9. Equations 7.1a and 7.1b, expressions for Burgers vectors for FCC and BCC crystal structures, are of theform(uvw)b =where a is the unit cell edge length. Also, since the magnitudes of these Burgers vectors may bedetermined from the following equation:²+v²+w-determine values of |b| for aluminum and chromium. You may want to consult Table 3.1 in (Callister).

For FCC,r=a22For aluminum metal,r=0.1431nma=22ra=22(0.1431)a=0.4047nmFor BCC,r=a34For chromium…

Answered: 9. Equations 7.1a and 7.1b, expressions…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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6. Within a cubic unit cell, sketch the following directions: ( :)) (a) [110] (e) [771] (b) [121] (f) [722] (c) [072] (d) [133] (g) [123] (h) [103]
4. Platinum (atomic radius = 1.38 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face- centered cubic unit cell and the density of platinum.
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3.45 Determine the Miller indices for the planes O shown in the following unit cell: +z A +y 1 2 +x Note: Both of these planes are displayed in Virtual Materials Science and Engineering (VMSE) (accessed through all digital versions of the text) under Metallic Crystal Structures and Crystallography/ Crystallographic Problem 2: Planes/Unit Cell Planes: Plane A Plane B Each image may be rotated (by click-and-drag), which allows you to obtain a better 3D perspective of the plane's orientation. 2/3
Sodium chloride (NaCl) has the rock salt crystal structure and a density of 2.17 g/cm³. The atomic weights of sodium and chlorine are 22.99 g/mol and 35.45 g/mol, respectively. (a) Determine the unit cell edge length. nm (b) Determine the unit cell edge length from the radii in the table below assuming that the Nat and Cl- ions just touch each other along the edges. nm Cation Mg2+ Fe2+ Na+ Ionic Radius (nm) 0.072 0.077 0.102 Anion Ionic Radius (nm) CI- 0²- 0.181 0.140
Consider two hypothetical metallic crystal structures A and B. A has an fcc and Ba bcc lattice structure. Both have the same unit cell volume. The Atomic Packing Factor for A is 0.740 and that for B is 0.680. What is the ratio of the fcc to bcc atomic radii? (This question has only one correct answer) O a. 0.92 Ob. 0.44 Ос. 0.63 O d. 0.82 е. 1.21
Tungsten (W) has the BCC crystal structure. The atomic radius is R = 0.137 nm and the atomic mass is M = 183.95 g/mole. %3D What is the density of Tungsten in g/cm³? Given: Avogadro's Number NA = 0.6023 × 1024 (atoms/mole) Select one: O a. 5.30 O b. 6.28 Oc. 2.534 Od. 6.77 Oe. 19.28
1. It is often convenient to represent a fcc lattice as simple cubic, with a cubic primitive cell of side a and a four-point basis. (a) Show that the structural factor is then either 4 or 0 at all points of the simple cubic reciprocal lattice. (b) Show that when points with zero structural factor are removed, the remaining points of the reciprocal lattice makes up a bcc cubic lattice with conventional cell of side 4л/a. Why is this expected?
Copper is a typical face-centered-cubic metal. In its pure state, it has extremely high electrical conductivity. Oxygen in copper causes a major reduction of conductivity. To investigate this issue further, I want to you to tell me whether oxygen atoms substitute for copper atoms, or if they enter interstitial sites.
Iron (Fe) has the Body-Centered Cubic (BCC) crystal structure. The edge length is a = 0.2864 nm. What is the linear density in (atoms/nm) along direction [011]? Select one: O a. 6.98 O b.7.48 Oc494 Od.349 O e 2.47
Lead and Aluminum both have FCC crystal structures. Given the following information and knowing that solid aluminum has a density of 2.71 g/cm3, determine the density of solid lead in g/cm3 Atomic Weight (amu) Atomic Radius (nm) Aluminum 26.98 0.143 Lead 207.2 0.175
Consider an fcc lattice with cube side length 1.2 Å. What are the planar densities of the (110) and (100) planes respectively (in units of 1020 atoms/m?): (This question has only one correct answer) O a. (100) = 2.08, (110) = 0.98 O b. (100) = 1.39, (110) = 0.98 O c. (100) = 1.39, (110) = 1.20 O d. (100) = 0.98, (110) = 1.39

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