All right. I guess we can we can get started now. Welcome back.
It's the 13th lecture of the condensed matter course.
When last we left off, we were talking about scattering the basic experiment.
If you have some sample, you have a wave at some way vector K that you send into the sample and then it's scattered off to some wave vector.
K prime. And by measuring how much and in what directions you have scattering, you're able to do something about what's actually in the sample.
A couple of things that we know about these type of scattering experiments.
First of all, in order to get scattering from chaotic prime, you must have K minus K prime equals reciprocal lattice vector g.
And we derived in the last lecture that's equivalent to the Bragg condition two d sine theta equals and lambda.
In other words, since a reciprocal lattice vector is orthogonal to a family of lattice planes,
what you're actually getting is diffraction off a diffraction grating with lattice plane spacing D orthogonal to that vector g.
The intensity of scattering at any reciprocal lattice vector is proportional to the structure factor of g squared,
where the structure factors the 48 transform of the scattering potential in the unit cell.
So this intensity gives you some amount of information about what's actually in the unit cell.
Now, how you actually do this experiment is a little more complicated.
There are several methods one can go about.
There is the first method, which is the best method, but also the hardest method, best bit hardest, which I'll discuss only very briefly.
The best method, but also the hardest method is to use a single crystal.
Single crystal. So a single crystals look like this.
This is a gorgeous single crystal of sodium chloride, a big single piece of the material without any defects.
Now what you'll find so you take a single crystal, you put it in the path of your of your wave, and you look for four waves being scattered off of it.
What you find is if you take your single crystal, you put it in some arbitrary orientation in front of your your beam.
Here, for most orientations, you won't get any scattering at all.
In other words, you will not manage to align the families of lattice planes just at the right angle,
such that the Bragg condition is actually satisfied. So in order to actually get any scattering, what you need to do is to rotate the crystal.
Rotate the crystal. Another way to get scattering is to vary the wavelength very lambda, do one or the other,
and then at some angle of the crystal or at some value of the wavelength, you get scattering.
You write down in which direction you get the scattering, you measure the intensity of the scattering,
and you can deduce a lot of information about the structure factor and therefore about the structure of the crystal.
Now, we're not actually going to go into how you analyse such data in any detail in this course.
And the reason we're not going to do it is because this method is very rarely used,
and the reason it's very rarely used is because it's actually the hardest thing to do.
The reason it's hard is because, almost without exception, materials don't like to form nice big crystals.
Sodium chloride is a suggestion. You have these beautiful big crystals of sodium chloride.
Quartz makes nice big crystals. You can probably make a nice big crystal of sugar something.
I mean, so there are a couple of materials that you can get to form nice big crystals, but most materials don't.
If you cook up some new material in your laboratory or you find some new material in nature, usually it's not crystalline.
It doesn't form big crystals. Frequently it's powder. Or even worse than that, you can have what's known as Polycrystalline.
So this is a picture of a Polycrystalline material where you have tiny little crystal lights, which are about on the size of a micron.
So there's a little crystal here, a little crystal here,
and those crystals have their axes miss oriented with respect to each other, and they're also smushed together.
You can think of it as like a powder of tiny little crystals all stuck together.
And typically materials look like this when you make them. It's very, very difficult.
It takes a very good chemist and very good material scientist to turn something that looks like this into something that looks like this.
So most of the time, when one is doing experiment, one is working with something that looks like this, it's probably crystal.
So to take data on a polycrystalline sample, you use what's known as the device Scherrer method to buy Scherrer.
The buy is the same guy from the Buy Theory of Vibrations and Solids share someone else obviously.
And the device share. This method is also known as powder diffraction.
Diffraction. Because you can use a powder of the material, lots of tiny, tiny little crystal lights in a powder.
And you can put that in your sample and get data out of it and actually make sense of it.
Now, the general principle of the idea of powder diffraction or doing diffraction on
polycrystalline samples is if you're sending a wave into your into your crystal here,
if you imagine for a second you got a big a big single crystal,
and you could rotate that single crystal around if there was any direction in which
you rotated that single crystal and you could get diffraction out at this angle.
Then somewhere in your powder, somewhere in your polycrystalline sample, the axes will be correctly oriented.
In other words, the poly crystal material or the powder represents every possible orientation of the material.
So somewhere in your material you'll have a situation where your family of lattice
planes is aligned just in the right way in order to get diffraction from K2 K Prime.
If there's any orientation that will do it, you'll find that crystal light with the right orientation somewhere in your sample.
That's the general principle we're going to live by.
Now, if if you can get diffraction in this direction with some lattice planes that are sort of lined up parallel this way,
you can also imagine that if you lined up your lattice planes this way, you could get diffraction out this way.
And in fact,
you can imagine rotating those crystal lights around the incoming axis and you realise that you can get a whole cone of diffraction outwards.
If there's any direction that you can scatter, you can rotate that crystal light around the incoming access and get a get a cone of outgoing waves.
So the way the experiment actually looks at in a setup is kind of like this.
There's an incident x ray beam, there's a polycrystalline or powder powder specimen here.
And then you get these cones of outgoing waves, which you then try to measure.
Remember that the it's marked here, so you don't forget the total deflection angle is two theta very important.
A common source of source of error and to theta can go anywhere from zero,
meaning it's not scattered at all to 180 degrees, meaning it's directly back scattered.
Now, the way one used to measure it is you would take this sort of round environment and you would line the inside with photographic paper,
and you would just expose the photographic paper to the x rays,
and you would see these nice conical signals coming out for these various different cones coming out like this from your polycrystalline sample.
But in modern days, photographic paper is not used.
It's much more it's much more convenient and actually much more accurate and much more sensitive to use semiconductor detectors.
So it's usually what you actually get out of your sample, out of your experiment now is a plot that looks like this,
where what's plotted across the bottom is the scattering angle and the intensity vertically.
So you'll see there's a peak here at 20 something degrees, another peak at 30 something degrees,
which might correspond to this ring and then that ring and so forth.
And this is the kind of data that we're going to spend our time analysing.
Okay. So this is sort of sort of a classic type of question for exams, analyse a powder diffraction pattern with the device sharing method.
So there's a couple of rules for following this method, so I'll write them down.
It's fairly difficult to find a book that explains this well, I think as explained well in my book,
but if you don't like the explanation of my book, you're kind of out of luck because there aren't a lot of other books to explain it.
So sorry about that, but that's just how it is. So rule number zero is no the wavelength, no your wavelength.
Lambda Why do you know you're wavelength lambda? Well, you're the experimentalist.
You put in the X-ray beam, you had the X-ray apparatus.
So, you know, presumably what the wavelength of the X-ray is. So it's a good thing to start with.
Know your wavelength, probably know it pretty accurately, actually, if you're a good experimentalist,
step one measure the angles, measure angles, angle of scattering, which is two theta, not theta.
Remember, that is going to be important. So here's some data we're actually going to analyse where they've done.
The first two steps for this is aluminium.
They're doing scattering off of aluminium, I guess, in this country and the wavelength is known extremely precisely in this setup.
This is done with a typical X-ray tube and you only have six digits of precision as to how how precisely they know the wavelength and
remember the reason they know their wavelength so precisely is because this is an electronic transition between two iron states in an atom.
It's an electron falling down from one eigen state to another eigen state.
So the the energies of the eigen states are very, very precisely defined.
So so you know,
the wavelength with very high precision then it you know you get this data all these peaks as a function of of scattering angle to theta.
And fortunately in this particular picture they've actually measured the angles for us 38.43 degrees, 44.6, seven degrees and so forth.
And I've also labelled the peaks with letters so we can keep track of them peak A, peak B peaks and so forth.
Just one hint that if you're ever given a powder diffraction pattern and you have to measure the angles yourself,
it will really help a lot if you get the angles very precise. If you get the angles off by a little bit, you can get very confused.
It won't be obvious what the result is supposed to be, so try to measure it as precisely as you can.
For each scattering angle you measure step two angle.
Sorry. Angle. So step two is calculate the lattice plane spacing.
So lattice plane spacing, plane spacing using Bragg's law and that is D equals lambda over two sine theta.
So for each scattering angle, you see that's representing a family of lattice planes with the spacing d equals lambda over two sine theta.
Now you might think it should be and lambda over to sine theta,
but in fact you don't need the end up here because the higher values of end will correspond to different theta is that you will also measure,
so you'll just be over counting if you put an end upstairs. So you just need to keep lambda over to sine theta.
Each angle corresponds to one one lattice plane spacing.
Okay. All right, so step three.
So this is for us, this is sort of not in the real world. We can assume cubic of some sort assume cubic.
It's actually not a bad assumption in the real world too. Meaning simple cubic, simple boxy or FCC.
I have never seen in 25 years of exams.
I have never seen a 30 year exam.
There's asked you to figure out a powder diffraction pattern for anything more complicated than simple cubic beaker FCC.
Those are the only lattices you'll run into on your 30 year exam. That's almost a promise.
I mean, they could do something for the first time, but. But you're pretty.
It's pretty good assumption that this is all you're ever going to run into in the real world.
It's you'll run into things that are more complicated, but it's not a bad thing to start with.
This is an assumption. And if you can't get the data to fit these, then you go on to more complicated possibilities.
All right. Assuming cubic we also have will remember that the lattice plane spacing this
is something we derived corresponding to Miller indices h k l is the lattice
constant a divided by the square root of h squared plus k squared plus l squared
is something we derive I think in the last lecture or maybe the one before. So what does that give us.
That gives us that a squared over D squared is h squared plus K squared plus L squared.
Now we don't know what A is yet. We're going to try to figure it out, but we know a bunch of D's.
So what is this going to tell us?
The ratio of D's are going to be these k's and L's are integers, so the ratio of one over D squared are going to be an integer ratios.
Okay. So we're going to declare this thing, this quantity h group plus case, but elsewhere we're going to call it N and step.
Let's see what up we step up to step four. We're going to look for integer ratios.
Four integer ratios. Ratios of N.
And then once we have the integer ratios of and we will look for selection roles, look for selection roles.
Rules. Rules. Okay, so let's, let's recall from the last lecture what the selection rules were given h can l if we had a simple cubic lattice.
Simple cubic lattice. All h can l are allowed.
All H can now correspond to reciprocal lattice vectors for the simple cubic layout lattice.
So these are all allowed for a basic lattice.
We derive the selection rule that h plus k plus l l must be even must be even for an FCC lattice.
We derive the selection rule that hk l all even or all odd, so we can actually start making a table of the possibilities.
Actually, maybe I'll put the table over here. So over first column we'll put h can l this is the family possible family's last place.
We'll write the number n which is h squared plus k squared plus l squared.
And then we'll write simple cubic boxy and FCC and ask whether they're allowed.
So what's the simple as possible family of last place we can be talking about?
We could be talking about 100. So n a square plus case proposal squared is one simple cubic yes that's allowed for simple cubic all a
squared plus k all h can l are allowed for simple cubic bc it's not allowed because one +00 is odd.
So this is no, the FCC is not allowed because they're not all odd or all even so it's not allowed for FCC.
How about the next simplest 1110?
A square plus l squared is two simple cubic well.
Yes, that's also allowed bc well one plus one plus zero is even one plus one plus zero is two.
So that's allowed. Yes, FCC is not allowed because they're not all odd or all even 111 what l squared is three.
It's allowed for simple cubic for BC. It's not allowed because the sum is odd, but it is allowed for FCC because they're all odd.
Then 200 keep going.
So two squared +00 squared is four. It's allowed for simple cubic.
The sum is even so it's allowed here and there all even so it's allowed here.
And you can make up a big table of these things. It's worth doing to keep track of which indices you should observe for which possible lattice type.
Now, from this type of table, you can make a list of the possible integers n that you would see for the different lattice types.
So for simple cubic cubic you get and the possible ends you get are one, two, three, four, five, six, eight.
So what happened to seven. So all the possible a's case ls are allowed, but you'll discover if you try there's no h can l that if you square them,
take eight squared plus k squared plus l squared you'll get seven. Just can't find integers.
It's sort of a number theory type of result that you just can't find three integers that when squared and added together give you seven.
So seven is missing and so forth. And I think the next one that's missing is 15 for BC, for BC and is two, four, six, eight, so forth and so on.
I think the first one that's missing is 28. Actually, FCC is the most interesting one.
FCC, the series is an equals three, three, four, eight, 11, 12, 16, 16 and so forth.
Actually, I think I wrote these all down already. Yeah.
So on this on this slide, so here's exactly building that whole table again, we wrote down the first four elements of this table.
Here's the ends. A square was case growth was L squared. One, two, three, four, five, six.
When you get to seven, there's no set of Miller indices that when squared together gives you seven.
And you look at FCC and the possible the possible indices that give you that are allowed for FCC.
The first the first one is 111. Those are all odd that that gives you a total of three.
The next one that's allowed is 200. Those are all even that gives you two squared +00 squared is four.
That's allowed. The next one that's allowed is 220. All even two squared plus two squared plus zero squared is eight.
The next one that's allowed is 311. All odd that gives you 11 and so forth.
So these are the series that we want to want to look for.
So actually let's do that for this aluminium data. Oh, and here's, here's.
A series written out. One, two, three, four, five, six, eight, nine.
Missing seven, 15 and 23. I I'll even integers excluding 28, 60 and so forth.
Actually, it's not a coincidence. The 28 happens to be four times seven and 60 happens to be four times 15.
You can convince yourself why that's true. And then FCC is three, four, eight, 11, 12 and and and so forth.
Okay, so here's the data again. Let's go through this procedure, all the steps, one by one.
So here we measured all of the the angles are someone measured it for us here.
It's very nice of them. So we make a table of all these angles.
We calculate D equals lambda over to sine theta.
So here is the lambda over to sine theta for all of these scattering peaks.
Then we have this statement that one or a over eight squared over squared is squared, plus K squared plus L squared.
And we want to find integer ratios of one over dx squared.
So how do we do that?
Well, let's declare this number here 2.3405 Angstroms we'll call it this of a and we'll make a table of this of a squared over dx squared where d is,
is all of these possible values of D. So the first slot is by definition 12.3 squared over 2.3 squared is one.
But the second 12.3 squared over 2.02 squared is 1.33 and so forth and so on.
We get this in this table here. Those aren't integers, but they're pretty close to integers.
If you multiply it by three, you get a set of things that are really, really close to integers,
and you'll see that that pattern is really three, four, eight, 11, 12, 16, 1920, which is the FCC series three, four, eight, 11, 12, 16, 1920.
So, so we've concluded from this that the the aluminium, the aluminium scattering pattern we found is corresponds to an FCC lattice.
We can actually go on a little bit further to calculate lattice constants.
This might be step five, calculate lattice constant the size of the units L constant a which would be d times the square root of k squared
plus l squared plus h squared because I did that the wrong order e squared plus case proposal squared usually.
All right, same thing. And if we do that there,
the Miller indices have written down for their for the peaks and here's the
A's and they're all pretty close to the same to about three or four digits.
They're not exactly the same. So there's probably some error in the measurement of the angles or something like that has gone wrong.
But within four digits of accuracy, we've measured the lattice constant, a four for aluminium.
Okay, so that's generally how it works. Now when you get good at this, you don't even have to really do this whole calculation.
You can sort of look at the pattern and have a pretty good idea what it is.
So this oh, there's aluminium there.
There's the Miller Indices labelled appropriately.
So I'm going to show you a pattern. And without actually doing any calculation, we can tell what the lattice type is.
Now, there's only three possibilities. If you give me a reason, just give me a reason.
If it's simple cubic, what's the reason? Sound very good.
All right, so that's a so No. Seven. That's a so that one's years.
So you can see here that the first peak so that the spacing of the peaks is not completely uniform and that's because of the sine theta factor.
That sine theta so expresses things out more when you're at low angles than it is that at at high angles.
But, but you can kind of see that one.
The spacing from 1 to 2 isn't so different from 2 to 3, 3 to 4, 4 to 5, 5 to 6.
But then there's a hole where seven is supposed to be, and then 8 to 9, ten, 11, 12.
And you see if they actually there's a hole where 15 is supposed to be.
And if you go all the way up to 23, you'll notice there's a hole where 23 is supposed to be.
And that's the series that you expect for a simple cubic.
So you can just look at this at the scattering pattern for barium Titan eight and know that barium type Nate is a simple cubic lattice.
That's how it works. Pretty cool. All right. It's only very tight. And it is a fairly important material for the Optoelectronics industry.
It's used for various things. Now, there's more information in this.
There is it. There's more information in this scattering pattern than we've used so far.
All we've used is the angles of scattering. We haven't used anything about the intensity of scattering.
And there's a lot of information in the intensity of scattering that we can make use of.
So let's see if we can understand what how much scattering we should get for Miller Indices H Candle.
Well, first of all, there's the structure factor, acid H candle, which gets squared now.
All right, let's even write it out. So the structure factor, remember this s equals s lattice times s basis.
Times s basis. And as Gladys does nothing more than enforce selection rules.
Enforce selection rules. So we've done that.
So that's not going to be interesting for us as a basis.
Well, there's only one atom in this basis because it's just one type of atom.
A single, single atom in the basis is just a pure FCC lattice with no, no interesting basis.
So all you get is the the form factor for aluminium and this form factor for aluminium
decays decays with increasing with increasing increasing g as we mentioned last time,
or increasing angle or increased increase theta.
So we expect that the total scattering is going to drop slowly as a function of increased scattering angle.
But there's actually that's not all that contributes to the the intensity.
There's two other factors which contribute to the intensity of first is the probability, probability of alignment.
So this is the probability that a crystal light will align in just the right way in order to get scattering.
It will come back to that in a moment. And the last factor is known as the Lorenz factor.
LORENZ Or sometimes. LORENZ Polarisation factor.
Polarisation factor, which is really a geometric factor.
We're not going to say too much about it, but I'll show you what it looks like.
So this function is a function which varies fairly rapidly as a function of angle, but then when you add in intermediate angles,
it's fairly flat between around 60 and 140, and then it goes back up eventually at around 140, it starts going back up.
It's the Lorenz polarisation factor. It's really a feature of how you set up your experiment.
And so if you're an experimentalist, you will know what this factor is in advance before you do any sort of wave,
before you do any sort of measurement.
Now, there's a there's sort of a glitch here, which is that sometimes when people present you with x ray patterns like this one,
they'll have already divided by the Lorenz polarisation factor here.
And other times they will not have divided by the range polarisation factor.
So really, if someone shows you data like this, you should ask Is that raw data?
Or have you divide it out by this Lorenz polarisation factor already?
And embarrassingly enough, on many exams they'll ask you a question and they won't tell you whether they've divided by this or not.
So if you want to be safe, you should say you should have told me whether you divide it by this factor or not, I'm assuming whatever.
So anyway, so that's that's this factor.
And it also for most angles. So here we're only scattering up to about 120 degrees here.
So this so this factor here is decaying as a function of angle up to 120 degrees.
This factor here is also decaying as a function of angle. So we'd sort of naively guess that we should always decay as a function of angle.
As you increase the angle and and more or less the scattering amplitude is decaying is a function of angle.
But we haven't looked at this piece here. Probability of alignment.
Well, you might think to yourself, you know, all the crystal lights are aligned completely randomly, so that that factor should be trivial.
It should be just one. But think about this for a second.
You remember when we talked about families of lattice planes.
If we're thinking about a lattice plane spacing, which is associated with the one one, one family of lattice planes,
it could have been the 111 bit family of large planes or the one one bar one or the one bar one one or the one by one bar one.
I mean, there's eight possibilities, which would all give you exactly the same lattice plane spacing that you can scatter off of.
So in fact, this probability of alignment is actually the multiplicity of a scale.
So if we go back to this table here to see where yeah.
Add the multiplicity on to this table.
If we're thinking about the 100 and the Miller indices, there's six possible directions which would present the same lattice.
Plane spacing is 100, the six possible axes of the cube.
But if we're looking at the 110, there would be 12 different phases for 12 different faith ways you could face the cube,
that you would present the same lattice plane spacing. And someone wanted a chocolate bar for realising that 3 to 1 has a multiplicity of 48.
So these are the factors that we want to keep track of if we want to understand the intensity of the scattering.
Now let's look at this.
Let's look at two peaks that are actually fairly close to each other, an angle.
Why do I want to look at two things that are fairly close to each other an angle?
Well, if you both F and L are decaying kind of slowly as a function of angle.
So here, if we're take if we take angles somewhere in the middle, you know, around 100, 120, then then this thing is pretty flat.
This thing is decaying only slowly, and we can only worry about the multiplicity.
So let's take these two peaks here. Three, one, one and two, two, two, three, one, one and two, two, two.
They're they they have these F's and L's in them.
But an L are pretty much the same because we're scattering at about the same angle for three, one, one and two, two, two.
Just a slight difference in the angle. But the multiplicity is let's back up to the multiplicity here.
311 has a where is it. 311 has a multiplicity of 24 where a two, two, two has a multiplicity of eight.
So we would guess at the peak, four, three, one, one should be three times higher than the peak for eight and indeed the peak four,
three, one one is pretty close to three times higher than the peak for 2 to 2.
Okay. So that's how the multiplicity is get in the game as well.
Okay. So these intensities of scattering can actually be very useful for analysing data as well.
I'm going to show you a sort of artificial example of where this might be useful.
So here is some fairly bad data for a taken on iron.
That was bad for a couple of reasons. The first reason is because the peaks are pretty broad.
The second reason is because you only see three peaks with three peaks.
It's pretty hard to tell which series you're talking about here.
You can't count out to the seventh and see if the seventh is missing because you only have three peaks.
So this is going to be a little bit difficult, but we have the amplitudes of the scattering and that actually might be might be helpful.
So we'll see if we can we can make use of that. So we go through the same exercise of measuring the the angles, calculate the latest plane spacing,
take the ratio of the first lattice plane spacing to the various last plane spacing da2
.03 squared over 1.4 squared is two is 22.13 squared over 1.17 squared is about three.
We realise those are in the integer ratios of one, two and three. So those could be two possibilities.
Either it's simple cubic and then these scattering peaks correspond to 100110 and 111,
meaning this integer and a squared plus k squared plus l squared is one, two or three or it could be and is two, four and six.
Those would still be in the ratio of one, two and three, but they would correspond to Miller Indices of 110200 and 211,
which when added squared and added up, give you two, four and six.
So we don't know which one it is just by looking at this data.
So what do we do? Well, there's a couple of approaches we could take. One approach is that we could try calculating the the lattice constant.
A So if we calculated that we could assume simple cubic calculate the lattice constant and we
would get d times square root of a square plus k squared plus l squared gives us 2.03 angstroms.
We did the same thing for BC. We'd get something that square root of two bigger because we're multiplying all the ends by two.
So this thing inside the square root is multiplied by two overall.
So instead of getting 2.3, we get 2.86 OC.
So we have a different lattice plane, a different size of the, of the unit.
So conventional units l for a BC case.
Now why does that help us?
Because we don't know what the lattice constant is for iron, but it does help us because we could try calculating the atomic density.
For the simple Cuba case, we would have one atom per 2.03 angstroms cubed, whereas for a BBC it's two atoms.
Two atoms in the conventional unit cell per 2.86 angstroms cubed.
And those densities differ from each other by a factor of square root of two.
So if we knew the atomic density of ion, we could tell which one it is, so we'd be done.
But suppose we don't happen to know the atomic density of ion. So what do we do then?
Okay. There are other. Other things we could invoke. We can look at the scattering amplitudes, which was what I was leading up to now, remember.
So at this at this scattering angle, the Lorentz factor is decreasing.
The form factor is always decreasing as a function of increased angle.
So you would sort of naively expect that A should be bigger than B should be bigger than C, but it's not C is bigger.
The only thing that could possibly explain that is if the multiplicity of the peak C is bigger than the multiplicity of B.
So if you look at the simple cubic, the simple cubic, we identify the first peak is 100.
The second peak is 110 and the third peak as 111.
The multiplicity of B is bigger than multiplicity of C. So there's no reason you would ever get the C peak to be bigger than B peak.
However, for BC, the multiplicity of the C peak to one one is 24 ways.
The multiplicity of peak is 200, so c should actually be bigger than B.
Now if you just look at the multiplicity as you would expect a factor of four,
that C should be four times bigger than B and it's not four times bigger than B, it's only about two, maybe two and a bit bigger times bigger than B.
But the reason it's not a four, four times bigger is because F is dropping,
the form factor is dropping and Lawrence Factor is dropping also at the same time.
So it's competing with this decreasing effect from these other two terms.
But at any rate, just by looking at this data, we can we can tell just from the amplitudes that this must be BC.
Of course, we also know that iron is not simple cubic because there's only one element that takes a simple cubic lattice,
and that's polonium and iron isn't polonium. And so that does it also in sort of a cheating way to go about it.
All right. So let's do one more example.
This is an example of neutron scattering. Neutron scattering, very similar to X-ray scattering.
But it's a little simpler because the form factor is just a constant.
It's just a nuclear scattering length. It's not a function of g. So what's the latest type?
Oh, my gosh. Oh, yes, yes.
Not so good at this. So. Yes.
So it's so you can tell it's it's this is FCC without doing any calculation because of the pattern that you can sort of imagine if this is three,
this is four, then there's a big spacing to eight, then a little spacing to 11, 12, close together.
So you can kind of see that the spacing is the right spacing for an FCC lattice.
Okay. So but if you want to go through the whole the whole the whole story, we can go through the whole story carefully, measure the angles,
calculate the lattice plane spacings, take the ratios of the one over the one over the square of the last plane spacings you get one, 1.3, 2.6.
We really multiply these by three. We realise these are integer ratios, three, four, pretty close to eight, pretty close to 11 actually.
The data isn't so great and if you back up well you'll realise that these peaks are pretty broad.
They're not really that accurately resolved.
It's not the greatest data. This was data taken in 1959.
I mean, modern, modern data is much better. But but okay,
it's you can definitely see that it's going to be FCC and you can also label the the Miller Indices and get an estimate of the lattice constant.
Okay. So the lattice constant, maybe you don't know it within a few percent, but you have a pretty good estimate of what it is.
But in fact, what we have here and we can write out the multiplicity is we can label the peaks as well.
Actually, I noticed that by doing this you'll notice that there's you would have predicted a slight peak here,
a peak of 40 zero peak here where the shoulder is.
It's not there you. But if you kind of look at it carefully, you see that there's kind of a weak shoulder.
You can imagine that it's just a very small peak that's sitting there now.
So this tells us where where the peaks are. But we there's more information that we would like to get.
We would like to know what the basis is. So we know that titanium carbide titanium carbide is F.C.C.
And it has to have a basis with the basis that it has two elements in it, with the basis of two atoms.
Of two atoms. And we'd like to know what is the form of the basis.
In other words, where do you put the titanium? Where do you put the carbon in the unit cell?
So can we figure that out? Well, okay, so arbitrarily we can we can always assign titanium to the position at position 000.
And the unit cell will just define the titanium position to be 000.
But then carbon we'll put at position u, v w which we don't know.
Incidentally, I should have mentioned Titanium Carbide is actually a a really important material.
Does anyone know? Here's another one. Anyone know why? Titanium Carbide is really important?
Use it for what? Yeah, yeah. Anything that needs to be really hard then.
Yeah. Okay. So. So Titanium Carbide is one of the hardest materials out there.
Short of diamond, I think there's only one or two others that are as hard as titanium carbide.
It's easy to, um. And it's easier to make titanium carbide.
It's not as expensive a diamond. It's so whenever you need anything that needs to be very tough, it's very frequently used for things like tools.
If you need a saw or a drill bit, that's going to cut through something very hard.
It's very frequently titanium carbide. So that's why they did this experiment way back in in 1959,
because it was an interesting material even back in 1959, and they didn't even know its structure.
So. Okay, so I think they did maybe they did know this structure from something else.
I don't I'm not sure, but I don't I don't think they knew it. So anyway, we don't know where the carbon is in the, in itself.
So we're going to see if we can use this intensity data from these peaks to figure out where in the units all it is.
So generally the piece we're going to be interested in is the basis structure factor, which is which ends up getting squared.
And that is, well, maybe I'll write this out here first as basis equals sum over alpha in the units l
and units l e to the i g dot are alpha times b alpha for the two atoms in the cell.
So space is squared here will be v titanium plus b carbon e to that will write it as either the two pi i h k l
dotted into u v w and then you take that thing and you square it and that will give you the basis structure factor.
Now if you knew what these nuclear scattering lengths were, b titanium in b b carbon, you might have a fighting chance of figuring out what you were.
But suppose we didn't know that. Can we still make any progress? Well, we can make a little bit of progress.
If I give you a hint. This is a kind of hint. You get an exam. The kind of hit you get on an exam is that F.C.C. with two atom basis.
Atom basis. There's only two common ones.
Only two common ones. Common possibilities.
In fact, I don't even know of any cases that are not one of these two.
There may be even a theorem that you can't have anything else. I'm not sure about that.
So don't quote me. But anyway, the two possibilities we've both we've seen them both before.
There's a so-called zinc blend structure, which is the same as the gallium arsenide structure,
which we talked about in class, which puts you BW at the position of one quarter, one quarter, one quarter.
And the other possibility we also discussed is the sodium chloride structure which has you've w at one half, one half, one half.
So can we figure out which one of these two possibilities it is.
Okay. Well, let's see if we can do it. So let's try first try one, try zinc blend structure.
Okay. So in this case s basis.
Squared equals well, we have b titanium plus B carbon either the two pi i h k l and then one quarter of one quarter,
one quarter, which we can also write this factor here.
This exponential factor here is also equivalent to E2 the pi over two times I
times h plus k plus l or equivalently this would be I to the h plus k puzzle.
So I'm out of room here. We're here so we have s basis squared equals b titanium plus v carbon times I to the h plus k plus l squared.
Okay. And what is that as a function of H canal? Well, there's three possibilities.
Case one, I mean, in case A is that H plus k plus l is a multiple of four equals four times an integer n.
So in that case we get SBC squared equals v titanium plus v carbon absolute value squared.
The second possibility case b e is h plus plus l is still even, but it's not a multiple of four.
So we'll call for n plus two, in which case we get spaces squared equals B titanium minus B carbon squared.
And the third case, Casey. Is H plus k plus l is odd.
Odd in which case s basis squared is B titanium plus or minus i times b carbon squared which is b titanium squared plus b carbon squared.
Good. All right. So with the stare at that for a second.
And what can you conclude from this? Well, no matter what B, titanium and B carbon are, as long as they're real,
which they are the largest of the three possibilities is always either case or case.
B, if they be titanium, be carbon to have the same sine in case A is the largest of the three possibilities
if they have opposite sign in case B is the largest of three possibilities.
So the largest peaks should definitely have even H plus K plus l.
Now let's look at this data and you'll see H plus K, plus L and you'll notice that.
The highest peaks always have od plus k plus l systematically always there.
So if 111 is high, 311 is high 331 is high, 333 is high.
All the other ones are small are even so this does not agree with the data does not agree but agree.
Well, we can try something else. We can try the sodium chloride structure, so try sodium chloride, in which case we have.
Well, I wrote it over here. Somewhere over there is one half, one half, one half.
So s basis squared is then b titanium plus.
B carbon times e to the to pi i h k now dotted into one half, one half, one half, and then all squared.
And that we can be right as v titanium plus v carbon times minus one to the H plus k plus l all squared.
So there's two cases we can have. Case one case or case a is h plus k plus l is odd.
In which case we have B titanium minus v carbon squared is the answer.
Whereas case B case v is h plus k plus l equals, even in which case we have b titanium plus b carbon squared.
Okay. So now what's which one is largest here?
Well, you can see that if vitamin B carbon have the same sign, then the even will always be larger.
Whereas if they have the opposite sign, then the odd will have larger.
Well, if you look at the data, the H plus Capossela Ott is clearly much larger.
And so that tells us it's it's consistent with this case that if B titanium be titanium has opposite sine it sine of be carbon.
So the large peaks are be titanium minus be carbon squared with be titanium and be carbon having
opposite side and the small peaks are the sum of the two with them having opposite sine as well.
Okay. Everyone happy with that? More or less.
Little bit. Yes, please. Yes. All right. Thank you.
All right. Good. So let's see.
So here's the summary. Summary slide. So we conclude it has to be sodium chloride structure.
Let's see. So we can actually go on and discuss.
Oh, here. Let's discuss how you make neutrons for a second. So.
We discussed how you make how you make ashtrays in the last in the last last lecture, I think.
But we haven't discussed how you make neutrons. This is a neutron facility escalation.
Neutron source, also right down the road next to the synchrotron source, rather large and expensive apparatus.
What we need is we need neutrons with a wavelength on the order of one angstrom.
How do we get that? Well, if the wavelength is one angstrom, the momentum is two pi h bar over lambda.
And from that we can calculate that the energy we need is p squared over two m equals approximately 80 million electron volts or about 800 kelvin.
You put in the appropriate factor Botswana's constant to make energy a temperature.
And this might sound to you like it's hot, but in fact, 800 Kelvin is really, really cold for neutrons.
Why is that? Well, how do you get the neutrons in the first place?
Making neutrons make neutrons.
Well, back in the old days, they didn't have neutrons.
And then along came the the Second World War and the Atomic Bomb Project.
And after the atomic bomb project, they had nuclear reactors. And nuclear reactors have neutrons as a by-product.
So possibly one is a by-product of vision.
And you can imagine that when a neutron comes off of a of a nucleus and some fission process,
it's coming out with millions of electron volts rather than miller electron volts.
So it's extremely, extremely energetic. So this 800 Kelvin is extremely low energy.
There's a more favoured method for making because people don't like to have atomic reactors in their backyard these days.
There's a more favourite method for making neutrons and in fact it's more efficient method as well, which is known as spoliation.
And that's what this source is. This neutron source is a separation source.
The idea of a separation source is not unlike the idea of an X-ray tube, where you take usually a proton,
a proton, and you accelerate it to about a giga electron volt and you hit it onto a target target.
And the idea is to smash some nucleus, some nuclei, and kick off some neutrons.
Now, again, since you're hitting it with such a high a high energy, you're going to kick off neutrons with some very, very high energy as well.
So whichever method you use to create neutrons, they're coming off with an incredibly high energy and you have to make them cold.
So to make them cold, what you use is you use a moderator and a moderator is basically just a big tank of some substance
where the neutrons can go in the substance and bounce around giving off their energy to the substance.
So frequently people use carbon graphite.
Graphite is carbon graphite, carbon water, heavy water, lots of materials,
anything that the neutrons can bump into and give off their energy to and eventually out.
The other end of this thing you get you get neutrons and you want to kill them down until they have about 800 kelvin worth of energy.
Now, once you have the the neutrons coming out,
you need to somehow figure out how to make them a monochromatic monochrome, meaning you just want to get one wavelength.
Remember, that was the first rule. Know your wavelength. So you have to monochrome eight.
So how do you monochrome one way method one is to distract.
So you can use a crystal. It's known we mentioned this in the case of X-rays as well.
Different from a known crystal. From known crystal.
And that sometimes works. But there's another method which is also used, which is very nice for neutrons,
which is a time of flight method, time of flight, which works in the following way.
So you have a beam of neutrons coming in over here, and then you have sort of a window which you can open and close,
open and close whenever you want, and then you have a long spacing.
L From here to here and another window which you can open and close.
So you open up and close the window very quickly to let in a little pulse of neutrons.
And then at some time Delta T later you open up the other window.
So then you know l you know T So if it made it from the first window to the second window in Delta T You know, it's velocity.
So L divided by delta t is its velocity.
If it's no, it's velocity, you know, it's momentum. If you know it's momentum, you know it's wavelength.
So you can select by knowing how by giving it a certain amount of time to get from
the first window to the second window you can select for a particular velocity,
therefore predict particular momentum, therefore a particular wavelength.
One more thing that I do want to mention about neutrons will say maybe just one or two more things about them in the.
Next lecture. But I do want to mention now is one of the reasons people like neutrons.
Is the neutrons actually have a spin. And that enables you to see things that have important where the spin is important.
If you have more spins in one region than another region that neutrons are sensitive to,
that they feel to the local magnetic environment as a potential, whereas photons do not.
They don't care about the local magnetic environment. Neutrons do because the neutrons have some spin.
Okay, we'll stop there and we'll pick up. Finish this off next next lecture.