The SCIENCE and ENGINEERING of MA TERIALS Second SI edition

Donald R. Askeland

Solutions manual Solutions supplied by Paul Porgess and Ian Brown Department of Polymers, Metals and Dental Technology, Manchester Polytechnic

Chapman & Hall LONDON NEW YORK. TOKYO. MELBOURNE. MADRAS

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Chapman & Hall, 2-6 Boundary Row, London SEI 8HN Chapman & Hall, 29 West 35th Street, New York NYlOOOI Chapman & Hall Japan, Thomson Publishing Japan, Hirakawacho Nemoto Building, 7F, 1-7-11 Hirakawa-cho, Chiyoda-ku, Tokyo 102 Chapman & Hall Australia, Thomas Nelson Australia, 102 Dodds Street, South Melbourne, Victoria 3205 Chapman & Hall India, R. Seshadri, 32 Second Main Road, CIT East, Madras 600 035

First edition 1991

1991 Chapman & Hall

The Ipswich Book Company

ISBN-13:978-0-412-39600-7 e-ISBN-13:978-94-009- 1842-9 DOl: 10.1 007/978-94-009- I 842-9

All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, or stored in any retrieval system of any nature, without the written permission of the copyright holder and the publisher, application for which shall be made to the publisher. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made.

British Library Cataloguing in Publication Data available

Chapter 1:

Chapter 2:

Chapter 3:

Chapter 4:

Chapter 5:

Chapter 6:

Chapter 7:

Chapter 8:

Chapter 9:

Chapter 10:

Chapter 11:

Chapter 12:

Chapter 13:

Chapter 14:

Chapter 15:

Chapter 16:

Chapter 17:

Chapter 18:

Chapter 19:

Chapter 20:

Chapter 21

TABLE OF CONTENTS

Solutions to Practice Problems

Introduction to Materials

Atomic Structure

Atomic Arrangement

Imperfections in the Atomic Arrangement

Atom Movement in Materials

Mechanical Testing and Properties

Deformation, Strain Hardening, and Annealing

Solidification and Grain Size Strengthening

Solidification and Solid Solution Strengthening

Solidification and Dispersion Strengthening

Dispersion Strengthening by Phase Transformation and Heat Treatment

Ferrous Alloys

Non-Ferrous Alloys

Ceramic Materials

Polymers

Composite Materials

Electrical Conductivity

Dielectric and Magnetic Properties

Optical and Thermal Properties

Corrosion and Wear

Failure - Origin, Detection, and Prevention

1

5

11

37

50

63

81

92

106

116

126

135

151

160

174

191

204

218

229

247

259

Chapter 1

INTRODUCTION TO MATERIALS

1. The Stealth aircraft is designed so that it will not be detected by radar. What physical property should the materials used in the Stealth plane possess to meet this design requirement?

The materials should be capable of absorbing radiation having the wavelength of radar. In addition, the materials must also have a good strength-to-weight ratio, appropriate corrosion resistance, and other properties that typical aircraft should possess.

2. Certain materials, such as tungsten carbide, are compounds consisting of both metallic and non-metallic elements. To which category of "materials does tungsten carbide belong?

Ceramic materials.

3. From which category of materials would you select a material best suited for building a vessel to contain liquid steel?

Ceramic materials, which often have high melting temperatures and chemical resistance to liquid metals. However, the ceramic must be chosen with care, because many ceramics will react with the metal.

4. Consider a cement wall reinforced with steel bars. Into which category of materials would you place this reinforced concrete?

This is a composite material.

5. Boron nitride (BN) and silicon carbide (SiC) are important materials in abrasive grinding wheels. In what category of materials do BN and SiC belong? BN and SiC are in the form of small particles and "are often embedded in a polymer to produce the grinding wheel. In what category of materials does the entire wheel belong?

The BN and SiC are ceramic materials; because the ceramic particles are embedded in a polymer, the entire wheel is a composite material.

6. Silicon carbide (SiC) fibers are sometimes mixed with liquid aluminum. After the mixture freezes, a fiber-reinforced composite results. Would you reinforce aluminum with high strength polyethylene fibers in the same manner? Explain your answer.

No. The polyethylene fibers are a polymer material and consequently have low melting and degradation temperatures. Introducing liquid aluminum at a temperature above 6000 C will destroy the polymer fibers.

1

7. The nose of the space shuttle is composed of graphite (carbon). Based on this information, what type of material would graphite be?

The nose of the shuttle will experience extremely high temperatures upon re-entering the atmosphere from orbit. Protection therefore requires a very high melting point material, such as a ceramic. Graphite is sometimes considered to be a ceramic, although it is not a combination of metallic and non-metallic elements.

8. Suppose we would like to make a porous metal filter to keep the oil in our automobile engine clean. Which one of the metal processing techniques listed in Table 1-3 might be used to produce these filters?

Powder metallurgy might be an excellent choice. We can compact spherical metal powder particles to a small degree and sinter just long enough so the powder particles are joined together. This will leave interconnected voids between the particles that will allow liquid oil to penetrate but will trap small solid impurities.

9. Sintering is listed in Table 1-3 as a ceramic processing technique. In which one of the metal processing techniques would you expect sintering also to be used?

Sintering is an integral portion of the powder metallurgy process, joining the powder particles together, reducing void space, increasing density, and providing good mechanical properties after the powder metallurgy part has been initially formed by compaction.

10. Which of the three ceramic processing methods mentioned in Table 1-3 do you think is used to produce glass bottles?

The bottles are normally produced by a "compaction" process, in which the glob of hot, viscous glass is introduced into a die and then formed, often using gas pressure.

11. By which one of the four methods of producing composite materials listed in Table 1-3 would you expect plywood to be made?

Plywood is produced by "joining"; the individual plies are joined by adhesive bonding, or glueing.

12. Injection molding to produce plastic parts most closely resembles which one of the metals processing methods?

Injection molding is very similar to die casting, in which pressure is used to force a molten material into a metal die to give the desired shape.

13. The Voyager is an experimental aircraft that flew around the world non-stop on a single tank of fuel. What type of material do you think made up most of the aircraft? Explain why this type of material was selected.

The Voyager was produced primarily from composite materials, including carbon fiber-epoxy and fiberglass materials. These materials provided both the exceptionally light weight and the high strength and stiffness required.

2

14. United Stages coinage, such as the quarter, appear silvery on the face, but close inspection reveals a reddish color at the edges. Based on your observations, to which one of the five categories of materials should a quarter belong? Explain.

The coinage is a composite material composed of high nickel sheets on the two flat surfaces surrounding a core sheet of high copper. the high nickel provides good corrosion resistance and the appropriate silvery appearance, while the high copper in the core minimizes cost. The edge appears reddish because when the coins are stamped from the original sheet, the copper core is partly revealed.

15. Relays in electrical circuits open and close frequently, causing the electrical contacts to wear. MgO is a very hard, wear resistant material. Why would this material not be suitable for use as the contacts in a relay?

The MgO is a ceramic material and consequently acts as an electrical insulator rather than as a conductor. Although it may not wear, it also will never allow current to flow through the relay. Also, if the relay closes very rapidly, the brittle MgO could fracture; however this latter point is not very important compared to the electrical properties.

16. What mechanical properties would you consider most important when selecting a material to serve as a spring for an automobile suspension? Explain.

The spring must have a high strength in order to support the automobile; it must have a high modulus of elasticity so little elastic deformation occurs; it must have sufficient ductility so that it can be formed in the first place; it should have reasonably good resistance to corrosion, particularly to salt that might be picked up from the highway during the winter.

17. The devices used for memory in personal computers typically contain an integrated circuit, electrical leads, a strong, non-conducting base, and an insulating coating. From what material should each of these four basic components be made? Explain your selections.

The heart of the integrated circuit should be a semiconducting material such as silicon or GaAs so that electrical signals can be properly processed and information can be stored. The electrical leads must have a high electrical conductivity and might be made of aluminum or gold. Because the base must be both strong and non-conducting, it should be made from a ceramic material. Finally, the insulating coating could be made from either ceramic or polymer material. Polymers are most often used.

3

18. Automobile bumpers might be made from a polymer material. recommend a thermoplastic of a thermosetting polymer application? Explain.

Would you for this

A thermosetting polymer, due to its network structure, is expected to be very brittle; even slight impacts of the bumper against another car, the end of the garage, or flying rock or gravel could cause it to break. The thermoplastic polymer has better ductility and impact resistance and consequently would be the better choice.

19. Sometimes a nearly finished part is coined. During coining, a force is applied to deform the part into its final shape. For which of the following could this be done without danger of breaking the part - brass, Al203 , thermoplastic polymers, thermosetting polymers, silicon?

In order to be coined, the material must possess at least some ductili ty. Of the materials mentioned, Al203 is a ceramic and is very brittle, thermosetting polymers are brittle, and silicon is brittle; none of these could easily be coined without a danger of introducing cracks or even. fracture. Both brass and thermoplastic polymers have good ductility and can be deformed.

20. A scrap metal processor would like to be able to identify different materials quickly, without resorting to chemical analysis or lengthy testing. Describe some possible testing techniques based on the physical properties of materials.

He could separate copper-base alloys from other metals by color -copper, brass, and bronze are yellow or red. A magnet could be used to identify most iron and steel alloys - with only a few exceptions these are magnetic while most other common alloys that a scrap yard might encounter are not. Austenitic stainless steels could be separated from other stainless steels by the same method. The weight or density might also be used; aluminum and magnesium are lightweight compared to iron, copper, or nickel. Chapter 21 will also describe a variety of non-destructive tests, some of which might be easily adapted by a scrap metal processor.

4

Chapter 2

ATOMIC STRUCTURE

1. Silicon, which has an atomic number of 14, is composed of three isotopes: 92.21% of the Si atoms contain 14 neutrons, 4.7% contain 15 neutrons and 3.09% contain 16 neutrons. Estimate the atomic mass of silicon.

MSi (0.9221)(14 + 14) + (0.047)(15 + 14) + (0.0309)(16 + 14)

28.1099 g/mol

2. Titanium, which has an atomic number of 22, is composed of five isotopes: 7.93% of the Ti atoms contain 24 neutrons, 7.28% contain 25 neutrons, 73.94% contain 26 neutrons, 5.51% contain 27 neutrons and 5.34% contain 28 neutrons. Estimate the atomic mass of titanium.

MTi = (0.0793)(24 + 22) + (0.0728)(25 + 22) + (0.7394)(26 + 22) + (0.0551)(27 + 22) + (0.0534)(28 + 22) 47.9305 g/mol

3. Bromine, which has an atomic number of 35 and an atomic mass of 79.909 g/mol, contains two isotopes - Br79and Br8l. Determine the percentage of each isotope of bromine.

Let "x" represent the fraction of the Br79 isotopes and "x-l" represent the fraction of the Br81 isotopes. Then

79.909 (x)(79) + (1 - x)(81) = 79x + 81 - 81x x = (81 - 79.909)/(81 - 79) = 0.5455

Therefore Br contains 54.55% Br79 and 45.45% Br8l.

4. Silver, which has an atomic number of 47 and an atomic mass of 107.87 g/mol, contains two isotopes - Agl07 and Agl09 . Determine the percentage of each isotope of silver.

Let "x" represent the fraction of AgI07 isotopes and "x-1" represent the fraction of Agl09 isotopes. Then

107.87 (x)(107) + (1 - x)(109) = 107x + 109 - 109x x = (109 - 107.87)/(109 - 107) = 0.565

Therefore Ag contains 56.5% Agl07 and 43.5% Agl09 .

5

5. Tin, with an atomic number of 50, has all of its inner energy levels filled except the 4f level, which is empty. From its electronic structure, determine the expected valence of tin.

First let's sum the electrons in the first four energy shells:

lS2 = 2 electrons 2s22p6 = 8 electrons

3s23p63dl0 18 electrons 4s24p64dl04fO 18 electrons

46 electrons There must be SO - 46 = 4 electrons in the outer energy shell:

5s2Sp2= 4 electrons = valence of tin

6. Mercury, with an atomic number of 80, has all of its inner energy levels filled except the Sf and 5g levels, which are empty. From its electronic structure, determine the expected valence of mercury.

First let's sum the electrons in the first five energy shells:

2 electrons 8 electrons

3s23p63dl0 = 18 electrons 4s24p64JO 4[4 32 electrons

5s25p6Sdl05f05g0 = 18 electrons 78 electrons

There must be 80 - 78 2 electrons in the outer energy shell:

2 electrons = valence of mercury

7. Calculate the number of atoms in 100 grams of silver. Assuming that all of the valence electrons can carry an electrical current, calculate the number of these charge carriers per 100 grams.

(a) The number of atoms in 100 grams of Ag can be calculated from the molecular weight 107.87 g/mol and Avogadro's number:

number of atoms = (100 g)(6.02 x 1023 atoms/mol) = 5.58 x 1023atoms 107.87 g/mol

(b) From Table 2-2, we expect silver to have a valence of 1. Therefore is all of the valence electrons can carry a current, the number of valence electrons equals the number of atoms in the 100 gram sample, or

number of charge carriers 5.58 X 1023 electrons

6

8. Suppose there are 8 x 1013 electrons in 100 grams of germanium that are free to move and carry an electrical current. (a) What fraction of the total valence electrons are free to move? (b) What fraction of the covalent bonds must be broken? (On average, there is one covalent bond per germanium atom and two electrons in each covalent bond. )

(a) First let's calculate the total number of valence electrons, using the molecular weight of 72.59 g/mole and the valence of germanium, which is 4.

(100 g)(6.02 x 1023 atoms/mol) number of atoms = ~--~~~~~~~~--------~ 72.59 g/mol 8.293 X 10

23

number of valence electrons = (4 electrons/atom) (8.293 = 3.7317 x 1024electrons

X 1023 atoms)

fraction that move 8 x 1013

= 2.41 X 10-11

(b)

3.317 X 1024

Because on the average there is one covalent bond per Ge atom, the number of covalent bonds in 100 g is 8.293 X 1023 Since each broken covalent bond frees two electrons, the number of broken bonds is half the number of free electrons, or 4 x 1013 bonds. The fraction of broken bonds is therefore

fraction = 4 X 1013 11

----------- = 4.82 x 10-8.293 X 1023

9. Compare the number of a toms in one gram of uranium wi th the number of atoms in one gram of boron. Then, using the densities of each (See Appendix A), calculate the number of atoms per cubic centimeter in uranium and boron.

(a)

(b)

The molecular weights of uranium and boron and 10.81 g/mol. In 1 gram of metal,

are 238.03 g/mol

(1 g)(6.02 x 1023 atoms/mol) U atoms = ~~~~~~~~~~~------~ 238.03 g/mol

B atoms (1 g)(6.02 x 1023 atoms/mol) 10.81 g/mol

2.53 X 1021

5.57 X 1022

The densities of uranium and boron are 19.05 Mg/m3 and 2.3 Mg/m3. The volumes of one gram of U and Bare

volume of U 1 g / 19.05 Mg/m3 = 5.25 x 10-8 3 m volume of B 1 g / 2.3 Mg/m3 = 4.35 x 10-8m3

U atoms/cm 3 (2.53 x 1021 atoms) / 0.0525 cm3 ) 4.82 B atoms/cm 3 (5.57 x 1022 atoms) / 0.435 cm3 ) 1. 28

7

x 1022 x 1023

10. Suppose you collect 5 x 1026 atoms of nickel. Calculate the mass in grams and volume in cubic centimeters represented by this number of atoms. See Appendix A for the density.

The molecular weight and density of 8.902 Mg/m .

mass (5 x 1026 atoms) (58. 71 g/moU

6.02 X 1023

nickel are 58.71 g/mol and

48.76 x 103 g

11. Calculate the volume in cubic centimeters occupied by one mol of gold. See Appendix A for the necessary data.

The mass of 1 mol of gold is equal to its atom\c weight, or 196.97 g. The density of gold is 19.302 Mg/m. The volume of one mol is therefore

12. Suppose you have 15 mols of iron. Calculate the number of grams and the volume in cubic centimeters occupied by the iron. See Appendix A for the necessary data.

The mass of 15 mols of iron, which has an atomic weight of 55.847 g/mol is

mass = (15 mols) (55. 847 g/mol) = 837.7 g The volume of 15 mols of iron which has a density of 7.87 Mg/m3, is

volume = (837.7 g) / (7.87 Mg/m3) (10-6 cm3/m3) = 106.4 cm3

13. A decorative steel item having a surface area of 93750 mm2 is plated with a layer of chromium 0.125 mm thick. Calculate the number of atoms required to produce the plating.

The number of atoms can be determined by calculating the volume of chromium (Cr), then calculating the mass of chromium from its density (7.19 Mg/m3 ) , and finally calculating the number of atoms from the atomic mass (51.996 g mol-i).

Volume area x thickness (93i~g cm2) (0'i~5)cm 11. 72 cm3

Mass volume x density = 84.26 g

mass Number of atoms = mass/mol x number of atoms/mol

( 84.26 g _1)(6.02 x 1023 atoms mol-i) = 9.76 x 1023 51. 996 g mol

8

14. Examine the elements in the IVB and VIIIB columns of the periodic table. As you go to a higher atomic number in each column (as from Ni to Pd to Pt), how does the melting temperature change? Would you expect this, based on the atomic structure?

IVB VB VIB VIIB VIIIB

Ti-1668 V-1900 Cr-1875 Mn-1244 Fe-1538 Co-1495 Ni-1453 Zr-1852 Nb-2468 Mo-2610 Tc-2200 Ru-2310 Rh-1963 Pd-1552 Hf-2227 Ra-2996 W-3410 Re-3180 Os-2700 Ir-2447 Pt-1769

For the elements listed, the melting temperature increases for each column as the atomic number increases.

15. Examine the elements in IA column of the periodic table. As you go to a higher atomic number, how does the melting temperature change? Would you expect this, based on the atomic structure? Is this behavior different from what was observed in the elements in Problem 14? Can you explain this difference?

H-gas Li-180.7 Na-97.8 K-63.2

Rb-38.9 Cs-28.6

The melting temperature decreases as the atomic number increases; this is opposite the behavior noted in Problem 14.

16. Determine the formulas of the compounds formed when each of the following metals react with oxygen. (a) calcium, (b) aluminum, (c) germanium, (d) potassium.

(a) Ca: ... 4s 2 0: ... 2s2p 4 ~ CaO (b) AI: ... 3s2p 1 0: ... 2s2p 4 Al 0 ~

2 3 (c) Ge: ... 4s2p 2 0: ... 2s2p 4 GeO ~

2 (d) K: ... 4s 1 0: ... 2s2p 4 ~ K20

17. Would you expect Al203 or aluminum to have the higher modulus of elasticity? Explain.

The ions in the ceramic Al203 are joined primarily by the particularly strong ionic bonds; aluminum atoms are joined by metallic bonding, which is normally less strong. We would expect a deeper energy trough in the alumina than in the aluminum, leading to a higher modulus of elasticity for Al203 . The actual values are

A1 203: 380 GPa aluminum: 69 GPa

9

18. Would you expect silicon or nickel to have the higher coefficient of thermal expansion? Explain.

Atoms in silicon are joined by covalent bonds, while atoms in nickel are joined by metallic bonds. We expect the covalent bonds to be stronger, leading to a deeper energy trough, a high modulus of elasticity, and a lower coefficient of thermal expansion in the silicon than in the nickel. the actual values are:

silicon: 3 x 10-6/ o C nickel: 13 x 10-6/ o C

19. The compound GaAs is an important semiconductor material in which the atoms are joined by mixed ionic-covalent bonding. What fraction of the bonding is ionic?

The electronegativity of Ga is about 1.8 and the electronegativity of As is about 2.2. From Equation 2-1, the fraction of bonding that is covalent is

fcovalent exp[(-0.25)(2.2 - 1.8)2] = exp [-0.04] = 0.961 The fraction of bonding that is ionic must be 0.039 or 3.9%.

20. The compound InP is an important semiconductor material in which the atoms are joined by mixed ionic-covalent boding. If the fraction of covalent bonding is found to be 0.914, estimate the electronegativity of indium. Does your calculated value compare well with what you might expect, based on Figure 2-3?

The electronegativity of P from Figure 2-3 is about 2.1. From Equation 2-1, letting E be the electronegativity of Indium,

2 fcovalent = 0.914 = exp[(-0.25)(2.1 - E) ] In(0.914) = -0.0899 = (-0.25)(2.1 - E)2 (2.1 - E)2= 0.3596

2.1 - E 0.6 E 1. 5

21. Would you expect bonding in the intermetallic compound Ca2Mg to be predominantly ionic or metallic? Explain.

The electronegativity of Ca is about 1. 1 and that of Mg is about 1. 3. The electronegativities are relatively the same, so we would expect that bonding might be predominantly metallic.

22. The electronegativities of both nickel and copper are 1. 8. Would you expect bonding in the intermetallic compound Ni Mg to be more or less metallic than in CuAI? Explain. 2

2

The electronegativities are Ni: 1.8, Mg: 1.3, Cu: 1.8, and AI: 1.5. There is a greater difference in electronegativies between Ni and Mg than there is between Cu and AI. Therefore we would expect bonding in CuAl2to be more metallic than in Ni 2Mg.

10

Chapter 3

ATOMIC ARRANGEMENT

1. How many lattice points are unique to the base-centered orthorhombic unit cell?

(1/8)(8 corners) + (1/2) (2 faces) = 2 points/cell 2. Why is there no base-centered tetragonal structure? Draw a lattice for

this structure, then determine what the actual unit cell is.

As the sketch indicates, the base centered tetragonal structure could be redrawn as a simple tetragonal structure.

3. Why is there no base-centered cubic structure? Draw a lattice for this structure, then determine what the actual unit cell is.

4.

As the sketch indicates, the base-centered cubic structure could be redrawn as a simple tetragonal structure.

A material has a cubic unit cell with one atom per lattice point. If a = 4.0786 A and r = 1. 442 A. determine the crystal structure. 0

For simple cubic, a = 2r = 2.884 ~ 4.0786 0

For BCC, a 4r/V3 = 3.33 ~ 4.0786 0

For FCC, a = 4r/Y2 = 4.079 9! 4.0786 Material is FCC! 0

5. A material has a cubic unit cell with one atom per lattice point. If a = 5.025 A and r = 2.176 A, determine the crystal structure.

o

For simple cubic, a = 2r 4.352 ~ 5.025 0

For BCC, a 4r/V3 5.025 5.025 Material is BCC! 0

For FCC, a 4r/Y2 6. 156 ~ 5.025

11

6. Using the atomic radius data in Appendix B, calculate the packing factor for crystalline polyethylene.

The lattice parameters for orthorhombic polyethylene are a = 7.41 A, b = 4.94 A and c = 2.55 A. The atomic radii are

o 0 0 rC = 0.77 A and r H= 0.46 A. From example 3-22, there are 8 H atoms and 4 C atoms per cell

(4 C atoms) [4n(0.77)3/31 + 8 H atoms) [4n(0.46)3/31 PF = (7.41}(4.94)(2.55) 0.117

The unusually low packing factor is due to the restrictions of covalent bonding and partly to the use of the particular atomic radii listed in the Appendix.

7. The density of lead is 11.36 Mg/m3 , its atomic mass is 207.19 g/mol and the crystal structure is FCC. Calculate (a) the lattice parameter and (b) the atomic radius for lead.

(a) Because lead is FCC, there are 4 atoms/unit cell and we can calculate the lattice parameter from the density equation

(b)

(4 atoms/cell) (207. 19 x 10-6 ) (a )3 (6.02 x 1023 )

p = 11.36

:. a o

o

121. 18 x 10-30

4.9485 X 10-10 m = 4.9485 A For FCC unit cells

r = iZ ao/4 = (iZ) (4.9485) /4= 1.749 A

8. The density of tantalum is 16.6 Mg/m3 , its atomic mass is 180.95 g/mol and the crystal structure is BCC. Calculate (a) the lattice parameter and (b) the atomic radius for tantalum.

(a) From the density equation, with 2 atoms per unit

p = 16.6 (2 atoms/cell) (180.95 x 10-6 Mg/mol) (a )3 (6.02 x 1023 )

o

(ao )3 = 36.215 x 10-30

:. a =3.3085 x 10- 10 m = 3.3085 A o

(b) For BCC unit cells r = 13 a /4 = (13) (3.3085)/4

o

12

1. 433 A

cell,

9. How many unit cells are present in a cubic centimeter of face-centered cubic nickel? The atomic radius of nickel is 1.243 A.

The lattice parameter and unit cell volume for FCC nickel are

a o

= 4r/V2 = (4)(1.243)/12 = 3.5163 A V = (a )3 = (3.5163)3 = 43.4768 A3 = 43.4768

o 3 -24

unit cells/cm = 1 / 43.4758 x 10 = 2.3 x

10. Calculate (a) the volume and (b) the mass of one million unit cells of body-centered cubic iron. The atomic radius of iron is 1.241 A.

The lattice parameter and unit cell volume for BCC iron are

a = 4r/13 = (4)(1.241)/13 = 2.866 A o

V = (a )3 = (2.866)3 = 23.541 A3 = 23.541 X 10-24 cm3 o

(a) The volume of one million unit cells is

(b) The mass of one million unit cells can be obtained from the density which is 7.87 Mg/m3.

mass = pV = (7.87 g/cm3)(23.541 x 10-18 cm3)

11. A material with a cubic structure has a density of 0.855 Mg/m3, an atomic mass of 39.09 g/mol, and a lattice parameter of 5.344 A. If one atom is located at each lattice point, determine the type of unit cell.

12.

We would like to find "x", the number of atoms per unit cell, using the density equation.

(x)(39.09 x 10-6 ) p = 0.855 = --~~~--------~---------(5.344 x 10-10 )3(6.02 x 1023 )

.. x = 2 a toms/unit ce 11

.. BCC structure

A material with a cubic structure has a density of 10.49 M$/m3, an atomic mass of 107.868 g/mol, and a lattice parameter of 4.0862 A. If one atom is located at each lattice point, determine the type of unit cell.

We would like to find "x", the number of atoms per unit cell, using the density equation.

-6 P = 10.49 = (x) (107.868 x 10 ) (4.0862 x 10-1)3(6.02 x 1023 )

.. x = 4 atoms/unit cell

.. FCC structure.

13

13. Antimony has a hexagonal unit cell with ao = 4.307 A and Co = 11.273 X. If its density is 6.697 Mg/m3 and its atomic mass is 121. 75 g/mol, calculate the number of atoms per cell.

v = a 2c cos30 = (4.307)2(11.273)cos30 =181.1 X3 = 181.1 x 10-30m3 o 0

We would like to find "x", the number of atoms per unit cell, using the density equation.

-6 P = 6.697 = (x) (121.75 x 10 ) (181.1 x 10-3)(6.02 x 1023 )

x = 6 atoms/unit cell

14. One of the forms of plutonium has a face-centered orthorhombic structure, with a = 3.159 X, b = 5.768 X, and c = 10.162 X. The density of Pu is 17.14 Mg/m3 and the atomic mass is 239.052 g/mol. Determine (a) the number of atoms per cell and (b) the number of atoms at each lattice point.

(a) From the density equation we can calculate the number of atoms per cell "x".

(x) (239.052 x 10-6 ) p = 17.14 = ----:.::..::.:.---=..:::..::..:..:....:..::.::.....::..::....=-:~------::-:,---(3.159) (5. 768) (10. 162) (10-30 ) (6. 023 x 1023 )

:. X = 8 a toms/unit ce 11

(b) There are 4 lattice points/unit cell in the face-centered orthorhombic crystal structure. Therefore there must be 2 atoms per lattice point.

15. Prasiodymium has a special hexagonal structure with 4 atoms per uni t cell; the lattice parameters are a = 3.6721 A and c = 11.8326 X, while

o the atomic radius is 1.8360 X. Calculate the packing factor of Pro

V (3.6721)3(11.8326)cos30 = 138.18 X3 unit cell

PF

Vatom 4nr3/3 = (4) (n) (1.8360)3/3 = 25.9243 X3 (4 atoms/cell) (25. 9243 X3/atom)

138.18 X3/ce11

14

0.75

16. Gadolinium has a HCP structure just below 12600 C with a = 3.6745 A and c = 5.825 A. Just above 12600 C, Gd transforms to a BCC structure with a = 4.06 A. Calculate the percent volume change when Gd cools from the BCC to the HCP structure. Does the metal expand or contract during cooling?

17.

Below 12600 C: 2 93 VHCP = (3.6745) (5.8525)cos30 = 68.4335 A Above 12600 C: 3 03 VBCC = (4.06) = 66.9234 A Both unit cells contain 2 atoms, so we can directly compare the two volumes.

%change 66.9234 - 68.4335 x 100 66.9234 -2.26% expansion

Lanthanum has a FCC structure just below 865C with a = 5.337 A, but has a BCC structure with a = 4.26 A just above 865C. Calculate the percent volume change when La heats from the FCC to the BCC structure. Does the metal expand or contract during heating?

Below 865C: Above 865C:

3 93 VFCC = (5.337) = 152.02 A 3 93 VBCC = (4.26) = 77.309 A

But the FCC structure contains 4 atoms/cell while the BCC structure contains only 2 atoms/cell. To compare the volume of equal numbers of atoms, we should use two BCC cells.

%change 152.02 - 2(77.309) x 100 152.02 -1.71% expansion

18. Lanthanum has a special HCP structure just below 325C and the FCC structure just above 3250 C. At 3250 C, the lattice parameters for the HCP structure are a = 3.779 A and c = 12.270 X; the lattice parameter for the FCC structure at this temperature is 5.303 X. Lanthanum has a density of 6.146 Mg/m3and an atomic mass of 138.9055 g/mol. (a) Calculate the number of atoms in the special HCP unit cell. (b) Calculate the percent volume change when the FCC form of La transforms to the HCP structure on cooling. Does the metal expand or contract during cooling?

(a) The volume of the special HCP structure is V (3.779)2(12.270)cos30 = 151.75 A3= 151.75 x 10-30 m3 p 6.146 Mg/m3 =

~ x = 4 atoms/cell

(b) Below 3250 C: Above 3250 C:

(x) (138. 9055 x 1(f6 Mg/moll (151.75 x 10-3)(6.02 x 1023 )

151. 75 X3 (5.303)3 = 149.13 X3

Both structures have 4 atoms/cell so the volumes can be compared directly.

%change = 149.13 - 151.75 x 100 149.13

15

-1.76% expansion

19. At 1450oC, thorium changes from one type of cubic unit cell to a different cubic cell, with a 0.5% decrease in volume during heating. Below 1450oC, the lattice parameter is 5.187 X while the lattice parameter of the higher temperature form is 4.11 X. What is the ratio between the number of atoms in the unit cell of the high-temperature from the the number of atoms in the unit cell of the low-temperature form of Th?

20.

v = (5.187)3 = 139.556 X3 low Vhigh= (4.11)3 = 69.4265 X3

We really do not need to consider the 0.5% volume difference. By inspection of the volumes of the unit cells, it is apparent that the high temperature form of thorium must contain half the number of atoms as the low temperature form.

atoms in high T form 1 atoms in low T form = 2

a-Mn has a cubic structure with a o

8.931A and a density of 7.47 Mg/m3 ~-Mn has a different cubic structure with ao = 6.326 AOand a density of 7. 26 ~/m3. -r-Mn has a tetragonal structure with a = 3.784 A and c = 9.40 X and a density of 7.21 Mg/m3 The atomic mass of manganese is 54.9380 g/mol. (a) Calculate the number of atoms in each of the three polymorphic forms of manganese. (b) Assuming that the radius of the Mn atom is 1.12 A in all three forms, calculate the packing factor for each of the three unit cells.

(a)

(b)

a-Mn: p = 7.47 Mg/m3 = (x)(54.938 x 10-6 ) (8.931 x 10-1)3(6.02 x 1023 )

:. x = 58

~-Mn: p 7.26 Mg/m3 (x) (54. 938 x 10-6 ) (6.326 x 10-1)3(6.02 x 1023 )

:. x = 20 (x)(54.938 x 10-6 )

-r-Mn: P 7.21 Mg/m3 (9.4)(3.784)2 X (10-3) (6. 02

:. x = 10

The packing factors are approximately

v = ~nr3 atom 3 4(n) (1. 12 x 10-1)3/3 5.8849 x 10-30m3

(58) (5. 8849 x 10-30 ) PF(a-Mn) = -'.....-....:....:---~-=---'-(8.931 x 10-1)3

16

0.479

x 102~)

20(5.8849 x 10-3 ) PF(/3-Mn) = = 0.465 (6.326 x 10-1)3

PF(7-Mn) (10)(5.8849 x 10-30 ) -'---'-'---'--'~"--'-:= = O. 437 (9.40)(3.784)2 X 10-30

21. Determine the Miller indices for the directions in the cubic unit cell shown in Figure 3-35

A: (1, 0, 0) - (0, 1, 1/2) B: (0, 112, 1) - (1, 1/2, 0) c: (1, 0, 1) - (0, 1, 0) D: (3/4, 1, 0) - (1, 0, 213)

= 1, -1, -1/2 = -1, 0, 1

1, -1, 1 = -1/4, 1, -2/3 =

[221] [101] [1111 [3 12 8]

22. Determine the Miller indices for the directions in the cubic unit cell shown in Figure 3-36.

A: 0, 112, 1) B: (0, 1/3, 1) C: (1/4, 1, 0) D: 0, 3/4, 0)

(0, 0, (0, 0,

1, 1) 1, 0) 0, 0) 1/4, 1)

1, -1/2, 0 -1, -2/3, 1 114, 1, 0

= 0, 1/2, -1

[210] [323] [140] [012]

23. Determine the Miller indices for the planes in the cubic unit cell shown in Figure 3-37.

A: x = -1, Y = 1/2, z 1 l/x -1 l/y = 2, liz 1

z = 2/3 liz = 3/2

(121)

B: x = "', l/x 0, l/x = 0,

y = "', l/y = 0, 1/y = 0, liz = 3 (003)

c: X = "', 1/x = 0, l/x = 0,

y = -2/3, z = -1 l/y = -3/2,l/z -1 l/y = -3, liz = -2 (032)

24. Determine the Miller indices for the planes in the cubic unit cell shown in figure 3-38.

A: x = -1, y = 3/4, z -1/2 l/x = -1, l/y = 4/3, liz -2 l/x = -3, l/y = 4. liz -6 (346)

B: x = 2, y = 1, z 1/4 l/x 1/2,l/y = 1, liz 4 l/x = 1, l/y = 2, liz 8 (128)

C: x = 01), Y = -1, z 114 l/x = 0, l/y = -1, liz 4 (014)

17

25. Determine the Miller indices for the directions in the hexagonal unit cell in Figure 3-39. using the three-digit system.

A: (1. o. 0) - (-1. -1. 0) 2. 1. a [210] B: (0. -1. 3/4) - (1. 1. 0) -1. -2. 3/4. -4. -8. 3 [483] c: (1. 1/2. 0) - (1. 1. II O. -112. -1 O. -1. -2 [012] D: (0. 1. 0) - (-1. -1. II 1. 2. -1 [121]

26. Determine the Miller indices for the directions in the hexagonal unit cell in Figure 3-40. using the three-digit system.

A: (0. -1. 1) - (1. 1. II B: (1. O. 1) - (0. o. 0) c: (0. 1. 112) - (0. -1. 0) D: (1. O. 0) - (112. 1. 0)

-1. -2. a 1. O. 1 O. 2. 112 112. -1. a

O. 4. 1 1. -2. a

[120] [lOll [04ll [120]

27. Determine the Miller-Bravais indices for the planes in the hexagonal unit cell in Figure 3-41.

28.

A: a

B:

C:

1 l/a

1

a 1

l/a 1

a 1

lIa 1

1

a =-1 2

lIa =-1 2

a 00 2

l/a a 2

a = 00 c = 1 3

l/a = a llc = 1 3

a -1 c = 3

l/a -1 llc -1 3

a -1/2 a c = 00 2 3

l/a -2 l/a lIc a 2 3

Determine the Mi ller-Bravai s indices for the planes cell in Figure 3-42.

A: a = 00 a = 1 a = -1 c = 1/3 1 2 3 l/a = a l/a l/a -1 llc = 3

1 2 3

B: a = 00 a a -1 c = 1/2 1 2 3

l/a a l/a 1/a -1 lIc 2 1 2 3

c: a = 00 a = 00 a 00 c = 2/3 1 2 3

lIa a l/a a l/a a lie 3/2 1 2 3

l/a a l/a a l/a a llc 3 1 2 3

18

(1101 )

(1011)

(1210)

in the hexagonal unit

(0113)

(0112)

(0003)

29. Sketch the following directions and planes within a cubic unit cell. z

a. [112] b. [310] c. [111] (111)

d. [101] e. [041] f. [203]

g. crOll h. (111) 1. (013)

J. (12ll k. (20ll l. (120) y

x

z z (201 )

/--=--+---7- y y x x

30. Sketch the following directions and planes within a cubic unit cell. 1 2

[123] b. [lID] [010] z 3'3,1 a. c.

d. [131] e. [1211 f. [134] g. (220) h. (301) i. (112) j. (011) k. (421) l. (141)

x

z z (301) (220) z

y y

x x

31. Draw the (111) plane and identify the six directions that lie in that plane in a cubic lattice.

z

~--+--+-y

x

19

[lOll

[011]

[lID]

[101]

[OIl]

[110]

32. Draw the (110) plane and identify the four directions of the form that lie in that plane in a cubic lattice.

z

x

y

[111] 611] [111] [111]

33. How many planes of the form {131} are found in a cubic system? Would you give the same answer if we used a tetragonal or orthorhombic system? Explain.

For the cubic system. there are 12 unique planes of the form {131}. or 24 planes if the negatives are included.

(13) (131) (311)

(113) (131) (311)

(13) (131) (311)

(113) (31) (311)

In the tetragonal system. planes of the form {131} would only include 2/3 of those listed above. giving 8 unique planes or 16 planes if the negatives are included.

(131) (311)

(131 ) (311 )

(131) (311)

(131) (311)

In the orthorhombic system. only 4 unique planes. or 8 if the negatives are included. belong to planes of the form {131}.

(31) (131 ) (31) (131)

34. How many planes of the form {123} are found in a cubic system? What are the indices of the planes of the form {123} in a tetragonal system?

In the cubic system. there are 24 planes of the form {123}. or 48 planes if the negatives are included.

(23) (123) (123) (123) (132) (132) (32) (132) (213) (213) (213) (213) (231) (231) (231 ) (231) (312) (312) (312) (312) (321) (321) (321) (321)

In the tetragonal system. there are 8 unique planes of the form {123}. or 16 planes if the negatives are included.

(123) (123) (23) (23) (213) (213) (213) (213)

20

35. How many directions of the form are found in a cubic system? Would you give the same answer if we used a tetragonal or orthorhombic system? Explain.

In a cubic system there are 48 directions of the form , including the following plus their negatives.

[123] [123] [123] [123] [132] [132] [132] [132] [213] [213] [213] [213] [231] [231] [231] [231] [312] [312] [312] [312] [321] [321] [321] [321]

In tetragonal systems, there are 16 directions of the form , including the following plus their negatives.

[123] [213]

[123] [213]

[123] [213]

In orthorhombic systems, there are 8 directions of the form , including the following plus their negatives.

[123] [123]

36. How many directions of the form are found in a cubic system? What are the indices of the directions of the form in a tetragonal system?

There are 24 directions of the form in cubic systems, including the following and their negatives.

[221] [122] [212]

[221] [122] [212]

[221] [122] [212]

[221] [122] [212]

In a tetragonal system, there are 8 directions of the form , including the following and their negatives.

[221] [221] [221] [221]

37. What are the indices of the planes of the form {412} in an orthorhombic system?

Planes of the form {412} in orthorhombic systems include the following 4 planes plus the 4 negatives.

(412) (412) 38. What are the indices of the directions of the form , in an

orthorhombic system? Directions of the form include the following 4 directions plus the 4 negatives.

[121] [121] [121]

21

39. Determine whether the [101] direction in a tetragonal unit cell with a c/a ratio of 1.5 is perpendicular to the (101). plane. If it is not perpendicular, calculate the angle between the direction and the plane.

The sketch shows the [101] direction and (101) plane in a tetragonal unit cell with a= 2 and c = 3 (giving a c/a ratio of 1.5). The second sketch shows the trace that the plane and direction make on the (010) face of the cell.

tan(et/2) et/2

CIt

1. 5/1. 0 56.30 112.60

1.5

Obviously the direction and plane are not perpendicular to one another, as they would be in a cubic system.

z

40. Determine whether the [110] direction in an orthorhombic unit cell with a = 3 A, b = 4 A, and c = 5 A is perpendicular to the (110) plane. If it is not perpendicular, calculate the angle between the direction and the plane.

The sketches show the plane and direction in the unit cell and a view of the (001) plane which contains the trace of the plane and direction.

tan(et/2) et/2

IX

2/1. 5 1. 3333 53.130 106.260

z

k--t--::::;r#-- Y

x

41. Draw the plane in a cubic system that passes through the coordinates I, I, 0; 0, I, 1; and 0, 0, 1. What are the Hiller indices of this plane?

z

r----I+-+-- Y

x

22

x l/x

1 Y 1 l/y

co Z o l/z

(101)

1 1

42. Draw the plane in a cubic system that passes through the coordinates 1, 1, 0; 0, 0, 1; and 0, 1, O. What are the Miller indices of this plane?

z

)-->',.--H'JIJ-- Y

x

x = m y = 1 z 1 l/x = ally = 1 liz 1

(all)

43. Draw the plane in a cubic system that passes through the coordinates 1, 0, 1; 1/2, 0, 1; and 1, 1/2, O. What are the Miller indices of the plane?

z

Jr-+-+---;f- y

x

x = m y = 1/2 z 1 l/x = ally = 2 liz 1

(021)

44. Draw the plane in a cubic system that passes through the coordinates

45.

1, 0, 0; 0, 0, 1; and 1/2, 1, 1/2. what are the Miller indices of this plane?

z

J----+hf-y

x

x l/x

1 y = m Z 1 lly a liz

(010)

1 1

In the four-digit system for finding the indices for a direction in HCP unit cells, is [110] equal to a [1120]? Show this by constructing the path from the tail to the head of the direction.

The sketch below shows the [110] Direction, which is actually the negative a axis. Let's let the origin be one of the p5ints on our [1120]direction,. If we start at the origin and move 1 lattice parameter in the a direction, 1 lattice parameter in the a direction, and -2 lattice parameters in th~ a direction, we have a second point on the d~rection. This point and the origin form a direction that also is the negative a direction and is identical in direction t5 the [110] direction.

23

c

+ I \-z \

\ ,

~ to]:[f t2.0]

46. In the four-digit system for finding the indices for a direction in HCP unit cells, is the [100] equal to a [2110]7 Show this by constructing the path from the tail to the head of the direction.

The [100] direction is the at direction. Let's let the origin be one point on the [2110] direction. We move 2 lattice parameters in the a l direction, -1 lattice parameters in the a2 direction, and -1 lattice parameter in the a 3 direction and produce a point 3 lattice parameters along the at direction. The [2110] direction also lies on the at axis and is identical to the [110] direction.

,

-I ,'.z r ---- #'

-I' ~OOl" c%.iiol

t

47. In the four-digit system for finding the indices for a direction in HCP unit cells, is the [011] equal to a [1213]7 Show this by constructing the path from the tail to the head of the direction.

The [011] direction is shown in the sketch and lies in the plane formed by the a2 and c axes. Let's let one point on the [1213] direction lie at the origin we move -1 lattice parameter in the a1 direction, 2 lattice parameters in the a2 direction, -1 lattice parameter in the a3 direction, and 3 lattice parameters in the c direction. This puts us at coordinates indicated by +3 in the a2 direction and +3 in the c direction and gives a direction that is identical to the [011].

,., 1

,

I

I }t--+:HlIIl-- - - _._l.. - --- - . :

-I \~

0..

24

48. Is the [1210] direction in an HCP unit cell perpendicular to the (1210) plane? Draw each and verify your answer.

First we need to construct the direction and plane in the unit cell. We let the origin be one point on [1210]. We move 1 lattice parameter in the a1 direction, -2 lattice parameters in the aa direction and 1 lattice parameter in the a3 direction, giving a second point that lies 3 lattice parameters in the negative aa direction. The [1210] direction is the same as the negative aa axis.

The intercepts for the plane will be a1 = 1; a = -1/2, a3 = 1; c = m. The plane shown in

a the sketch satisfies these four intercepts. It shol!.1d be aeparent by simple inspection that [1210] ~ (1210).

c

ct.

49. Calculate the linear density of a line in the [111] direction in (a) simple cubic, (b) body-centered cubic, and (cl face-centered cubic unit cells, assuming a lattice parameter of 4.0 A in each case.

(al In simple cubic, atoms are located at corners of the cube. If we start at the origin and move in the [111] direction, or body diagonal, we do not encounter another lattice point until coordinates 1, 1, 1.

repeat distance = iJa = (iJl(4 Al = 6.9282 x 10-8 cm o

linear density = l/repeat distance = 1.443 x 107 points/cm

(bl When we start at the origin of BCC and move in the [111] direction we encounter the body centered atom at 1/2. 1/2. 1/2.

repeat distance = iJa /2 = (iJl(4 Al/2 = 3.4641 x 10-8cm o

linear density = l/repeat distance = 2.887 x 107 points/cm

(cl In FCC, we again do not encounter a second lattice point until coordinates 1, 1, 1, just as in simple cubic.

repeat distance = 6.9282 x 10-8 cm

linear density = 1.443 x 107 points/cm

25

50. Calculate the linear density of a line in the [110] direction in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells, assuming a lattice parameter of 4.0 X in each case.

(a) As we move in a simple cubic structure from the origin along the [110], or face diagonal, we do not encounter another lattice point until coordinates 1, 1, O.

repeat distance = iZa = (iZ)(4 X) o

5.6569 X 10-8 cm

linear density = l/repeat distance 1.768 X 107 points/cm

(b) In BCC, the situation is identical to that in simple cubic along the [110] direction.

repeat distance = iZa = (iZ)(4 X) linear density = l/repeat distance

5.6569 X 10-8 cm

1.768 x 107 points/cm

(c) As we move along the [110] direction in FCC, we encounter another lattice point at 1/2, 1/2, o.

repeat distance = iZa/2 = (iZ)(4 X)/2 = 2.8284 x 10-8 cm linear denSity = l/repeat distance = 3.536 x 107 points/cm

51. Calculate the packing fraction in the [111] direction in (a) simple cubic (b) body-centered cubic, and (c) face-centered cubic unit cells. In

which, if any, of these structures is the [111] direction a close-packed direction?

(a) Simple cubic: There are two atomic radii along the [111] Length iJa = iJ(2r) = 3. 464r or PF = 2r / 3. 464r

o

(b) BCC: There are four atomic radii along the [111] Length = iJ(4r/iJ) = 4r or PF = 4r / 4r = 1.00

(c) FCC: There are two atomic radii along the [111]

0.577

Length = iJ(4riZ) 4.899r or PF = 2r / 4. 899r = 0.408 The [111] direction is a close packed direction in the BCC structure

52. Calculate the packing fraction in the [110] direction in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells. In which, if any, of these structures is the [110] direction a close-packed direction?

(a) Simple cubic: There are two atomic radii along the [110] Length iZa = iZ(2r) = 2. 828r or PF = 2r / 2. 828r

o

(b) BCC: There are two atomic radii along the [110]

26

0.707

Length = iZ(4r/iJ) = 3.266r or PF = 2r I 3.266r = 0.612 (c) FCC: There are four atomic radii along the [110]

Length = iZ(4riZ) = 4.r or PF = 4r I 4r = 1.00 The [111] direction is a close packed direction in the FCC structure

53. Calculate the packing fraction of a (111) plane in (a) a simple cubic, (b) a body-centered cubic, and (c) a face-centered cubic unit cell. In which, if any, of these structures is the (111) plane a close-packed plane?

Sketches of the atoms centered on the (111) plane in each unit cell are shown.

11

(a) Simple cubic: Only 1/6 of each corner atom is included in the plane within the cell.

(iZa 12)(iJa liZ) = 0.866a2 = 0.866(2r)2 = 3. 464r2 000

Aatoms = (3 corners) (1/6 atomlcorner)nr2 = nr2/2 2 2 PFSC = [nr 12] 1 3.464r = 0.453

(b) Body-Centred cubic: (Note that the (111) plane does not pass through the center atom! ) A = 0.866(4r/iJ) = 4.619r2 111

2 2 Aatoms = (3 corners) (1/6)nr = nr 12 PF = (nr2/2) I 4. 619r2 = 0.34

(c) Face-centered cubic: (Note that the (111) plane also bisects an atom along each of the three edges)

_'" 2 2 Alll = 0.866(4r/v2) = 6.928 r Aatoms = [(3 corners)(1/6) + (3 2 2 edges)(1/2)]nr = 2nr PF = 2nr2 I 6. 928r2 = 0.907

The (111) plane is a close packed plane in FCC unit cells.

27

54. Calculate the packing fraction of a (110) plane in (a) a simple cubic, (b) a body-centered cubic, and (c) a face-centered cubic unit cell. In which, if any, of these structures is the (110) plane a close-packed plane?

(a) Simple cubic ""fio.'

A = iZa Z = iZ(2r)z = 5. 657rZ 110 0

Z Z Aatoms = (4 corners)(1/4 atom/corner)nr = nr PF = nrZ / 5.657rZ = 0.555

(b) Body-centered cubic: A = iZa Z = iZ(4r~)Z = 7. 542rz

110 0

A = [(4 corners)(1/4) + 1 centre]nrz = nrz atoms

PF =2nrz / 7. 542rz = 0.833 (c) Face-centered cubic:

Z A110 = iZao = iZ(4r/iZ)z = 11.314rz

Aatoms = [(4 corners)(1/4) + (2 edges) (1/2)]nrZ 2nrZ

PF = 2nrZ / 11.314rZ = 0.555

55. Calculate the planar density on a (111) plane in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells assuming a lattice parameter in each case of 4.0 A.

Referring to the sketches in problem 53.

A = 0.866aZ = 0.866(4 x 10-8 )z = 13.856 x 10-16 111 0

Z Cia

(a) Simple cubic: The number of atoms located on the (111) plane is (3 corners)(1/6 atom/corner) = 0.5 atom.

PO = 0.5 atom / 13.586 x 10-16 = 3.61 X 1014 atoms/cmZ

(b) BCC: The number of atoms located on the (111) plane is (3 corners)(1/6 atom/corner) = 0.5 atom

PO = 0.5 atom / 13.856 x 10-16 = 3.61 X 1014 atoms/cmZ

(c) FCC: The number of atoms located on the (111) plane is (3' corners) (1/6 atom/corner) + (3 edges)(1/2 atom/edge) = 2 atoms

PO = 2 atoms / 13.856 x 10-16 = 14.43 x 1414 atoms/cmZ

28

56. Calculate the planar density of a (110) plane in (a) simple cubic, (b) body-centered cubic, and (c) face-centered cubic unit cells assuming a lattice parameter in each case of 4.0 A.

Referring to the sketches in problem 54:

A = .f2a'2. = .f2(4 x 10-8 )'2. = 22.627 x 10-16 cm'2. 110 0

(a) Simple cubic: There is one atom located on the (110 plane) PD = 1 atom / 22.627 x 10-16 = 4.42 X 101' atoms/cm'2.

(b) BCC: There are two atoms located on the (110) plane. PD = 2 atoms / 22.627 x 10-16 = 8.84 X 101' atoms/cm'2.

(c) FCC: There are two atoms located on the (110) plane. PD = 2 atoms / 22.627 x 10-16 = 8.84 X 101' atoms/cm'2.

57. Calculate the llnear densities in the [110] and [101] directions in a face-centered tetragonal unit cell with a = 4.0 A and 6 = 6.0 A.

(110): The [110] passes through a face-centered lattice point repeat distance = (1/2) (.f2)a = (1/2)(.f2)(4) = 2.828 x 10-8 LD = 1 / repeat distance = 3.54 x 107 points/cm

[101]: The [101] passes through a face-centered lattice point

repeat distance = (1/2)(..4'2. + 6'2.) = 3.6055 X 10-8 cm

LD = 1 / repeat distance = 2.774 x 107 points/cm " 58. Calculate the planar densities in the (110) and (101) planes in a

face-centered tetragonal unit cell with a = 4.0 A and c = 6.0 A. (110): There are (4 corners) (1/4) +(2 faces) (1/2) or two atoms

located on the (110) plane A110 = .f2ac = (.f2) (4) (6)

33.94 x 10-16 cm'2.

PD 2 atoms / 33.94 x 10-16 5.89 X 1014 atoms/cm'2.

"

(101): There are (4 corners)(1I4) + (2 faces)(1/2) or two atoms located on the (101) plane

A101 = a./a'2. + c'2. = (4) (/4'2. + 6'2. = 28.84 X 10-16 cm'2.

PD 2 atoms/ 28.84 x 10-16 = 6.93 X 1014 atoms/cm'2.

29

(101)

59. Calculate the linear densities in the [100], [010], [110] and [0011 directions of a base-centered orthorhombic unit cell with a = 3.0 A, b = 5.0 A, and c = B.O-A.

[100]: repeat distance = a = 3 x 10-8 cm

LD = l/repeat distance = 3.33 x 107 points/cm

[010]: repeat distance = b = 5 X 10-8 cm

LD = l/repeat distance = 2.0 x 107 points/cm

[110]: There are two points along the [110]

repeat distance (1/2)/a2 + b 2 = (1/2) ,h2 + 52 2.9155 X 10-8 cm

LD = l/repeat distance = 3.43 x 107 points/cm

[001]: repeat distance = c B X 10-8 cm

LD = l/repeat distance = 1.25 x 107 points/cm

f

60. Calculate the planar densities in the (100), (010), (110), and (00l) planes of a base-centered orthorhombic unit cell with a = 3.0 A , b = 5.0 A and c = B.O A.

(100): There is one atom located on the plane (only corners

(010):

(110)

contribute) \00 = bc = (5) (B) = 40 x 10-16 cm2

PO -16 1 atom /ti0 x 10 2 2.5 x 10 atoms/cm

There is one atom located on the plane (only corners contribute)

A010 = ac = (3) (B) = 24 x 10-16 cm2

1 atom / 24 x 10-16 PO 4.17 X 1014 atoms/cm2

!l~>.: .- ... 8 ,:, . '.,-' .

,', .

3

There are two atoms located on the plane (corners and half of each base centered atom contribute)

~ ~ A = cVa- + b- = (B)"'3- + 5-110

46.65 x 10-16 cm2

PO = 2 atoms / 46.65 x 10-16 = 4.29 X 1014 atoms/cm2

30

(001) : There are two atoms located on the plane (corners and one base centered atom contribute) AOO1 = ab = (3) (5) = 15 x 10-16 cm2

PD 2 atoms I 15 x 10-16 13.33 X 1014 atoms/cm2

61. Calculate the interplanar spacing between the following planes in gold (see Appendix A for the lattice parameter).

(a) (12i) (b) (201 (c) (li2) (d) (~21) The lattice parameter for gold is 4.0786 A

(a) d 4.0786 I ~2 + 22 + 42 0.890 A (b) d = 4.0786 I /z2 + 02 + 12 = 1. 824 A 201 (c) d 112 4.0786 I ~2 + 12 + 22 1. 665 A (d) d 4.0786 I /32 + 22 + 12 1. 090 A 321

62. The interplanar spacing between (231) planes is found to be 0.89 A. Calculate the lattice parameter if the material has a cubic crystal structure.

d 231 a I /z2 + 32 + 12 = 0.89 or a = 0.89V14 = 3.33 A

63. Show that the radius ratio for an atom or ion that just fits into a tetrahedral interstitial site without disturbing the surrounding atoms or ions is 0.225.

Let "R" be the radius of the normal atoms and "r" be the radius of the interstitial site. The atoms in a tetrahedron actually touch one another along the face diagonal. Along the body diagonal or the tetrahedral cube

2r + 2R = v'3a = v'3VZR = v'6R

2r v'6R - 2R = 0.4495R

rlR = 0.225

64. Show that the radius ratio for an atom or ion that just fits into a triangular interstitial site without disturbing the surrounding atoms or ions is o. 155

cos30 = R I r (r + R) 0.866 R = 0.866r + 0.866R

rlR = 0.134 I 0.866 0.155

31

65. List the coordinates for all six of the octahedral sites that lie in a BCC unit cell. How many of these sites belong uniquely to one BCC cell?

The six octahedral sites are the faces of the cube.

1/2, 1/2, 0 0, 112, 1/2 112, O. 112 1, 112, 1/2 1/2, 1, 112 112, 112, 1

Each is shared with another cell, so only three sites belong uniquely to one cell.

66. Using the ionic radii given in Appendix B, determine the coordination number expected for the following compounds.

67.

(a) FeO (b) CaO (c) SiC (d) PbS From Appendix B, we find the ionic radii.

(a) r Fe 0.74, rO 1. 32, rFe/rO = (b) rCa 0.99, rO .1. 32, rCa/rO = (c) r Si 0.42, rC 0.16, rC/rSi (d) rpb 0.84, rS 1. 84, rpb/rS (e) r B 0.23, rO 1. 32, rB/rO

(e)

0.56

0.75

0.38

0.46

0.17

BO 2 3

CN = 6

CN 8

CN 4

CN 6

CN 3

Using the ionic radii given in Appendix B, determine the number expected for the following compounds.

(a) Al 0 (b) 2 3

no (c) 2

MgO (d) SiO 2

(e) CuZn From Appendix B, we find the ionic radii:

(a) r AI 0.51, rO 1. 32, rAl/rO 0.39 CN 4 (b) rn 0.68, rO 1. 32, rn/rO 0.52 CN 6 (c) r Mg 0.66, rO 1. 32, rMlro 0.50 CN 6 (d) rCu 0.96, r = Zn 0.74, rcu/rZn 0.77 CN 8

coordination

68. Based on the ionic radius ratio and the necessity for charge balance, which of the cubic structures discussed in the text would you expect CdS to possess?

From the ionic radii,

rCd = 0.97, rs = 1.84, rCd/rS = 0.527 CN = 6

The two ions have equal but opposite charges, and with the coordination number of 6, a sodium chloride structure seems likely.

32

69. Based on the ionic radius ratio and the necessity for charge balance which of the cubic structures discussed in the text would you expect CoO to possess?

70.

From the ionic radii,

rCo = 0.72, rO = 1.32, rCo/rO = 0.545 CN = 6

The ions have equal but opposite charges and the coordination number is six, making the sodium chloride structure a possibility.

The compound NiO has the sodium chloride crystal structure. data in Appendix B, calculate (a) the lattice parameter, density, and (c) the packing factor for NiO.

The ionic radii are r Ni = 0.69 and rO = 1.32

(a) the ions touch along the [100] direction, so ao = 2rNi + 2rO = 2(0.69) + 2(1.32) = 4.02 A

Based on the (b) the

(b) MNi = 58.71 g/mol, MO = 16 g/mol, and there are 4 nickel and 4 oxygen ions per unit cell.

p = (4)(58.71) + (4)(16) = 7.64 g/cm3= 7.64 Mg/m3 (4. 02 x 10-8 )3(6. 02 x 1023 )

(cl PF

PF

3 3 4(4n/3)rNi + 4(4n/3)rO (4.02)3

4(4n/3) (0. 69)3 + 4(4n/3) (1. 32)3 (4.02)3

71. The compound Te02 has the fluorite structure. Determine (a) the number of tellurian and oxygen ions in each unit cell, (b) coordination number based on the radius ratio, (c) parameter, and (d) the packing factor for the compound.

the expected the lattice

(a) From inspection of the fluorite crystal structure and the formula of the compound, there must be 4 tellurium ions and 8 oxygen ions.

(b) The ionic radii are

r Te = 2.11, rO = 1.32, rO/rTe = 0.626 CN = 6

Note, however, that the expected coordination number is not achieved by the fluorite crystal structure due to the requirements of the charge balance.

33

72.

(c) The ions touch along the body diagonal. There are 4 tellurium ionic radii and 4 oxygen ionic radii along the body diagonal, al though not all of the positions are actually occupied by ions. Therefore

(d) PF

PF

v'3a o

4rTe + 4rO 4(2.11) + 4(1.32) = 13.72 13.72 I v'3 7.92 A

3 3 4(4nI3)rTe + 8(4nI3)ro (7.92)3

4(4nI3) (2. 11)3 + 8 (4nI3) (1. 32)3 (7.92)3

0.472

One of the forms of BeO has the zinc blende temperatures Determine (a) the lattice parameter, and (c) the packing factor for the compound.

structure at high (b) the density,

From Appendix B, r Be = 0.35 A, ro = 1.32 A, MBe = 9.01 g/mol, and MO = 16 g/mol.

(a) The ions touch along the body diagonal of the zinc blende unit cell. There are 4 beryllium and 4 oxygen ionic radii along this distance.

(b) p

(c) PF

PF

v'3a o

a o

4rBe + 4rO = 4(0.35) + 4(1.32) 6.68 I v'3 = 3.857 A

6.68 A

4(9.01) + 4(16) (3.857) x 10-6 )3(6.02 x 1023 )

3 4(4nI3)rBe3 + 4(4nI3)ro

2.896 g/cm3

(3.857)3 4(4nI3)(0.35)3 + 4(4nI3)(1.32)3

(3.857)3 0.684

73. Which of the cubic structures discussed in the text would you expect CsBr to possess, based on its expected coordination? Calculate (a) the lattice parameter, (b) the packing factor, and (c) the density of the compound.

o From Appendix B, rcs 1. 67 A and r Br 1. 96 A.

rCs/rBr = 1.67 I 1.96 = 0.852 CN 8

(a) The ions touch along the body diagonal. there are 2 Cs ionic radii and 2 Br ionic radii along this distance. v'3a

o

a o

2rCs + 2rBr = 2(1.67) + 2(1.96) = 7.26 7.26 I v'3 = 4.1917 A

34

(b) There is one Cs ion and one Br ion per unit cell. (4n/3)r Cs3 + (4n/3)r Br3

PF '" ---=-=------==-

PF

(4.1917)3 (4n/3)(1.67)3 + (4n/3)(1.96)3

(4.1917)3 0.693

132.91 g/mol and MBr 79.909 g/mol

p 132.91 + 79.909 '" 4.S g/cm3 4.S Mg/m3 (4.1917 x 10-8 )3(6.02 x 1023 )

74. Germanium has the diamond cubic structure with a lattice parameter of 5.6575 A. Calculate the size of the germanium atom in the unit cell. Does this best match up with germanium's atomic radius or ionic radius?

The atoms in diamond cubic touch along the body diagonal. there are S atomic radii (although not all are actually occupied by atoms) along this distance.

V3a '" Sr o

r = V3a /S = V3(5.6575) / S = 1.2249 A o

The atomic radius listed in Appendix B is 1. 225 A and the ionic radius is 0.53 A. It appears that we should use the atomic rather than the ionic radius.

75. Calculate the fraction of the [111] direction and the fraction of the (111) plane actually covered by atoms in the diamond cubic cell.

There is space along the [111] direction for S atomic radii. although only 2 atoms (4 atomic radii) are actually located along this distance. Therefore

fraction [111] = 4r/Sr = 0.50

The area of the (111) plane is 0.S66a 2 o

a = Sr / V3 o

A = O. S66(Sr / V3)2 = lS.4747r2 111

In diamond cubic,

Only the corners and face-centered atoms lie on the (111) plane: A = [(3 corners) (1/6) + (3 edges) (1/2)]nr2 = 2nr2

atoms

fraction (111) = 2nr2 / lS.4747r2 = 0.34

35

76. Calculate the fraction of the [111] direction and the fraction of the (111) plane actually covered by sodium ions in the NaCI unit cell.

Let's assume that the Na + ions are located at normal FCC lattice points. In the NaCI structure, ions touch along the edge of the cell, or

ao 2rNa + 2rCI = 2(0.97) + 2(1.81) = 5.56 X There are only two Na+ionic radii along the [111] direction of the unit cell.

fraction = 2r / V3a = 2(0.97) / V3(5.56) = 0.201 o

Note: The [111} also passes through a CI ions at the center of the cell so the total fraction covered by ions is

The area of the (111) plane' is

A = 0.866a 2 = 0.866(5.56)2 = 26.77 X2 111 0

~a ions [(3 corners)(1/6) + (3 edges) (1/2)]rrrNa2 2 92 2rrrNa2 = 2rr(0.97) = 5.912 A

fraction = 5.912 / 26.77 = 0.221

77. How many oxygen and silicon ions are present in cristobalite? Using the ionic radii in Appendix B, calculate (a) the lattice parameter, (b) the packing factor, and (c) the density of cristobalite.

Silicon ions are located at face centered positions and at four tetrahedral locations, giving 8 Si ions per cell. There must therefore be 16 0 ions per cell, since the formula is Si02.

(a) The ionic radii are rSi = 0.42 X and rO = 1.32 X. The ions touch along a body diagonal, with 8 silicon radii and 8 oxygen radii along that length.

(b)

(c)

V3a o

13.92 a = o

8.037 A

PF

= 8(rSi + rO) = 8(0.42 + 1.32) (8 Si ions) [4rr(0.42)3/3 ] + (160 ions) [4rr( 1. 32)3/3 ]

(8.037)3

PF = 156.628 / 519.1369 = 0.302 (8 Si ions)(28.08) + (16 0 ions) (16) p = ~------~~--~--~--------~~

(8.037 X 10-8 )3 (6.02 X 1023 )

36

1. 538 g/cm3

1. 538 Mg/m3

1.

Chapter 4

IMPERFECTIONS IN TIlE ATOMIC ARRANGEMENT

Calculate, using the data in Appendix A, the cadmium, rhenium and beryllium. In which expected to occur easily on the basal planes? to occur on the prismatic planes?

c/a ratios for magnesium, of these metals is slip In which is slip expected

Mg: c/a Cd: c/a Rh: c/a Be: c/a

5.209 / 3.2087 5.6181 / 2.9793 4.458 /2.760 3.5842/2.2858

1.623 basal plane slip 1.886 basal plane slip 1.615 prismatic plane slip 1.568 prismatic plane slip

2. Determine the Miller indices of the slip directions on the (111) plane in an FCC unit cell.

[101] [110] [011]

[101] [110] [011]

3. Determine the Miller indices of the slip directions on the (101) plane in a BCC unit cell.

z

4. Determine the Miller indices of the slip planes in FCC unit cells that include the (110) slip direction.

x

37

(111) (111)

5. Determine the Miller indices of the {110} slip planes in Bee unit cells that include the [111] slip direction.

(110) (101) (011)

(110) (101) (011)

6. Does the [111] slip direction lie in the (112) slip plane in the BCC unit cell? Show by suitable drawings .

. ~, Yes Note that the direction drawn in the sketch is: 0, 0, 1/2, -1/2, 112, 0 = -1/2, -1/2, 112 or -I, -I, I, = [111] 7. Does the [111]slip direction lie in the (123) slip plane in the Bee unit

cell? Show by suitable drawings. 1:

No

Simple inspection reveals that the direction does not lie in the plane.

8. Using the three-digit form for indices in the HCP unit cell, determine the slip directions the (0111) slip plane.

[100] [100]

9. Using the thr~e-digit form for indices in the HCP unit cell, determine the slip directions in the (1010) slip plane.

[010] [010]

38

10. Verify that \he planar density of the {112} planes in BCC iron is 9.94 x 10 atomslcm, as used in Example 4-3.

The (112) plane is shown in the cubic cell and is also isolated to better illustrate the dimensions and atom locations. The three corners of tpe plane contribute a total of 1/2 atom to the plane; one edge contributes 1/2 atoms to the plane; and the center contributes 1 atom to the plane.

atoms = 1/2 + 1/2 + 1 = 2 atoms.

A = (112) (2v'2a Hv'3a ) = v'6a 2 112 0 0 0

a = 2.866 A o

A = v'6(2.866 x 10-1)2 = 2. 012 X 10-19 m = 2. 012 X 10-15 cm2 112

PD = 2 atoms I 2. 012 X 10-15 cm2 = 9.94 X 1014 atoms/cm2 112

11. Determine the length of the Burger's vector in BCC tungsten.

3.1652 A The Burger's vector is the repeat distance rd111 along the direction, which is half of the body diagonal.

b rd = (1/2)v'3a = (1/2) (v'3) (3.1652 x 10-10) 111 0 b 2.7411 X 10-10 m

12. Compare the planar density and interplanar spacings for the Ull} and {1l0} planes in FCC aluminum. Based on your calculations, on which planes would slip occur?

a o

= 4.04958 A For (110): atoms = (4 corners)(1/4) + (2 edges) (1/2) 2

A = (v'2a )( (a ) = v'2a 2 110 0 0 0

PD = 2 I v'2a 2 = 2 I v'2(4. 04958 x 10-1)2 = 8.625 X 1018 pOints/m2 110 0

For (111) atoms = (3 corners) (1/6) + (3 edges) (1/2) = 2 A (1/2)(v'3a 1v'2)(v'2a) 0.866a 2

111 0 0 0

39

PD = 2 / 0.866a 2 = 2 / (0.866) (4. 04958) x 10-1)2 111 0

= 14.083 X 1018 points/m2

Slip will occur on the (111) planes due to the higher planar density.

13. When BCC iron is in the softest condition, the dislocation density is about 106cm/cm3; large amounts of deformation of the iron increase the dislocation density to 1012cm/cm3. How many grams of iron are necessary to produce 1600 kilometers of dislocation in (a) soft iron and (b) deformed iron?

Firstly, find the volume of iron containing the length of dislocation, and hence the mass of iron.

length of dislocation 1 = 1600 km = 1600 x 1000 m = 1600 x 1000 x 100 cm = 1.6 x 108 cm

density of iron, p = 7.87 Mg m3 = 7.87 g cm-3

(a) mass [1.6 X 108 cm/106 cm cm-3] [7.87 g cm-3] 1259 g

(b) mass [1.6 x 108 cm/1012 cm cm-3] [7.87 g cm-3] = 0.00126 g

14. The dislocation density in an aluminum sample is found to be 5 x 107 cm/cm3. Calculate the total length of dislocations in 100 g of aluminum (see Appendix A for the necessary data). How many kilometers of dislocation are present in the sample?

p 2.669 g/cm3

1 (100 g)(5 x 107 cm cm3 ) 2.669 g/cm3

18730 km

1. 87 X 109 cm

15. The circumference of the earth is roughly 38,600 kilometers. If dislocations totalling this length were placed into one cubic centimeter, what would be the dislocation density?

Dislocation density = length of dislocations volume

(38.600 km)(1000 m/km) (100 cm/m) 1 cm3

3.86 X 109 cm/cm3

40

16. Suppose that a single crystal of an FCC metal is oriented so that the [0011 direction is parallel to an applied stress of 20 HPa. Calculate the resolved shear stress acting on the (111) slip plane in the [110], [011] and [101] slip directions. Which slip system(s) will become active first?

~ = 54. 76cos ~ = 0.577 T = ~ cos ~ cos A [110]: A = 90, cos A = 0

T = (20)(0)(0.577) = 0 HPa [011]: A 45, cos A = 0.707

T = (20)(0.707)(0.577) = 8.16 HPa [101]: A 45, cos A = 0.707

T = (20)(0.707)(0.577) = 8.16 HPa Slip begins first on [011] and [101]; no slip occurs on [110].

17. Suppose that a single crystal of an FCC metal is oriented so that the [001] direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 1.5 HPa, calculate the magnitude of the _appliec!. stress reguired to cause slip to begin on the (111) plane in [110], [011] and [101] slip directions.

~ = 54.76 cos ~ = 0.577 ~ = T/cos ~ cos A [lio]: A = 90 cos A = 0

~ = 1.5/(0)(0.577) = ~ HPa

[011]: A = 45, cos A = 0.707 ~ = 1.5/(0.707)(0.577) 3.67 HPa

[101]: A = 45, cos A = 0.707 ~ = 1.5/(0.707)(0.577) = 3.67 HPa

Slip never occurs in the [110] direction.

18. Suppose that a single crystal of a BCC metal is oriented so that the [001] is parallel to an applied stress of 80 HPa. Calculate the resolved shear stress acting on the (110), (011) and (101) planes in the [111] slip directions. Which slip system(s) will become active first?

A = 54.76, cos A = 0.577 T = ~ cos ~ cos A (110) ~ = 90, cos ~ = 0

T = (80)(0.577)(0) = 0 HPa

(010): ~ = 45, cos ~ = 0.707 T = (80)(0.577)(0.707) 32.6 HPa

(101): ~ = 45, cos ~ = 0.707 T = (80)(0.577)(0.707) = 32.6 HPa

Slip begins first on the (011) and (101) slip planes.

41

19. Suppose that a single crystal of a BCC metal is oriented so that the [001] direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 58.6 MPa, calculate _the magnitude of the applied stress to cause slip to begin in the [1111 direction on the (110), (Olll and (10ll slip planes.

A = '54.76 cos A = 0.577 u = cos ~ cos A (11)): ~ = 90, cos ~ = 0,

u = 58.6/(0.577)(0) = m MPa

(011): ~ = 45, cos ~ = 0.707, u =-58.6/(0.577)(0.707)

(101): ~ = 45, cos ~ = 0.707, 143.6 MPa

u = 58.6/(0.577)(0.707) = 143.6 MPa 20. Suppose that a single crystal of an FCC metal is oriented so that the

[0011 direction is parallel to an applied stress. When the applied stress is 48 MPa, a dislocation on the (111) plane just begins to move in the [011]direction. Calculate the critical resolved shear stress in this material. Based on this result, do you suspect that there are numerous lattice imperfections in the metal?

cos ~ cos A

0.577 0.707

T = TCRSS U cos ~ cos A = 48(0.577)(0.707) = 19.6 MPa

Because TCRSS 0.34 - 0.69 MPa

there must be many lattice imperfections which make it more difficult for the dislocations to move.

21. Suppose that a single crystal of a BCC metal is oriented so that the [001] direction is parallel to an applied stress. When the applied stress is 93 MPa, a dislocation on the (101) plane begins to move in the [111] direction. Calculate the critical resolved shear stress in this material. Based on this result, do you suspect that there are numerous lattice imperfections in the metal?

cos A cos ~

0.577 0.707

T = TCRSS U cos ~ cos A 93(0.707)(0.577) = 37.9 MPa

This is a typical value for the critical resolved shear stress in BCC metals; there are probably few lattice imperfections that are capable of interfering with slip.

42

22. Suppose a single crystal of a hexagonal~close packed metal is oriented so that the [001] direction is parallel to an applied stress. Can slip occur in the basal plane? Can slip occur in the prismatic planes such as (1010)? Explain your answer to both questions.

In the basal plane. A = 90; in the prismatic planes. + = 90. Consequently slip will not occur in either plane.

23. FCC aluminum has a density of 2.695 Mg/m3 and a lattice parameter of 4.04958 A. Calculate (a) the fraction of the lattice points that contain vacancies and (b) the. total number of vacancies in a cubic centimeter of aluminum.

Using the density equation. 2.695 x 106 /m3 = (x atoms/cell) (26.981 g/mol)

g (4.04958 x 10-10 m)3(6.02 x 1023 atoms/mol)

x = 3.9933 Al atoms/cell

vacancies/cell = 4 - 3.9933 = 0.0067 vacancies/cell

(a)

(b)

0.0067 vacancies/cell fraction = 4.0000 lattice points/cell = 0.001675

vacancies/cm3 = (0.0067 vacancies/cell) =.1 x 1020 vacancies/cm3 (4.04958 x 10- 8 cm)3

24. HCP magnesium has a density of 1.735 Mg/m3 and lattice parameters of a = 3.2087 A and c = 5.209 A . Calculate (a) the average number of

o 0 atoms per lattice point in the unit cell and (b) the total number of vacancies in a cubic centimeter of magnesium.

Using the density equation: 1.735 x 106 = (x atoms/cell) (24.312 g/mol)

(46.4455 x 10-29m3)(6.02 x 1023 atoms/mol) x = 1.99535 Mg atoms/cell

(a) average Mg atoms/lattice point = 1.99535 / 2 0.99767

(b) vacancies/cell = 2 - 1.99535 = 0.00465 vacancies/cm3 = ____ 0_._0_0_4_6_5 _____ = 1 x 1026 m3

4.64455 x 10-24m3

43

25. The lattice parameter of BCC caesium is 6.13 A. from one out of 800 unit cells, calculate (a) the cubic centimeter and (b) the density of caesium.

(a) vacancies/cm3 = (1 vacancy / 8000 cells) (6.13 x 1O-Bcm )3

If one atom is missing number of vacancies per

5.43 X 101B

(b) In 800 cells of a BCC metal, there are 1600 lattice points. Since one atom is missing in 800 cells, the average number of Cs atoms per cell is

(1599 / 1600)(2 lattice points/cell) = 1.99875 Cs atoms/cell p = ~9875 atoms/cell) (132.91 g/mol) 1.9157 g/cm3

(6.0849 x 10- B cm)3(6.02 x 1023 atoms/mol)

26. The lattice parameter of FCC strontium is 6.0849 A. If one atom is missing for each 1,500 strontium atoms, calculate (a) the density of strontium and (b) the number of vacancies per gram.

(a) The number of Sr atoms/cell is

(b)

(1499/1500)(4 lattice points/cell) = 3.99733 Sr atoms/cell

= (3.99733 atoms/cell) (87.62 g/mol) = 2.582 M /m3 p -B 3 23 g (6.0849 x 10 cm) (6.02 x 10 atoms/mol)

vacancies/m3 = (1 vacancy/1500 atoms) (4 atoms/cell) (6.0849 x 10- 10m)3

vacancies/gram

0.11836 x 1026

(0.11836 X 1026 vacancies/m3 ) 2.582 x 106 g/m3

4.552 X 101B

27. A BCC alloy of tungsten containing substitutional atoms of vanadium has a density of 16.912 Mg/m3 with a lattice parameter of 3.1378 A. Calculate the fraction of vanadium atoms in the alloy

Let Fy be the fraction of Y atoms; then (1 - fy) is the fraction of tungsten atoms. The molecular weight of Y is 50.941 g/mol and that of W is 183.85 g/mol. From the density equation

6 3 (2 atoms/cell)[(~ )(50.941) + (1 - fy )(183.85) 16.912 x 10 g/m

(3.1378 x 10-10 m)3(6.02 x 1023 atoms/mol)

fy 0.2

44

28. An FCC alloy of co~per containing substitutional atoms of tin has a density of 7.17 Mg/m and a lattice parameter of 3.903 A. Calculate the fraction of tin atoms in the alloy.

29.

Let fSn be the fraction of Sn atoms; then (1 - fSn) is the fraction of copper atoms. The molecular weight of tin is 118.69 g/mol and that of copper is 63.54 g/mol. From the density equation

(4 atoms/cell) [(fSn)(118.69) + (1 - f Sn)(63.65) 7.727 x 106 g/m3 ; ------------~~~~------~~----------~-

(3.903 X 10- 10 m)3(6.02 x 1023 atoms/mole)

Suppose that when one-third of the atoms in RCP magnesium are replaced by cadmium atoms, the lattice parameters of the alloy are a ; 3.133 A and

o c

o 5.344 A. Calculate the expected density of the alloy.

MWMg ; 24.312 g/mol MWCd ; 112.4 g/mol -10 3 -10 -29 3 V cell; (3.133 x 10 m) (5.344 x 10 m)cos30; 4.54274 x 10 m

p ; (2 atoms/cell) [(0.667)(24.312) + (0.333)(112.4) = 3.924 Mg/m3 (4.54274 x 10- 29 m3)(6.02 x 1023 atoms/mol)

30. Body centered cubic iron has a lattice' parameter of 2.868 A after a carbon atom enters one interstitial site in every twentieth unit cell. Estimate (a) the density of the iron-carbon alloy and (b) the packing factor for the structure, assuming r Fe = 1.241 A and rC = 0.77 A.

(a) When we use the density equation, we must add onto the usual 3 Fe atoms per cell a factor of 1/20 of a carbon atom per cell.

MW = 55.847 g/mol Fe

MWC 12 g/mol

p (2 Fe/cell) (55. 847) + (1/20 C/cell)(12) (2.868 x 10-10 m)3(6.02 x 1023 atoms/mol)

7.907 Mg/m 3

(b) In each unit cell, there are 2 Fe atoms and 0.05 C atoms.

PF 2(4/3)(n)(1.241 A)3 + (0.05)(4/3)(n)(0.77 A)3 (2.868 A)3

45

0.6828

31. Carbon atoms enter interstitial positions in FCC nickel. producing a lattice parameter of 3.5198 A and a density of 8.955 Mg/m3 Calculate (a) the atomic fraction of carbon atoms in the nickel and (b) the number of unit cells you would have to examine to find one carbon atom.

(a) The unit cell will contain 4 nickel atoms and "x" carbon atoms. From the density equation. using MWNi = 58.71 g/mol and

MWC = 12 g/mol.

8.955 x 106 = (4 Ni atoms/cell)(58.71) + (x C atoms/cell) (12) (3.5198 x 10- 10 m)3(6.02 x 1023 atoms/mol)

235.08 = 234.84 + 12x or x = 0.2 carbon atom per cell

fraction carbon atoms/cell 0.02 atoms/cell 4 Ni atoms/ cell 0.005

(b) Since there is 0.02 carbon atoms/cell. then 1/0.02 cells would be required to find one carbon atom.

50 unit

32. Would you expect a Frenkel defect to change the lattice parameter or density of MgO? Explain.

Yes; the misplaced ion will strain the lattice. causing the lattice parameter to increase and therefore reducing the density.

33. Suppose one Schottky defect occurred in every tenth unit cell of NaCI. producing a lattice parameter of 5. 57A.I Calculate the density of the sodium chloride.

Although normally there are 4 sodium ions and 4 chlorine ions per cell. the Schottky defects in every tenth unit cell produce. on the average. (39/40) (4) sodium ions and (39/40) (4) chloride ions per cell.

MWNa 22.99 g/mol MWCI = 35.453 g/mol

p (39/40)(4)(22.99 + 35.453) 2.191 Mg/m3 (5.57 X 10- 10 m)3(6.02 x 1023 ions/mol)

46

34. Suppose th3e lattice parameter of CsCI is 4.0185 A and the density is 4.285 Mg/m. Calculate the number of Schottky defects per unit cell.

Let "x" be the average number of Cs and CI ions in the unit cell.

MWCs = 132.91 g/mol MWCI = 35.453 g/mol

4.285 X 106 = ____ -'x:.c..:..:(1:...:3c.::2c:.. ..:..9.:..1 ::-+---=3.::.5.:.....4.:..5:...:3:....:)---=-=--__ _ (4.0185 x 10 10 m)3(6.02 x 1023 ions/mol)

x 0.99424 ions/cell

There should be 1 of each ion in the caesium chloride structure. The number of Schottky defects is therefore

defects/cell = 1 - 0.99424 = 0.00576

35. Suppose the (111) plane is parallel to the surface of an FCC metal. What is the coordination number for each atom at the surface?

At the surface, 6 atoms would touch a central atom at the surface, 3 more atoms from beneath the surface would contact the central atom, but the other 3 atoms are not present. Therefore the coordination number would be 9.

36. The ASTM grain size number for a metal is 6. How many grains would be observed per square inch in a photograph taken at a magnification of 100? How many actual grains are present per square inch?

N = 2n- 1 = 26- 1 = 25 = 32 grains at 100x

total grains = 32(100)2 = 320,000 grains

37. Twelve grains per square inch are counted in a photograph taken at a magnification of 500. Calculate the ASTM grain size number. Is this a coarse, medium, or fine grain size?

At 100x, the number of grains N would be

N = 12(500/100)2 = 300 = 2n-1

In(300) (n-l)(ln(2) or 5.7038 = (n-1)(0.693) or n 9.23 This is a fine grain size.

47

38. Eighteen grains per square inch are counted in a photograph taken at a magnifIcation of 75. Calculate the ASTM grain size number. Is this a coarse, medium or fine grain size?

At 100x, the number of grains N would be

N = 18(75/100)2 = 10.125 = 2n-1

In(10.125) = (n-l)ln(2) or 2.315 = (n-l)(0.693) or n = 4.34 This is medium grain size.

39. Figure 4-23 shows the microstructure of a material at a magnification of 100. Estimate the ASTM grain size number.

From the photograph, there are approximately 7 grains/in2.

7 = 2n- 1

In(7) = (n-l)ln(2) or 1.95 (n-l)(0.693) or n 3.8

40. Figure 4-24 shows the microstructure of a material at a magnification of 500. Estimate the ASTM grain size number.

There are approximately 25 grains/in2 at 500x. Therefore at 100x

N = 25 (500/100) 2 = 625 = 2n-1

In(625) = (n-l)ln(2) or 6.44 = (n-l)(0.693) or n 10.3

41. Calculate the angle 9 of a small angle grain boundary in BCC iron when the dislocations are 7500 A apart.

The lattice parameter for BCC iron is 2.866 A. The repeat distance of Burger's vector is half of the body diagonal, or

b = (1/2)(13)(2.866) = 2.482 A sin(9/2) = 2.482/7500 = 0.0003309 9/2 = 0.01896 or 9 = 0.038

42. A small angle grain boundary is tilted 0.75 in FCC copper. the average distance between the dislocations.

Calculate

The lattice parameter is 3.6151 A. The repeat distance of Burger's vector is half of the face diagonal, or

b = (1/2)(i2)(3.6151) = 2.556 A If "x" is the distance between dislocations, then

sin(0.7512) 2. 556/x 0.006545 2.556/x or x 390 A

48

43. Suppose that 2 Fe3+ ions are substituted for normal Fe2+ ions in FeO. What other changes in the atom arrangement (creation of vacancies etc) would be required to maintain the proper charge balance?

We must create a Fe2+ vacancy. The addition of two Fe3+ ions introduces a charge of +6; removal of the two Fe2 + ions only removes a charge of +4. However. if three Fe2+ ions are removed. then a total of +6 charge is removed~+the charge is balanced. but a vacancy is created where the third Fe ion was removed.

44. Suppose an Fe2+ ion is substituted for an Na+ ion in NaCl. Explain why you would expect that the charge balance would be maintained by forming a vacancy rather than by adding another Cl- ion to the lattice.

The large Cl- ion would be added as an interstitial ion and would create a larger amount of lattice strain than would be caused by the formation of a vacancy.

49

Chapter 5

ATOM MOVEMENT IN MATERIALS

1. When a force is applied at 300C, each inch of a metal stretches at the rate of 0.025 mmlmin; at 400C, the rate of stretching is 0.09375 mmlmin; and at 500C, the rate is 0.25 mmlmin. Calculate (a) the activation energy for the stretching process and (b) the constant co' What are the units for each?

(a) From the equation rate = coexp(-QlRT): 0.025 _ coexp[-Q/(8. 314) (300 + 273)] 0.25 - c exp[ 01(8.314)(500 +2731)]

o

0.1 = exp [(-Q) (0.0002099 - 0.0001556)] 0.1 = exp [-Q(0.0000543)] In(O.l) = -2.302585 =.-0.00005430 :. Q = 42,405 J/mol

0.025 = coexp[-42,405/(8.314) (573)] 0.025 = coexp(-8/9012792) 0.025 = co (0.0001362) c = 183.55 mmlminute

o

2. The fraction of the lattice points containing vacancies in copper is 2.24 x 10-15 at 100C but is 2.42 X 10-6 at 700C. Calculate (a) the activation energy required to form a vacanc~, (b) the fraction of the lattice points that contain vacancies 5 C below the melting point (l085C), and (c) the number of vacancies per unit cell 5C below the melting point.

(a) 2.24 x 10-15 _ exp!-Q/(8.314)(373)] 2.42 x 10-6 - exp[-Q/(8.314)(973)]

(b)

9.256 x 10-10 = exp!-Q(0.0003224 - 0.0001236)] In(9.256 x 10-19 ) = -20.80056 = -0.0001988Q Q = 104,631 J/mol To find the constant c

o

2.42 X 10-6 = c exp[-104,631)/(8.314) (973)] o

c exp(-12.934143 = c (2.42 x 10-6 ) o 0

.. c 1. 0 o

50

At 5C below the metling point, or 1080C, the fraction 'f' is

f 1.0 exp [-104,631/(8.314)(1353)]

1.0 exp(-9.3014908) 9. 12 X 10-5 vacancies/ lattice point.

(c) In FCC copper there are 4 lattice points/cell :. vacancies/cell (9.12 x 10-5 ) (4)

36.48 x 10-5

3. The diffusion coefficient for Al in Al 0 is 7.48 x 10-23m2 /s at 1000C 2 3

and is 2.48 x 10-14 m2/s at 1500C. Calculate the activation energy and the diffusion constant D .

o

7.48 X 10-23 D o[exp -Q/(8.314)(1273)] 2.48 x 10-14 Do[exp - Q/(8.314)(1773l]

3.016 x 10-9 = exp[-Q(0.0000944 x 0.0000678)] = exp[-Q(0.0000266)]

In(3.016 x 10-9 ) = -19.619334 = -Q(0.0000266)

Q = 737,569 J/mol

2.48 x 10-14 = D Jxp[-737,569/(8.314)(1773)] = Doexp(-50.036152)

2.48 x 10-14 = D (1. 8602 x 10-22 ) o

4. The diffusion coefficient for Ni in MgO is 1.23 X 10-16 m2/s at 1200C and 1.45 x 10- 14 m2/s at 1800C. Calculate the activation energy and the diffusion constant D .

o

1.23 X 10-16 D J xp[-Q/(8.314)(1473)] 1.45 x 10-14 Doexp[-Q/(8.314) (2073)]

8.427 x 10-3 = exp[-Q(0.0000816 - 0.000058)]

= exp[-Q(0.0000236)]

In(8.4827 x 10-3 ) = -4.7697 = -Q(0.0000236)

Q = 202, 107 J/mol

1.23 x 10-16 = D exp[-202,107/(8.314)(1473)] o

51

1.23 X 10-16 = (6.8036 x 10-8)D o

-8 2 :. DO = 1. 808 x 10 m Is

5. Estimate the activation energy for self-diffusion in titanium.

From Figure 5-S the activation energy is approximately 293,000 to 314,000 J/mol because the melting point is 166SoC.

6. Would you expect the activation energy for self-diffusion in silicon, which has the covalent bond, to follow the curve in Figure 5-8? Explain.

No. We would expect a higher activation energy for silicon due to the strong covalent bonds.

7. Suypose a 0.1 mm thick wafer of germanium