APPENDIX 3

Solutions to selected end-of-chapter problems (marked in text by an asterisk).

CHAPTER 1

5)

Ra(608) ¼

1 0 0

0 cos 60 � sin 60

0 sin 60 cos 60

2664

3775 ¼

1 0 0

012�

ffiffiffi3p

2

0

ffiffiffi3p

212

266664

377775

Rb(608) ¼

cos 60 0 � sin 60

0 1 0

sin 60 0 cos 60

2664

3775 ¼

12

0 �

ffiffiffi3p

20 1 0ffiffiffi3p

20

12

266664

377775

Principles of Inorganic Materials Design, Second Edition. By John N. Lalena and David A. ClearyCopyright # 2010 John Wiley & Sons, Inc.

569

9) The rotation matrix for the 2608 rotation about the [1 1 1] direction, which in the

1ffiffiffi3p ,

1ffiffiffi3p ,

1ffiffiffi3p

� �axis is: R[1 1 1](�608)

¼

0:666667 0:666667 �0:333333

�0:333333 0:666667 0:666667

0:666667 �0:333333 0:666667

264

375

The rotation matrix for the 2908 rotation about the [1 0 0] direction, which in thek1 0 0l axis, is:

R[1 0 0](908) ¼1 0 00 0 10 �1 0

24

35

The product matrix, J, of the two rotations is:

J ¼

0:666667 0:666667 �0:333333

0:666667 �0:333333 0:666667

0:333333 �0:666667 �0:666667

264

375

The rotation angle is

cos�1([0:6667� 0:3333� 0:6667� 1]=2) ¼ u ¼ 2:300 radians

2:3000� 180=p ¼ 131:88

On inspection it can be seen that J is a nonsymmetric matrix ( jij = jji). Therefore,the components of the rotation axis are:

u0x ¼ [0:6667� (�0:6667)]=2 sin(131:8) ¼ 0:8943

u0y ¼ [0:3333� (�0:3333)]=2 sin(131:8) ¼ 0:4470

u0z ¼ [0:6667� 0:6667]=2 sin(131:8) ¼ 0

From vector algebra, it is known that any ordered set of three numbers that can beobtained from kux, uy, uzl by multiplying all of them by the same positive constantk is also a set of direction numbers for the vector r, in that they define the directionof the vector. Hence choosing k to be (1/0.4470) gives: [0.8943/0.4470, 0.4470/0.4470, 0] or k2 1 0l. Therefore, the equivalent axis angle pair is rotation by131.88 about k2 1 0l.

APPENDIX 3570

10) The rotation angle is given by Eq. 1.19 as:

J11 þ J22 þ J33 ¼ 1þ 2 cosf

0:36þ 0:60þ 0:60 ¼ 1þ 2 cosf

0:56 ¼ 2 cosf

0:28 ¼ cosf

þ73:7 ¼ f

The rotation axis is given by Eq. 1.20:

u0x ¼J23 � J32

2 sinf¼

0� 0:642 sin 73:7

¼ �13

u0y ¼J31 � J13

2 sinf¼

0:48� (�0:80)2 sin 73:7

¼23

u0z ¼J12 � J21

2 sinf¼

0:48� (�0:80)2 sin 73:7

¼23

So, the axis is denoted as: � 13 , 2

3 , 23

� �, which has direction indices of the general

form [1 2 2].

CHAPTER 2

1) Using Eq. 2.4 gives 6.5 � 103 distinct boundaries.

4) Annealing is the term applied to the process by which a solidification product isheld at an elevated temperature for an extended time period and then slowlycooled for the purposes of relieving internal stress, increasing plasticity, andproducing a specific microstructure through grain growth. Sintering refers tothe heating of a polycrystalline aggregate at a temperature below its solidus (melt-ing point), but high enough that grain coalescence occurs via solid-statediffusion.

7) The three principle methods of strengthening materials are: grain-size reduction,solid-solution strengthening, and plastic-deformation processes, like strain (work)hardening.

CHAPTER 3

3) VSEPR theory indicates nothing about the nature of the chemical bond (loca-lized Heitler–London versus delocalized Hund–Mulliken). It simply predictsthe geometrical shape, specifically, the X–X, M–X and/or X–M–X bondangles in the molecule.

6) Based on Eq. 3.1, the bond energies would be predicted to increase with increas-ing ion charges.

APPENDIX 3 571

9) A solid-solution alloy consists of random substitution of one element for anotherin the same sublattice or into interstitial sites. An IMC is an ordered placement oftwo or more elements into different sublattices.

CHAPTER 4

5) Utilization of Gibbs’ relations (Eqs. 4.21–4.23) yields:

a� ¼2pffiffiffi

3p

axþ

2pa

y b� ¼ �2pffiffiffi

3p

axþ

2pa

y c� ¼2pc

z

Thus the triangular (hexagonal) lattice is its own reciprocal, but the reciprocallattice is rotated with respect to the crystal lattice.

7) If the three vectors a, b, and c are written in terms of unit vectors, the determinantof [a b c] is ja b cj, which is expressed as:

ax ay az

bx by bz

cx cy cz

��������

��������and [a b c] ¼ axbycz þ bxcyaz þ cxaybz � axcybz � bxaycz � cxbyaz.

CHAPTER 5

1) Equation 5.52 is:

E(k) ¼+ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(1þ e�ik�a1 þ e�ik�a2 )(1þ eik�a1 þ eik�a2 )

pE(10)

This can be expanded by multiplying the two trinomials to obtain:

E(k)¼+ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þeika1þeika2þe�ika1þ1þe�ika1 eika2þe�ika2þe�ika2 eika1þ1

pE(10)

The equivalency between Eqs. 5.52 and 5.53 is then made evident byuse of the following two relations: eika¼cos(ka)þ i sin(ka) and e�ika¼

cos(ka)� i sin(ka). Making the substitutions and simplifying leads to:

E(k)¼+E(10)ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi{3þ2 cos(k �a1)þ2 cos(k �a2)þ2 cos[k �(a2�a1)]}

p

4) The guiding principles can be listed as follows.

1) Combining atomic orbitals must have the same symmetry about the inter-nuclear axis.

APPENDIX 3572

2) The strength of the interactions generally decreases in going from s to p to d

symmetry.

3) Orbitals of very different energies give small interactions.

4) The many-electron wave function in a crystal forms a basis for some irredu-cible representation of the space group. This means that the wave function,with a wave vector k, is left invariant under the symmetry elements of thecrystal class (e.g. translations, rotations, reflections) or transformed into anew wave function with the same wave vector k.

CHAPTER 6

4) None. From Eq. 6.6, it is seen that the conductivity tensor is isotropic in the abplane perpendicular to the c axis.

7) Using the values given for ktot and s in the relation ktotal ¼ kph þ selLT, kph canbe readily calculated as:

kph ¼ ktotal � selLT

kph ¼ 2:2� 40,000(0:0000000165)(298) ¼ 2:0 W m�1 K�1

CHAPTER 8

1) 16 terms, 252 microstates.

7) All cases have 1808 M–X–M angles. Hence, the sign of the superexchangecomes directly from the middle portion of Table 8.5. The Mn4þ–O–Mn3þ

superexchange interaction gives a d3 and a d4 cation, and the sign is positive(ferromaganetic). The Fe3þ–O–Fe3þ superexchange interaction, gives two d5

cations, so the sign is negative (antiferromagnetic). The Co3þ–O–Co4þ

superexchange interaction gives a d6 and a d5 ion and the sign is negative(antiferromagnetic).

CHAPTER 10

2) EVRH ¼ 201.1; ER ¼ 397.1; GVRH ¼ 78.7; BVRH ¼ 150.6.

5) Following Example 10.3 in the text:

50,000N ¼ F ¼ sA0 ¼ sd0

2

� 2

p ¼ (97,000� 106)1x

0:34

� 0:02 m

2

� 2

p

1x ¼ 5:57� 10�4 mm

1z ¼0:000557

0:34

� ¼ 0:00164 mm

8) 18.4 MPa.

APPENDIX 3 573