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Lecture

ENG 2002 Lecture Notes - Diamond Cubic, Monochrome, Diffractometer

97 pages128 viewsWinter 2014

Department
Engineering
Course Code
ENG 2002
Professor
Hany E.Z.Farag

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CHAPTER 3
THE STRUCTURE OF CRYSTALLINE SOLIDS
PROBLEM SOLUTIONS
Fundamental Concepts
3.1 What is the difference between atomic structure and crystal structure?
Solution
Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the
number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the
arrangement of atoms in the crystalline solid material.
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Unit Cells
Metallic Crystal Structures
3.2 If the atomic radius of aluminum is 0.143 nm, calculate the volume of its unit cell in cubic meters.
Solution
For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC
crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.4 as
V
C
= 16R32 = (16)(0.143 10-9 m)3(2)= 6.62 10-29 m3
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3.3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic
radius R are related through a =4R/3.
Solution
Consider the BCC unit cell shown below
Using the triangle NOP
(NP)2= a2+ a2=2a2
And then for triangle NPQ,
(NQ)2=(
QP)2+(
NP)2
But
NQ = 4R, R being the atomic radius. Also,
QP = a. Therefore,
(4R)2 = a2 + 2a2
or
a = 4R
3
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