as caps. Let's come to our last case the body centered cubic. In BCC A effective is equal to 2 but what is the R by A ratio? How do you calculate the R by A ratio in BCC? Well the atoms along the corners do not touch each other so we won't be able to calculate R by A from here and if you look across this face diagonal you'll realize that this atom doesn't lie on this face but it's actually inside right. So this distance is definitely not twice R. So the face diagonal cannot be written as 4R. So how do we find out R by A? Well we need to think of some line which can be written in terms of R right? Pause this video and see if you can find out such a line. It can be an edge or a face diagonal or anything else. Now in a BCC if you look closely we'll realize that the atoms along this line they touch each other. So this line is called the body diagonal and atoms along this body diagonal they touch each other. So the length of this line can be written as R plus twice R and R. So this line can be written as 4R. Now to find out R by A we need to be able to write this R in terms of A right? So we need to be able to write this line in terms of A. So how can we do that? Well in this cube even this line is the edge length A right? And now if I join these two points I'll have a right angle triangle out here. Let me name this triangle A, B and C. Now in this right triangle ABC, AB is A, SC is 4R but I don't know the value of BC. If I can somehow figure out the value of BC I might be able to relate A and R using Pythagoras theorem right? So how do we figure out BC? Hmm well if you look at the base of this cube this is a square right? And this BC is the diagonal of this square. Now we have seen that for a square of sides A the diagonal can be written as root 2A right? So in triangle ABC, AB and BC are the sides of the triangle while SC is the hypotenuse. So using Pythagoras theorem we can write 4R whole thing square equal to A square plus root 2A whole thing square. So this implies that 4R whole thing square will be equal to A square plus 2A square. So this will be equal to 3A square right? So we can write 4R to be equal to root 3 times of A and this implies that R by A will come out to be equal to root 3 by 4. So in BCC the R by A ratio will come out to be equal to root 3 by 4 and it does not matter if these atoms are big or small as long as they're arranged in the form of BCC the atoms across the body diagonal will touch each other. So the body diagonal can always be written as 4R and this body diagonal can be written in terms of A as root 3A. So R by A will always come out to be equal to root 3 by 4. Now if I plug in these values then the pecking fraction of BCC will come out to be equal to 0.68. So in BCC the pecking efficiency is 68%. So to summarize the pecking fraction of FCC is 0.74 that of BCC is 0.68 and that of simple cubic is 0.52. So FCC has the highest pecking efficiency amongst all the cubic unit cells.