(a) Diamond is an electrical insulator and is one of the hardest substances known. Interpret these properties in terms of its bonding.

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1 MSE 200A Survey of Materials Science Fall, 2008 Problem Set No. 1 Problem 1: In the diamond modification of solid carbon each atom has exactly four nearest neighbors that are configured in a tetrahedral coordination around it. In the graphite modification of carbon the carbon atoms form planar sheets that are relatively widely separated from one another. Each carbon atom is bonded to three near neighbors in the plane, and has a weak interaction with atoms in adjacent planes. (a) Diamond is an electrical insulator and is one of the hardest substances known. Interpret these properties in terms of its bonding. Since each carbon atom has four carbon neighbors, its four sp 3 hybrid bonds are saturated. All electron states in the bonds are filled, so there can be no net electron motion from bond to bond. The only mechanism available to conduct electricity is the excitation of valence electrons into excited orbitals. However, this requires a significant activation energy and is, consequently, a rare event at all normal temperatures. The conductivity is very low; diamond is an insulator. The hardness of diamond is largely due to the rigidity of its bonding. In order for the carbon to deform its atoms must move with respect to one another. The local distortion of the atom configuration requires some reconfiguration of the valence electron distribution. However, these electrons are in very strong, saturated bonds that are difficult to distort. (b) Graphite is an anisotropic conductor. It has high conductivity in the plane of the sheet, and very low conductivity perpendicular to the sheet plane. Explain this behavior in terms of a simple chemical bonding model. The structure of graphite is one in which carbon atoms form sheets in which the atoms are hexagonally configured as shown in Fig The sheets are stacked in a configuration in which successive sheets are weakly bonded to one another. Assuming that the valence bonding of carbon is in the plane of the sheet only, each carbon atom has three neighbors, and hence its bonds are unsaturated. Graphite is a good conductor in the plane of its sheet. On the other hand, graphite is a very poor conductor in the direction perpendicular to the sheet. Since the sheets are relatively far apart the bonds between them are nearly of the Van der Waals (polarization) type, and do not permit electrical conduction. Page 1

2 Fig. 1.1: Atom configuration in a plane of graphite. (c) In its usual microstructure, small three-dimensional particles made of stacked planar sheets, graphite deforms easily, and is used in pencil lead and lubricants. Given that the carbon sheets themselves are very rigid, explain why graphite is so easily deformed. The flakes of graphite slide easily over one another, much like cards in a deck of playing cards, so that small particles that are stacks of planar sheets are relatively easy to deform. (d) Graphite fiber is made by modifying the microstructure into long, thin ribbons in which the carbon sheets parallel the plane of the ribbon. The ribbons are then rolled, somewhat as one would roll a newspaper, to form the graphite fiber. While normal graphite is weak, graphite fiber is extremely strong when pulled along the fiber axis. Explain in terms of the microstructure and the bonding in graphite. When graphite sheets are rolled into tubes to create fibers they are very strong when pulled along the fiber axis. To deform a rolled sheet one must stretch the bonds in the sheet or reconfigure the atoms. Since the carbon atoms in graphite are tightly bound in the sheet plane, this is very difficult to do. Problem 2: Let two atoms bond together so that the potential energy of the pair is described by equation 2.1 of the notes. (a) Find the equilibrium separation of the atoms, r 0, in terms of the constants A, B, m and n. From eq. 2.1 the potential energy of the bond is Ï(r) = A r m - B r n 2.1 page 2

3 The equilibrium separation, r 0, is at the minimum of the potential, Ï(r). The condition dï dr = 0 (at r = r 0 ) 2.2 yields the result r 0 = ma nb 1 m-n 2.3 (b) Show that the molecule behaves like a simple spring for small displacements, r = r - r 0, away from the equilibrium separation; that is, show that the force that acts to restore the equilibrium separation is given approximately by the relation Evaluate the spring constant, K. F = - K r 2.4 The force on the molecule is related to the potential by the equation F = - dï dr = ma nb rm+1 - r n It follows that F vanishes when r = r 0, is positive when r < r 0, and is negative when r > r 0. Hence the molecule behaves as a spring; a displacement away from r 0 induces a force that acts to restore the equilibrium separation. To find the "spring constant" of the molecule we compute the restoring force for a small displacement, r = r - r 0. A Taylor expansion of the force, F(r), about the point, r 0, gives F( r) = F(r 0 ) + df dr ( r) r0 where the subscript on the derivative indicates that it is to be evaluated at r 0, and we neglect higher order terms in the expansion since they depend on ( r) 2 or higher powers of r, which is taken to be small. Since F(r 0 ) = 0 we have F( r) = df dr ( r) = - d 2 Ï r0 dr 2 ( r) 2.7 r 0 The force on a spring that is displaced from equilibrium by the amount, r, is F = - K r 2.8 page 3

4 where K is the spring constant. It follows that, for small displacements, the molecule behaves like a spring with a spring constant Since K = d 2 Ï dr r 0 d 2 Ï dr 2 = m(m+1)a r m+2 - n(n+1)b r n the spring constant is found to be, after some algebra, K = d 2 Ï dr 2 = 1 r 0 r n+2 [(m-n)b] (This is the simplest expression I can find.) Note that K is positive, as it should be for a stable molecule. (c) Find the force that would have to be applied to the molecule to break it in two. This is called the ultimate tensile strength of the molecule. If the potential and force are plotted against (r) in the same coordinate frame the plot appears as shown below. The force is the negative gradient of the potential. When r < r 0 the force is positive, when r > r 0 it is negative, so the force that develops between the particles always acts to restore their equilibrium configuration. However, as shown in the diagram, the force has a minimum value that is reached at the separation, r m. For r > r m the force increases again (decreases in magnitude) and asymptotes to zero at large interatomic distances. Let an external force, F a, be applied to the molecule, and let the force be positive (tensile) so that it tends to separate the atoms. The equilibrium separation in the presence of the force, F a, is the value of r at which F(r) = - F a 2.12 so that the total force is zero. Now suppose that F a is gradually increased. The atoms separate further until F a exceeds - F(r m ). Since this is the minimum value of F(r), it is not possible to find a solution to equation 2.12 when F a > - F(r m ), and the atom flies apart. Hence the tensile strength of the atom is - F(r m ). page 4

5 Fig. 1.2: Plot of the potential, Ï(R), and the force, F(R), required to separate a binary molecule to the distance, R. The maximum of this force, F(max), is the ultimate tensile strength of the molecule. We find r m from the condition that the derivative of F(r) vanishes there. Then which gives the result df dr = - d2 Ï dr 2 = r m = m(m+1)a n(n+1)b m-n = r 0 m+1 n+1 m-n 2.14 which shows that r m > r 0. The tensile strength is F = - F(r m ) = 1 r n+1 ma rm-n - nb 1 = 1 r n+1 nb 1 - n+1 m page 5

6 Problem 3: (a) Consider a two-dimensional crystal made up of spherical atoms that are packed together as tightly as possible in the plane. Draw the Bravais lattice for this crystal, and the primitive cell defined by the unit vectors of the Bravais lattice. Fig. 1.3 shows the planar distribution of close-packed atoms, and shows the primitive unit cell that is a parallelogram with atoms at each of its corners. Since each of the corner atoms is shared by four cells, there is one atom/cell. A unit cell that contains only one atom is called a primitive cell. Fig. 1.3: Close packing of atoms with a unit cell in the form of a parallelogram The Bravais lattice can also be chosen as the hexagonal cell shown in Fig. 1.4, wich reveals its hexagonal symmetry. As you can show by geometric construction, or can simply reason from the number of atoms/cell, that there are three cells of the parallelogram type in each hexagonal unit cell. Fig. 1.4: Close-packed planar configuration of atoms with hexagonal unit cell. (b) Show that the unit cell can be drawn as a hexagon with an atom in the center and atoms at each of the six corners. How many atoms are there per cell? page 6

7 The hexagonl cell is shown in Fig Recalling that the structure is twodimensional, the hexagonal cell has 1 atom in its interior plus 6 corner atoms that are shared between three cells. Hence the total number of atoms/cell is 3 (c) Consider a two-dimensional crystal that has a hexagonal unit cell with an empty center and atoms located at the corners of the hexagon. Draw the atom positions generated by several neighboring cells. Show that the unit cell used in part (b) is a satisfactory cell for this structure, but the cell used in part (a) is not. The hexagonal configuration is drawn in Fig It is not an unphysical configuration. The graphite modification of carbon, which is, in fact, the stable form of carbon at normal temperature and pressure, is a stacking of atom planes that have this configuration. If the cell used in Fig. 1.3 is drawn in the structure shown in Fig. 1.5, it has an empty corner. It is not an acceptable unit cell since the lattice points it defines do not identify identical groups of atoms. However, the hexagonal cell drawn in Fig. 1.4 is a satisfactory choice since all atoms can be located by repeating identical hexagons through space. Fig. 1.5: A two-dimensional hexagonal configuration of atoms (d) Draw a proper set of lattice vectors for the structure in (c). Indicate the basis vectors that are needed to complete the description of the structure. Given your choice of lattice vectors, what atom group is repeated at each lattice point? One correct choice for the unit cell is shown in Fig of the notes, which is repeated in Fig. 1.6 below. With this choice the repeated atom group is a two-atom arrangement in which the atoms are displaced from the lattice point by the vectors τ 1 and τ 2. The lattice point is vacant. [Note that this problem does not have a unique answer. You can consistently choose a different set or lattice vectors and/or a different origin for the lattice vectors. In this case the basis vectors will change as well. The critical thing is that the lattice vectors must generate a periodic lattice of points, each of which has exactly the same atom group associated with it.] page 7

8 a 2 a 2 2 a 1 a 1 1 Fig. 1.6: A simple hexagonal arrangement of atoms showing the hexagonal unit cell and location and basis vectors of the two atoms in the cell. Problem 4: (a) Prove that there is a [ 101] direction in the (111) plane. The vector [111] is perpendicular to the (111) plane. It is also perpendicular to the vector [ 101] since [ 101] [111] = 1-1 = It follows that the [ 101] direction is in the (111) plane. (b) Prove that there is no [101] direction in the (111) plane. Since [101] [111] = the [101] direction does not lie in the (111) plane. (c) Find the angle at which the (111) and ( 111) planes intersect. [Hint: this is a very difficult problems in geometry unless you remember that the [hkl] direction is a vector perpendicular to the (hkl) plane and recall that the scalar product of two vectors, a and b, obeys the relation a^b = a b cos(œ) With this relation the problem becomes simple.] Since [ 111] [111] = 1 = 3 cos(œ) 4.3 page 8

9 we have œ = cos -1 (1/3) = 1.23 radians = 71º 4.4 Problem 5: (a) Let the lattice sites of the FCC bravais lattice be filled with spherical atoms that are close-packed in the sense that they just touch one another. What is the radius of the largest sphere that will fit inside an octahedral void without distorting the structure? What is the radius of the largest sphere that will fit in a tetrahedral void without distortion? [Hint: the geometry of the tetrahedral void is much easier to treat if one first shows that the center of the void lies at the center of a cube that is 1/8 the size of the FCC unit cell.] Let the radius of a spherical atom be R and let the edge of the unit cell have length, a. If the lattice atoms just touch, then they touch along the face diagonal. Hence 2 a = 4R, and a = 2.83R. The diameter of the octahedral void is equal to the gap between lattice atoms along the cube edge. Hence d = 0.83R, or r, the radius of the largest atom that will fit in the octahedral hole, is r oct = 0.414R 5.1 To obtain the radius of a tetrahedral void note that it lies at the center of an octet of the cube, that is, a small cube of edge (a/2) centered on the corner, as shown in Fig of the notes. An atom in the tetrahedral hole would first touch a lattice atom along the diagonal of the octet. Hence, letting d' be the length of the diagonal of the octet, d'/2 = R+r tet. Since d' = 3 2 a =.866a = 2.45R 5.2 r tet = ( )R = 0.225R 5.3 (b) Let a crystal have the hexagonal close-packed structure, and consist of spherical atoms that just touch one another. Find the number of octahedral and tetrahedral voids per atom, and compute the sizes of the largest spheres that will fit in each type of site without distortion in terms of the radius, R, of the atoms on the HCP lattice sites. [Hint: assuming that you have solved parts (a) and (b), if you have to compute anything at all to answerthis question you are approaching it wrong.] As one can see by considering the stacking of close-packed planes, the coordination of voids in fcc and hcp are the same. Hence the numbers and sizes of the voids are identical. page 9

10 Problem 6: (a) Let the lattice sites of the BCC bravais lattice be filled with spherical atoms that are close-packed in the sense that they just touch one another. What is the radius of the largest sphere that will fit inside an octahedral void without distorting the structure? What is the radius of the largest sphere that will fit in a tetrahedral void without distortion? Let a BCC unit cell be made of spherical atoms that are packed as densely as possible. In that case the atoms touch along the cube diagonal. If the atom radius is R and the cube edge is a, a =(4/ 3)R 6.1 There is an octahedral void in BCC at the center of each edge. If there is a spherical atom or radius r in that void, the largest that will fit has R oct = (a/2) R = (2/ 3-1)R = 0.158R 6.2 There is a tetrahedral void at positions like (0. 1/2, 1/4) that has neighboring atoms at (0,0,0) and (0,1,0). The largest sphere that can fit in this void has radius, r, such that (r+r) is the hypotenuse of a right triangle with sides a/2 and a/4. It follows that from which (r+r) 2 = (a/2) 2 + (a/4) 2 = (5/16)a 2 = (5/3)R 2 r tet =.291R 6.3 (b) The common interstitial solutes in BCC metals, in particular, C and N in Fe, are found in the octahedral interstices rather than the tetrahedral ones. Given your answer to part (a), how do you explain this? While the radius of the octahedral site is smaller than that of the tetrahedral site, its volume is larger. In fact, since there are 6 octahedral voids and 12 tetrahedral voids per unit cell, the free volume available per octahedral void is twice as large (roughly 0.16V vs. 0.08V, where V is the volume of a lattice atom.) Since the whole electron cloud of the interstitial atom must be squeezed into the interstitial hole, the volume of the hole is more important than its shape, and interstitials prefer the octahedral site in BCC. Problem 7: page 10

11 (a) The perovskite crystal structure, which is a common structure of ferroelectric oxides like BaTiO 3, is usually drawn in a cubic cell with Ti atoms at the corners, Ba at the center, and O at the centers of each of the edges. Draw the structure and show that it has the stoichiometric formula BaTiO 3. Ti Ba O Fig. 1.7: The perovskite structure of BaTiO 3 The unit cell is drawn in Fig The unit cell contains 1 Ba atom at the cell center, a net of 1 Ti atom (8 corner atoms, each 1/8 in the cell), and a net of 3 O atoms (12 edge atoms, each 1/4 in the unit cell). Hence the formula is BaTiO 3, and is an electrically neutral ionic compound if Ti has valence +4, Ba has +2, and O has -2. These are common valences for these elements. Note that the structure is such that all the nearest neighbors of an ion of given charge are ions of opposite charge. Referring to Fig. 3.1, the Ti +4 cations are located at the corners of the unit cell. Their closest neighbors are the O -2 ions at the centers of the edges of the cell. The Ba +2 ion is located in the center of the cell. It is closer to the O -2 ions on the cell edges than to the Ti +4 ions at the cell corners. The closest neighbors of the O -2 ions are the Ti +4 ions on the corners. The Ba +2 ions in the cell centers are only slightly further away. O -2 ions are not neighbors of one another. It follows that this is a suitable structure for an ionic crystal. (b) To relate this structure to FCC most simply we consider lattice vacancies as a component. By placing a vacancy at the center of each face of the cube show that BaTiO 3 can be regarded as the compound (Ti 3)(BaO 3 ) where is a vacancy. The FCC lattice is filled by Ti and vacancies while the octahedral interstitial voids are filled by Ba and O. Show that this structure is the NaCl structure if we ignore the difference between Ti and, and between Ba and O. Show that Ti and are distributed over the FCC sites in the Cu 3 Au pattern, while Ba and O are distributed over the FCC lattice of octahedral voids in a similar pattern. The perovskite structure is re-drawn to show the positions of the vacancies in Fig If we assume a structure in which the Ti atoms and vacancies are joined into the pseudo-atom Ti 3 and the Ba and O are joined into the pseudo-atom BaO 3, the structure is just the NaCl structure, as shown in Fig page 11

12 Fig. 1.8: The perovskite structure re-drawn to show the vacant FCC sites. = Ti 3 = BaO 3 Fig. 1.9: The perovskite cell redrawn as an NaCl lattice of pseudo-atoms. Fig shows the distribution of Ti atoms and vacancies in a perovskite unit cell from which the Ba and O atoms have been removed. The distribution is precisely a Cu 3 Au ordering over the FCC lattice sites. = Ti = vacancy Fig. 1.10: The Cu 3 Au distribution of Ti atoms and vacancies in BaTiO 3. (c) Show that the perovskite structure of part (a) can also be understood as a BCC lattice filled by Ba and Ti with O atoms in one-half of the octahedral voids. Show that the Ba and Ti atoms are substitutionally ordered in a CsCl pattern. The possibility of this description should be obvious on inspection of Fig The octahedral voids on the edges of the BCC cell are filled. Those in the center of the faces are page 12

13 vacant. If we simply remove the O atoms from Fig. 1.8 the result is a unit cell of BaTi in the CsCl ordered structure. (d) While this representation may seem to be a simpler description of the structure, in at least one important respect it is not. Show that the O atoms do not have a simple BCC pattern, even when vacancies are included. [To treat the structure as a set of BCC sublattices with CsCl order on all of them one must assume that the oxygen is distributed over three separate sublattices, one associated with each type of octahedral void (O x, O y, and O z ). The mixture of oxygen and vacancies on each of the oxygen sublattices is ordered in a CsCl pattern. However, this leads to a much more complicated picture of the structure.] If we place vacancies in the centers of each face in the perovskite structure shown in Fig. 1.8, which is equivalent to placing them in the face-centered octahedral voids of the BCC-like CsCl lattice of Ba and Ti, move the origin to the center of one of the oxygen atoms, and erase the Ba and Ti atoms, the resulting structure shows the distribution of oxygen and vacancies with respect to the BCC reference frame, Fig = O = vacancy Fig. 1.11: The distribution of oxygen atoms and vacancies in the perovskite structure referred to a BCC cell. This can be seen to be a set of three interpenetrating BCC lattices, each of which contains oxygen atoms and vacancies in the CsCl pattern. One lattice includes the O x interstitials in BCC, one the O y, and one the O z. page 13