In this example we're going to calculate linear density
and planar density for iron.
Iron has the body centered cubic structure. Calculate the linear density
along the (1 0 0) direction
and the planar density for the (1 0 0) plane
So I have indicated the (1 0 0) direction
which axis I call "x" is arbitrary when we have this cubic structure
I get the same answer if I picked one of the other axis as "x"
Linear density, atoms per length, whose centers are on the direction vector
So if I look at the
(1 0 0) direction then this atom and this atom are on
this vector and so for the distance
of the unit cell. So let's call this
distance "a" which of course then
is the length of all the sides of this
unit cell. Distance "a" I have
half of this atom and half of this atom
so I have a total of one atom per distance "a"
and so if I can determine the distance "a"
relative to the radius of the atoms and
keep in mind this is not a space-filling model, I have
indicated the lighter color blue atoms as the corner atoms
and this darker blue one is the in the center
atom in the body centered cubic structure. And of course in this structure
in a space-filling model the center atom
touches all these other atoms I have indicated here
So let's look at using geometry to determine
distances. I can look at the diagonal going from this atom to this atom
and I could represent that as, in a space-filling model,
three atoms. Here's the corner here
this is the atom on this corner. This now is this center atom
and so this distance "R" "R" "R" where "R"
is the atomic radius
this distance is four "R". So what I wanna do is use geometry and draw
a triangle that will allow me to calculate
"a" from this value "R". I have a diagonal going from these corners
I can now have one length of the triangle as this
side, which is length "a". And then
the last would be this diagonal. And I will call
that for now "b". Draw this so it is a little clearer
So I know a^2 plus b^2
is equal to (4R)^2
I have to first calculate "b" so if you look at the
triangle that relates "a" and "b"
so just look at this front face where
"a", side "a" the diagonal "b"
a^2 plus a^2 is equal to
b^2. So I can then solve
this to determine "b". So what I am going to do is substitute "b"
back into this equation and so I have just reduced this to solve for "a"
in terms of "R" and then I can look up "R"
property, the atomic radius of iron
from a periodic table which means I have now a
value for "a". So once I have the dimension for "a"
there's one atom per distance "a" which means the linear density
is 2.52 atoms per nanometer So this is the linear density
for iron along the (1 0 0) direction
Second part of the problem is to calculate
the planar density for the (1 0 0) plane. So if I look at
the x-axis again in this direction
then the plane we're talking about is this plane here
Intercept x is one and it doesn't intercept the y and z-axis
and we want the number of atoms per area. So we look at this plane
each of these atoms is shared with the
cube in the plane now. So it means we have four atoms but
only one quarter is associated with this particular
square that I have drawn here and the area is just
a^2. So I have one atom and the area
is 0.397 nanometers squared
so this is 6.34 atoms per nanometer squared
and that's the planer density for iron
for the (1 0 0) plane