CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
2.1 (a) When two or more atoms of an element have different atomic masses, each is termed an
isotope.
(b) The atomic weights of the elements ordinarily are not integers because: (1) the
CHAPTER 18
ELECTRICAL PROPERTIES
PROBLEM SOLUTIONS
18.5 (a) In order to compute the resistance of this copper wire it is necessary to employ Equations (18.2)
and (18.4). Solving for the resistance in terms of the conductivity,
R=
l
l
=
A
A
7
-1
From Table
1.
(a) Compare the atomic packing factor of Simple cube, BCC and FCC.
(b) Explain linear density and planar density.
(c) Discuss x-ray diffraction technique to determine the crystal structure
2.
Strontium (Sr) has an FCC crystal structure, an atomic radiu
Diffusion
Diffusion is a time dependent process
How do we quantify the amount or rate of diffusion?
moles (or mass)diffusing mol
kg
J Flux
or
2
surface areatime
cm s m2s
Measured empirically
Make thin film (membrane) of known surface area
Impose conc
EGN 3364 Exam
C x C0
= 1 erf
C s C0
Equation Sheet
x
2 Dt
=E
l
l0
0 = y k y d -1/2
z z1 erf ( z ) erf ( z1)
=
z 2 z1 erf ( z 2)erf ( z1)
= 2(1 + )
0
% = (
= 1 [
= 0 ( )
= ( )
z =
) 100
% = (
x x
B
D = J A
CA CB
0
0
) 100
0
1 2 + 1 2 + 1 2
(
1. Calculate the fraction of atom sites that are vacant for copper at its melting temperature
of 1084 C (1357K). Assume energy for vacancy formation of 0.90 eV/atom.
Solution:
= ( )
= (
0.90
)
8.62 105 1357
= 4.55 10-4
1357 4.55 10-4.
2. The purification
Homework 3 Questions
1. A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile
stress is applied parallel to the [100] direction. If the critical resolved shear stress for this
material is 0.5 MPa, calculate the mag
1. (a) Electronic configuration
Fe 1s2 2s2 2p6 3s2 3p6 3d6 4s2 or [Ar] 3d6 4s2
S2- - 1s2 2s2 2p6 3s2 3p6 or [Ar]
Cu2+ - 1s2 2s2 2p6 3s2 3p6 3d9
(b)
2. a. When the properties of a material vary with different crystallographic orientations,
the material is
10.7 Compute the rate of some reaction that obeys Avrami kinetics, assuming that the constants n and k have values
of 2.0 and 5 104, respectively, for time expressed in seconds.
10.23 Make a copy of the isothermal transformation diagram for a 1.13 wt% C i
EGN 3365 Exam 1
Name:_
4
Volume of a sphere = 3 3
Lattice parameter a for FCC is 22
Lattice parameter a for BCC is
Theoretical density: =
4
3
nAV
VC N A
NA = 6.02 1023
ave =
nAave
VC N A
n = 2 dhkl Sin
Nv
Q
= exp v
N
kT
1
1. What does each of the four quantum numbers specify with respect to electrons and electronic states?
Solution
The n quantum number designates the electron shell.
The l quantum number designates the electron subshell.
The ml quantum number designates the
CHAPTER 16
COMPOSITES
PROBLEM SOLUTIONS
16.3 The elastic modulus versus volume percent of WC is shown below, on which is included both upper
and lower bound curves;
these curves were generated using Equations (16.1) and (16.2),
respectively, as well as th
CHAPTER 14
POLYMER STRUCTURES
PROBLEM SOLUTIONS
14.4 We are asked to compute the number-average degree of polymerization for polypropylene, given
that the number-average molecular weight is 1,000,000 g/mol.
The mer molecular weight of
polypropylene is jus
CHAPTER 13
APPLICATIONS AND PROCESSING OF CERAMICS
PROBLEM SOLUTIONS
13.5
(a)
From Figure 12.25, the maximum temperature without a liquid phase corresponds to the
temperature at the MgO(ss)-[MgO(ss) + Liquid] boundary at this composition, which is approxi
CHAPTER 3
THE STRUCTURE OF CRYSTALLINE SOLIDS
PROBLEM SOLUTIONS
3.9 We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure.
For FCC, n = 4 atoms/unit cell, and V = 1 6R 3 2 [Equation (3.4)]. Now,
C
=
n AIr
VC N
CHAPTER 4
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
4.4 In this problem we are asked to cite which of the elements listed form with Cu the three possible solid
solution types.
For complete substitutional solubility the following criteria must be met: 1) t
CHAPTER 5
DIFFUSION
PROBLEM SOLUTIONS
5.12 This problem asks that we determine the position at which the carbon concentration is 0.25 wt%
after a 10-h heat treatment at 1325 K when Co = 0.55 wt% C. From Equation (5.5)
C x Co
C s Co
=
x
0 .25 0.55
= 0 .545
CHAPTER 6
MECHANICAL PROPERTIES OF METALS
PROBLEM SOLUTIONS
6.9
This problem asks that we calculate the elongation l of a specimen of steel the stress-strain
behavior of which is shown in Figure 6.24. First it becomes necessary to compute the stress when
CHAPTER 7
DISLOCATIONS AND STRENGTHENING MECHANISMS
PROBLEM SOLUTIONS
7.2 When the two edge dislocations become aligned, a planar region of vacancies will exist between the
dislocations as:
7.7 Below is shown the atomic packing for a BCC cfw_110 type plan
CHAPTER 8
FAILURE
PROBLEM SOLUTIONS
8.31 (a) The fatigue data for this alloy are plotted below.
(b) The fatigue limit is the stress level at which the curve becomes horizontal, which is 290 MPa
(42,200 psi).
(c) From the plot, the fatigue lifetimes at a s
CHAPTER 9
PHASE DIAGRAMS
PROBLEM SOLUTIONS
9.24 (a) We are given that the mass fractions of and liquid phases are both 0.5 for a 40 wt% Pb-60 wt
% Mg alloy and asked to estimate the temperature of the alloy.
Using the appropriate phase
diagram, Figure 9.1
CHAPTER 10
PHASE TRANSFORMATIONS IN METALS
PROBLEM SOLUTIONS
10.11 We are called upon to consider the isothermal transformation of an iron-carbon alloy of eutectoid
composition.
(a) From Figure 10.13, a horizontal line at 550 C intersects the 50% and reac
CHAPTER 11
APPLICATIONS AND PROCESSING OF METAL ALLOYS
PROBLEM SOLUTIONS
11.3 Ferritic and austenitic stainless steels are not heat treatable since "heat treatable" is taken to mean
that martensite may be made to form with relative ease upon quenching aus
CHAPTER 12
STRUCTURES AND PROPERTIES OF CERAMICS
PROBLEM SOLUTIONS
12.5 This problem calls for us to predict crystal structures for several ceramic materials on the basis of
ionic charge and ionic radii.
(a) For CsI, from Table 12.3
r
Cs +
r
=
I
0.170 n m
Energies of Orbitals
As the number of electrons
increases, though, so does
the repulsion between
them.
Therefore, in many-electron
atoms, orbitals on the same
energy level are no longer
degenerate.
SURVEY OF ELEMENTS
Most elements: Electron configurati