MSEG302 HOMEWORK 2 (Chapters 3 and 4) 3.2 (10 points, completion) 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 calculate the volume of a unit unit cell of aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed computed from Equation 3.4 as
(16) (0.143 V C = 16 R 3 2 = (16)
×
10 -9 m) 3( 2 ) = 6.62 6.62
×
10 -29 m3
3.14 ( 20 20 points, detailed grading) Below are listed the atomic weight, density, and atomic radius for
three hypothetical alloys. For each determine whether whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in Figure 3.24.
Alloy
Atomic Weight ( g/mol )
A B
C
Density 3 ( g/cm )
Atomic Radius nm)) (nm
77.4
8.22
107.6
13.42
127.3
9.23
0.125 0.133 0.142
Solution For each of these three alloys we need, by trial and error, to calculate the density using Equation 3.5, and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, structures, the respective values of n are 1, 2, and 4 R 4, whereas the expressions for a (since V C = a3) are 2 R, and 2 R 2 . 3 For alloy A, let us calculate
ρ assuming a simple cubic crystal structure.
ρ
=
=
nAA V C N A
nAA
(2 R)3 N A
(1 atom/unit cell)(77.4 g/mol)
=
[(2)(1.25 × 10− )] /(unit cell) (6.022 × 10 8
3
23
atoms/mol)
= 8.22 g/cm3
Therefore, its crystal structure is simple cubic.
For alloy B, let us calculate
ρ assuming an FCC crystal structure.
ρ
=
nAB
=
(2 R 2) 3 N A
(4 atoms/unit cell)(107.6 g/mol)
[(2 2 )(1.33 × 10
-8
]
3
cm) /(unit cell) (6.022
×
10 23 atoms/mol)
= 13.42 g/cm3
Therefore, its crystal structure is FCC.
For alloy C, let us calculate
ρ assuming a simple cubic crystal structure. =
=
nAC
(2 R)3 N A
(1 atom/unit cell)(127.3 g/mol)
[(2)(1.42 × 10
-8
]
3
cm) /(unit cell) (6.022
×
10 23 atoms/mol)
= 9.23 g/cm3 Therefore, its crystal structure is simple cubic.
3.30 (10 points, completion) Within a cubic unit cell, sketch the following directions: (a)
[1 10],
(e) [1 1 1] ,
(b)
[1 2 1],
(f)
[1 22] ,
(c)
[01 2] ,
(g) [12 3 ],
(d)
[13 3] ,
(h) [1 03] .
Solution The directions asked for are indicated in the cubic unit cells shown below.
3.40 (20 points, detailed grading) Sketch within a cubic unit cell the following planes: (a) (01 1 ) ,
(e) (1 11 ) ,
(b) (112) ,
(f) (12 2) ,
(c) (102) ,
(g) (1 23),
(d) (13 1) ,
(h) (01 3)
Solution The planes called for are plotted in the cubic unit cells shown below.
4.2 (10 points, detailed grading) 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 3
are 7.65 g/cm and 55.85 g/mol, respectively. Solution Determination of the number of vacancies per cubic meter in iron at 850°C (1123 K) requires the utilization of Equations 4.1 and 4.2 as follows:
N v = N exp
Qv − = kT
N A ρFe AFe
Q exp − v kT
And incorporation of values of the parameters provided in the problem statement into the above equation leads to
N v =
(6.022 ×
10 23 atoms/ mol)( 7.65 g/cm3) 55.85 g/mol
exp
− (8.62 ×
1.08 eV/atom 10−5 eV/atom − K ) (850 °C +
273 K)
= 1.18 × 1018 cm-3 = 1.18 × 1024 m-3
4.7 (10 points, detailed grading) What is the composition, in atom percent, of an alloy that consists of
30 wt% Zn and 70 wt% Cu? Solution In order to compute composition, in atom percent, of a 30 wt% Zn-70 wt% Cu alloy, we employ Equation 4.6 as
'
C Zn =
=
C Zn ACu C Zn ACu + C Cu AZn
×
100
(30)(63.55 g / mol) (30)(63.55 g / mol) + (70)(65.41 g / mol)
= 29.4 at%
'
C Cu =
C Cu AZn C Zn ACu + C Cu AZn
×
100
×
100
=
(70)(65.41 g / mol) (30)(63.55 g / mol) + (70)(65.41 g / mol)
×
100
= 70.6 at%
4.8 (10 points, completion) 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
'
C Pb =
C Pb APb '
'
100
C APb + C ASn
Pb
=
×
Sn
(6)(207.2 g / mol) (6)(207.2 g / mol) + (94)(118.71 g / mol)
×
100
= 10.0 wt%
'
C Sn
=
=
C Sn ASn '
'
×
100
C Pb APb + C Sn ASn
(94)(118.71 g / mol) (6)(207.2 g / mol) + (94)(118.71 g / mol)
×
100
= 90.0 wt%
EXTRA (not assigned): 4.14 Calculate the number of atoms per cubic meter in aluminum. Solution In order to solve this problem, one must employ Equation 4.2,
N
=
N A ρ Al AAl
The density of Al (from the table inside of the front cover) is 2.71 g/cm 3, while its atomic weight is 26.98 g/mol. Thus,
N =
(6.022 × 10 23 atoms/mol)( 2.71 g/ cm3) 26.98 g / mol
= 6.05 × 1022 atoms/cm3 = 6.05 × 1028 atoms/m3
4.16 (10 points, completion) Determine the approximate density of a high-leaded brass that has a
composition of 64.5 wt% Cu, 33.5 wt% Zn, and 2.0 wt% Pb. Solution In order to solve this problem, Equation 4.10a is modified to take the following form:
ρ ave
=
100 C Cu
ρCu
+
C Zn
+
ρ Zn
C Pb
ρPb
And, using the density values for Cu, Zn, and Pb—i.e., 8.94 g/cm3, 7.13 g/cm3, and 11.35 g/cm3 —(as taken from inside the front cover of the text), the density is computed as follows:
ρ ave
=
64.5 wt% 8.94
g /cm3
+
100 33.5 wt% 7.13
g/cm3
+
2.0 wt% 11.35 g / cm3
= 8.27 g/cm3
NOT ASSIGNED 4.29 (a) For a given material, would you expect the surface energy to be greater
than, the same as, or less than the grain boundary energy? Why? Solution (a) The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary.
