Hello, today’s screencast is all about miller indices of planes in the unit sells.
Some monies points from students are. Why are the planes important determining the new origin, how to draw the planes what families of planes are planar packing density diagrams and the whole bracket issue? so if there is a particular muddiest point that you wish to see, there are a bunch of links down here. So, let’s get started crystal planes are important for a variety of reasons: for example, metals deform on close packed planes and close-packed directions, which is useful for industry and making things? Lasers are grown by paper, depositing layer by layer of atoms on a specific plane, as shown right there, vapor deposition, my personal favorite is that gems grow on planes that reflect the crystal structure. So all of our shiny gemstones reflect crystal structure. So that’s why crystal planes are important so now that we know why planes are important. We need to be able to determine the indices given a picture of the plane. So let’s just get some notation down. Planes are given in moon brackets like that, and these would be notated as h k l like so also here we have a diagram of the unit cell. The edges of the unit cell are labeled, so this edge in the x-direction would be a this one in the y-direction would be b, and this one in the z direction would be c? Now that we have some notation down, let’s look at the four step method to determine the indices given a plane.
First, we choose the origin l in the unit cell or a nuo prime with axes x, prime y, prime and z prime, depending on where the plane is.
We will talk more about this.
Later too, we find intercepts from our origin which, at whichever corner it is to the a intercept? Then we go back to the origin. Then we go from the origin to be intercept and back to the origin, and then we go from the origin to the see, intercept the big difference between directions and planes. Is that with directions? we go back to the origin when we take our intercepts step three.
Is we take reciprocals of the intercept position so, whatever numbers we came up with here, to keep the indices within the unit cell and within proper form and then number four? is we clean up? we produce multiples, we eliminate fractions, we put bars above negatives and place round brackets around the integers round is important! Some common mistakes are listed here. This one? We need moon brackets around them to make sure we’re talking about a plane in this one. We have commas and we do not use commas when we’re talking about planes or directions or anything like that. With this one, we have negatives in front of the numbers we need bars over them. That is proper notation bars over them like. So this would be the fixed one problem with this. One is that we have a fraction, so we need to multiply by two to get rid of this fraction, so this would give us one four six. Finally, this one we have a common factor in all of them, so we can multiply by one half and still get whole numbers, so this would end up being one two three. Finally, we have made this table to make finding the indices easier. Let’s do an example problem step. One is we choose the origin now here are some tips for choosing the origin! The plane cannot touch the origin so that corner that corner in that corner or out, and we choose the closest corner to the plane so out of the remaining five corners, this corner down here is the closest to the plane!
So this would be our origin.
Oh and it is at zero, zero, zero?
Next we find the a b and c intercepts, so the plane intersects the a edge here at one back to our origin, the plane intersects, the b edge one.
Finally, the plane intersects the sea at one also, that would be a 1. Next, we take the reciprocal of our found, a b and c intercepts so 1 over 1 1 over 1 + 1 over 1. Then we give the value for the reciprocal, so f, of something other than 1. This would be difficult, but for us it’s just 1 1 1! Next, we give the final miller indices and we do this by making everything pretty taking away negatives, putting bars over numbers multiplying through fractions that kind of thing, but for us it’s an easy one? 1 1 and we finished the entire thing off with brackets. Let’s do a more challenging example. First, we need to choose our origin, remembering the rules we used last time.
The origin cannot intersect the plane.
So this point- and this point are out the origin- is- should also be the closest corner to the plane that doesn’t touch it. So this would be our new origin, o prime, and what is the value for that we went. It is at the point 1 1 0 is our new origin. After finding the origin, we draw our new axis. So then up so here is our new x prime y prime z prime. Our new axis step 2 is be fine, the intercepts so the plane intersects the a edge back here at negative 1, the plane, intersects the b edge right here at negative 1/2 and the plane intersects the c edge up here at positive 1.
Next we take the reciprocal x’, so this would be 1 over negative 1! This would be 1 over negative 1/2 and this would be 1 over 1? Then we find our reciprocal values, so this would be negative, 1, negative, 2 and 1! Finally, we make our indices pretty we get rid of fractions, we put bars over negatives, so ours becomes bar 1 bar 2 and 1 all in parenthesis. Now we need to be able to draw the plane given the indices. Now we know how to give the indices having a drawn plane. We need to go in reverse, there’s a similar, four-step method to do this and we are going to use this indicee right here. As our example remember, we have h k l for our miller indices, notation our step one is we select the origin o at 0, 0 0 or if there’s negatives at oh, prime with x prime y, prime z prime? however i’m talking about our example, there are no negative, so our origin is here at 0, 0 0, step 2. Is we take the reciprocals of the indices to get 1 over h 1 over k and 1 over l, which will be b a bnc intercepts? so let’s do that! We’ve got 1 of 0 1 of 1 and 1 of 1. Now we evaluate the reciprocals?
The 1 over 0 thing is equal to infinity and that’s kind of odd, but what it means is that the plane is parallel to the a for our case, and that means it will never intersect this a edge. So our other two are fairly easy, 1 and 1. Now we go to our unit cell? Now we have our a b and c intercepts. We mark them, so ok, starting here a at infinity! So nobody and we go b at 1 over here, b at 1 and then go back to the origin and then see it 1 also, so we go up 1 there we go so we marked our intercepts next. We draw the plane by connecting the intercepts. So here we go back.
You know we only have one diagonal line.
However, we can drag this line in order to get it to the front. So it would be the same.
We drag it over here tada, because the plane continues this way and this way so there we have it we’ve drawn it and we have done what was asked of us! Let’s do another example: we can use our table and our steps in order to draw this plane so step one. We choose our origin following the same rules as we did with directions to repeat: if there is a negative, you put a 1 in the origin at the same place and if there isn’t a negative, you put a 0.
All of these are positive, so our origin is at 0 0 0, which would be right there origin.
Next, we take our given indices and we take the reciprocal of them.
So we’ve got 1 over 1 1 over 0 and 1 over 0?
The reciprocal values would be 1, infinity and infinity. Remember that in affinity just means that the plane does not touch the unit cell if the reciprocal is infinity!
So now that we’ve found our reciprocal values, we mark so marking our a b and c intercepts. We’ve got a is up here at 1! We don’t have a b because of the infinity and we don’t have a c, so i’ll check, and now we draw now. This might be a little bit confusing, because we only have one point. However, remember this one point would continue in all directions so continue this way that way up up and down. So it continues all around so this plane that we’re talking about is the front face of the unit cell! So there we go check now that we know how to draw planes given indices and how to give indices from a drawn plane. We need to now learn about families of planes. Families of planes are equivalent, they have the same packing density, environment, etc.
They are denoted by capital integers in curly brackets, like so so notice.
These are different than the moon brackets? Given here is the family of 1 1 1 planes and over here are the individual indices for each member of the family is important to note for a cubic system. Indices of families of planes are given by all the permutations, plus and minus of 3 integer indices and again, as an example, we can see that here we are now going to look at some common families of planes!
Here we have the 1 1 1 family of planes, i’ve drawn the first one right here in blue over here, i’ve also labeled, the new origins for the other three members of the plane. However, i did not draw them because that would be really messy on one cube here we have the 100 family! All of these planes have the origin, 0 0, 0 notice that there’s only 3 of them, because parallel planes are not unique. So, for example, this one would be the same thing as this one etc. Finally, we have the 1 1 oh family i have drawn the first plane here and i’ve labeled, the origins for the other ones.
The three positive ones have the same origin, there’s no real special name for this plane, but it is, and importantly because there are so many variations of it. As you can see, we need to be able to visualize what different planes on different types of unit cells would look like. So one way to do this is by taking the planar packing density, which is the number of atoms within the plane, divided by the unit area of the plane again for cubics. All of this, the cube edges are equal to a an important thing. To note is that we count fractions of circle area on the plane only when the atom centers sit exactly on the plane. Let’s go through an example of finding this planet, packing density, they’re asking us to give it for the 100, the 1 100 and the 1 1 1 planes, so i’ve already gone in and put in what the planes would look like using that cell up there! We need to count the number of atoms within the plane, so how we do this is we know that there are 360 degrees in the circle, so we can take the angle that is here and use it to count the number of atoms! So this is a full circle, so this would be 1. This is a 90 degree angle, so this would be a quarter because 90 is a quarter of 360 again? All of these would be 1/4 1/4! All of these would be 1/4! Then we take the number of atoms and divide it by the unit area as a plane which was given to us for each plane? So for the first one, we’ve got 2 divided by a squared and again we’re doing these in terms of a sub-zero! Now, let’s do the next one. So this is a 180 degree angle, so this would be 1/2, and this down here would also be 1/2, because one half of 360 is 180 and again these are all 90 degrees, so all of the corners would be 1/4, so we can see that one half plus one half would be one plus four one fourths would be two total atoms divided by the area, and the area is given here so 2 to the one-half is the same thing as the square root of 2 times a sub 0 squared. Finally, let’s take the last plane: let’s count the atoms, as these are all 180 degrees, so this be 1/2, 1/2 and 1/2 and using the geometry, knowing that all of the side. Lengths of this triangle are the same. We know that each of these angles is 60 degrees and that would be 1/6 of 360. So here we go one six, one six one six, so, let’s add them up, that’d be one 1/2, plus 1/2 would be 1 and then 1/2 plus 3 6 would also be once we’ve got 2 divided by i’m, going to say square root of 3/2 times a sub 0. Squared and again we can make all of these prettier, but we were just asked to calculate the planar density? It’s another good fact to note that this would be the close, close packed plane for an fcc unit cell, because all of the atoms are touching notice that they’re spaced here and there’s space here? So those are not close. Packed. This one is close, packed one of the most difficult things about drawing the planar packing density would be telling if the atom is on the plane or, if it’s not so, in order to better explain this, we’re going to draw the bcc on the 1 1 1 plane? Now, if the atom is 1/2 above the plane and 1/2 below in the plane, we count it if anything else, if it’s atom, center above the plane or if the atom center is below the plane, then we don’t count it so, let’s draw the plane here would look like this. Drawing in between would be a triangle, that’s a horrible triangle, but you get the idea. It’d be this triangle, and the origin would be at the very back left corner notice! We have the planar packing density drawn here and notice? There is no center atom? The reason is is because the center in here is actually above the plane, so it would look like this, but we only draw atoms that centers are on the plane, and so this is a very common mistake that many people make so be sure that you understand this is bcc on the 1 1 1 plane after we’ve learned all that new information, it’s good to review! So let’s go over what the different brackets mean to review. We have here our square brackets, which are used for individual directions. Here we have caret brackets and these are used for families of directions? The halfmoon brackets over here are used for individual planes and the curly brackets are used for families of planes. Now, there’s no real cute fancy schmancy way to memorize this, but you just need to burn this table into your brain.
To be sure that you understand what you’re asked and what you’re giving this screencast has now successfully addressed the menus points given at the beginning, we learned why crystallographic planes are important!
We learned how to determine the new origin and how to draw the planes. We learned all about families of planes, we learned about planar, packing density and how to calculate it, and we learned when to you certain types of brackets. So if you have any questions, please leave them in the comment section and good luck. Indexing. .