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

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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 m−3

5.85 × 1028 m−3

= 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

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