3.16 Crystals with 2 Atoms per Lattice Point

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Del curso dictado por Georgia Institute of Technology

Material Behavior

256 calificaciones

Georgia Institute of Technology

Material Behavior

256 calificaciones

Have you ever wondered why ceramics are hard and brittle while metals tend to be ductile? Why some materials conduct heat or electricity while others are insulators? Why adding just a small amount of carbon to iron results in an alloy that is so much stronger than the base metal? In this course, you will learn how a material’s properties are determined by the microstructure of the material, which is in turn determined by composition and the processing that the material has undergone. This is the first of three Coursera courses that mirror the Introduction to Materials Science class that is taken by most engineering undergrads at Georgia Tech. The aim of the course is to help students better understand the engineering materials that are used in the world around them. This first section covers the fundamentals of materials science including atomic structure and bonding, crystal structure, atomic and microscopic defects, and noncrystalline materials such as glasses, rubbers, and polymers.

De la lección

Crystalline Structure [Level of Difficulty: Medium || Student Effort: 2hrs 30mins]

This module covers how atoms are arranged in crystalline materials. Many of the materials that we deal with on a daily basis are crystalline, meaning that they are made up of a regularly repeating array of atoms. The "building block" of a crystal, which is called the Bravais lattice, dtermines some of the physical properties of a material. An understanding of these crystallographic principles will be vital to discussions of defects and diffusion, which are covered in the next module.

3.16 Crystals with 2 Atoms per Lattice Point6:24

3.17 Crystals with 2 Ions or 2 Different Atoms per Lattice Point7:02

3.18 Crystals with Several Atoms per Lattice Point9:42

3.19 Polycrystalline Materials and Liquid Crystals10:20

3.20 X-Ray Diffraction and Crystal Structure14:26

Conoce a los instructores

Thomas H. Sanders, Jr.
Regents' Professor
School of Materials Science and Engineering

In this lesson we're going to be talking about slightly more complex structures,

namely what happens when we have additional atoms per lattice point.

So we'll begin first by talking about two atoms per lattice point.

An example of a material that has 2 atoms per lattice point.

Even though there are 2 atoms per lattice point,

the atoms are identical to one another, and

we'll see which of the materials of interest to us has this type of structure.

But first we're going to describe it.

We have here a face centered cubic structure.

And what we have in terms of a face center of cubic structure,

if you recall, there are a total of four lattice points per unit cell.

Now we said we're dealing with 2atoms per lattice point.

So, now that means that we're going to be looking at a total of 8 atoms

in the unit cell.

Ultimately, we'd like to be able to develop a relationship between

the axes a0 and the body diagonal of the cube.

So we can go back and forth between the radius and

the dimensions of the unit cell.

So we can ultimately calculate things like the packing factor as well as the density.

So, first of all, what we're going to do, is start on the plane, z = 0.

And I'm going to go through, and I'm going to add atoms subsequently as I go

through the structure, all the way down to z = 0.

So we put our first atom at the corner position.

And now we're going to go down to the plain that is three-quarters

of the way up or a quarter of the way down, and that's our plain.

And on that plain I'm going to put the next atom.

And so now I have two of my atoms so far included into this structure.

And now we'll go down to 0.50.

And when I do that, my position, according to the structure to the left,

the atom position actually sits on the face.

But now what we're really trying to do, is to develop a relationship that will take

us back and forth between the radius of the atoms and the edges of the unit cell.

So in order to do that, we need to calculate how many atoms we actually have

moving from z down to the front corner of the cell.

So as a consequence, what I'm going to do is, I'm going I'm going to imagine that

atom actually sits here and it's going to be right there along that body diagonal.

So it doesn't exist there.

But there is in fact a space that would be equal to

the dimensions of the atom that is at the face centering operation.

Now what I'm going to do is, I'm going go down to 25% up the z-axis or

a quarter the way up the z-axis.

And that's my plan of interest.

And the position of the atom that I have on that particular plane lies

where the atom has just appeared.

And in order to get it a long the body diagonal of the cube,

what I'm going to do is just displace it.

And so now I have a body diagonal for that.

Now remember, the atoms don't actually exists at this points,

but I've included those atoms there, so

that they actually are controlling the spacing between the blanks.

Now, the last plane that I'll look at is the z = 0,

so now I have an atom at that front corner.

So I'll draw my line that goes between the position at z = 1,

all the way down to the position at z = 0.

And now what you see is, those atoms are sitting at the particular positions that

are equivalent to those positions that lie along the body diagonal of the cube.

And I'm doing this primarily because I'd like to be able

to develop this relationship between a0 and r.

If I look at the atoms, and those atoms are, in effect,

tightly packed along that body direction, I would have a total of eight of those.

And when I look at those 8 radii,

I can relate the line a0 onto the square root of 3,

which is that magnitude of the body diagonal to the value 8r.

So now what I've done, I've come up with a relationship between a0 and r.

I can use that relationship in the following way.

I can describe all of the structures that are of the type of diamond cubic.

Where I have 2 atoms per lattice point and they're the same, and

let's consider what materials they are.

The important group of group IVB contain carbon, silicon, and germanium.

They all have the identical

description in terms of their crystal structures as diamond cubic.

So, carbon exists as a diamond cubic structure, so does silicon, and so

does germanium.

And they all form that structure that we have illustrated on the left.

I can take this information and I can determine what the packing factor for

the diamond cubic structure is.

Now remember how we did it in the past, what we did was to take the volume of

the spheres that are inside of the unit cell, which is represented by a cube.

And here you can begin to see what I'm going to do,

is to substitute in one of those expressions, either a or r.

And I'll do that using the expression, that a onto the square root of 3 = 8r.

And now I can put in the value for a in terms of r.

And when I do that, I come up and I get a packing factor of 0.34.

So this is the packing factor for the diamond cubic structure.

And if you recall the packing factors that we have previously described, that is,

face center cubic and body center cubic.

What you see is that the packing factor for

the FCC is the greatest followed by the BCC and then last by the diamond cubic.

So these are all structures in which we're dealing with the same sphere diameter,

or the same atoms.

And we can see how the packing factors change as we

go from FCC all the way down to diamond cubic.

Thank you.

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