14-5 Crystal Stuctures: Body centered cubic bcc


Uploaded by JUSTANEMONERD on 18.12.2012

Transcript:
PROFESSOR CIMA: The body centered cubic crystal structure starts with
the cubic lattice with lattice parameter a. The atoms are again at the corners of the
cube, one at that back corner. But there's also an atom in the center of
the unit cell shown right there. Now the way I've drawn the atoms, it's clear
that they are not drawn to scale.
Now all the atoms are exactly the same. And as such, it turns out that they're going
to be touching along a diagonal rather than the edge that we saw on the simple
cubic. To see that, let's draw the following triangle.
That's going to be a right triangle. Here you can see the angle of that.
I've drawn a similar triangle down here. It connects these two atoms at these two corners
with the atom at that corner.
And you can see this, the hypotenuse of the triangle, will go right along
the diagonal. It'll go right through that atom at the middle
of this body centered cubic structure.
Let's look at some of the dimensions. This one is obviously a.
This dimension here is radical 2 a. Why?
Because this is a and this is a. So this hypotenuse of this right triangle
here has to be the square root of 2 times the length of this side.
So I'll just reproduce these dimensions down here.
Now we can ask the question how long is this hypotenuse for this triangle,
the blue triangle that's shown there? Well this has to be a^2 + 2a^2, has to equal
this length. So you see that that's going to be 3 * a^2
or this distance is radical 3 * a. Now if we superimpose the atoms as they're
in their real size on this triangle--
that is, this atom, this atom, this atom, and this atom.
You can see that they will actually touch along the diagonal.
So now I can see since there are three atoms touching along this diagonal to
the cube, that if this distance has to equal 4 * r, this length, this length,
this length, and this length. And that'll allow us to derive a relationship
between the size of an atom and the lattice parameter.
So that requires that 4 * r = radical 3 * a. So now let's calculate the atoms per unit
volume for the BCC structure. Well there's one atom in the center and there
are eight atoms on the corners.
Now if you think about it, each one of these atoms at the corners here is
shared with eight other unit cells. The volume of course is just a cubed.
So the atoms per unit volume is just 2 over a cubed.
And obviously, given the numerator of this expression, the number of atoms
per unit cell is just 1 + 8 * (1 / 8), or 2 for the BCC structure.