Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-answers GTBL042-Callister-v3 October 11, 2007 13:37

Revised Pages

Answers to

Selected Problems

Chapter 2

2.4For P^5 +:1s^22 s^22 p^6

2.8FA= 5. 89 × 10 −^10 N

2.1273.4% for MgO

Chapter 3

3.1VC= 1. 213 × 10 −^28 m^3

3.4R= 0 .138 nm

3.7Metal B: simple cubic

3.8 (a)n=4 atoms/unit cell;

(b)ρ= 7 .31 g/cm^3

3.13APF=0.79

3.15 (a)a= 0 .437 nm;

(b)a= 0 .434 nm

3.22Cl−point coordinates: 000, 100 , 110 , 010 ,

001 , 101 , 111 , 011 ,^12120 ,^12121 , 11212 , 01212 ,

1

20

1

2 ,and

1

21

1

2.

3.26[102]

3.28Direction B: [ 40 3];

Direction D: [ 11 1]

3.29 (b)[100],[010],and [010]

3.31Direction A: [4 2 23]

3.36Plane A: (111) or ( 1 11)

3.37Plane B: (021)

3.41 (a)(2 1 10)

3.43 (a)LD 100 =

1

2 R

√

2

3.44 (b)PD 110 (Mo)= 1. 014 × 1019 m−^2

3.47 2 θ= 45. 88 ◦

3.48d 111 = 0 .1655 nm

3.49 (a)d 211 = 0 .1348 nm;

(b)R= 0 .1429 nm

3.50d 200 = 0 .2455 nm,d 311 = 0 .1486 nm,

a= 0 .493 nm

Chapter 4

4.3 (a)Mn=49,800 g/mol; (c)DP= 498

4.5L=2682 nm;r= 22 .5nm

4.89333 of both acrylonitrile and butadiene

repeat units

4.9Vinyl chloride

4.14 (a)ρa= 0 .841 g/cm^3 ;ρc= 0 .946 g/cm^3

Chapter 5

5.1Nv/N= 4. 56 × 10 −^4

5.2Qv= 1 .10 eV/atom

5.4 (a)Qs= 7 .70 eV

5.6 (a)Li+vacancy; one Li+vacancy for every

Ca^2 +added

5.13NMo= 1. 73 × 1022 atoms/cm^3

5.15CGe= 11 .7 wt%

5.21 (b)NM=320,000 grains/in.^2

5.D1CLi= 2 .38 wt%

Chapter 6

6.4M= 4. 1 × 10 −^3 kg/h

6.6t= 31 .3h

6.8t=135 h

6.10T=901 K (628◦C)

6.13T=900 K (627◦C)

Chapter 7

7.2l 0 =475 mm (18.7in.)

7.4 (a)F=44,850 N (10,000 lbf);

(b)l= 76 .25 mm (3.01 in.)

7.6

(

dF

dr

)

r 0

=−

2 A

(

A

nB

) 3 /(1−n)+

(n)(n+1)B

(

A

nB

)(n+2)/( 1 −n)

7.8F=7,800 N (1785 lbf)

7.11 (a)l= 0 .15 mm (6. 0 × 10 −^3 in.);

(b)d=− 5. 25 × 10 −^3 mm (− 2. 05 × 10 −^4 in.)

7.12Steel and brass

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