Calister-concept check questions answers ch01-21 fundamental of materials science and engineering
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Calister-concept check
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Calister-concept Check
Calister-concept check questions answers ch01-21
fundamental of materials science and engineering (Orta Doğu Teknik Üniversitesi)
Calister-concept check questions answers ch01-21
fundamental of materials science and engineering (Orta Doğu Teknik Üniversitesi)
QUESTIONS AND ANSWERS
Cha...
Calister-concept check questions answers
ch01-21
fundamental of materials science and engineering (Orta Doğu
Teknik Üniversitesi)
Calister-concept check questions answers
ch01-21
fundamental of materials science and engineering (Orta Doğu
Teknik Üniversitesi)
QUESTIONS AND ANSWERS
Chapter 2
Atomic Structure and Interatomic Bonding
Concept Check 2.1
Question: Why are the atomic weights of the elements generally not integers? Cite
two
reasons.
Answer: The atomic weights of the elements ordinarily are not integers because: (1)
the
atomic masses of the atoms normally are not integers (except for 12C), and (2) the
atomic weight
is taken as the weighted average of the atomic masses of an atom's naturally
occurring isotopes.
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Concept Check 2.2
Question: Give electron configurations for the Fe3+and S2- ions.
Answer: The Fe3+ ion is an iron atom that has lost three electrons. Since the
electron
configuration of the Fe atom is 1s 22s 22p 63s 23p 63d 64s 2 (Table 2.2), the
configuration for Fe3+ is
1s 22s 22p 63s 23p 63d 5 .
The S2- ion a sulfur atom that has gained two electrons. Since the electron
configuration
,of the S atom is 1s 22s 22p 63s 23p 4 (Table 2.2), the configuration for S2- is 1s 22s
22p 63s 23p 6 .
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Concept Check 2.3
Question: Explain why covalently bonded materials are generally less dense than
ionically or metallically bonded ones.
Answer: Covalently bonded materials are less dense than metallic or ionically
bonded
ones because covalent bonds are directional in nature whereas metallic and ionic
are not; when
bonds are directional, the atoms cannot pack together in as dense a manner,
yielding a lower mass
density.
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The Structure of Crystalline Solids
Concept Check 3.1
Questions:
(a) What is the coordination number for the simple-cubic crystal structure?
(b) Calculate the atomic packing factor for simple cubic.
Answers:
(a) For the simple cubic crystal structure the coordination number (the number of
nearest
neighbor atoms) is six. This is demonstrated in the figure below.
Consider the atom labeled A, which is located at the corner of the reduced-sphere
simple cubic
unit cell. It has three nearest neighbors located in this unit cell—labeled 1, 3, and 5.
In addition,
the three shaded atoms, labeled 2, 4, and 6, are also nearest neighbors that belong
to adjacent unit
cells.
(b) The atomic packing factor is the total sphere volume-cell volume ratio (Equation
3.3). For total sphere volume it is necessary to compute the number of atoms per
unit cell using
,Equation 3.2, realizing that, for simple cubic there are eight corner atoms and no
face and no
interior atoms—i.e., Nc = 8 and
. Therefore,
N = Ni + N
f
2
+ N
c
8
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lOMoARcPSD|47166518
Ni = N f = 0= 0 + 0
2
+ 8
8
= 1 atom/per unit cell
Therefore, the total sphere volume in terms of the atomic radius R is equal to
Because the unit cell is cubic, its volume is equal to the edge length (a) cubed—that
is
And because a = 2R
Finally, using Equation 3.3, the APF is computed as follows:
= 0.52
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lOMoARcPSD|47166518
VS = (1 atom/unit cell)
4
3
R 3
æ
è
ç
ö
, ø ÷ = 43R 3
VC = a 3
VC = (2R) 3 = 8R 3
APF = VS
VC
=
4
3 R
3
8R
3 lOMoARcPSD|47166518
Concept Check 3.2
Question: What is the difference between crystal structure and crystal system?
Answer: A crystal structure is described by both the geometry of, and atomic
arrangements within, the unit cell, whereas a crystal system is described only in
terms of the unit
cell geometry. For example, face-centered cubic and body-centered cubic are
crystal structures
that belong to the cubic crystal system.
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Concept Check 3.3
Question: For cubic crystals, as values of the planar indices h, k, and l increase,
does
the distance between adjacent and parallel planes (i.e., the interplanar spacing)
increase or
decrease? Why?
Answer: The interplanar spacing between adjacent and parallel planes decreases as
the
values of h, k, and l increase. As values of the planar indices increase, the
magnitude of the
denominator in Equation 3.22 increases, with the result that the interplanar spacing
(dhkl)
decreases.
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