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E 344  Stevens Study Resources
 Stevens
 Staf
 Steel Industry: Price and Policy Issues, Mechanical Behavior of Materials, Foundations of Electromagnetic Theory (4th Edition), Elements of Continuum Mechanics and Conservation Laws, Magnetic Nanostructures in Modern Technology: Spintronics, Magnetic MEMS and Recording (NATO Science for Peace and Security Series B: Physics and Biophysics)

2011 Summer PS 1 Solutions
School: Stevens
Course: Materials Processing
1) Core 2.04 2) Callister 2.6 3) Callister 2.12 4) Callister 2.14 5) Callister 2.15 6) Core 2.21 7) From test 1, spring 2011 8) From test 1, fall 2010 9) Callister 2.23 10) Callister 3.3 11) Callister 3.5 12) Callister 3.7 13) From midterm exam, summer 20

2010 Spring Final Solutions
School: Stevens
Course: Materials Processing
(0.001m) 2 RA 8 0.2m 10m 1.6 x10 m m L 4 (B) The resistance would increase because: (1) A would get smaller; and (2) would increase slightly because of the defects introduced by cold drawing. (C) The resistivity of pure Nickel wire would be lower than

2011 Spring Final Solutions
School: Stevens
Course: Materials Processing
C) i . Al, because of need for low density, strength and toughness. i i. Al2 O3, because it is a good thermal insulator. Also, i t is in oxide form, so it is stable at high temperatures. i ii . Al, because fracture cannot be tolerated and Al is much tough

2011 Spring Test 1 Solutions
School: Stevens
Course: Materials Processing
This test consists o f 8 pages. There are 5 problems. Questions 614 are multiple choice questions. Some constants are given on page 7, and a periodic table is given on page 8. L\~ E 344 Spring 2011 Test 1 3 March, 2011 Name: Honor Pledge: Room:  ~

Final Exam Sheet
School: Stevens
Course: Materials Processing
Rubber has a low elastic modulus, cant be easily recycled due to its crosslinks, typically cant go much plastic deformation prior to failure, is a type of polymer Sharp notches & cracks on the surface of a material loaded in tension are potentionally ver

2011 Summer PS 3 Solutions
School: Stevens
Course: Materials Processing
1. 2.Callister 14.3 3.Callister 14.9 4.Callister 14.10 5.Callister 14.16 6.Callister 14.24 7.Callister 15.15 8.Callister 15.31 9.Callister 15.32 10.From test 1 spring 2011 11.Core 5.21 12.Core 5.18 13.Core 5.19 14.From test 2 spring 2007 15.b 16.b 17.b 18

2011 Summer PS 4 Solutions
School: Stevens
Course: Materials Processing
6.5 A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lbf), and experiences an elongation of 0.10 mm (4.0 103 in.). Assuming that the deformation is entire

2013 Spring PS 3 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 3 1. Sodium has the BCC crystal structure. Its density is 0.971 g/cm 3, and its atomic weight is 22.99 g/mole. Use this information to show that the atomic radius of a sodium atom is 0.186 nm. Solution ASSUMPTION: Assuming th

2013 Spring PS 10 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 10 1. 4.4 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated. Cu Atomic Radius (nm) 0.127

2013 Spring PS 12 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 12 1. (From spring 2006 final) The band gap for pure silicon is 1.12 eV. The donor level for Sb doped into silicon is 0.039 eV below the bottom of the conduction band. If you have samples of pure Si and Sbdoped Si at room te

2013 Spring PS 11 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 11 Solution (A) Cementite is much harder but more brittle than ferrite. Increasing the carbon concentration increases the fraction of Fe3C in a steel alloy and will result in a harder and stronger material. (B) Martensite is

2013 Spring PS 2 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 2 1. 2.14 The net potential energy between two adjacent ions, E N, may be represented by the sum of Equations 2.8 and 2.9; that is, EN = AB + r rn Calculate the bonding energy E0 in terms of the parameters A, B, and n using t

2013 Spring PS 1 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 1 1. Wafers of silicon (Si) are used to make semiconductor devices. The density of Si is 2.33 g/cm 3 and its atomic weight is 28.086 g/mole. Estimate how many Si atoms are contained in a wafer 500 m thick and 50.8 cm in diame

Test 1
School: Stevens
Course: Materials Processing
Metals & alloys shaped by casting, rolling, forging, extrusion Cement hardens by a chemical reaction with water, is a component in both mortar & concrete, can be considered a ceramic material, forms by an exothermic reaction ; doesnt form a drying reacti

Test 1 Sheet
School: Stevens
Course: Materials Processing
Metals & alloys can be shaped by casting, rolling, forging, extrusion Cement hardens by a chemical reaction with water, is a component in both mortar & concrete, can be considered a ceramic material, forms by an exothermic reaction; doesnt form a drying

2013 Spring PS 6 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 6 1. Callister 14.23 For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible,

2013 Spring PS 5 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 5 1. 14.2 Compute repeat unit molecular weights for the following: (a) poly(vinyl chloride), (b) poly(ethylene terephthalate), (c) polycarbonate, and (d) polydimethylsiloxane. Solution (c) For polycarbonate, from Table 14.3,

2013 Spring PS 7 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 7 1. (From test 2 Fall, 2011) A plaincarbon steel has a yield strength of 35 ksi, a tensile strength of 52 ksi and a modulus of 29,000 ksi. Suppose a 3 inch diameter rod of this steel with a length of 10 feet is loaded in te

2013 Spring PS 9 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 9 1. 10.1 Name the two stages involved in the formation of particles of a new phase. Briefly describe each. Solution The two stages involved in the formation of particles of a new phase are nucleation and growth. The nucleati

2013 Spring PS 8 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 8 1. 8.4 A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/m 2 (2.86 103 in.lbf

2013 Spring PS 4 Solutions
School: Stevens
Course: Materials Processing
2013 Spring E344 Solution Set 4 1 . From Summer 2011 midterm: The SiC ubit cell is given in the diagram below. A) Show that the SiC unit cell contains 4 Si atoms and 4 C atoms. B) Use the Miller index notation to describe the positions of the 4 C atoms i

Test 2 Sheet
School: Stevens
Course: Materials Processing
Rubber has a low elastic modulus, cant be easily recycled due to its crosslinks, typically cant go much plastic deformation prior to failure, is a type of polymer Sharp notches & cracks on the surface of a material loaded in tension are potentionally ver

Test 2
School: Stevens
Course: Materials Processing
Rubber has a low elastic modulus, cant be easily recycled due to its crosslinks, typically cant go much plastic deformation prior to failure, is a type of polymer Sharp notches & cracks on the surface of a material loaded in tension are potentionally ver

Callister7e_sm_ch02_03
School: Stevens
Course: Materials Processing
23 2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as 1 g / mol 1 mol # g/amu = 23 atoms 1 amu /atom 6.023 x 10 = 1.66 x 10

Callister7e_sm_ch02_04
School: Stevens
Course: Materials Processing
24 2.4 (a) Two important quantummechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells. (b) Two important refinements resulting fr

Callister7e_sm_ch02_05
School: Stevens
Course: Materials Processing
25 2.5 The n quantum number designates the electron shell. The l quantum number designates the electron subshell. The ml quantum number designates the number of electron states in each electron subshell. The ms quantum number designates the spin moment o

Callister7e_sm_ch02_06
School: Stevens
Course: Materials Processing
26 2.6 For the L state, n = 2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and 1; and possible ms values are . Therefore, for the s states, the quantum numbers are 1 1 1 1 1 1 200 ( ) and 200 ( )

Callister7e_sm_ch02_07
School: Stevens
Course: Materials Processing
27 2.7 The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6). P5+: 1s22s22p6 P3: 1s22s22p63s23p6 Sn4+: 1s22s22p63s23p63d104s24p64d10 Se2: 1s22s22p63s23p63d104s24p6 I: 1s22s22p63s23p63d104s24p64d105s25p6 Ni2+: 1s22s22

Callister7e_sm_ch02_08
School: Stevens
Course: Materials Processing
28 2.8 The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon (Figure 2.6). The I ion is a iodine atom that has acquired one extra electron; therefore, it has an electron configurati

Callister7e_sm_ch02_09
School: Stevens
Course: Materials Processing
29 The Periodic Table 2.9 Each of the elements in Group IIA has two s electrons. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in course

Callister7e_sm_ch02_10
School: Stevens
Course: Materials Processing
210 2.10 From the periodic table (Figure 2.6) the element having atomic number 112 would belong to group IIB. According to Figure 2.6, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the rightmost column of group VIII. Movi

Callister7e_sm_ch02_02
School: Stevens
Course: Materials Processing
22 2.2 The average atomic weight of silicon (ASi ) is computed by adding fractionofoccurrence/atomic weight products for the three isotopes. Thus ASi = f 28 A28 + f 29 A29 + f 30 A30 Si Si Si Si Si Si = (0.9223)(27.9769) + (0.0468)(28.9765) + (0.0309)(

Callister7e_sm_ch02_01
School: Stevens
Course: Materials Processing
21 CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS Fundamental Concepts Electrons in Atoms 2.1 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally

Ss_8
School: Stevens
1. Callister 6.3 A specimen of copper having a rectangular cross section 15.2 mm x 19.1 mm (0.60 in. x 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. 2. Callister

Ss_4
School: Stevens
1. Callister 4.28 For a BCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why? (Note: You may want to consult the solution to Problem 3.54 at the end of Chapter 3.) 2. Callis

Ps_2
School: Stevens
Problem Set 2 Due: Wednesday, 30 May 2007 at beginning of lecture E344 Summer 2007 1. Callister 3.56 Explain why the properties of polycrystalline materials are most often isotropic. 2. Core 3.27 PZT 3. Callister 13.13 viscosity of sodalime glass

Ss_2
School: Stevens
1The atomic weight of vanadium (V) is 50.94 g/mole. Its density is =5.96 g/cm3. It has the BCC crystal structure. Calculate the atomic radius of vanadium. Confirm that this matches the value given in Appendix A. 3a = 4 R a 2a V has the BCC crysta

Ss_3
School: Stevens
1. Core 8.01 0.75 0.05082 A0 = (0.75in)(0.05082in) = 0.03816in 2 FMAX = 10,000 lbs 10,000lbs ! MAX = = 262ksi 0.03816in 2 ! MAX = 262 > ! T = 65 Thus the tensile testing machine has the capacity to break tensile bars of this steel. 2. Core 8.05

Ss3
School: Stevens
3.31 Determine the indices for the directions shown in the following cubic unit cell: +z A 1 2 1 1 , 2 2 1 2 B +y +x Direction A is the 110 direction. the direction vector. [ ] To identify it, we position the origin at the tail of Then, in t

Ps_9
School: Stevens
Problem Set 9 Due: Tuesday, 20 November, 2007 at beginning of lecture E344 Fall 2007 1. Callister 10.4 Nucleation during nickel solidification 2. Core 9.11 critical size for cubeshaped nucleus 3. Core 9.02 Boron diffusion profiles in silicon 4. C

Ss_11
School: Stevens
1. Callister 9.8 ae Cite the phases that are present and the phase composition for the following alloys: (a) 15wt% Sn85 wt% Pb at 100C (212F) (b) 25wt% Pb75 wt% Mg at 425C (800F) (c) 85wt% Ag15 wt% Cu at 800C (1470F) (d) 55wt% Zn45 wt% Cu at 600

Callister7e_sm_ch02_11
School: Stevens
Course: Materials Processing
211 2.11 (a) The 1s22s22p63s23p5 electron configuration is that of a halogen because it is one electron deficient from having a filled p subshell. (b) The 1s22s22p63s23p63d74s2 electron configuration is that of a transition metal because of an incomplete

Callister7e_sm_ch02_12
School: Stevens
Course: Materials Processing
212 2.12 (a) The 4f subshell is being filled for the rare earth series of elements. (b) The 5f subshell is being filled for the actinide series of elements. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit bas

Callister7e_sm_ch02_23
School: Stevens
Course: Materials Processing
225 Secondary Bonding or van der Waals Bonding 2.23 The intermolecular bonding for HF is hydrogen, whereas for HCl, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting tempera

Callister7e_sm_ch04_04
School: Stevens
Course: Materials Processing
44 Impurities in Solids 4.4 In this problem we are asked to cite which of the elements listed form with Ni the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic rad

Callister7e_sm_ch04_01
School: Stevens
Course: Materials Processing
41 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and SelfInterstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 4.1. As stated in the problem, Qv = 0.90 eV/atom. Thus

Callister7e_sm_ch04_02
School: Stevens
Course: Materials Processing
42 4.2 Determination of the number of vacancies per cubic meter in gold at 900C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: Q Q N A Au N v = N exp v = exp v AAu kT kT = (6.023 10 23atoms / mol)(18.63 g / cm 3) exp 196.9

Callister7e_sm_ch04_03
School: Stevens
Course: Materials Processing
43 4.3 This problem calls for the computation of the activation energy for vacancy formation in silver. Upon examination of Equation 4.1, all parameters besides Qv are given except N, the total number of atomic sites. However, N is related to the density

Test 3 Sheet
School: Stevens
Course: Materials Processing
Lithography is used in semiconductor device processing for patterning; To create ntype Ge, one could dope using P Consider a parallel plate capacitor where the electrodes are separated by barium titanate. If the barium titanate is removed so that the ga

Callister7e_sm_ch02_22
School: Stevens
Course: Materials Processing
21 CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS Fundamental Concepts Electrons in Atoms 2.1 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally

Callister7e_sm_ch02_21
School: Stevens
Course: Materials Processing
223 2.21 For silicon, having the valence electron structure 3s23p2, N' = 4; thus, there are 8 N' = 4 covalent bonds per atom. For bromine, having the valence electron structure 4s24p5, N' = 7; thus, there is 8 N' = 1 covalent bond per atom. For nitrogen,

Callister7e_sm_ch02_13
School: Stevens
Course: Materials Processing
213 Bonding Forces and Energies 2.13 The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation 2.8, which is just A d A r = = dr r2 FA = dE A dr The constant

Callister7e_sm_ch02_14
School: Stevens
Course: Materials Processing
214 2.14 (a) Differentiation of Equation 2.11 yields B A d d r rn = + dr dr dE N dr = nB A  = 0 r (1 + 1) r ( n + 1) (b) Now, solving for r (= r0) A 2 r0 = nB r0( n + 1) or A 1/(1  n) r0 = nB (c) Substitution for r0 into Equation 2.11 and solving fo

Callister7e_sm_ch02_15
School: Stevens
Course: Materials Processing
215 2.15 (a) Curves of EA, ER, and EN are shown on the plot below. (b) From this plot r0 = 0.24 nm E0 = 5.3 eV (c) From Equation 2.11 for EN A = 1.436 B = 7.32 x 106 n=8 Thus, A 1/(1  n) r0 = nB 1/(1  8) 1.436 = 0.236 nm (8)(7.32 106 ) and Excerp

Callister7e_sm_ch02_16
School: Stevens
Course: Materials Processing
217 2.16 This problem gives us, for a hypothetical X+Y ion pair, values for r0 (0.38 nm), E0 ( 5.37 eV), and n (8), and asks that we determine explicit expressions for attractive and repulsive energies of Equations 2.8 and 2.9. In essence, it is necess

Callister7e_sm_ch02_17
School: Stevens
Course: Materials Processing
219 2.17 (a) Differentiating Equation 2.12 with respect to r yields r C dD exp d dE r = dr dr dr C r2 Der / = At r = r0, dE/dr = 0, and (r0 /) C 2 r0 = De (2.12b) Solving for C and substitution into Equation 2.12 yields an expression for E0 as (r0 /)

Callister7e_sm_ch02_18
School: Stevens
Course: Materials Processing
220 Primary Interatomic Bonds 2.18 (a) The main differences between the various forms of primary bonding are: Ionicthere is electrostatic attraction between oppositely charged ions. Covalentthere is electron sharing between two adjacent atoms such that

Callister7e_sm_ch02_19
School: Stevens
Course: Materials Processing
221 2.19 The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation 2.10. The electronegativities of the elements are found in Figure 2.7. For MgO, XMg = 1.2 and XO = 3.5, and therefore, 2 % IC = 1

Callister7e_sm_ch02_20
School: Stevens
Course: Materials Processing
222 2.20 Below is plotted the bonding energy versus melting temperature for these four metals. From this plot, the bonding energy for molybdenum (melting temperature of 2617C) should be approximately 7.0 eV. The experimental value is 6.8 eV. Excerpts fro

Problem Core 3.28
School: Stevens
The bodycentered tetragonal (BCT) crystal structure is just like the bodycentered cubic structure except that it is elongated in one direction. For BCT: a = b c and = = = 90o. The body diagonal is a closepacked direction. Assume c=1.25a. Show t