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CHAPTER 3
THE STRUCTURE OF CRYSTALLINE SOLIDS
PROBLEM SOLUTIONS
3.9
We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure.
For FCC,
n
= 4 atoms/unit cell, and
V
C
=
16
R
3
2
[Equation (3.4)].
Now,
ρ
=
nA
Ir
V
C
N
A
And solving for
R
from the above two expressions yields
R =
nA
Ir
16
ρ
N
A
2
1/3
=
4 atoms/unit cell
(
29
192.2 g/mol
(
29
2
( 29
16
(
29
22.4 g/cm
3
(
29
6.023 x 10
23
atoms/mol
(
29
1/3
= 1.36 x 10
8
cm = 0.136 nm
3.15
For each of these three alloys we need to, by trial and error, calculate the density using Equation
(3.5), and compare it to the value cited in the problem.
For SC, BCC, and FCC crystal structures,
the respective values of
n
are 1, 2, and 4, whereas the expressions for
a
(since
V
C
=
a
3
) are 2
R
,
2
R
2
, and 4
R
/
3
.
For alloy A, let us calculate
ρ
assuming a BCC crystal structure.
ρ
=
nA
A
V
C
N
A
11
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(2 atoms/unit cell)(43.1 g/mol)
(4) 1.22 x 10
8
cm
(
29
3
3
/(unit cell)
6.023 x 10
23
atoms/mol
(
29
= 6.40 g/cm
3
Therefore, its crystal structure is BCC.
For alloy B, let us calculate
ρ
assuming a simple cubic crystal structure.
ρ
=
(1 atom/unit cell)(184.4 g/mol)
2
( 29
1.46 x 10
8
cm
(
29
3
/(unit cell)
6.023 x 10
23
atoms/mol
(
29
= 12.3 g/cm
3
Therefore, its crystal structure is simple cubic.
For alloy C, let us calculate
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This note was uploaded on 08/30/2011 for the course EGN 3365 taught by Professor Staff during the Spring '11 term at University of Florida.
 Spring '11
 Staff

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