 Next write down hexagonal primitive unit cell, hexagonal is not that much important the unit cell is a hexagon in this okay two three points write down the atoms in the corner the atoms at atoms in the corner is distributed among six unit cell distributed among six unit cell and in this three atoms are present inside the unit cell inside the unit cell okay. So, all these are well corners here. What? So, how do you just say there is. Drawing it. At the corners we have atoms present at these corners since the unit cell is a hexagon that is why it is distributed among six different unit cells okay. So, three atoms present at the within the unit cell like this. We will discuss this again in the arrangement of atoms. You understand how these atoms are arranged okay. Just to draw the structure now. Within the unit cell. Contribution is completely one for this within the unit cell okay. We will see the arrangement of atoms then you can understand this structure. Is it arranged in the same way or we just arrange it in the same way? No, it is like a triangle you can see in the plane triangle plane. It is like this it is one atom here and here and here. So, those other two are on the edges. This one? No sir. The triangle. It is like edge like this. But they are all inside. You are not touching the face right. They are not touching any of the faces. Are you just touching inside? No, no, no. Randomly we cannot say. It is an arrangement. We will see that at CT arrangement we call it as. We will see that you must have heard ABC, ABC type, AB, AB type. So, when we discuss that you will understand how these three atoms are arranged. This just you draw the structure now. We will see the arrangement of atoms and then you can correlate the structure with this. Okay. So, just give me some time for that. Now, the contribution of these atoms are like you see how many hexagon you can draw with it. One hexagon is like this you can draw. Right? And one hexagon is this. So, in this plane you see this atom is distributed around 1, 2 and 3 hexagon. Okay. Similarly, you can place 3 above this also. 1, 2 and 3. The distribution is 1, 6, 9, 7. Right? 1, 6 unit 7. So, what is the rank of this? Number of the 50 atoms we need to tell you. Q5. 1 by 6 all 7 and 1 by 6 all 7. 1 by 1 plus 3 inside. No. So, I said those three have contribution of 1, right? So, it should be 3. No. Each of them have contribution of 1. Inside those three, the triangle one, each of them have contribution of 1. Yes. So, that is 3. Yes, correct. And it is 5. Yes. So, you can draw 6 of them on each side. Oh, I missed one thing. One face center also. Then it should be 6. 6. Okay. Face center also. We will understand this structure. Okay. So, 12 atoms at the corner and the contribution is 1 by 6. Okay. 2 at the face center. So, 2 into 1 by 2 and 3 in the unit cell 3 into 1. So, it is 6. Okay. Coordination number write down 12. Coordination number is 12. This atom is in touch with this atom 6 and this 6 atoms. Okay. Coordination number again we will discuss when we see the arrangement. Just write down here coordination number 12. We will come to this structure again. How do we get hexagonal packing? Okay. We will discuss that. Triangular spathir is one unit. Wait. No. This is not tri-flap. These 3 atoms present within the unit cell. You just said it is 12. You said 12. So, if you take each atom inside then it should be 12 plus 2. No, no, no. It is not. It is give me some 5 minutes to give me. You understand this. Until 5 minutes you will give me. I will draw this structure. The arrangement will see two types of arrangements there. You will understand this or the coordination number is 6. From this it is difficult to understand the coordination. Just you draw this. We will see the packing fraction of this and then we will move on to the arrangement. Okay. So, rank is 6. Coordination number you write down 12. Packing fraction you find out. Packing fraction. Here the relation is A is equals to 2 R. A is equals to 2 R we have here. These two coordinate atoms are in touch. A is equals to 2 R. Packing fraction you find out. That is all arrangement. Yes. I will discuss that. Arrangement we have not done. We will do that in arrangement. Just give me 5 minutes. Simple. Simple. Simple what? Simple hexagonal. Simple hexagonal. We will see how this hexagon forms when arrangement of atoms 6. We will find the packet ring of the simple hexagonal. This one. This one. Or this one. See here the rank is 6. So packing fraction is what? 6 into 4 by 3 pi r cube divided by the volume of this hexagon. Right? Now volume of this hexagon is what? If it is cube it is A cube. For volume of this what we can find out the cross sectional area. This area of this hexagon into the height. Right? This height suppose I am assuming h. Right? Regular hexagon. Regular hexagon. All this h length is t. There is a certain thing. A certain thing. Yes. See one relation we have here I will just give you one relation. The length of the body diagonal fcc divided by root 3. This relation we can also derive this by geometry but that derivation is not required. It is equals to the height of the hexagon divided by 2. This relation you should memorize. Sometimes they will give you this value A of fcc. You can find out length of the body diagonal fcc and then you can find out h. Sometimes they ask you to find out the height of the hexagon. Okay? So length of the body diagonal fcc is what? Root 3A. Here A is equals to what we have? Root 2R for hexagon. So we can write root 3A fcc by root 3 is equals to h by root 3. That is what you are going to get. That is right. Okay? So what is the height h by root 3? Once you know the h length fcc you can find out the height of the hexagon. Okay? Now what is the relation of, because you see, what is the area of this hexagon? We have 6 what? Triangle, equivalent triangle. Okay? 6 equivalent triangle. So its area is what? 6 into root 3 by 4A square. Right? This area into height is the volume. A into h. Right? So A is in terms of A. If as you know in terms of A you can substitute here. Then you get volume. Right? Volume of this is what? 6 into 4 by 3 pi r cube. Okay? So volume of the unit cell is what? 6 root 3 by 4A square into h is what? 2 by root 3A of fcc. See this A of fcc and this A is different. All of you, here. This A of fcc we cannot substitute. A is equals to 2r here. Because it is the h length of fcc unit cell. So what is the relation of A and r in fcc unit cell? 3A is equals to 4r. Right? No. 2A is equals to 4r. Right? So this fcc is equals to what we can write? 4r by root 2. We cannot write it as 2r. But this A is what? This A is 2r. Did you understand this? This is the border diagonal of fcc. So this becomes the h length of fcc. And h length of fcc is what? 4r by root 2. So this A and this A is different. It is not the same. So now you can find out volume in terms of r. And then the packing efficiency is what? Volume of the atom, of the volume occupied by the atoms divided by the volume of the unit cell. So packing efficiency is the same. Take my right down here. The packing fraction is the same. So packing efficiency, next take my right down here. The packing fraction phi is equals to, there are 6 atoms, rank as 6 into 4 by 3 pi r cube divided by the volume of unit cell is this. Which is 6 root 3 root 3 gets cancelled. 3A square is 2r square. And A of fcc is 4r by root 2. When you solve this, we will again get 0.74 by root 2. Same as fcc. Did you understand this? Here the calculation is a bit typical because you do not know the formula of this hexagon directly. Volume of this hexagon. But how do you eliminate h? Actually there are 2 A hcc by root 3 from this variation. Done. So packing efficiency of both hexagon, hcp and what we say fcc is same that is 0.74. Next right down, density. Density of cubic unit cell. Ecc is nothing. Just we have, you would not get any question on Ecc. We have corners and any opposite face center. So we will have Ecc and hcp. Not important. Just you should know the arrangement of atoms. Coronets will have atoms. And any of the 2 opposite face centers. For example, any center is this. You would not get any question on this. Next right down, density. Can you find out the formula of density for unit cell or crystal lattice? Density is what, mass by volume. And mass means mass of any atoms. How many atoms? Any atoms you have got to know to write. So how many atoms here we have? That is suppose rank is z. So z is the atom we have become. For other different unit cells. The number of effective atom is different. Correct? Yes. Like for pcc it is 2, xcc 4. For hcp it is 6. So we need to find out the mass of the given number of atoms for a unit cell. So density, it is very basic formula we have. Density d is equals to mass by volume. Atomic mass, suppose of an atom is m. Means it is a mass of any atoms. Mass of one atom is what? If the rank is z, what is the mass of this atom? M by any into z. This is what, this is the rank of the unit cell. So density is what? D is equals to mass by volume, mass is this? M z by any into volume of the unit cell. If it is a cubic unit cell, M z by any into a cube where a is the edge length of the unit cell. This is the formula.