In this module, we're going to be introducing the concept of homogeneous nucleation. Now, remember what I said at the end of the last lesson. Homogeneous nucleation is the process that occurs anywhere inside of the phase that's transforming. So since we're talking about a liquid transforming to a solid, we're looking at the possibility of the nucleation event to be occurring at any atom that happens to be in that volume. So we'll start out with a liquid. And that liquid is going to be at a certain temperature. And I'm going to choose that temperature that is below. So I'm going to have positive undercooling. So I have super cooled in effect the liquid below its equilibrium melting temperature. Then what I'm going to do is in that liquid I'm going to examine what happens at the same temperature as the previous illustration. But now what we're going to be looking at is the fact that to that liquid I have added a solid which has a particular free energy associated with it. And depending upon how big that solid is, it's going to lower the free energy of the system. But in addition, notice that I have created an interface. And that interface separates the liquid phase from the solid phase. Here I'm making a simple assumption that the interfacial energy, because I've described the material that's forming as a solid as a sphere. I'm assuming that my liquid phase and the boundary that separates it from the solid, that behavior is isotropic. So this surface energy then becomes isotropic. So we start off with a liquid that's below the equilibrium melting temperature. Then what we do is, we generate just a little bit of solid. And the solid that forms is a sphere. And what we're going to examine are the various characteristics of the sphere that are important to solving this problem. So we start out with a sphere. It has a specific radius. And the other important term that we need to know. Because of surface energy behaviors, we need to look at the surface area of the sphere, which is nothing more than pi r square. And now we'll also be interested in the volume term. Because, remember, as the volume of liquid increases, we're going to have more and more free energy that is being released as a consequence of the liquid phase transforming to the solid. So these three parameters, the radius, the surface area and the volume. Will be used in the calculation of the driving force for this process of homogeneous nucleation. So we'll start out with the free energy of the solid is less than that of the liquid. And what that's telling us is that the liquid wants to become a solid phase. So the volume energy that is contributing to this process is negative. That is, supplying energy for the continued solidification process. And how much delta G sub V is released will depend upon the volume of the new phase that's being formed. So, what I can do is, I can plot the radius of the sphere that's growing out of my liquid, and I can calculate what that delta G is as a function of r. And what I see is that it is four-thirds pi r cubed, that's the volume of the sphere. And it's going to be modified by the term delta G sub V, which is a negative term. And it is increasing the bigger the particle gets. So I'm looking at a single particle that's nucleated. And it is nucleated randomly. And it's growing as a sphere. Now the next thing I need to consider is the surface energy. So the surface energy that separates the solid phase from the liquid phase is going to be greater than 0. That's going to help consume some of the free energy that we are getting as a result of the solidification. So the interface requires energy. So we're going to be taking away some of that Gibbs free energy. And of course then the surface area contribution is going to be positive. And the magnitude will depend upon the actual surface area of the material. And it's going to look like 4 pi r square times the surface energy of the solid and the liquid phase. So we look at a small particle to the left and that particle has gotten bigger. And as a result of the dimensional change, what we're seeing is an increase in the associated free energy. With the particle getting larger, and therefore increasing the surface area. So now we have those two behaviors. The one that increases with radius, and the other one that decreases with radius. One of them is the square of the radius, and the other one is the cube of the radius. Now what we know is, when the particles are small, the square term dominates, and when the particles become larger, then the cube term dominates. And when we look at the net curve, this is what we have. So, the net curve is given in red. And that shows us the contributions of both the volume term and the surface area term. We'll turn to this and examine more of it as we go into subsequent discussions in the upcoming modules. Thank you.