0:04

We're beginning a new lesson here and finally after all these time we

Â are going to be introducing Fick's First Law, which is really the first time we

Â are going to begin to talk about the process of diffusion in solid state.

Â We're going to introduce here Fick's First Law.

Â In the case of Fick's First Law, it's an empirical relationship, and

Â we start out by looking at a cubic element.

Â We're going to be describing something we refer to as the flux, and

Â we're looking at in this cubic element the flux in x, y, and z directions.

Â And what we mean by that idea of flux, it is a measure of the number of

Â atoms that we have that cross a particular plane that's in a unit area.

Â 0:53

Now, if we're in a situation and we're going to be in this as an approximation

Â as we go through the lectures that talk about solutions of problems in diffusion.

Â We're going to make an assumption that we're going to reduce the problem

Â to one dimension.

Â And the way we do, of course,

Â is we're going to assume that the flux in two of the directions,

Â namely in the y and z direction they're going to be equal to one another, and

Â we have a flux consequently in the x direction, which is different.

Â Now when we have a situation where the two fluxes in the y-z plane are equal,

Â it means that the composition in that plane is the same,

Â unlike the composition along the x direction, which is different.

Â 1:43

So when we look at a one dimensional problem then we look at the flux coming

Â out of the plane and as a result of the uniform composition we can

Â reduce fixed law rather than considering fixed law in three dimensional space.

Â We're just going to be looking at a one-dimensional problem here,

Â where we're only looking at the diffusion in the plane that is

Â given by J of X, which represents the flux in the X direction.

Â 2:17

Okay, so let's look at an example.

Â We're going to be looking at a plane which contains two different types of atoms,

Â a blue atom and a light blue atom.

Â We're looking at the A atoms and we're looking at the B atoms.

Â And what's happening here is that, as we go from left to right,

Â we are seeing a concentration gradient, associated with the A atoms.

Â In other words, the number of atoms on the plane at one.

Â A blue is going to be greater than that of plane one, or

Â plane two containing the blue atoms.

Â So now what it tells us is we have a concentration gradient between

Â plane one and plane two.

Â 3:07

Now, what's going to determine how quickly the atoms are going to move

Â from that plane one throughout the microstructure

Â will depend on the concentration gradient, how large the difference is between

Â the atoms that are on plane one and the atoms on plane two.

Â We're also interested in the concept of a term that we refer to as the diffusivity.

Â The diffusivity is how fast are the atoms moving at a given temperature.

Â We're therefore, also going to be interested in the temperature.

Â The higher the temperature, what we expect is that the process of diffusion,

Â which will occur much more readily as the temperature goes up.

Â The other thing that is important is how far the atoms have to move or

Â the jump distance, and of course the crystal structure itself.

Â And in this case,

Â we're just looking at it from a point of view of a simple two dimensional crystal.

Â 4:11

And all of these factors combine to lead to an expression that was developed

Â again in curricula by Fick.

Â And what he said was that the flux of atoms in the X direction

Â is going to be proportional to the diffusivity D and the difference

Â in composition and the difference between the distance between the planes.

Â And when we write this in the form of a differential what we have

Â is that the flux in the x direction is going to be equal to

Â the negative of the diffusion coefficient times the C-D-X.

Â The reason that the negative is included here is because we're moving down

Â a concentration gradient.

Â We'll see in more advanced courses that the flux of atoms in

Â a particular direction more generally is related to.

Â The concentration gradient, so down a concentration gradient.

Â But we're going to find in more advanced courses that's not a necessary condition.

Â We just simply need to have a concentration gradient and

Â it doesn't necessarily mean that it is down the concentration gradient.

Â 5:40

equal to some consonant D0 times the exponential of minus Q,

Â and here we're talking about an activation energy divided by rt.

Â What we mean by the activation energy of the defusion process is how much energy

Â is necessary in order to start the atoms moving from one place to another place.

Â So how much energy do we need to put in into the structure in order for

Â that defusion process to occur.

Â Again we have the gas constant and we have t as the temperature.

Â The expression that we have here is referred to as an erroneous expression,

Â and we have what's on the left hand side related to

Â this exponential dependence of the activation energy and D0.

Â What becomes important is, what is D0 and what are Q?

Â And depending upon the particular diffusion process and

Â the particular system D0 and Q will be specified.

Â So if we make this particular behavior linear so now we have the log of D

Â is going to be related to the log of D zero which becomes our intercept.

Â Now we have the linear relationship where we have negative q over r time one over t.

Â And I've put it that way that one over t because when we

Â starting looking at these erroneous plots the way we

Â usually plot them is based upon 1 over the temperature,

Â or the reciprocal temperature in units of kelvin to the minus 1.

Â So if we have our equation of Fick's First Law and

Â we look at this as a function of temperature, what we can do is

Â we can measure the diffusivity, and these can be done experimentally,

Â and we can measure the diffusivity at one temperature,

Â and the diffusivity of another temperature.

Â And over this entire relationship the expressions that I have written here hold.

Â And we're assuming that at these two different temperatures

Â the process of atomic motion is exactly the same.

Â So, therefore, we're going to have the same activation energy.

Â 8:07

And, when we write the equation then,

Â knowing the diffusivity at two different temperatures, we can actually

Â calculate the activation energy cue, because everything else is known.

Â We have the diffusivity at one temperature,

Â the diffusivity of another temperature.

Â And, hence, we can solve for the value of Q.

Â So, let's go ahead and use an example calculation here, where,

Â what we are going to be looking at, is zinc dissolved in copper.

Â 8:37

The concentration of zinc is going to be relatively low.

Â And we're interested in the diffusivities of the zinc and the copper,

Â and what we're going to be given is how much at this two different temperatures

Â will be the diffusivity and by doing that, then we can come up with a value for

Â the activation energy for Zinc diffusing inside of Copper.

Â So the required data that we need to have is, what is the diffusivity

Â at one temperature, and the diffusivity of another temperature?

Â And if we measure these diffusivities, we can use those values or we

Â can use tabulated values of diffusivities at two different temperatures.

Â Once we have the diffusivities at two different temperatures,

Â we can put those back into our relationship and as a result of that,

Â we can write the expression that I have indicated up on the screen.

Â And we have again the constant which is our gas constant

Â in the appropriate units and the temperature again used in Kelvin.

Â 9:51

And we can come up with a value of the activation energy.

Â And it's on the order of 190.8 kilo calories or kilo joules per mole.

Â Now, it is important for you to take this calculations and actually carry them out.

Â The reason that you want to do this is because you're looking at numbers or

Â you're using numbers that are extremely small like numbers like ten to the 11 and

Â ten to the 16 and

Â you're not maybe necessarily use to seeing that dimension number.

Â So it's important for you to actually try some of these calculations out and

Â make sure that you understand and

Â come up with what are orders of magnitude and typical values that you would have.

Â Now we're going to be doing these types of calculations throughout the remainder of

Â this module so you want to make sure that you understand this particular process.

Â Thank you.

Â