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ENGR-1210-HW-4-solutions.pdf
ENGR 1210 HW 4
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Homework 4 on chapter 4; (Total points: 60)
Reading assignments: Chapter 4 from textbook, notes and slides
Student Name: _______________________________ MTSU ID: ____________________
Honor pledge: I _______________________________ acknowledge that I have neither received nor given any unauthorized help/aid during this assignment.
Part 1 Key Engineering Terms: fill in the blanks – 20 points
1. Crystalline defect: a __________lattice___________________________ irregularity
having one or more of its dimensions on the order of an atomic diameter.
2. Grain: a single crystal of a __________polycrystalline____________ material.
3. Polycrystalline material: a material made of many _____crystals________.
4. Solid solution: a ____single___ phase atomic structure of an alloy of two metals.
5. Substitutional solid solution: solute atoms of one element replace those of solvent atoms
of other element occupying _____regular_________ lattice positions.
6. Interstitial solid solution: when solute atoms occupy ______interstitial___________ sites
or holes inside the solvent crystal lattice.
7. Vacancy: a ______point_________ defect/imperfection in a crystal lattice where an atom
is missing from a ______regular_________ lattice site in a crystal structure.
8. Self-interstitial: is an atom from the crystal that is ___crowded__ into an interstitial site.
9. Frenkel defect: a point defect in an ionic crystal in which a ____cation____ vacancy is
associated with an interstitial cation.
10. Schottky defect: a point defect in an ionic crystal in which a cation vacancy is associated
with ____anion_______ vacancy.
11. Dislocation: a ______line______ defect around which some of the atoms are misaligned.
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12. Edge dislocation: a _____line______ defect that centers on the line that is defined along
the end of ______extra________ half plane of atoms.
13. Screw dislocation: a dislocation produced by ______skewing_______ a crystal by one
atomic spacing so that a spiral ramp is produced.
14. Grain boundary: a surface or two dimensional defect that ____separates____ grains or
crystals of different orientation.
15. Twist boundary: an array of ________screw________ dislocations creating mismatch
inside a crystal.
16. Twin boundary: is a special type of grain boundary across which there is a specific
_____mirror____ lattice symmetry.
17. Phase boundary: exists in multiphase materials in which a _____second________ phase
exists on each side of boundary.
18. Stacking fault: a surface defect formed due to improper (out of plane) stacking of atomic
____planes____.
Part 2 Learning concepts/reflections/problems – 30 points
1. What are the four factors affecting the formation of a Substitutional solid solution?
Explain in details effect of each factor. 10 points
Atomic size factor. Appreciable quantities of a solute may be accommodated in this type of solid
solution only when the difference in atomic radii between the two atom types is less than about
15%. Otherwise the solute atoms will create substantial lattice distortions and a new phase will
form.
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Crystal structure. For appreciable solid solubility the crystal structures for metals of both atom
types must be the same.
Electronegativity. The more electropositive one element and the more electronegative the other,
the greater the likelihood that they will form an intermetallic compound instead of a
substitutional solid solution.
Valences. Other factors being equal, a metal will have more of a tendency to dissolve another
metal of higher valency than one of a lower valency.
2. Calculate equilibrium number of vacancies per m3 of copper at 1000 º C. assume energy
required to create one vacancy = 0.9 eV/atom, atomic weight of copper = 63.5 g/mol,
density of copper = 8.4 g/cm3 and Avogadro’s number = 6.022 x 1023 atoms/mol.
5 points
𝑁𝑁𝑣𝑣 = 𝑁𝑁 exp (− 𝑄𝑄𝑣𝑣 𝑅𝑅𝑅𝑅
)
Here, 𝑁𝑁 = 𝜌𝜌𝜌𝜌𝑁𝑁𝐴𝐴 𝐴𝐴
= 8.4∗10 6∗6.022∗1022
63.5 = 8 ∗ 1028𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎/𝑎𝑎3
Substituting the values,
𝑁𝑁𝑣𝑣 = 8 ∗ 1028 exp �− 0.9
8.62 ∗ 10−5 ∗ 1273 �
𝑁𝑁𝑣𝑣 = 2.2 ∗ 1025𝑣𝑣𝑎𝑎𝑣𝑣𝑎𝑎𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑎𝑎/𝑎𝑎3
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3. What is the composition in atomic % of an alloy that consists of 92.5 wt. % Ag and 7.5
wt. % Cu. Atomic weights of Ag and Cu are 107.87 g/mol and 63.55 g/mol respectively.
5 points
Using equation 4.6 (a and b);
𝐶𝐶𝐴𝐴𝑔𝑔 ′ =
𝐶𝐶𝐴𝐴𝑔𝑔𝐴𝐴𝐶𝐶𝐶𝐶 𝐶𝐶𝐴𝐴𝑔𝑔𝐴𝐴𝐶𝐶𝑢𝑢 + 𝐶𝐶𝐶𝐶𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴
𝑋𝑋100
Substituting the values,
𝐶𝐶𝐴𝐴𝑔𝑔 ′ = 87.9%
𝐶𝐶𝐴𝐴𝑔𝑔 ′ =
𝐶𝐶𝐶𝐶𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝐴𝐴𝑔𝑔𝐴𝐴𝐶𝐶𝑢𝑢 + 𝐶𝐶𝐶𝐶𝐶𝐶𝐴𝐴𝐴𝐴𝐴𝐴
𝑋𝑋100
Substituting the values,
𝐶𝐶𝐶𝐶𝐶𝐶 ′ = 12.1%
4. Differentiate between grains and grain boundary in terms of following:
a. Energy level
b. Accumulation of second phase (foreign atoms) 5 points
Grains Grain boundary
Energy level Low High
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Accumulation of second
phase
Not uniformly distributed Higher accumulation
Part 3 Materials design – 10 points
1. Which of the following systems (i.e., pair of metals) would you expect to exhibit
complete solid solubility? Explain your answers. Use following elemental data.
Metal Atomic Radius (nm) Crystal Structure Electronegativity Valence
Cr 0.125 BCC 1.6 +3
V 0.132 BCC 1.5 +5 (+3)
Mg 0.160 HCP 1.3 +2
Zn 0.133 HCP 1.7 +2
Al 0.143 FCC 1.5 +3
Zr 0.159 HCP 1.2 +4
Ag 0.144 FCC 1.4 +1
Au 0.144 FCC 1.4 +1
Pb 0.175 FCC 1.6 +2
Pt 0.139 FCC 1.5 +2
(a) Cr-V (b) Mg-Zn (c) Al-Zr (d) Ag-Au (e) Pb-Pt 10 points
In order for there to be complete solubility (substitutional) for each pair of metals, the four Hume-Rothery rules must
be satisfied: (1) the difference in atomic radii between Ni and the other element (∆R%) must be less than ±15%; (2)
the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the
same.
(a) A comparison of these four criteria for the Cr-V system is given below:
Metal Atomic Radius (nm) Crystal Structure Electronegativity Valence
Cr 0.125 BCC 1.6 +3
V 0.132 BCC 1.5 +5 (+3)
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For chromium and vanadium, the percent difference in atomic radii is approximately 6%, the crystal structures are the
same (BCC), and there is very little difference in their electronegativities. The most common valence for Cr is +3;
although the most common valence of V is +5, it can also exist as +3. Therefore, chromium and vanadium are
completely soluble in one another.
(b) A comparison of these four criteria for the Mg-Zn system is given below:
Metal Atomic Radius (nm) Crystal Structure Electronegativity Valence
Mg 0.160 HCP 1.3 +2
Zn 0.133 HCP 1.7 +2
For magnesium and zinc, the percent difference in atomic radii is approximately 17%, the crystal structures are the
same (HCP), and there is some difference in their electronegativities (1.3 vs. 1.7). The most common valence for both
Mg and Zn is +2. Magnesium and zinc are not completely soluble in one another, primarily because of the difference
in atomic radii.
(c) A comparison of these four criteria for the Al-Zr system is given below:
Metal Atomic Radius (nm) Crystal Structure Electronegativity Valence
Al 0.143 FCC 1.5 +3
Zr 0.159 HCP 1.2 +4
For aluminum and zirconium, the percent difference in atomic radii is approximately 11%, the crystal structures are
different (FCC and HCP), there is some difference in their electronegativities (1.5 vs. 1.2). The most common valences
for Al and Zr are +3 and +4, respectively. Aluminum and zirconium are not completely soluble in one another,
primarily because of the difference in crystal structures.
(d) A comparison of these four criteria for the Ag-Au system is given below:
Metal Atomic Radius (nm) Crystal Structure Electronegativity Valence
Ag 0.144 FCC 1.4 +1
Au 0.144 FCC 1.4 +1
For silver and gold, the atomic radii are the same, the crystal structures are the same (FCC), their electronegativities
are the same (1.4), and their common valences are +1. Silver and gold are completely soluble in one another because
all four criteria are satisfied.
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(e) A comparison of these four criteria for the Pb-Pt system is given below:
Metal Atomic Radius (nm) Crystal Structure Electronegativity Valence
Pb 0.175 FCC 1.6 +2
Pt 0.139 FCC 1.5 +2
For lead and platinum, the percent difference in atomic radii is approximately 20%, the crystal structures
are the same (FCC), their electronegativities are nearly the same (1.6 vs. 1.5), and the most common
valence for both of them is +2. Lead and platinum are not completely soluble in one another, primarily
because of the difference in atomic radii.
Bonus questions: 10 points
1. Explain differences between optical microscope and scanning electron microscope. Use
following points to differentiate. 5 points
Optical Microscope Scanning Electron Microscope
Source Visible light Electron Beam
Magnification 2000 X 300000 X
Image contrast Color Greyscale
EDS capabilities NO YES
2. In scanning electron microscope; why there is a need for sample to be always
conductive? 5 points
The source of SEM is a high energy electron beam; which interacts with the sample to generate
secondary electrons that makeup the image. If the sample is nonconductive, the electron beam
will create charging effect on the sample surface thus, making it impossible to distinguish
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contrast and a formation of image. The screen will be seen as a bright light with no contrast.
Hence, sample must be conductive to complete the electron path through the stage at ground.
Note: Please use extra sheets wherever necessary to answer the questions. All HW must be uploaded via D2L Dropbox system. Students are highly encouraged to refer textbook, notes and slides.
