A crystalline solid is one which has a crystal structure in which atoms or ions are arranged in a pattern that repeats itself in three dimensions.


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1 CHAPTER ATOMIC STRUCTURE AND BONDING. Define a crstalline solid. A crstalline solid is one which has a crstal structure in which atoms or ions are arranged in a pattern that repeats itself in three dimensions..2 Define a crstal structure. Give eamples of materials which have crstal structures. A crstal structure is identical to a crstalline solid, as defined b the solution of Problem.. Eamples include metals, ionic crstals and certain ceramic materials.. Define a space lattice. A space lattice is an infinite threedimensional arra of points with each point having identical surrounding points..4 Define a unit cell of a space lattice. What lattice constants define a unit cell? The unit cell of a space lattice represents a repeating unit of atomic spatial positions. The cell is defined b the magnitudes and directions of three lattice vectors, a, b, and c: aial lengths a, b, and c; interaial angles α, β,and γ..5 What are the 4 Bravais unit cells? The fourteen Bravais lattices are: simple cubic, bodcentered cubic, facecentered cubic, simple tetragonal, bodcentered tetragonal, simple orthorhombic, basecentered orthorhombic, bodcentered orthorhombic, facecentered orthorhombic, simple rhombohedral, simple heagonal, simple monoclinic, basecentered monoclinic, and simple triclinic..6 What are the three most common metal crstal structures? List five metals which have each of these crstal structures. The three most common crstal structures found in metals are: bodcentered cubic (BCC), facecentered cubic (FCC), and heagonal closepacked (HCP). Eamples of metals having these structures include the following. BCC: α iron, vanadium, tungsten, niobium, and chromium. FCC: copper, aluminum, lead, nickel, and silver. HCP: magnesium, α titanium, inc, berllium, and cadmium..7 How man atoms per unit cell are there in the BCC crstal structure? Smith Foundations of Materials Science and Engineering Solution Manual 2
2 A BCC crstal structure has two atoms in each unit cell..8 What is the coordination number for the atoms in the BCC crstal structure? A BCC crstal structure has a coordination number of eight..9 What is the relationship between the length of the side a of the BCC unit cell and the radius of its atoms? In a BCC unit cell, one complete atom and two atom eighths touch each other along the cube diagonal. This geometr translates into the relationship a= 4 R..0 Molbdenum at 20ºC is BCC and has an atomic radius of 0.40 nm. Calculate a value for its lattice constant a in nanometers. Letting a represent the edge length of the BCC unit cell and R the molbdenum atomic radius, a = 4 R or a= R= (0.40 nm) =0.2 nm. Niobium at 20ºC is BCC and has an atomic radius of 0.4 nm. Calculate a value for its lattice constant a in nanometers. For a BCC unit cell having an edge length a and containing niobium atoms, a = 4 R or a= R= (0.4 nm) =0.0 nm.2 Lithium at 20ºC is BCC and has an atomic radius of nm. Calculate a value for the atomic radius of a lithium atom in nanometers. For the lithium BCC structure, which has a lattice constant of a = nm, the atomic radius is, R= a = ( nm) =0.52 nm. Sodium at 20ºC is BCC and has an atomic radius of nm. Calculate a value for the atomic radius of sodium atom in nanometers. For the sodium BCC structure, with a lattice constant of a = nm, the atomic radius is, Smith Foundations of Materials Science and Engineering Solution Manual 22
3 R= a = ( nm) =0.86 nm.4 How man atoms per unit cell are there in the FCC crstal structure? Each unit cell of the FCC crstal structure contains four atoms..5 What is the coordination number for the atoms in the FCC crstal structure? The FCC crstal structure has a coordination number of twelve..6 Gold is FCC and has a lattice constant of nm. Calculate a value for the atomic radius of a gold atom in nanometers. For the gold FCC structure, which has a lattice constant of a = nm, the atomic radius is, 2 2 R = a = ( nm) =0.44 nm.7 Platinum is FCC and has a lattice constant of nm. Calculate a value for the atomic radius of a platinum atom in nanometers. For the platinum FCC structure, with a lattice constant of a = nm, the atomic radius is, 2 2 R = a = (0.929 nm) =0.9 nm.8 Palladium is FCC and has an atomic radius of 0.7 nm. Calculate a value for its lattice constant a in nanometers. Letting a represent the FCC unit cell edge length and R the palladium atomic radius, 2a = 4 R or a= R= (0.7 nm) =0.87 nm Strontium is FCC and has an atomic radius of 0.25 nm. Calculate a value for its lattice constant a in nanometers. For an FCC unit cell having an edge length a an containing strontium atoms, 2a = 4 R or a= R= (0.25 nm) =0.608 nm 2 2 Smith Foundations of Materials Science and Engineering Solution Manual 2
4 .20 Calculate the atomic packing factor for the FCC structure. B definition, the atomic packing factor is given as: volume of atoms in FCC unit cell Atomic packing factor = volume of the FCC unit cell These volumes, associated with the fouratom FCC unit cell, are 4 6 Vatoms = 4 πr = πr and Vunit cell = a 4R where a represents the lattice constant. Substituting a =, 2 64R Vunit cell = a = 2 2 The atomic packing factor then becomes, πr 6 2 π 2 APF (FCC unit cell) = = 2R 6 = How man atoms per unit cell are there in the HCP crstal structure? The heagonal prism contains si atoms..22 What is the coordination number for the atoms in the HCP crstal structure? The coordination number associated with the HCP crstal structure is twelve..2 What is the ideal c/a ratio for HCP metals? The ideal c/a ratio for HCP metals is.6; however, the actual ratios ma deviate significantl from this value..24 Of the following HCP metals, which have higher or lower c/a ratios than the ideal ratio: Zr, Ti, Zn, Mg, Co, Cd and Be? Cadmium and inc have significantl higher c/a ratios while irconium, titanium, magnesium, cobalt and berllium have slightl lower ratios..25 Calculate the volume in cubic nanometers of the titanium crstal structure unit cell. Titanium is HCP at 20ºC with a = nm and c = nm. For a heagonal prism, of height c and side length a, the volume is given b: Smith Foundations of Materials Science and Engineering Solution Manual 24
5 V = (Area of Base)(Height) = [(6 Equilateral Triangle Area)(Height)] 2 = (a sin 60 )( c) 2 = ( nm) (sin 60 )(0.468 nm) = 0.06 nm.26 Rhenium at 20ºC is HCP. The height c of its unit cell is nm and its c/a ratio is.6 nm. Calculate a value for its lattice constant a in nanometers. The rhenium lattice constant a is calculated as, c nm a = = =0.27 nm c/ a.6.27 Osmium at 20ºC is HCP. Using a value of 0.5 nm for the atomic radius of osmium atoms, calculate a value for its unitcell volume. Assume a packing factor of From the definition of the atomic packing factor, volume of atoms in HCP unit cell HCP unit cell volume = APF Since there are si atoms in the HCP unit cell, the volume of atoms is: V atoms 4 = 6 πr = 8 π(0.5) = nm The unit cell volume thus becomes, nm HCP unit cell volume = = nm How are atomic positions located in cubic unit cells? Atomic positions are located in cubic unit cells using rectangular,, and aes and unit distances along the respective aes. The directions of these aes are shown below Smith Foundations of Materials Science and Engineering Solution Manual 25
6 .29 List the atom positions for the eight corner and si facecentered atoms of the FCC unit cell. The atom positions at the corners of an FCC unit cell are:, (, 0, 0), (,, 0), (0,, 0), (0, 0, ), (, 0, ), (,, ), (0,, ) On the faces of the FCC unit cell, atoms are located at: (,, 0), (, 0, ), (0,, ), (,, ), (,, ), (,, ).0 How are the indices for a crstallographic direction in a cubic unit cell determined? For cubic crstals, the crstallographic direction indices are the components of the direction vector, resolved along each of the coordinate aes and reduced to the smallest integers. These indices are designated as [uvw].. Draw the following directions in a BCC unit cell and list the position coordinates of the atoms whose centers are intersected b the direction vector: (a) [00] (b) [0] (c) [] (,, ) (, 0, 0) (,, 0) (a) Position Coordinates: (b) Position Coordinates: (c) Position Coordinates:, (, 0, 0), (,, 0), (,, ).2 Draw direction vectors in unit cubes for the following cubic directions: (a) [ ] (b) [0] (c) [2] (d) [] (a) (b) = + =  =  = + =  = 0 [ ] [ 0] Smith Foundations of Materials Science and Engineering Solution Manual 26
7 (c) (d) [] Dividing b 2, =  = =  [2] Dividing b, = = =. Draw direction vectors in unit cubes for the following cubic directions: (a) [ 2] (d) [02] (g) [0] (j) [0 ] (b) [2 ] (e) [22] (h) [2 ] (k) [2 2] (c) [] (f) [2] (i) [2] (l) [22] (a) Dividing [2] b 2, =, =, = 2 2 (b) Dividing [2] b, 2 =, =, = (c) Dividing [] b, =, =, = (d) Dividing [02] b 2, = 0, =, = 2 (e) Dividing [2 2] b 2, =, =, = 2 (f) Dividing [2] b, 2 =, =, = Smith Foundations of Materials Science and Engineering Solution Manual 27
8 (g) For [ 0], =, = 0, = (h) Dividing [2] b 2, =, =, = 2 2 (i) Dividing [2] b, 2 =, =, = (j) Dividing [0] b, =, = 0, = (k) Dividing [22] b 2, =, =, = 2 (l) Dividing [22] b, 2 2 =, =, =.4 What are the indices of the directions shown in the unit cubes of Fig. P.4? (a) a ¾ d c b (b) e f g ¾ h ¾ Smith Foundations of Materials Science and Engineering Solution Manual 28
9 a / 6 b c New 0 New 0 a. Vector components: = , =, = 0 Direction indices: [0] b. Moving direction vector down, vector components are: =, = , = Direction indices: [44] c. Moving direction vector forward, vector components are: =  / 6, =, = Direction indices: [66] ¾ d e New 0 f  New 0 d. Moving direction vector left, vector components are: =, =, = Direction indices: [22] New 0 e. Vector components are: = ¾, = , = Direction indices: [44] New 0 f. Moving direction vector up, vector components are: = , =, =  Direction indices: [] ¾ ¾ g ¾ h ¾ g. Moving direction vector up, vector components are: =, = , =  Direction indices: [44] h. Moving direction vector up, vector components are: = ¾, = , = ¾ Direction indices: [4] Smith Foundations of Materials Science and Engineering Solution Manual 29
10 .5 A direction vector passes through a unit cube from the ( ¾, 0, ) to the (,, 0 ) positions. What are its direction indices? The starting point coordinates, subtracted from the end point, give the vector components: = = = 0= = 0 = 2 The fractions can then be cleared through multiplication b 4, giving =, = 4, =. The direction indices are therefore [4]..6 A direction vector passes through a unit cube from the (, 0, ¾ ) to the (,, ) positions. What are its direction indices? Subtracting coordinates, the vector components are: = = = 0= = = 2 Clearing fractions through multiplication b 4, gives =, = 4, = 2. The direction indices are therefore [ 4 2]..7 What are the crstallographic directions of a famil or form? What generalied notation is used to indicate them? A famil or form has equivalent crstallographic directions; the atom spacing along each direction is identical. These directions are indicated b uvw..8 What are the directions of the 00 famil or form for a unit cube? [0 0], [00], [00], [00], [00], [00].9 What are the directions of the famil or form for a unit cube? [], [], [], [], [], [ ], [], [] [ 0] [0 ].40 What 0 tpe directions lie on the () plane of a cubic unit cell? [0], [0], [0], [0], [0], [0 ] [ 0 ] [ 0] [0] [0] Smith Foundations of Materials Science and Engineering Solution Manual 0
11 .4 What tpe directions lie on the (0) plane of a cubic unit cell? [], [], [ ], [] [] [] [] [].42 How are the Miller indices for a crstallographic plane in a cubic unit cell determined? What generalied notation is used to indicate them? The Miller indices are determined b first identifing the fractional intercepts which the plane makes with the crstallographic,, and aes of the cubic unit cell. Then all fractions must be cleared such that the smallest set of whole numbers is attained. The general notation used to indicate these indices is (hkl), where h, k, and l correspond to the, and aes, respectivel..4 Draw in unit cubes the crstal planes that have the following Miller indices: (a) ( ) (d) (2) (g) (20) (j) () (b) (0 2) (e) (2) (h) (22) (k) ( 2) (c) ( 2 ) (f) (0 2) (i) (22) (l) ()   ( ) (02)  + ( 2 )  a. For ( ) reciprocals are: =, = , = (2) + b. For (02) reciprocals are: =, =, =   (2) + + c. For ( 2 ) reciprocals are: =, = , =  + (02)  d. For (2)reciprocals e. For (2) reciprocals f. For (02) reciprocals are: =, =, =  are: =, = , = are: =, =, =  Smith Foundations of Materials Science and Engineering Solution Manual
12 +  (20) g. For (20)reciprocals are: =, =, = (22) h. For (22)reciprocals are: =, =, = (22) i. For (22) reciprocals are: =, =, = () j. For () reciprocals are: =, =, =   (2) + k. For (2) reciprocals are: =, = , =  () k. For ()reciprocals are: = , =, = .44 What are the Miller indices of the cubic crstallographic planes shown in Fig. P.44? (a) a d c ¾ ¾ (b) a b c d b Smith Foundations of Materials Science and Engineering Solution Manual 2
13 Miller Indices for Figure P.44(a) Plane a based on (0,, ) as origin Plane b based on (,, 0) as origin Planar Intercepts = =  = 4 Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts 0 = =  = = 5 2 = = = = 0 = The Miller indices of plane a are (04). The Miller indices of plane b are (5 2 0). Plane c based on (,, 0) as origin Plane d based on as origin Planar Intercepts = =  = Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts 0 = = = = = = = 2 = = 2 The Miller indices of plane c are (0 ). The Miller indices of plane d are (22). Miller Indices for Figure P.44(b) Plane a based on (, 0, ) as origin Plane b based on (0,, ) as origin Planar Intercepts =  = = Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts = = = 0 = = = 2 = = 2 The Miller indices of plane a are (0). The Miller indices of plane b are (22). Smith Foundations of Materials Science and Engineering Solution Manual
14 Plane c based on (0,, 0) as origin Plane d based on (0,, 0) as origin Planar Intercepts = 5 = 2 = Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts = = = 2 = =  5 = 0 = = 2 2 = The Miller indices of plane c are ( 5 2 0). The Miller indices of plane d are ( 2)..45 What is the notation used to indicate a famil or form of cubic crstallographic planes? A famil or form of a cubic crstallographic plane is indicated using the notation {hkl}..46 What are the {00} famil of planes of the cubic sstem? (00), (00), (00), (00), (00), (00).47 Draw the following crstallographic planes in a BCC unit cell and list the position of the atoms whose centers are intersected b each of the planes: (a) (00) (b) (0) (c) () a. (, 0, 0), (, 0, ), (,, 0), (,, ) b. (, 0, 0), (, 0, ), (0,, 0), (0,, ), (,, ) c. (, 0, 0), (0, 0, ), (0,, 0).48 Draw the following crstallographic planes in an FCC unit cell and list the position coordinates of the atoms whose centers are intersected b each of the planes: (a) (00) (b) (0) (c) () Smith Foundations of Materials Science and Engineering Solution Manual 4
15 a. (, 0, 0), (, 0, ), (,, 0), (,, ) (,, ) b. (, 0, 0), (, 0, ), (0,, 0), (0,, ), (,, 0), (,, ) c. (, 0, 0), (0, 0, ), (0,, 0), (, 0, ) (,, 0), (0,, ).49 A cubic plane has the following aial intercepts: a =, b = , c =. What are the Miller indices of this plane? Given the aial intercepts of (, , ), the reciprocal intercepts are:,, 2. = = 2 = Multipling b 2 to clear the fraction, the Miller indices are (64)..50 A cubic plane has the following aial intercepts: a = , b = , c =. What are the Miller indices of this plane? Given the aial intercepts of (, , ), the reciprocal intercepts are: 2, 2,. = = = 2 Multipling b 2, the Miller indices are (44)..5 A cubic plane has the following aial intercepts: a =, b =, c = . What are the Miller indices of this plane? Given the aial intercepts of (,, ), the reciprocal intercepts are:,, 2. = = 2 = Multipling b 2, the Miller indices are (24)..52 Determine the Miller indices of the cubic crstal plane that intersects the following position coordinates: (, 0, 0 ); (,, ); (,, 0 ). First locate the three position coordinates as shown. Net, connect points a and b, etending the line to point d and connect a to c and etend to e. Complete the plane b Smith Foundations of Materials Science and Engineering Solution Manual 5
16 connecting point d to e. Using (,, 0) as the plane origin, = , =  and =. The intercept reciprocals are thus =, =, = 2. The Miller indices are ( 2). (, 0, 0) (,, ) (,, 0) a b d c e (,, 0).5 Determine the Miller indices of the cubic crstal plane that intersects the following position coordinates: (, 0, ); ( 0, 0, ); (,, ). First locate the three position coordinates as shown. Net, connect points a and b and etend the line to point d. Complete the plane b connecting point d to c and point c to b. Using (, 0, ) as the plane origin, = , = and =. The intercept reciprocals are thus =, =, =. The Miller indices are (). (0, 0, ) (,, ) (, 0, ) a b c (, 0, ) d.54 Determine the Miller indices of the cubic crstal plane that intersects the following position coordinates: (,, ); (, 0, ¾ ); (, 0, ). After locating the three position coordinates, connect points b and c and etend the line to point d. Complete the plane b connecting point d to a and a to c. Using (, 0, ) as the plane origin, = , = and = . The intercept reciprocals then become =, = 2, = 2. The Miller indices are (22). Smith Foundations of Materials Science and Engineering Solution Manual 6
17 d (, 0, ¾) b (, 0, ) (, 0,,) c (,, ) a.55 Determine the Miller indices of the cubic crstal plane that intersects the following position coordinates: ( 0, 0, ); (, 0, 0 ); (,, 0 ). After locating the three position coordinates, connect points b and c and etend the line to point d. Complete the plane b connecting point d to a and a to b. Using as the plane origin, =, = and =. The intercept reciprocals are thus =, = 2, = 2. The Miller indices are therefore (2 2). a d (0, 0,,) c (, 0, 0) (,, 0) b (0,, 0).56 Rodium is FCC and has a lattice constant a of nm. Calculate the following interplanar spacings: (a) d (b) d 200 (c) d nm nm d = = = nm (a) nm nm d = = = 0.90 nm (b) nm nm d = = = 0.5 nm (c) 220 Smith Foundations of Materials Science and Engineering Solution Manual 7
18 .57 Tungsten is BCC and has a lattice constant a of nm. Calculate the following interplanar spacings: (a) d 0 (b) d 220 (c) d nm nm d = = = nm (a) nm nm d = = = 0.2 nm (b) nm nm d = = = 0.00 nm (c) 0.58 The d 0 interplanar spacing in a BCC element is nm. (a) What is its lattice constant a? (b) What is the atomic radius of the element? (c) What could this element be? (a) (b) a= d0 h + k + l = (0.587 nm) =0.502 nm a (0.502 nm) R = = =0.27 nm (c) The element is barium (Ba)..59 The d 422 interplanar spacing in a FCC metal is nm. (a) What is its lattice constant a? (b) What is the atomic radius of the metal? (c) What could this metal be? (a) (b) a= d422 h + k + l = ( nm) =0.408 nm 2a 2(0.408 nm) R = = =0.44 nm (c) The element is gold (Au). Smith Foundations of Materials Science and Engineering Solution Manual 8
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