CHAPTER 4
IMPERFECTIONS IN SOLIDS
4-2, 4-19, AND 4-35
4.2
Calculate the number of vacancies per cubic meter in
iron at 850C.
The energy for vacancy formation is 1.08
eV/atom. Furthermore, the density and atomic weight for Fe
are 7.65 g/cm3 and 55.85 g/mol, respectively.
Solution
Determination of the number of vacancies per cubic
meter in iron at 850
Equations 4.1 and 4.2 as follows:
 Q  N A Fe
 Q 
N v = N exp  v =
exp  v 
 kT 
 kT 
AFe
And incorporation of values of the parameters provided in

the problem statement into the above equation leads to

Nv =
(6.022

 10 23atoms / mol)(7.65 g / cm3)
exp 
55.85 g / mol
 (8.62  10 5
= 1.18  1018 cm-3 = 1.18  1024 m-3
4.8 What is the composition, in weight percent, of an alloy
that consists of 6 at% Pb and 94 at% Sn?
Solution
In order to compute composition, in weight percent, of
a 6 at% Pb-94 at% Sn alloy, we employ Equation 4.7 as
CPb =
=
' A
CPb
Pb
C'
Pb APb
 100
Sn ASn
(6)(207.2 g / mol)
 100
(6)(207.2 g / mol)  (94)(118.71 g / mol)
= 10.0 wt%

CSn =


C'
=
' A
CSn
Sn
C'
Pb APb
C'
 100
Sn ASn
(94)(118.71 g / mol)
 100
(6)(207.2 g / mol)  (94)(118.71 g / mol)
= 90.0 wt%
For a solid solution consisting of two elements
4.19
(designated as 1 and 2), sometimes it is desirable to
determine the number of atoms per cubic centimeter of one
element in a solid solution, N1, given the concentration of
that
element
specified
in
weight
percent,
C1.
This
computation is possible using the following expression:
N1 
N A C1
C1 A1
A
 1 100  C1
1
2
(4.18)
where
NA = Avogadro’s number

ρ1 and ρ2 = densities of the two elements
A1 = the atomic weight of element 1
Derive Equation 4.18 using Equation 4.2 and expressions
contained in Section 4.4.
Solution
This problem asks that we derive Equation 4.18, using
other equations given in the chapter. The concentration of
component 1 in atom percent
(C1' )
is just 100 c1' where
atom fraction of component 1. Furthermore,



c1'
c1'
is the
is defined as

= N1/N where N1 and N are, respectively, the number of
c1'

atoms of component 1 and total number of atoms per cubic
centimeter. Thus, from the above discussion the following
holds:
N1 =
C1' N
100
Substitution into this expression of the appropriate form of N

from Equation 4.2 yields
C1' N A ave
N1 =
100 Aave
And, finally, substitution into this equation expressions for

'
C1

ave (Equation 4.10a), Aave (Equation
4.11a), and realizing that C2 = (C1 – 100), and after some
algebraic manipulation we obtain the desired expression:
N1 =
N AC1
C1 A1
1


A1
2
(100
 C1)
4.35 Determine the ASTM grain size number if 25 grains per
square inch are measured at a magnification of 600.
Solution
This problem asks that we determine the ASTM grain
size number if 25 grains per square inch are measured at a
magnification of 600.
In order to solve this problem we
make use of Equation 4.17:
2
 M 
N M    2n  1
100 
where NM = the number of grains per square inch at
magnification M, and n is the ASTM grain size number.
Solving the above equation for n, and realizing that NM = 25,
while M = 600, we have
 M 
log N M  2 log  
100 
n
1
log 2

 600 
log 25  2 log 

 100  110.8

log 2