CHEM 123: Assignment # 1
page 1 of 10
1.
Simple cubic packing.
The crystal lattic and unit cell for simple cubic packing are shown
below.
a) # spheres
=
()
(
1
8
sphere/corner
8 corners
)
=
1 sphere
The coordination number of each sphere is 6.
To see this, focus on an “interior” sphere
such as sphere “A” in the figure on the left. That sphere is in contact with 6 other spheres
– the one above it, the one below, the one in front, the one behind, the one to its left and
the one to its right.
b
) If the radius of each sphere is “
R
” then the edge length of the unit cell is
a
= 2
R
c)
The volume of the unit cell is
V
cell
=
a
3
=
8
R
3
There is
one full sphere
contained within the unit cell so the volume of spheres
contained within the unit cell is
3
spheres
4
3
VR
=
π
.
3
4
3
3
0.5236
6
8
spheres
cell
V
R
Packing
Efficiency
V
R
π
π
==
=
=
2.
Show that the following data are consistent with the fact that silver metal
crystallizes in a facecentred cubic lattice:
Edge length of unit cell,
a
= 408 pm
Density of silver, 10.6 g cm
−
3
Atomic weight of silver, 107.9 g mol
−
1
,
Avogadro's number,
N
A
= 6.022
×
10
23
If silver crystallizes in a facecentred cubic lattice, then
1
23
1
3
10
3
107.9 g mol
6.022 10
atoms mol
(4atoms)
10.6 g cm
( 408 10
cm)
cell
cell
m
V
d
−
−
−
−
×
×
⎛⎞
⎜⎟
⎝⎠
=
A
The calculated density agrees well with the observed density.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentCHEM 123: Assignment 1
page 2 of 10
2
3
.
A tetrahedral site ican be generated by placing four spheres of radius
R
at alternate
corners of a cube.
Since the spheres are in contact at the centre of each cube face, the
length of the face diagonal of this cube is equal to 2
R
.
Since the centre of the
tetrahedral site is located at the centre of the cube, the radius of the tetrahedral hole (
r
t
)
is equal to the difference between half the body diagonal and
R
.
What is the radius of
the tetrahedral hole?
At left is an “exploded view” of a small sphere in a tetrahderal hole. In reality the
larger spheres (of radius
R
) are in contact along the cube faces.
Let
a
be edge length,
FD
the length of the facediagonal and
BD
the length of the body
diagonal of the cube above.
The following rightangled triangles can be identified:
FD = 2
BD
a
a
FD = 2
From the first triangle:
22
2
2
2
(2 )
2
4
2
aa
R
a
R
a
R
+=
⇒
=
⇒=
Using the second triangle:
( )
2
2
2
2
2(
2
)
6
6
B
DaF
D
R
R
R
B
D R
=+
=
+
=
⇒
=
The distance from the center of the small sphere (at the centre of the cube) to the centre of
the larger sphere (at the corner of the cube) is
1
2
6
2
BD
R
=
. This distance is also equal to
r
t
+
R
where
r
t
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 Oakley
 pH, Cubic crystal system, Rate equation, Atomic packing factor

Click to edit the document details