Uploaded by JUSTANEMONERD on 18.12.2012

Transcript:

PROFESSOR CIMA: The face-centered cubic crystal structure has atoms at

the corners of the cubic unit cell and at the center of each face.

So let's draw them here. Here's the corners that are in toward the

front of the cube, and also this one up here that you can see.

And I'll draw the one in the back slightly grayer color.

The centers of the face are shown here. And of course there's one at the center of

the three back faces also. So that's back here, and one back here.

Now we want to derive the relationship between the lattice parameter and the

size of the atoms. Again, the atoms as I've drawn here are not

to scale. They actually touch.

And in this case, they touch across the face diagonal.

To get a better picture of this, let's draw one face of the FCC

crystal down here. If I put the atoms in their real size, it'll

look like this. Here's this atom here in the center of the

face. That's shown here above.

And here's the four atoms at the corners, this one, this one, this one,

and this one. It's clear that if I were to draw this diagonal

on the lower diagram, it would look something like this.

This is a, as is this. And this is a, as is that.

The dimension of this diagonal across the face, of course, if this is a and

this is a and this is a right triangle, means that this hypotenuse

is radical 2 times a. Given that the radius of the atom is r, we

can see that 1, 2, 3, 4 times r has to equal the length of this hypotenuse,

or radical 2a. So that's the relationship between the radius

of the atoms and the unit cell dimension for an FCC crystal.

Now let's calculate the atoms per unit volume. Well, this is a little more complicated.

The reason is that I have to count, again, the atoms on the corners.

So there's eight of them, each shared with eight other unit cells.

So each contributes 1/8 of an atom to this unit cell.

But then I have six atoms at the centers of the faces.

Those atoms are shared between two unit cells. So each contributes 1/2.

The volume is a cubed. So the atoms per unit volume for a FCC crystal

is 1 plus 3, or 4 (over) (a cubed).

The atoms per unit cell is just the numerator up here, or 4.

So these are the important relationships for the FCC crystal.

the corners of the cubic unit cell and at the center of each face.

So let's draw them here. Here's the corners that are in toward the

front of the cube, and also this one up here that you can see.

And I'll draw the one in the back slightly grayer color.

The centers of the face are shown here. And of course there's one at the center of

the three back faces also. So that's back here, and one back here.

Now we want to derive the relationship between the lattice parameter and the

size of the atoms. Again, the atoms as I've drawn here are not

to scale. They actually touch.

And in this case, they touch across the face diagonal.

To get a better picture of this, let's draw one face of the FCC

crystal down here. If I put the atoms in their real size, it'll

look like this. Here's this atom here in the center of the

face. That's shown here above.

And here's the four atoms at the corners, this one, this one, this one,

and this one. It's clear that if I were to draw this diagonal

on the lower diagram, it would look something like this.

This is a, as is this. And this is a, as is that.

The dimension of this diagonal across the face, of course, if this is a and

this is a and this is a right triangle, means that this hypotenuse

is radical 2 times a. Given that the radius of the atom is r, we

can see that 1, 2, 3, 4 times r has to equal the length of this hypotenuse,

or radical 2a. So that's the relationship between the radius

of the atoms and the unit cell dimension for an FCC crystal.

Now let's calculate the atoms per unit volume. Well, this is a little more complicated.

The reason is that I have to count, again, the atoms on the corners.

So there's eight of them, each shared with eight other unit cells.

So each contributes 1/8 of an atom to this unit cell.

But then I have six atoms at the centers of the faces.

Those atoms are shared between two unit cells. So each contributes 1/2.

The volume is a cubed. So the atoms per unit volume for a FCC crystal

is 1 plus 3, or 4 (over) (a cubed).

The atoms per unit cell is just the numerator up here, or 4.

So these are the important relationships for the FCC crystal.