Fundamentals of Materials Science and Engineering 5th ed - Solutions PDF

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CH A PTE R 2 ATOMIC STRU CTU RE A ND INTERATOMIC BOND ING

2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as #g/amu =



1 mol 6.023 × 1023 atoms



1 g/mol 1 amu/atom



= 1.66 × 10−24 g/amu 2.14 (c) This portion of the problem asks that we determine for a K + -Cl− ion pair the interatomic spacing (ro ) and the bonding energy (E o ). From E quation (2.11) for E N A = 1.436 B = 5.86 × 10−6 n=9 Thus, using the solutions from Problem 2.13 ro =



=



A nB

1/(1−n)

1.436 (9)(5.86 × 10−6 )

1/(1−9)

= 0.279 nm

and Eo = −

1.436 1.436 (9)(5.86 × 10−6 )

1/(1−9) + 

5.86 × 10−6 9/(1−9) 1.436 (9)(5.86 × 10−6 )

= −4.57 eV 2.19 The percent ionic character is a function of the electronegativities of the ions X A and X B according to E quation (2.10). The electronegativities of the elements are found in Figure 2.7. For TiO 2 , X Ti = 1.5 and X O = 3.5, and therefore,  2 % IC = 1 − e (−0.25)(3.5−1.5) × 100 = 63.2%

1

CH A PTE R 3 STRU CTU RES OF META LS A ND CERA MICS

3.3

For this problem, we are asked to calculate the volume of a unit cell of aluminum. A luminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from E quation (3.4) as √ √ V C = 16R 3 2 = (16)(0.143 × 10−9 m) 3 2 = 6.62 × 10−29 m 3

3.7

This problem calls for a demonstration that the A PF for H CP is 0.74. A gain, the A PF is just the total sphere-unit cell volume ratio. For H CP, there are the equivalent of six spheres per unit cell, and thus   4␲R 3 VS = 6 = 8␲R 3 3 Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACD E shown below.

D

C

a = 2R 30 60

A

E B a = 2R

a = 2R The area of ACD E is just the length of CD times the height BC. But CD is just a or 2R, and √ 2R 3 ◦ BC = 2R cos(30 ) = 2 Thus, the base area is just √  √ 2R 3 = 6R 2 3 A R E A = (3)(CD)(BC) = (3)(2R ) 2 

and since c = 1.633a = 2R (1.633) √ √ √ V C = (A R E A )(c) = 6R 2 c 3 = (6R 2 3)(2)(1.633)R = 12 3(1.633)R 3

2

Thus, A PF =

VS 8␲R 3 = √ = 0.74 VC 12 3(1.633)R 3

3.12. (a) This portion of the problem asks that we compute the volume of the unit cell for Z r. This volume may be computed using E quation (3.5) as VC =

nA Z r ␳N A

Now, for H CP, n = 6 atoms/unit cell, and for Z r, A Zr = 91.2 g/mol. Thus, VC =

(6 atoms/unit cell)(91.2 g/mol) (6.51 g/cm 3 )(6.023 × 1023 atoms/mol)

= 1.396 × 10−22 cm 3 /unit cell = 1.396 × 10−28 m 3 /unit cell (b) We are now to compute the values of a and c, given that c/a = 1.593. From the solution to Problem 3.7, since a = 2R, then, for H CP √ 3 3a 2 c VC = 2 but, since c = 1.593a VC =

√ 3 3(1.593)a 3 = 1.396 × 10−22 cm 3 /unit cell 2

Now, solving for a

a=



(2)(1.396 × 10−22 cm 3 ) √ (3)( 3)(1.593)

1/3

= 3.23 × 10−8 cm = 0.323 nm A nd finally c = 1.593a = (1.593)(0.323 nm) = 0.515 nm 3.17 In this problem we are given that iodine has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.479, 0.725, and 0.978 nm, respectively. (a) G iven that the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively we are to determine the number of atoms in each unit cell. From the definition of the A PF

A PF =

VS = VC

3

n



4 ␲R 3 3 abc



we may solve for the number of atoms per unit cell, n, as n=

=

(A PF)abc 4 ␲R 3 3 (0.547)(4.79)(7.25)(9.78)(10−24 cm 3 ) 4 ␲(1.77 × 10−8 cm) 3 3

= 8.0 atoms/unit cell (b) In order to compute the density, we just employ E quation (3.5) as ␳ = =

nA I abcN A (8 atoms/unit cell)(126.91 g/mol) [(4.79)(7.25)(9.78) × 10−24 cm 3 /unit cell](6.023 × 1023 atoms/mol)

= 4.96 g/cm 3 3.22 This question asks that we generate a three-dimensional unit cell for AuCu 3 using the Molecule Definition File on the CD-R OM. One set of directions that may be used to construct this unit cell and that are entered on the Notepad are as follows: [DisplayProps] R otatez=−30 R otatey=−15 [A tomProps] G old =LtR ed,0.14 Copper=LtYellow,0.13 [BondProps] SingleSolid =LtG ray [A toms] Au1=1,0,0,G old Au2=0,0,0,G old Au3=0,1,0,G old Au4=1,1,0,G old Au5=1,0,1,G old Au6=0,0,1,G old Au7=0,1,1,G old Au8=1,1,1,G old Cu1=0.5,0,0.5,Copper Cu2=0,0.5,0.5,Copper Cu3=0.5,1,0.5,Copper Cu4=1,0.5,0.5,Copper Cu5=0.5,0.5,1,Copper Cu6=0.5,0.5,0,Copper

4

[Bonds] B1=Au1,Au5,SingleSolid B2=Au5,Au6,SingleSolid B3=Au6,Au2,SingleSolid B4=Au2,Au1,SingleSolid B5=Au4,Au8,SingleSolid B6=Au8,Au7,SingleSolid B7=Au7,Au3,SingleSolid B8=Au3,Au4,SingleSolid B9=Au1,Au4,SingleSolid B10=Au8,Au5,SingleSolid B11=Au2,Au3,SingleSolid B12=Au6,Au7,SingleSolid When saving these instructions, the file name that is chosen should end with a period followed by mdf and the entire file name needs to be enclosed within quotation marks. For example, if one wants to name the file AuCu3, the name by which it should be saved is “AuCu3.mdf”. In addition, the file should be saved as a “Text Document.” 3.27 In this problem we are asked to show that the minimum cation-to-anion radius ratio for a coordination number of six is 0.414. Below is shown one of the faces of the rock salt crystal structure in which anions and cations just touch along the edges, and also the face diagonals.

r A

r C

H

G

F

From triangle FGH, G F = 2r A

and

FH = G H = r A + r C

Since FGH is a right triangle (G H ) 2 + (FH ) 2 = (FG ) 2 or (r A + r C ) 2 + (r A + r C ) 2 = (2r A ) 2 which leads to 2r A rA + rC = √ 2

5

Or, solving for rC /rA rC = rA



 2 √ − 1 = 0.414 2

3.29 This problem calls for us to predict crystal structures for several ceramic materials on the basis of ionic charge and ionic radii. (a) For CsI, from Table 3.4 0.170 nm r Cs+ = = 0.773 − rI 0.220 nm Now, from Table 3.3, the coordination number for each cation (Cs+ ) is eight, and, using Table 3.5, the predicted crystal structure is cesium chloride. (c) For KI, from Table 3.4 r K+ 0.138 nm = 0.627 = 0.220 nm r I− The coordination number is six (Table 3.3), and the predicted crystal structure is sodium chloride (Table 3.5). 3.36 This problem asks that we compute the theoretical density of diamond given that the C ––C distance and bond angle are 0.154 nm and 109.5◦ , respectively. The first thing we need do is to determine the unit cell edge length from the given C ––C distance. The drawing below shows the cubic unit cell with those carbon atoms that bond to one another in one-quarter of the unit cell.

a

y

x

From this figure, ␾ is one-half of the bond angle or ␾ = 109.5◦ /2 = 54.75◦ , which means that ␪ = 90◦ − 54.75◦ = 35.25◦ since the triangle shown is a right triangle. A lso, y = 0.154 nm, the carbon-carbon bond distance. Furthermore, x = a/4, and therefore, x=

a = y sin ␪ 4

6

Or a = 4y sin ␪ = (4)(0.154 nm)(sin 35.25◦ ) = 0.356 nm = 3.56 × 10−8 cm The unit cell volume, V C , is just a3 , that is V C = a 3 = (3.56 × 10−8 cm) 3 = 4.51 × 10−23 cm 3 We must now utilize a modified Equation (3.6) since there is only one atom type. There are eight equivalent atoms per unit cell (i.e., one equivalent corner, three equivalent faces, and four interior atoms), and therefore ␳ = =

n′A C VC NA (8 atoms/unit cell)(12.01 g/g-atom) (4.51 × 10−23 cm 3 /unit cell)(6.023 × 1023 atoms/g-atom)

= 3.54 g/cm 3

The measured density is 3.51 g/cm 3 . 3.39 (a) We are asked to compute the density of CsCl. Modifying the result of Problem 3.4, we get a=

2r Cs+ + 2r Cl− 2(0.170 nm) + 2(0.181 nm) √ = √ 3 3

= 0.405 nm = 4.05 × 10−8 cm From E quation (3.6) ␳=

n ′ (A Cs + A Cl ) n ′ (A Cs + A Cl ) = VC NA a3NA

For the CsCl crystal structure, n′ = 1 formula unit/unit cell, and thus ␳ =

(1 formula unit/unit cell)(132.91 g/mol + 35.45 g/mol) (4.05 × 10−8 cm) 3 /unit cell(6.023 × 1023 formula units/mol)

= 4.20 g/cm 3 (b) This value of the density is greater than the measured density. The reason for this discrepancy is that the ionic radii in Table 3.4, used for this computation, were for a coordination number of six, when, in fact, the coordination number of both Cs+ and Cl− is eight. U nder these circumstances, the actual ionic radii and unit cell volume (V C ) will be slightly greater than calculated values; consequently, the measured density is smaller than the calculated density. 3.45 We are asked in this problem to compute the atomic packing factor for the CsCl crystal structure. This requires that we take the ratio of the sphere volume within the unit cell and the total unit cell volume. From Figure 3.6 there is the equivalent of one Cs and one Cl ion per unit cell; the ionic radii of these two ions are 0.170 nm and 0.181 nm, respectively (Table 3.4). Thus, the sphere volume, V S ,

7

is just VS =

4 (␲)[(0.170 nm) 3 + (0.181 nm) 3 ] = 0.0454 nm 3 3

For CsCl the unit cell edge length, a, in terms of the atomic radii is just 2r Cs+ + 2r Cl− 2(0.170 nm) + 2(0.181 nm) √ √ = 3 3 = 0.405 nm

a=

Since V C = a3 V C = (0.405 nm) 3 = 0.0664 nm 3 A nd, finally the atomic packing factor is just A PF =

0.0454 nm 3 VS = = 0.684 VC 0.0664 nm 3

3.50 (a) We are asked for the indices of the two directions sketched in the figure. For direction 1, the projection on the x-axis is zero (since it lies in the y-z plane), while projections on the y- and z-axes are b/2 and c, respectively. This is an [012] direction as indicated in the summary below.

Projections Projections in terms of a, b, and c R eduction to integers E nclosure

x

y

z

0a 0 0

b/2 1/2 1 [012]

c 1 2

3.51 This problem asks for us to sketch several directions within a cubic unit cell. The [110], [121], and [012] directions are indicated below.

_ [012]

z __ [121]

_ [110]

y x 3.53 This problem asks that we determine indices for several directions that have been drawn within a cubic unit cell. Direction B is a [232] direction, the determination of which is summarized as follows.

8

We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x Projections Projections in terms of a, b, and c R eduction to integers

2a 3 2 3 2

E nclosure

y −b −1 −3

z 2c 3 2 3 2

[232]

Direction D is a [136] direction, the determination of which is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system

Projections Projections in terms of a, b, and c R eduction to integers

x

y

a 6 1 6 1

b 2 1 2 3

E nclosure

z −c −1 −6

[136]

3.56 This problem asks that we determine the Miller indices for planes that have been drawn within a unit cell. For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis, and one unit cell distance parallel to the x axis; thus, this is a (112) plane, as summarized below. x

y

Intercepts

−a

−b

Intercepts in terms of a, b, and c

−1

−1

R eciprocals of intercepts

−1

E nclosure

−1

z c 2 1 2 2

(112)

3.58 For plane B we will leave the origin at the unit cell as shown; this is a (221) plane, as summarized below.

Intercepts Intercepts in terms of a, b, and c R eciprocals of intercepts E nclosure

9

x

y

a 2 1 2 2

b 2 1 2 2 (221)

z c 1 1

3.59 The (1101) plane in a hexagonal unit cell is shown below. z

a 2

a

3

a 1

_ (1101)

3.60 This problem asks that we specify the Miller indices for planes that have been drawn within hexagonal unit cells. (a) For this plane we will leave the origin of the coordinate system as shown; thus, this is a (1100) plane, as summarized below.

Intercepts Intercepts in terms of a’s and c R eciprocals of intercepts E nclosure

a1

a2

a3

a 1 1

−a ∞a −1 ∞ −1 0 (1100)

z ∞c ∞ 0

3.61 This problem asks for us to sketch several planes within a cubic unit cell. The (011) and (102) planes are indicated below.

z

__ (011)

y

x

_ (102)

10

3.63 This problem asks that we represent specific crystallographic planes for various ceramic crystal structures. (a) A (100) plane for the rock salt crystal structure would appear as

+ Na Cl

3.64 For the unit cell shown in Problem 3.21 we are asked to determine, from three given sets of crystallographic planes, which are equivalent. (a) The unit cell in Problem 3.21 is body-centered tetragonal. Only the (100) (front face) and (010) (left side face) planes are equivalent since the dimensions of these planes within the unit cell (and therefore the distances between adjacent atoms) are the same (namely 0.40 nm × 0.30 nm), which are different than the (001) (top face) plane (namely 0.30 nm × 0.30 nm). 3.66 This question is concerned with the zinc blende crystal structure in terms of close-packed planes of anions. (a) The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not H CP) because the anion packing is FCC (Table 3.5). (b) The cations will fill tetrahedral positions since the coordination number for cations is four (Table 3.5). (c) Only one-half of the tetrahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion. 3.70* In this problem we are to compute the linear densities of several crystallographic planes for the face-centered cubic crystal structure. For FCC the linear density of the [100] direction is computed as follows: The linear density, LD , is defined by the ratio LD =

Lc Ll

where Ll is the line length within the unit cell along the [100] direction, and Lc is line length passing through intersection circles. Now, Ll is just √ the unit cell edge length, a which, for FCC is related to the atomic radius R according to a = 2R 2 [E quation (3.1)]. A lso for this situation, Lc = 2R and therefore LD =

2R √ = 0.71 2R 2

11

3.73* In this problem we are to compute the planar densities of several crystallographic planes for the body-centered cubic crystal structure. Planar density, PD , is defined as PD =

Ac Ap

where A p is the total plane area within the unit cell and A c is the circle plane area within this same plane. For (110), that portion of a plane that passes through a BCC unit cell forms a rectangle as shown below.

R

4R 3

4R 2 3

In terms of the atomic radius R, the length of the rectangle base is 4R a= √ . Therefore, the area of this rectangle, which is just A p is

√ 4R 2 √ , 3

whereas the height is

3

Ap =



√  √  4R 2 16R 2 2 4R = √ √ 3 3 3

Now for the number equivalent atoms within this plane. One-fourth of each corner atom and the entirety of the center atom belong to the unit cell. Therefore, there is an equivalent of 2 atoms within the unit cell. H ence A c = 2( ␲R 2 ) and PD =

2␲R 2 √ = 0.83 16R 2 2 3

3.80* U sing the data for aluminum in Table 3.1, we are asked to compute the interplanar spacings for the (110) and (221) sets of planes. From the table, aluminum has an FCC crystal structure and an atomic radius of 0.1431 nm. U sing E quation (3.1) the lattice parameter, a, may be computed as √ √ a = 2R 2 = (2)(0.1431 nm)( 2) = 0.4047 nm

12

Now, the d110 interplanar spacing may be determined using E quation (3.11) as 0.4047 nm a √ = 0.2862 nm = d 110 =  2 (1) 2 + (1) 2 + (0) 2 3.84* From the diffraction pattern for ␣-iron shown in Figure 3.37, we are asked to compute the interplanar spacing for each set of planes that has been indexed; we are also to determine the lattice parameter of Fe for each peak. In order to compute the interplanar spacing and the lattice parameter we must employ E quations (3.11) and (3.10), respectively. For the first peak which occurs at 45.0◦ d 110 =

n␭ (1)(0.1542 nm)   = 0.2015 nm = 45.0◦ 2 sin ␪ (2) sin 2

A nd   a = d hkl (h) 2 + (k) 2 + (l) 2 = d 110 (1) 2 + (1) 2 + (0) 2 √ = (0.2015 nm) 2 = 0.2850 nm Similar computations are made for the other peaks which results are tabulated below: Peak Index

2␪

d hkl (nm)

a (nm)

200 211

65.1 82.8

0.1433 0.1166

0.2866 0.2856

13

CH A PTE R 4 POLYMER STRU CTU RES

4.4 We are asked to compute the number-average degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol. The mer molecular weight of polypropylene is just m = 3(A C ) + 6(A H ) = (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol If we let nn represent the number-average degree of polymerization, then from E quation (4.4a) nn =

Mn 106 g/mol = = 23,700 m 42.08 g/mol

4.6 (a) From the tabulated data, we are asked to compute Mn , the number-average molecular weight. This is carried out below. Molecular wt R ange

Mean M i

xi

xi M i

8,000–16,000 16,000–24,000 24,000–32,000 32,000–40,000 40,000–48,000 48,000–56,000

12,000 20,000 28,000 36,000 44,000 52,000

0.05 0.16 0.24 0.28 0.20 0.07

600 3200 6720 10,080 8800 3640

Mn =



xi M i = 33,040 g/mol

(c) Now we are asked to compute nn (the number-average degree of polymerization), using the E quation (4.4a). For polypropylene, m = 3(A C ) + 6(A H ) = (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol A nd nn =

33040 g/mol Mn = 785 = 42.08 g/mol m

4.11 This problem first of all asks for us to calculate, using E quation (4.11), the average total chain length, L, for a linear polytetrafluoroethylene polymer having a number-average molecular weight of 500,000 g/mol. It is necessary to calculate the number-average degree of polymerization, nn , using E quation (4.4a). For PTFE , from Table 4.3, each mer unit has two carbons and four fluorines. Thus, m = 2(A C ) + 4(A F ) = (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol and nn =

500000 g/mol Mn = 5000 = 100.02 g/mol m

14

which is the number of mer units along an average chain. Since there are two carbon atoms per mer unit, there are two C ––C chain bonds per mer, which means that th...