Introduction to Materials Physics P Solutions 1, Fall 2017
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1 Introduction to Materials Physics P Solutions 1, Fall To find the primitive cell of a crystal structure, we should start by finding the smallest repetitive unit (=the basis). When each basis is replaced by a lattice point, the corresponding lattice is formed. If we have chosen the correct basis, the locations of corresponding lattice points are expressed by a linear combination of lattice vectors a1, a2, and a. If some extra points are produced or all lattice points are not produced by the linear combination, then the chosen basis is wrong and the second smallest repetitive unit should be tried next in order to get the proper lattice. When we have the correct lattice, the shortest possible lattice vectors form the edges of the primitive cell. The unit cell is formed by more symmetrical vectors (perpendicular to each other) if possible, and is thus in general larger than the primitive cell. (a) ccording to the Fig. 1.1, the smallest repetitive unit (=the basis) is the circled group of atoms, that consists of one -atom and one -atom. s the respective lattice points are formed (Fig. 1.1 right), all of them (and only them) can be expressed by the linear combination of the vectors a1 and a2. ecause a1 and a2 are the shortest vectors of this kind, they are the primitive lattice vectors of this lattice. b=a 2 a=a 1 Figure 1.1 Crystal structure and the smallest repetitive unit (=the basis) (left). The corresponding lattice points are formed by setting a lattice point to replace each repetitive unit (right). The primitive cell edges are formed now by these vectors (see Figure 1.2). We can see that the primitive cell contains atoms (corner atoms are only partially inside the cell). This can be also deduced, because it is known that the primitive cell always includes only the atom group that positioned around each and every lattice point (=the basis). In this case, the basis involves = 2 atoms, which is the repetitive unit. Since the primitive lattice vectors already are perpendicular to each other (symmetrical), the same vectors can also be chosen as crystallographical axes a and b (the dimensions of the unit cell). In this case, the unit cell is the same as the primitive cell and they have the same atomic content. 1
2 Koska hilan alkeisvektorit ovat kohtisuorassa, niin ne muodostavat myös yksikkökopin. Koska hilapiste voidaan sijoittaa mihin tahansa toistuvan rakenteen kohtaan, niin alkeiskoppeja voi olla useita erilaisia. alkeiskoppi = yksikkökoppi a 2 =b Kaikki ovat mahdollisia alkeiskoppeja ja niiden atomisisältö on sama!! a 1 =a Figure 1.2 The primitive cell formed by vectors a1 and a2 as well as the unit cell formed by axes a and b (left). Depending on the choice of lattice point location, there can be many choices for a primitive cell, but they are all equivalent and fill the space in regular manner (right). (b) b a 2 a 1 lkeisvektorit a Kideakselit Figure 1. Crystal structure and the smallest repetitive unit (left). The corresponding lattice is formed by replacing each repetitive unit by lattice point (right). Respective to the previous case, each lattice point (and only them) in Fig. 1. can similarly be expressed by vectors a1 and a2. s these vectors are the shortest vectors of this kind, they are the primitive lattice vectors of this lattice. These vectors (see Figure 1.4) form the edges of the primitive cell. The unit cell vectors are chosen to be perpendicular to each other, which in this case are the vectors a and b in Figure 1.. These vectors form the edges of a unit cell (see Figure 1.4) and they are called crystallographical axes. The lengths of axes are called lattice parameters (or lattice constants). 2
3 b a 2 a 1 a Figure 1.4 The primitive cell formed by vectors a1 and a2 as well as the unit cell formed by axes a and b (left). Depending on the choice of lattice point location, there can be many choices for a primitive cell, but they are all equivalent and fill the space in regular manner (right). We can see from the figure that the primitive cell contains one -atom and 2 -atoms (corner atoms are only partially inside the cell). This can be also deduced, because it is known that the primitive cell always includes only the atom group that positioned around each and every lattice point (=the basis). In this case, the basis involves = atoms, which is the repetitive unit. The unit cell has now 2 -atoms and 4 -atoms. This can also be deduced from the lattice, because the unit cell volume in Figure 1.4 involves two lattice points (one in the middle and ¼ from each corner). ecause one lattice point involves the basis (1+2), two lattice point includes twice the amount of atoms (i.e. 2+4). This means that the unit cell volume is twice the volume of the primitive cell in this crystals structure. Figure 1.4 also shows that the primitive cell can be drawn to begin from different locations depending on the choice of lattice point location. Usually it is easiest to put the lattice point to a position occupied by atom. Whatever the choice of lattice point location, the primitive cell volume and atom content should be similar.
4 Introduction to Materials Physics P Solutions 1, Fall The periodically repeating crystal structures are divided into seven -dimensional crystal systems, which include in total of 14 different lattice types, i.e. ravais lattices. Thus, all periodically repeated crystal structures has one of the ravais lattices as its lattice type. The unit cell of the crystalline structure is formed then the required atoms are placed to the lattice points of the respective lattice type. (a) Copper crystallizes to a face centered cubic (fcc) structure, and the ravais lattice is face centered cubic lattice. This indicates that every lattice point includes one atom as a basis. So the unit cell includes 1/8-part of each corner point and ½-part of each point sitting on cube faces. Thus. 1 1 Fcc-lattice contains n lattice points. 2 8 Since the lattice type contains 4 lattice points and each of them includes one copper atom, the unit cell contains 4 copper atoms. However, remember that the primitive cell always includes only on lattice point, meaning one copper atom (= the basis) in this case. (b) Silicon has a diamond structure, which forms from two overlapping fcc-structure that are separated along the cubic diagonal by the distance of ¼ from its length. The ravais lattice of diamond structure is still face centered cubic lattice, but now one lattice point involves 2 atoms (=basis). Following the previous example and taking the basis into account, we get 1 1 n ( 6 8 ) 4 lattice points in unit cell 2 8 2n = 8 silicon atoms a Now the primitive unit cell still has a one lattice point, which now involves 2 atoms as a basis. So the primitive unit cell involves two atoms. a/4 a/4 a/4 (c) Cesium chloride crystallizes into CsCl-structure, which has a cubic unit cell with Cs + -ions in the corners and bigger Cl - in the middle. The corresponding ravais lattice is primitive cubic and every lattice point includes Cs + Cl - ion-pair. The atoms are separated along the cubic diagonal by the distance of ½ from its length. Thus only one of the Cl-ions involved in the each lattice point remains inside the unit cell. (Note! Respectively, we could place the lattice point on top of the Cl-ion, which would give a unit cell with Cl-ions on the corners and Csion in the middle. So, the unit cell can be formed in two different manner. ) 4
5 Kiderakenne Hila Cl - Cs + The unit cell involves now one lattice point so it has to be equivalent to the primitive cell. oth cells have two ions (=the basis), which is one Cs + and one Cl -. 5
6 Introduction to Materials Physics P Solutions 1, Fall The density of copper can be deduced from the density of its unit cell, since a material consists of unit cells. ased on the exercise 2a, the unit cell includes 4 atoms, so the mass of the unit cell is same as the mass of 4 copper atoms. The density of the unit cell is m munit cell 4 M. V V N a unit cell The distance d between the nearest neighbouring atoms (corner atom and face center atom) in fcc-structure is ½ of the face diagonal a 2 d 2 a d 2 2 y substituting the lattice constant with the above-solved expression, we get the density 4M N 4M 2 a 2 2N d N 2 6,54 g/mol 2 6, (2,56 10 mol M d 8 cm) 8,894 g/cm 8,89 g/cm 6
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