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Exam (elaborations) TEST BANK FOR Materials Science and Engineering an
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, SOLUTIONS TO PROBLEMSPREFACE
This section of instructor's resource 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 recommend 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 E and F replace script E (electric field in Chapter
18) and script F (Faraday's constant in Chapter 17), 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.
1
, 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 masses
12
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
⎛ 1 mol ⎞ ⎛ 1 g / mol ⎞
# g/amu = ⎜ ⎟⎜ ⎟
⎝ 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 10
23
)
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.
2.5 The n quantum number designates the electron shell.
2
, 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
1
possible ml values are 0 and ±1. Therefore, for the s states, the quantum numbers are 200 ( ) and
2
1 1 1 1 1
200 (− ) . For the p states, the quantum numbers are 210 ( ) , 210 (− ) , 211 ( ) , 211 (− ) ,
2 2 2 2 2
1 1
21(-1 )( ) , and 21(-1 )(− ) .
2 2
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2;
1
possible ml values are 0, ±1, and ±2; and possible ms values are ± . Therefore, for the s states,
2
1 1 1 1 1
the quantum numbers are 300 ( ) , 300 (− ) , for the p states they are 310 ( ) , 310 (− ) , 311 ( ) ,
2 2 2 2 2
1 1 1 1 1 1
311 (− ) , 31(-1 )( ) , and 31(-1 )(− ) ; for the d states they are 320 ( ) , 320 (− ) , 321 ( ) ,
2 2 2 2 2 2
1 1 1 1 1 1 1
321 (− ) , 32(-1 )( ) , 32(-1 )(− ) , 322 ( ) , 322 (− ) , 32(-2) ( ) , and 32(-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 Na ion is just a sodium atom that has lost one electron; therefore, it has an electron
configuration the same as neon (Figure 2.6).
-
The Cl ion is a chlorine atom that has acquired one extra electron; therefore, it has an
electron configuration the same as argon.
2.9 Each of the elements in Group IIA has two s 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.
3
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