5.1 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
5.1. As stated in the problem, Qv = 0.55 eV/atom. Thus,


 Q 
Nv
0.55 eV / atom
= exp  v = exp 

 kT 
N
 (8.62  10 5 eV / atom- K) (600 K) 

= 2.41  10-5
5.6 Calculate the number of Frenkel defects per cubic meter in zinc oxide at 1000°C. The energy for defect
formation is 2.51 eV, whereas the density for ZnO is 5.55 g/cm3 at 1000°C.
Solution
This problem asks that we compute the number of Frenkel defects per cubic meter in zinc oxide at 1000C.
Solution of this problem is possible using Equation 5.3. However, we must first determine the value of N, the
number of lattice sites per cubic meter, which is possible using a modified form of Equation 5.2; thus
N =


N A
AZn + AO
(6.022  10 23 atoms/mol)(5.55 g/cm 3 )(10 6 cm 3/m3)

65.41 g/mol + 16.00 g/mol
= 4.11  1028 lattice sites/m3
And, finally the value of Nfr is computed using Equation 5.3 as
 Q fr 
N fr = N exp 

 2kT 


2.51 eV
3
= (4.11  10 28 lattice

 sites/m ) exp 
-5
 (2)(8.62  10 eV/K)(1000 + 273 K) 

= 4.43  1023 defects/m3
5.11
Atomic radius, crystal structure, electronegativity, and the most common valence are tabulated in the
following table for several elements; for those that are nonmetals, only atomic radii are indicated.
Cu
Atomic Radius
(nm)
0.1278
C
0.071
H
0.046
O
0.060
Ag
0.1445
FCC
1.9
+1
Al
0.1431
FCC
1.5
+3
Co
0.1253
HCP
1.8
+2
Cr
0.1249
BCC
1.6
+3
Fe
0.1241
BCC
1.8
+2
Ni
0.1246
FCC
1.8
+2
Pd
0.1376
FCC
2.2
+2
Pt
0.1387
FCC
2.2
+2
Zn
0.1332
HCP
1.6
+2
Element
Crystal Structure
Electronegativity
Valence
FCC
1.9
+2
Which of these elements would you expect to form the following with copper:
(a) A substitutional solid solution having complete solubility
(b) A substitutional solid solution of incomplete solubility
(c) An interstitial solid solution
Solution
In this problem we are asked to cite which of the elements listed form with Cu the three possible solid
solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic
radii between Cu and the other element (R%) must be less than ±15%, 2) the crystal structures must be the same,
3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are
tabulated, for the various elements, these criteria.
Element
R%
Cu
C
H
O
Ag
Al
Co
Cr
Fe
Ni
Pd
Pt
Zn
–44
–64
–53
+13
+12
-2
-2
-3
-3
+8
+9
+4
Electronegativity
Crystal
Structure
FCC
Valence
2+
FCC
FCC
HCP
BCC
BCC
FCC
FCC
FCC
HCP
0
-0.4
-0.1
-0.3
-0.1
-0.1
+0.3
+0.3
-0.3
1+
3+
2+
3+
2+
2+
2+
2+
2+
(a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete
solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure,
and thus display complete solid solubility at these temperatures.
(b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals
have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are
greater than ±15%, and/or have a valence different than 2+.
(c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly
smaller than the atomic radius of Cu.
5.43 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 8 grains per square inch are measured
at a magnification of 600. In order to solve this problem we make use of Equation 5.20:
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 = 8, while M = 600, we have
 M 
log N M  2 log  
100 
n
1
log 2

600 
log 8  2 log  
100 

 1  9.2
log 2
5.2FE What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?

The atomic weights for Pb and Sn are 207.19 g/mol and 118.71 g/mol, respectively.
(A) 2.6 at% Pb and 97.4 at% Sn
(B) 7.6 at% Pb and 92.4 at% Sn
(C) 97.4 at% Pb and 2.6 at% Sn
(D) 92.4 at% Pb and 7.6 at% Sn
Solution
We are asked to compute the composition of an alloy in atom percent. Employment of Equation 5.9 leads
to
CPb
 =
=
CPb ASn
 100
CPb ASn  CSn APb
(4.5)(118.71 g/mol)
 100
(4.5)(118.71 g/mol)  (95.5)(207.19 g/mol)
= 2.6 at%
CSn
 =
=
CSn APb
 100
CSn APb  CPb ASn
(95.5)(207.19 g/mol)
 100
(95.5 )(207.19 g/mol)  (4.5)(118.71 g/mol)
= 97.4 at%
which is answer A (2.6 at% Pb and 97.4 at% Sn).