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Lecture

# ENG 2002 Lecture Notes - Jmol, Linear Interpolation, Carburizing

6 pages34 viewsWinter 2014

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

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PROBLEM SOLUTIONS
1. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume
an energy for vacancy formation of 0.55 eV/atom.
Solution
In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation
4.1. As stated in the problem, Qv = 0.55 eV/atom. Thus,
N
v
N= exp Q
v
kT
= exp 0.55 eV/atom
(8.62 × 10
5
eV/atom- K)(600 K)
= 2.41 × 10-5
2. Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of
vacancies at 500
°
C (773 K) is 7.57
×
1023 m-3. The atomic weight and density (at 500
°
C) for aluminum are,
respectively, 26.98 g/mol and 2.62 g/cm3.
Solution
Upon examination of Equation 4.1, all parameters besides Qv are given except N, the total number of
atomic sites. However, N is related to the density, (ρAl), Avogadro's number (NA), and the atomic weight (AAl)
according to Equation 4.2 as
N = N
A
ρ
Al
A
Al
=
(
6.022 × 10
23
atoms /mol
)(
2.62 g /cm
3
)
26.98 g /mol
= 5.85 × 1022 atoms/cm3 = 5.85 × 1028 atoms/m3
Now, taking natural logarithms of both sides of Equation 4.1,
ln N
v
= ln N Q
v
kT
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and, after some algebraic manipulation
Q
v
= kT ln N
v
N
=
(
8.62 × 10-5 eV/atom- K
)
(500°C + 273 K) ln 7.57 × 1023 m3
5.85 × 1028 m3
= 0.75 eV/atom
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.