3.17 Crystals with 2 Ions or 2 Different Atoms per Lattice Point

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Del curso dictado por Georgia Institute of Technology

Material Behavior

269 calificaciones

Georgia Institute of Technology

Material Behavior

269 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 talk about structures in which we have two ions or

two different atoms per lattice point.

What I've illustrated up here is the Cesium Chloride Structure.

And when you first look at the cesium chloride structure on the left,

what it appears to be is body center cubic.

However, we need to go back and

consider exactly what we mean by the Concept of a Lattice Point.

Remember if it's body-centered cubic, then we have two equivalent lattice points.

All those that lie on the corners and

the one that lies wholly in the interior of the unit cell.

Now, if you stand at a lattice point and you move to another lattice point,

the environment that you see around that new lattice point

has to be exactly the same as the environment that you saw in the first.

However, in the case of the cesium chloride structure,

when you're standing on the corner, what you see is the open circle whereas when

you're sitting at the open circle, what you see is that filled in black circle.

So consequently, those are two different environments, so

rather than referring to this as a Body Center Cubic unit,

what we do is we refer to it as a Simple Cubic Structure

in which we have two ions, one cesium ion and one chlorine ion.

And those two ions move together as we translate the unit cell in the X,

Y, and Z directions.

Now there's an alternative way that we can look at the ‎cesium chloride structure.

And we can look at it in terms of having two different

simple cubic lattices that interpenetrate one another.

So for example, if we look at the filled in circles,

we're going to refer to those as the B ions and the open circles as the A ions.

And each of those on the corners has eight

corner positions in one eighth therefore a total of one lattice point for the blue.

And one lattice point to the associated black.

And so consequently we're dealing with two ions per lattice point.

Now, as I say, cesium chloride is the characteristic that we

usually refer to these structures by, by calling them cesium chloride.

But it turns out that there are other systems.

For example, in the metallurgy community, when we look at copper zinc alloys,

which are the bases of brasses, what we have is a structure where

when the material goes through a particular type of transformation,

the copper sits at one of those positions and the zinc sits at the other position.

So once again, it's a simple cubic structure.

Now what we can do is also because we have the relationship between the radius and

the body diagonal of the unit cell, we can actually therefore go through and

make a calculation for the density for example.

So here is the cesium chloride structure.

We have two different atoms or two different ions.

In this case, one at the positions 0,0,0.

And the other one at one-half, one-half, one-half.

Now what we need to do is to turn our attention to what is the diameter or

what is the radius of these two ions?

So what we're going to do is look up the values and it turns out that we

find the value for the cesium and it's written in terms of nanometers.

We now look for the radius of the chlorine and so

now we have little r and big R and we can come up with the radius ratio, of course.

And we can also come up with a relationship with a0 in r.

Now in this particular case because we have two different ions, a cesium and

a chlorine, we want those two ions to touch one another.

So they're going to actually determine

the dimensions of the unit cell as we described in an earlier module.

So what I can then do is,

based upon the fact that I am looking at a structure of cesium chloride.

I can look at the number of ions per unit cell times the mass of the ions and

in this case, I have two.

One of them is cesium and one of them is chlorine.

So then I need to come up with the atomic mass of each of those ions.

And once I know the atomic mass,

I know that I have one ion per unit cell and I can then calculate

the mass of that particular structure of cesium chloride in the unit cell.

Knowing the relationship between r and a0, I can come up with the value for a0.

And once I have a0, I now am able to calculate the density of cesium chloride.

So we go through these steps and each one of those then becomes highlighted.

And ultimately we come up with the answer of 3.99 grams per centimeter cubed.

We can turn our attention to another common structure,

namely that of Sodium Chloride.

This time sodium chloride is, again, it's 2 ions per lattice point,

but now what we're going to be dealing with is a face centered cubic structure.

So in this particular case we have a sodium ion and a chlorine ion.

And they're located at two types of positions.

One of them is on the corner positions, and

the other ones are on the face centering operations.

So we can look at the unit cells in alternate ways.

We can either put the black filled in circles as our corner position, or

alternatively we could put the blue filled in positions as our corner.

When we look at the structure, where I have the blue chosen as my alternate,

I see that I have the sodium ion and the chlorine ion together.

The last structure that we want to look at and examine in this particular lesson

is the structure called Zinc Blend, or Zinc Sulfide.

And here we have two different ions per lattice point,

we have a zinc ion and a sulfur ion and they're coupled together.

And of course what has to happen, because of the material has

to have the stoichiometry of the compound, what we see is

once we make a calculation of how many sulfur atoms, how many zinc atoms.

We know that there has to be an equal number of those.

So it turns out then, since we're dealing with FCC and

two ions per lattice point, we now have a total of

four zinc and four sulfur, giving us eight ions per unit cell.

In the next lesson we'll be looking at slightly more complicated structures.

But again, what we want to do in this case is to illustrate some very

important points regarding the development of crystalline materials.

Thank you.

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