Exam (elaborations)

TEST BANK FOR Engineering Materials Science By Milton Ohring (Solution Manual)

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TEST BANK FOR Engineering Materials Science By Milton Ohring

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Harvard University

Exam (elaborations) TEST BANK FOR Engineering Materials Science By Milton Ohring (Solution Manual) Exam (elaborations) TEST BANK FOR Engineering Materials Science By Milton Ohring (Solution Manual) SOLUTIONS TO PROBLEMS PREFACE This section of instructors materials contains solutions and answers ...

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milton ohring
test bank
engineering materials science

Institution
Harvard University
Course
TEST BANK FOR Engineering Materials Science By Milton Ohring

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, SOLUTIONS TO PROBLEMS

PREFACE

This section of instructors materials contains solutions and answers to all
problems and questions that appear in the textbook. My penmanship leaves
something to be desired; therefore, I generated these solutions/answers using
computer software so that the resulting product would be "readable." Furthermore, I
endeavored to provide complete and detailed solutions in order that: (1) the
instructor, without having to take time to solve a problem, will understand what
principles/skills are to be learned by its solution; and (2) to facilitate student
understanding/learning when the solution is posted.

I would recommended that the course instructor consult these
solutions/answers before assigning problems and questions. In doing so, he or she
ensures that the students will be drilled in the intended principles and concepts. In
addition, the instructor may provide appropriate hints for some of the more difficult
problems.

With regard to symbols, in the text material I elected to boldface those symbols
that are italicized in the textbook. Furthermore, I also endeavored to be consistent
relative to symbol style. However, in several instances, symbols that appear in the
textbook were not available, and it was necessary to make appropriate substitutions.
These include the following: the letter a (unit cell edge length, crack length) is used in
place of the cursive a. And Roman F and E replace script F (Faraday's constant in
Chapter 18) and script E (electric field in Chapter 19), respectively.

I have exercised extreme care in designing these problems/questions, and then
in solving them. However, no matter how careful one is with the preparation of a work
such as this, errors will always remain in the final product. Therefore, corrections,
suggestions, and comments from instructors who use the textbook (as well as their
teaching assistants) pertaining to homework problems/solutions are welcomed. These
may be sent to me in care of the publisher.

, 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 atomic
12
masses of the atoms generally are not integers (except for C), and (2) the atomic weight is
taken as the weighted average of the atomic masses of an atom's naturally occurring isotopes.

2.2 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 occurring isotopes.

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

 
#g/amu =
1 mol
( 1 g/mol
 6.023 x 1023 atoms 1 amu/atom )
= 1.66 x 10-24 g/amu

(b) Since there are 453.6 g/lbm ,

1 lb-mol = (453.6 g/lbm )(6.023 x 1023 atoms/g-mol)

= 2.73 x 1026 atoms/lb-mol

2.4 (a) Two important quantum-mechanical concepts associated with the Bohr model of the atom
are that electrons are particles moving in discrete orbitals, and electron energy is quantized into
shells.
(b) Two important refinements resulting from the wave-mechanical atomic model are that
electron position is described in terms of a probability distribution, and electron energy is
quantized into both shells and subshells--each electron is characterized by four quantum
numbers.

1

, 2.5 The n quantum number designates the electron shell.
The l quantum number designates the electron subshell.
The m quantum number designates the number of electron states in each electron subshell.
l
The m quantum number designates the spin moment on each electron.
s

2.6 For the L state, n = 2, and eight electron states are possible. Possible l values are 0 and 1,
while possible m l values are 0 and ±1. Therefore, for the s states, the quantum numbers are
1 1 1 1 1
200( ) and 200(- ). For the p states, the quantum numbers are 210( ), 210(- ), 211( ), 211(-
2 2 2 2 2
1 1 1
), 21(-1)( ), and 21(-1)(- ).
2 2 2
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2;
1
possible m l values are 0, ±1, and ±2; and possible m s values are ± . Therefore, for the s
2
1 1 1 1
states, the quantum numbers are 300( ), 300(- ), for the p states they are 310( ), 310(- ),
2 2 2 2
311(\F(1,2)), 311(-\F(1,2)), 31(-1)(\F(1,2)), and 31(-1)(-\F(1,2)); for the d states they are
1 1 1 1 1 1 1 1 1 1
320( ), 320(- ), 321( ), 321(- ), 32(-1)( ), 32(-1)(- ), 322( ), 322(- ), 32(-2)( ), and 32(-2)(- ).
2 2 2 2 2 2 2 2 2 2

2.7 The electron configurations of the ions are determined using Table 2.2.

2+ 2 2 6 2 6 6
Fe - 1s 2s 2p 3s 3p 3d
3+ 2 2 6 2 6 5
Fe - 1s 2s 2p 3s 3p 3d
+ 2 2 6 2 6 10
Cu - 1s 2s 2p 3s 3p 3d
2+ 2 2 6 2 6 10 2 6 10 2 6
Ba - 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p
- 2 2 6 2 6 10 2 6
Br - 1s 2s 2p 3s 3p 3d 4s 4p
2- 2 2 6 2 6
S - 1s 2s 2p 3s 3p

+
2.8 The Cs ion is just a cesium atom that has lost one electron; therefore, it has an electron
configuration the same as xenon (Figure 2.6).
-
The Br ion is a bromine atom that has acquired one extra electron; therefore, it has an
electron configuration the same as krypton.

2.9 Each of the elements in Group VIIA has five p electrons.

2 2 6 2 6 7 2
2.10 (a) The 1s 2s 2p 3s 3p 3d 4s electron configuration is that of a transition metal because
of an incomplete d subshell.

2

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