Homework # 1  Solution
Problem #1
Simple cubic structure:
The distance between nearestneghbor atoms is
a
. Therefore the radius
r
of the touching spheres
located at lattice sites is
/ 2
=
r
a
. Since in the simple cubic structure there is one atom in the unit cell,
the volume coved by the spheres is
3
3
4
4
3
3
2
π
π
°
±
=
=
²
³
´
µ
sc
a
V
r
, whereas the total volume of the unit cell is
3
=
cell
V
a
. The packing ratio is, therefore,
3
3
4
3
2
0.52
6
π
π
°
±
²
³
´
µ
=
=
=
≈
sc
sc
cell
a
V
p
V
a
.
Bcc structure:
The distance between nearestneghbor atoms is
3 / 2
a
and therefore
3 / 4
=
r
a
. In the bcc structure
there are 2 atoms in the conventional unit cell so that the volume coved by the spheres is
3
4
3
2
3
4
π
°
±
=
²
³
²
³
´
µ
bcc
a
V
. The packing ratio is
3
0.68
8
π
=
=
≈
bcc
bcc
cell
V
p
V
.
Fcc structure:
The distance between nearestneghbor atoms is
2 / 2
a
, so that
2 / 4
=
r
a
. In the fcc structure there
are 4 atoms in the conventional unit cell, and therefore the volume coved by the spheres is
3
4
2
4
3
4
π
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 Spring '08
 WORMER
 Crystallography, Mineralogy, Work, Cubic crystal system, Diamond cubic, primitive cell, Wigner–Seitz cell

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