We'll continue our description of these hard sphere models and we'll turn our attention now to the body center cubic hard sphere model. When we look at the body center cubic lattice, what we have is, positions of each one of the corners and a position that lies wholly inside of the center. Giving us a non-primitive unit cell in which we have two lattice points per unit cell. And what we can do is, if we examine each one of those points. For example, if you pick the center position which lies in the center of the cube. What we see is, that we have a total of 8 equivalent positions that are adjacent to one another at each one of those corners. Those represent our nearest neighbors. So we have a total of eight nearest neighbors. Like we had a total of 12 nearest neighbors in the case of FCC. Now if I continue to distribute these in three dimensional space, then what I can do, is to come up with an alternative description. Again, using the vectors that take us from certain position at an origin in the back, going to positions in the center, looking at three different unit cells. And so, these can represent, just as we did in the lecture on the Brave lattice of the FCC, we can describe three alternative vectors that will give us a primitive rhombohedral lattice, just as we did with the FCC structure. An alternative way of looking at the BCC structure is that we can consider the BCC as two interpenetrating lattices, an A lattice and a B lattice. But in order for our points to be lattice points, A and B have to be equivalent. In other words, whatever the surrounding environment around A is, it has to be the same around B. So those are equivalent points, and we've described those as our lattice points. So, we look at our structure. We have our lattice. And then we have the atoms that are, rather in this case were putting in hard spheres, we're putting on top of each one of those lattice points. And what we're developing is a hard sphere model for the BCC structure just like we did with the FCC. It turns out that the location where the hard spheres touch is different in the BCC than it was in the FCC. So, if we look at the unit, we will have in each one of those corners, one-eighth of a sphere, and in the center we have a complete sphere. Giving us a total of two spheres in that unit. For the body centered cubic hard sphere to develop, what we are required to have is that the spheres must touch along the body diagonal of the cube. Which again, permits us to write a relationship between the dimensions of the cube. As well as using the dimensions of the radius of the sphere. So, the radius of the sphere then is related directly to the edge of the cube by the edge of the cube dimension a0 times the square root of 3. And that gives us then that total length. Or alternatively, we can come up with the relationship which leads to how we can relate a zero and r. When we do our hard sphere calculations in the body center cubic case, here is our picture here, and we have a total of 2 spheres that lie inside of this cube. And so, what we would write for the packing factor, again, is we have 2 spheres times the volume of each sphere. And that's going to be divided by the volume of the cube containing those 2 spheres. And we know that for the case of the touching, they touch along the body diagonals. And so we can relate a and r through the expression, a onto the square root of 3 = 4r. And therefore a = 4r over square root of 3. And if we substitute the r for a in the denominator, what we find is that we develop a packing factor of 0.68. And if we consider that with respect to the FCC structure, what we see is that the packing factor is lower in the body centered cubic case than it is in the face center cubic case. This is a very important point, and we're going to use it over and over again to describe a number of important ideas and concepts that we'll use as we go through this course. This then concludes our discussion of the packing associated with the FCC and the BCC hard sphere distributions. Thank you.