Homework5_Solutions - Homework#5 Solutions Spring 2014 1.a...

Homework #5 Solutions Spring 2014
1.a fcc lattice:
The distance between nearest neighbors in the fcc lattice is just
a 2
2
. Thus the radius of
spheres that just touch will be half this value:
a 2
4
. The volume of such a sphere is
4
3
π r 3 =
4
3
π
a 2
4
⎛
⎝
⎜
⎞
⎠
⎟
3
. There are 4 such spheres per fcc unit cell so that the ratio of the
volume occupied by the spheres to the volume of total fcc cubic cell ( a 3 ) is:
4
4
3
π r 3
a 3
=
4
4
3
π
a 2
4
⎛
⎝
⎜
⎞
⎠
⎟
3
a 3
=
π 2
6
b. bcc lattice:
The distance between nearest neighbors in the bcc lattice is just
3 a
2
. Thus the radius of
spheres that just touch will be half this value:
a 3
4
. The volume of such a sphere is
4
3
π r 3 =
4
3
π
a 3
4
⎛
⎝
⎜
⎞
⎠
⎟
3
. There are 2 such spheres per bcc unit cell so that the ratio of the
volume occupied by the spheres to the volume of total bcc cubic cell ( a 3 ) is:
2
4
3
π r 3
a 3
=
2
4
3
π
a 3
4
⎛
⎝
⎜
⎞
⎠
⎟
3
a 3
=
π 3
8
c. Simple Cubic Cell:

The distance between nearest neighbors in the simple cubic lattice is just a . Thus the
radius of spheres that just touch will be half this value:
a
2
. The volume of such a sphere
is
4
3
π r 3 =
4
3
π
a
2
⎛
⎝
⎜
⎞
⎠
⎟
3
. There is one sphere per simple cubic lattice unit cell so that the ratio
of the volume occupied by the spheres to the volume of total simple cubic cell ( a 3 ) is:
4
3
π r 3
a 3
=
4
3
π
a
2
⎛
⎝
⎜
⎞
⎠
⎟
3
a 3
=
π
6
d. diamond lattice:
The diamond lattice is an fcc lattice with a basis of two atoms located at (0,0,0) and
a/4(1,1,1). The basis atoms are nearest neighbors separated by a distance
a 3
4
. Thus
the radius of spheres that just touch will be half this value:
a 3
8
. The volume of such a
sphere is
4
3
π r 3 =
4
3
π
a 3
8
⎛
⎝
⎜
⎞
⎠
⎟
3
. There are 8 such spheres per fcc unit cell so that the ratio of the
volume occupied by the spheres to the volume of total diamond (fcc) cubic cell ( a
3
) is:
8
4
3
π r 3
a 3
=
8
4
3
π
a 3
8
⎛
⎝
⎜
⎞
⎠
⎟
3
a 3
=
π 3
16

2. We know that for a lattice with a density of lattice points, n,
nv = 1
where v is the volume of the primitive cell.
If a
2
and a
3
are primitive lattice vectors, they form a plane with a planar density of lattice
points:
σ =
1
a
2
× a
3
since there is only one lattice point per primitive cell.
The volume of the lattice primitive cell is
a
1
i ( a
2
× a
3
) = a
2
× a
3
a
1
cos θ = a
2
× a
3
( a
1
i
ˆ
g ) = v , where
ˆ
g is a unit vector normal to
the plane formed by a
2
and a
3
. However, a
1
i
ˆ
g is just the spacing between lattice planes, d.
So we find that:
σ =
1
a
2
× a
3
=
a
1
i
ˆ
g
v
=
d
v
b. Given the expression above for the density of lattice points in a plane, to maximize this
density, we must maximize the spacing between lattice planes. If we think of the spacing
between lattice planes as being determined by the magnitude of a reciprocal lattice vector,
d =
2 π
G
, to maximize d we have to find the shortest reciprocal lattice vector of the lattice.