Complete Solutions to Selected Problems

1 Complete Solutions to Selected Problemsto accompanyMATERIALS SCIENCEAND ENGINEERINGAN INTRODUCTIONS ixth EditionWilliam D. Callister, University of UtahJohn Wiley & Sons, 2003 John Wiley & Sons, Inc. All rights part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or byany means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permittedunder Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior writtenpermission of the Publisher, or authorization through payment of the appropriate per-copy fee to theCopyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508)750-8400, fax (508)750-4470.

2 Requests to the Publisher for permission should be addressed to the Permissions Department, JohnWiley & Sons, Inc., 605 Third Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008,E-Mail: order books or for customer service please call 1(800) Complete Solutions to Selected Problems has been developed as asupplement to the sixth edition of Materials Science and Engineering: AnIntroduction. The author has endeavored to select Problems that are representativeof those that a student should be able to solve after having studied the relatedchapter topics.

3 In some cases problem selection was on the basis of illustratingprinciples that were not detailed in the text discussion. Again, Problems havingsolutions in this supplement have double asterisks by their numbers in both Questions and Problems sections at the end of each chapter, and in the Answersto Selected Problems section at the end of the printed Solutions begin with a reiteration of the problem , the author has sought to work each problem in a logical andsystematic manner, and in sufficient detail that the student may clearly understandthe procedure and principles that are involved in its solution.

4 In all cases,references to equations in the text are cited. The student should also keep in mindthat some Problems may be correctly solved using methods other than , the course instructor has the option as to whether or not to assignproblems whose Solutions are provided here. Hopefully, for any of these solvedproblems, the student will consult the solution only as a check for correctness, oronly after a reasonable and unsuccessful attempt has been made to solve theparticular problem . This supplement also serves as a resource for students, to helpthem prepare for examinations, and, for the motivated student, to seek additionalexploration of specific author sincerely hopes that this Solutions supplement to his text will be auseful learning aid for the student, and to assist him/her in gaining a betterunderstanding of the principles of materials science and engineering.

5 He welcomesany comments or suggestions from students and instructors as to how it can D. Callister, John Wiley & Sons, 2 ATOMIC structure AND INTERATOMIC BONDINGPROBLEM (a) In order to determine the number of grams in one amu of material, appropriate manipulation ofthe amu/atom, g/mol, and atom/mol relationships is all that is necessary, as# g/amu = 1 x 1023 atoms 1 g /mol1 amu / atom = x 10-24 g/amu(b) Since there are g/lbm,1 lb - mol = g/lbm() x 1023 atoms/g - mol()= x 1026 (c) This portion of the problem asks that, using the Solutions to problem , we mathematicallydetermine values of ro and Eo.

6 From Equation ( ) for ENA = = x 10-6n = 8 Thus, ro = AnB 1/(1 - n)= (8) x 10-6() 1/(1 - 8)= nmandCopyright John Wiley & Sons, (8) x 10 6 1/(1 8) + x 10 (8) x 10 6 8/(1 8)= - The percent ionic character is a function of the electron negativities of the ions XA and XBaccording to Equation ( ). The electronegativities of the elements are found in Figure MgO, XMg = and XO = , and therefore,%IC = 1 e( ) ( )2 x 100 = CdS, XCd = and XS = , and therefore,%IC = 1 e( ) ( )2 x 100 = John Wiley & Sons, 3 THE structure OF CRYSTALLINE SOLIDSPROBLEM For this problem , we are asked to calculate the volume of a unit cell of aluminum.

7 Aluminum has anFCC crystal structure (Table ). The FCC unit cell volume may be computed from Equation ( )asVC = 16R32 = (16) x 10-9 m()32= x 10-29 (a) The volume of the Ti unit cell may be computed using Equation ( ) asVC=nATi NANow, for HCP, n = 6 atoms/unit cell, and for Ti, ATi = g/mol. Thus,VC = (6 atoms/unit cell)( g/mol) g/cm3() x 1023 atoms/mol()= x 10-22 cm3/unit cell = x 10-28 m3/unit cell(b) From the solution to problem , since a = 2R, then, for HCPVC = 33a2c2but, since c = = 33(1 . 5 8 ) a32= x 10-22 cm3/unit cellNow, solving for aCopyright John Wiley & Sons, = 2() x 10-22 cm3()3()3() () 1/3= x 10-8 cm = nmAnd finallyc = = ( )( nm) = (a) From the definition of the APFAPF = VSVC= n43 R3 a2cwe may solve for the number of atoms per unit cell, n, asn = (APF)a2c43 R3= ( )( )2( ) 10-24 cm3()43 x 10-8 cm()3= atoms/unit (a) We are asked for the indices of the two directions sketched in the figure.

8 For direction 1, theprojection on the x-axis is zero (since it lies in the y-z plane), while projections on the y- and z-axesare b/2 and c, respectively. This is an [012] direction as indicated in the summary belowxyzProjections0ab/2cProjections in terms of a, b,and c01/21 Copyright John Wiley & Sons, to integers012 Enclosure[012](b) This part of the problem calls for the indices of the two planes which are drawn in the 1 is an (020) plane. The determination of its indices is summarized ab/2 cIntercepts in terms of a, b,and c 1/2 Reciprocals of intercepts020 Enclosure(020) Direction B is a [4 03 ] direction, which determination is summarized as follows.

9 We first of allposition the origin of the coordinate system at the tail of the direction vector; then in terms of thisnew coordinate systemxyzProjections- 2a30b- c2 Projections in terms of a, b,and c- 230- 12 Reduction to integers- 40- 3 Enclosure[4 03 ]Direction D is a [1 11 ] direction, which determination is summarized as follows. We first ofall position the origin of the coordinate system at the tail of the direction vector; then in terms of thisnew coordinate systemxyzProjections- a2b2- c2 Copyright John Wiley & Sons, in terms of a, b,and c- 1212- 12 Reduction to integers- 11- 1 Enclosure[1 11 ] For plane B we will leave the origin of the unit cell as shown; thus, this is a (122) plane, assummarized in terms of a, b,and c11212 Reciprocals of intercepts122 Reductionnot necessaryEnclosure(122) (a) For this plane we will leave the origin of the coordinate system as shown.

10 Thus, this is a (12 11)plane, as summarized a2acIntercepts in terms of a's and c1- 1211 Reciprocals of intercepts1- 211 Reductionnot necessaryEnclosure(12 11) (a) The unit cell in problem is body-centered tetragonal. Only the (100) (front face) and (01 0)(left side face) planes are equivalent since the dimensions of these planes within the unit cell (andCopyright John Wiley & Sons, the distances between adjacent atoms) are the same (namely nm x nm), whichare different than the (001) (top face) plane (namely nm x nm). (a) In the figure below is shown a [100] direction within an FCC unit this [100] direction there is one atom at each of the two unit cell corners, and, thus, there isthe equivalent of 1 atom that is centered on the direction vector.