Okay. Let's get started. Welcome back.
Now that this is the 12th lecture of the condensed matter course.
Hello. Hello. Is it Monday still? So when we left off, we were talking about scattering the standard scattering experiment.
If you have some sample, you send in some way.
That way vector k and something comes out at wave vector.
K prime. And by measuring what comes out at various wave vectors.
K prime, we're supposed to be able to deduce the structure, the microscopic structure of the sample.
I want to emphasise again that this type of experiment is the way to understand the
microscopic structures of things on a scale smaller than you can actually see with your eye.
The number of Nobel Prizes that have been awarded for people doing this type of experiment is enormous.
There's like 20, 25, 30 IG Nobel Prizes, just a couple of famous ones.
You know, Oxford's own Dorothy Hodgkin, Somerville's favourite daughter, won a Nobel Prize for discovering the structure of penicillin and insulin.
Using X-ray scattering, Watson and Crick very famously figured out the DNA double helix structure,
which they figured out by looking at X-ray scattering data taken by Rosalind Franklin.
Rosalind Franklin sadly died extremely young and was therefore denied a Nobel Prize as well as the rest of her life.
It is very sad, but everyone should know her name nonetheless.
Anyway, this is an extremely important experiment and that's why we're spending so much time on it.
At least this lecture, next lecture, and maybe some of the lecture thereafter.
One of the things we derived in the last lecture is that in order to get scattering from K to prime,
we must satisfy in a crystal the Laue condition that k minus K prime must be a reciprocal lattice vector g.
Now, at the very end of of the last lecture, I claimed that this condition,
this Laue condition is very similar to a condition that you have already thought about when you considered just simple diffraction off of planes.
If you have light or x rays or something, or neutrons coming into a set of parallel planes, a family of lattice planes and you have are different.
We understand how you can get diffraction of this sort at certain angles off of parallel family of planes.
And I'm going to claim that this this condition here is actually equivalent to that.
So first, what we're going to do is we're going to review some things about diffraction off of planes.
Then we're going to show that the LAUE condition and the diffraction condition the same.
So let's abstract this picture a little bit to just a set of planes like this.
And we'll imagine we have some wave coming in like this.
It reflects off like this at an angle theta, but also another parallel wave coming in like this.
Exactly the same wave just goes down to one layer further.
The fracture off like this, the same angle theta like this.
And I'm just going, I'm going to actually emphasise this and maybe I'll even write it down because it's an extremely common source of error.
Total deflection angle. Deflection angle equals to theta.
Why is that while there's theta there and another theta there.
So the difference, the difference in angle between the incoming wave and the outcome wave is actually two theta.
This this causes all sorts of grief to a lot of people.
So make sure, you know, when you're talking about theta and when you're talking about two theta.
So as you did this this type of calculation in prior years,
you know that the what you need to do is you need to calculate the additional distance travelled by the second wave.
So if I drop a perpendicular here and a perpendicular here, the additional distance here is d sine theta if d is the distance between the the layers.
So in order to get constructive interference, constructive, maybe, maybe I'll write this first.
The additional distance. Additional distance distance is then two d sine theta equals two d sine theta.
So de sine theta on this side and then another d sine theta on that side.
So the additional distance taken by the by the wave that reflected off the
second plane is two d sine theta to get constructive interference constructive.
Interference into Ference.
I think as the term goes on, my writing gets worse. So we get is end lambda equals two sine theta.
So you have to have an integer a number of wavelengths contained in that extra distance.
Does that all look familiar from previous years? Yeah. Okay, so I'll put a box around it because it's an important equation.
This is known as the Bragg condition, Bragg condition, after William Henry and William Lawrence Bragg,
the father and son team who pioneered X-ray scattering.
William Lawrence Bragg winning a Nobel Prize at the age of 25 for his work on X-ray, scattering for a chocolate bar.
Does anyone know who the second youngest Nobel laureate was? It's a name you all know.
So those are all really good guesses. But those are all wrong.
Really close to all of them. Same era. Heisenberg.
He said it up there. Oh, my gosh. Close enough. Yes, Heisenberg.
Exactly. Dirac was just a few months older than then, then Heisenberg.
All right, so now what I claim to you was that this condition, this Bragg condition, is actually forgetting constructive interference.
You only get constructive interference coming off on certain angles from your crystal
that that condition is actually the same condition as this lousy condition over here,
which is the conservation of crystal momentum. So let's see if we can prove that you write this down, because this is important proof equivalence.
Of Bragg, Bragg and Laue.
Okay, so same diagram. We have two parallel planes in a family of lattice plane.
You can actually you can kind of understand why it is that these two are going to be related.
Remember from last lecture that reciprocal lattice vectors are associated with families of lattice
planes so that gee up there is going to be telling you something about this family of lattice planes.
So the fact that we have parallel planes is because we have a reciprocal lattice vector.
Those are sort of equivalence statements. So here we have our family of lattice planes separated by some distance.
D We have an incoming wave at wave vector.
K And outgoing at wave vector. K Prime that.
And then since we have a family of lattice planes, there is a orthogonal vector g which is a reciprocal lattice vector.
That was one of the things we derive last time that every time we have a family of last planes,
there's a reciprocal lattice vector perpendicular to it.
Since we have since we have elastic scattering, the magnitude of K and the magnitude of K prime are the same.
And let's actually define some unit vectors.
Unit vectors, vacc k hat, k prime hat and g hat to be pointing in the the appropriate directions and then with a little bit of geometry.
K hat dot jihad equals sine theta.
Just take a second. Make sure you're convinced of that. If K hat and jihad were pointing in the same direction.
Then K had a g hat would be one sine theta would also be one in that case.
Whereas K prime hat which is going in the opposite direction, jihad is minus sine theta.
Okay. And we can also use we're going to need that, that k vector.
We can write as the magnitude of K which is two pi over lambda with lambda the way vec with the wavelength times k hat.
Everyone happy with this so far. Yeah. Okay, good.
So we're going to assume the Laue condition. So assume Laue that is k minus K prime equals g.
So assume that satisfied and I can rewrite that then as two pi over lambda times k hat minus k prime hat equals g
then dot with g hat to get to pi over lambda g hat okay hat minus g hat okay prime hat equals g hat to g vector.
Okay then we'll substitute in. We just had the jihad k hat is sine theta so we'll substitute that in so we get to pi over lambda.
So this is sine theta here. The second term is minus sine theta and then g hat g vector is just the absolute value of g vector and one.
Still happy. Okay. No, not happy.
Yeah. Still happy. Okay, so just a little bit of rearrangement.
This becomes two pi over absolute g vector times two sine theta equals lambda.
And now remember what we said, what we derived last time about reciprocal lattice vectors.
The spacing between lattice planes D is two pi over the shortest reciprocal lattice vector in a given direction.
G min. So any reciprocal lattice vector g vector has to be some integer times the minimum lattice vector in that given direction.
So PN is some integer integer.
So substituting this equation in this equation into the equation just above gives us two d sine theta equals and lambda the bragg condition.
So what we've shown is that the Laue condition,
this conservation of crystal momentum is the same thing as requiring constructive interference or having the Bragg condition satisfied.
Okay. Good. All right. So supposing that we have the bad condition or the lousy condition satisfied?
One of the things we derived last time was that the intensity of scattering.
Intensity of scattering is proportional to the structure factor.
At G came out as K prime squared structure factor s equals the four eight transform of the scattering potential v of.
Ah. Does that sound familiar from last lecture? I hope.
Okay. But that still leaves up in the question what is V of ah and what v of are is what the potential your wave feels as it goes into the crystal.
Depends on well, depends on the crystal, but it also depends on what kind of wave you're using.
They the wave interacts with the crystal very differently depending on the type of wave that you're using.
So the easier case to think about is neutrons.
So case one, neutrons, neutrons interact.
Mainly mainly are via nuclear forces by a short range short range nuclear forces and nuclear forces.
So what does that mean? It means the neutrons go into the sample and they see very short range potentials associated with the nuclei.
And we can write that we of our equals or is proportional to, I guess, some of our atoms,
all atoms alpha in the whole system, some coefficient B alpha times a delta function potential at r minus R alpha.
So basically saying that every nucleus.
So our alpha is position of atom alpha. Position of atom alpha.
And B Alpha is in the in the magnitude of the interaction with that nucleus.
It's known as the nuclear scattering length. Scatter length, length.
And so the essence of this equation is basically saying that the the neutrons coming in,
they see very sharp peaks in potential at the positions of the nuclei.
And and that's all they don't see anything else. Now these coefficient B, they have to do with nuclear physics.
And nuclear physics can be extremely complicated.
They can be positive or they can be negative depending on the particular nucleus we're talking about.
Positive means you have a repulsive interaction with the nucleus. Negative means you have an attractive interaction with the nucleus,
and you can have the either and the size of these B's can vary by lots, depending on the particular nucleus you're thinking about.
And really the only way to know what B is for a particular nucleus is to just look it up on a table or measure it.
You know, you could have two atoms on the periodic table that are right next to each other, like carbon, nitrogen and the nucleus.
Scattering length is completely different. Some can be positive, some can be negative.
It just goes all over the place. So it's actually something that we can't actually predict very easily.
Even nuclear physicist, even good nuclear physicists have a hard time predicting these things.
So we just have to resort to a table and assume that these things are known or they somehow we're going to be able to measure them.
But given that we have this sort of functional form of the interaction with the nuclei,
we can then calculate the structure factor, structure factor SMG,
which remember the the definition is that it's actually deeper and therefore you transform in the unit
cell the for you transform in the unit cell of e to the i g r times v of r something we derive last time.
And if we actually plug in the, the form that we just wrote down above,
we have some of our alpha b alpha delta, three dimensional delta function R minus hours of alpha.
Now we'll let the Delta Function Act and we get the fairly simple outcome that the sum over Adams alpha Adams Alpha in unit
cell of B sub alpha e to the i g dot our sub alpha where r sub alpha is the position of position of atom alpha and unit cell.
Adam Alpha. So there we have it.
So that's the structure factor for scattering, scattering of neutrons against nuclei.
Now that's versus for neutrons. Neutrons being fairly simple.
But there's a second case that we need to worry about, which is x rays and x rays and more complicated x rays.
X-rays. Well, they're electromagnetism X-rays, you know, like light.
They mainly scatter. Mainly scatter from electrons.
Scatter from electrons. From from electrons.
Electrons. Via Thompson scattering in space.
Thompson scattering last year. So. Sound familiar? Yeah. So it may be scattering from electrons via Thompson scattering.
There's a little bit of Thompson scattering off the nuclei, but really very, very,
very small amount because the nuclei are very heavy compared to the electrons.
And so they're very hard to scatter off of. It's hard to push them around. It's easy to push the electrons around because they're light.
So mainly the of our four x rays, the effective interaction is proportional to the density of electrons.
The density of electrons. So what's the density of electrons?
Well, we can write sort of an approximate expression for the density of electrons,
which will rate as some sum over atoms alpha Z, sub alpha Z here being the atomic number, atomic number of atom, alpha,
atomic number being the number of electrons in the in the atom,
also the number of protons in the atom times some function which I'll call g sub alpha of R minus R alpha,
which I guess we can call something like a shape function. Which is normalised.
So it's integral as one and we can compare that to what we have over here.
Looks kind of similar, except here it's a delta function and here it's not a delta function.
So the difference being that when you if you're a neutron, you you scatter off of only the nucleus.
Whereas if you're an X-ray, you scatter off of the whole atom.
So you can kind of think of this G as being a very, very fat delta function.
It's spread out, spreading out the delta function into some functionality or just takes into account the entire size of all the electrons in the atom.
So think of it as a sort of a fat helper function.
So now I should say, however, that this this form where we're summing up over the shapes of all these different atoms,
times the number of electrons in each atom is a little bit approximate.
And the reason it's a little bit approximate is because there's an assumption,
a hidden assumption here, that the shape of the atom is independent of its environment.
If an atom happens to be bonded to one particular type of atom, it will actually distort the electron cloud of the atom a little bit.
So the shape will change a little bit, and we've ignored that.
And if it's bonded to a different type of atom, its shape will will change a little bit and maybe another direction.
It will depend on which direction it's bonded and so forth and so on.
But to a very good first approximation, this is a fairly good way to describe the the shape of the density of electrons in any crystal,
just you only have a shape of for a particular type of atom and you add it up over
all atoms and you ignore the fact that neighbouring atoms will distort the shape.
Okay, good. All right.
So we can take this form of the, of the density of electrons and again construct the structure factor as a g which is now sum over all atoms.
Atoms alpha integral D three are very similar to what we did just over here, over a unit.
So then we have the I got R, we have Z Sub Alpha and we have g sub alpha.
Of our ways are alpha. If we do that for you, transform.
What we get is extremely similar to what we get over here, very similar to this equation here.
What we get is of the following form sum over atoms, alpha and unit cell of each of the.
I got our efforts of alpha use different letter as a function of g.
This is known as the atomic form factor. Form factor.
It depends on the type of atom we're talking about.
And if you want to know what the atomic form factor actually is, so actually maybe just compare it to over here.
The atomic form factor takes the place of the nuclear scattering length over here of this takes the place of the B alpha.
But the overall structure of that of the of the structure factor here out of this equation stays pretty much the same.
The only difference is the atomic form factor is also a function of the, of the the reciprocal lattice factor g.
Whereas over here it's not in the reason it's not over here is because we have a nice simple
delta function interaction over here and we have a more complicated interaction over here.
If you want to know the actual form of F of G, it's actually an integral d3r, it's just the 48 transform g r r g sub alpha.
I guess maybe this about for the z sub alpha first, then g of alpha of R.
So it's just the for the transform of the shape function g.
The surprise maybe is this integral is actually the overall space, not just over the moon itself.
And that I'm not going to go through the whole derivation.
I think there's a revision homework assignment that asks you to try to work through it maybe.
Or there was last year or maybe it's not there this year. But it's worth trying to do.
It's in the book. But to try to convince you that the integrals should be over all space, let me make the following argument.
So these the shape functions g have long tails.
I mean, this tails are small, but they are. But they go out a long way.
If you're only adding up over atoms within the unit cell,
you would never feel the tail of one of those those shape functions because you're only adding up over alphas within the unit cell.
What you should be including the tails of all your neighbours.
So there's atoms way out there and other unit cells. Then you should be feeling the tails of those of those shape functions.
And so the fact that you're integrating all overall space is keeping track of both the
short range part of G and the tails of the of the GS from other atoms farther away.
So they all sort of get re some and come into the the form factor here.
Okay. So I'm not going to do the derivation. It's probably not worth doing.
But. Might be worth doing for fun.
Anyway, the the form of this form factor more or less looks like this.
So we'll draw alpha here.
We'll draw G over here for g equal to zero for g equals zero equals zero f sub alpha is just z sub alpha just basically counting.
You can kind of see if you insert zero into that integral all over space.
G is a normalised shape function, so you just get Z back out.
So, so if g zero you get z sub alpha up here,
but then it sort of decays slowly down to something much smaller and the decay length is more or less one over the radius of the atom.
Radius of atom. Very frequently, people just make the approximation that F of alpha is a constant,
and the constant is just the number of electrons because you know, if you have more electrons, you scatter more off of it, off of the atom.
So just let's assume that the form factor is independent of G and it's basically just proportional to the number number of electrons in G.
It's not a horrible approximation. It isn't completely right, but it's not too bad and makes your life a lot simpler.
It makes the the x ray problem look a lot more like the the neutron problem where there's this form factor.
This B doesn't depend on G anymore. It doesn't depend on the reciprocal lattice factor.
But strictly speaking, F should decay as a function of g as well.
So generally, whether we're talking about X-rays or we're talking about neutrons,
we will always have s of g of the following form some over atoms, alpha atoms, alpha in units.
So each of the i g are alpha and then well I'll write it as f of g but or or b sub alpha.
If we're talking about neutrons, it's convenient frequently to rewrite this in terms of Miller Indices.
So if we rewrite these statements in terms of Miller Indices, we would have some of our alpha and units all.
E to the two pi i h times you alpha plus k times v alpha plus l times w alpha f sub alpha hk l so here hk
l hk l is the Miller Indices Miller Indices of g g vector and you alpha v alpha w alpha equals position.
Position of atom alpha alpha in unit cell.
Okay. That make people happy. Good so far.
All right. Let's actually do an example to make make some of this more clear.
So example, a material we talked about before, caesium chloride.
So remember, caesium chloride is simple cubic with basis.
With basis. And the basis has a caesium at 000000 and a chlorine at one half.
One half. One half. And I think I may have a picture of that.
Yeah, I hope this looks familiar. We spent some time discussing it.
So using that expression for the structure factor, we can we can write s at a scale equals.
So it's going to be a sum over the two atoms in the unit cell.
So the first atoms at 000. So each of the two pi, then we have a bunch of zeros upstairs.
So the first time will just be f caesium because either the two pi is zero is just going to give you one good.
And the second term will be f chlorine times e to the two pi.
And then we have h times one half plus k times one half, plus L times one half.
Plugging in the position of chlorine, UVA and W both being one half for chlorine and I can rewrite that as F strictly speaking.
Strictly speaking, I should have given f it caesium an f chlorine h can l dependencies as I just ranted about up up there.
So they're strictly speaking, they're also functions of h can't l weak functions of h candle as well.
They depend slightly on on the magnitude of G two, but I typically won't write that right, that dependence.
But you have to remember that it's actually there so we can rewrite that as f caesium in here.
I'm going to drop the dependence on h can't l for convenience times f plus f chlorine and this factor of each of the
two pi i h times one half plus k times one half was l times one half is actually minus one in the H plus plus l.
Good. So this is the structure factor. And you square this thing.
This will give you the equity squared. It will give you the amplitude of scattering to O with k minus k prime given by Miller Indices.
H can l. Okay. All right. Let's do another example.
Example 2x2. Pure caesium.
Pure caesium. Now you remember that that's BC,
but PCC can be thought of as simple cubic with a basis with basis where the basis is caesium at 000 and caesium at one half, one half, one half.
So in other words, we're just replacing the chlorine by another caesium.
We discussed this before. And so there's a picture of it.
So you just replace the chlorine with the caesium. So we get exactly the same result, except now chlorine.
This is the chlorine from the centre of the unit, the conventional unit cell.
It's going to be the same as if caesium. So now we'll get s h cl equals F caesium plus F caesium times minus one to the H plus k plus l.
And we want to we can factor all the F caesium and F caesium times one plus minus one, the eight plus plus L.
Now you'll notice that this vanishes unless write this down vanishes.
Unless. H plus k plus l is even age plus k plus l is even.
This is what's known as a selection rule or a systematic absence.
Maybe I'll write those down. Selection rule. Rule or systematic absence.
Systematic absence. Absence. Absence.
Sense. Systematic absence.
The idea being that you will get no scattering unless h plus k plus l is even.
So all of the Miller indices with H plus people to L being odd have no scattering at all.
They are systematically absent from this, from the scattering or another way words.
In order to get scattering, you must satisfy this selection rule.
And this is generally true for BC lattices, for BC lattices, for BC lattices.
These are the selection rules now.
So whenever you do a scattering experiment, if you discover that all the Miller indices scattering by these wave vectors with H plus K plus L,
any time if you discover that H was odd that there's no scattering at those Miller indices, but you do find scattering, for instance.
Kate Purcell even. It's a very, very good bet that what you have is a BC lattice.
However, you can be fooled. How can you be fooled? You can be fooled.
For example, let's consider the case of caesium chloride.
Suppose just coincidentally, F caesium and F chlorine were equal to each other by some coincidence,
in which case you'd get the same vanishing over here. You'd have the same selection wrong.
But if they're different, maybe by a little bit, then the selection rule is almost satisfied.
Meaning that you have to look really, really carefully to see that there was some amplitude because these would almost exactly cancel.
So caesium chloride does not have this property of caesium and chlorine are not very similar,
but there's a very, very similar material for which these two are extremely close.
And you might be mistaken, you might if you looked at the data, you might make a mistake.
And unless you looked at it really carefully, you would think that you would have the same selection role.
Anyone want to guess what that material is? Well reduction potassium chloride so that we have there.
I'll give this to you anyway, but it's not the right answer. Potassium chloride is a very good gas.
And the reason the reason thallium chloride is a good gas is because remember that these aff these form factors are
basically proportional to the atomic number and the atomic number of caesium and chlorine are extremely different.
Chlorine is what is chlorine? Chlorine is number 17 and caesium is number number 55.
So these two numbers will be extremely different. But if you replaced the chlorine, the caesium with potassium, potassium is right above it.
On the periodic table it's chemically very similar. Then potassium has as atomic number 19 and chlorine has atomic number 17.
So these would be pretty close to each other and you'd have a near cancellation.
Potassium chloride is not the right answer because in fact,
potassium chloride does not take this crystal structure and it takes a different crystal structure.
It's FCC. It's like it's like sodium chloride. It's FCC with a basis.
There's another material, but that's a hint. What's the other material?
That would be more like caesium chloride. No.
No, no. Magnesium. No, no, no.
All right. I'm going to eat this one myself. It's a caesium iodide.
If you replace chlorine with iodine, which is which is to two spaces under it on the periodic table,
that iodine has has atomic number 53 and caesium has atomic number 55.
So the cancellation between these two would be almost exact.
I mean, only within a few percent you would, you would have a slight non cancellation.
So you're very easy to make the mistake and look at the data and think that you have a BSC lattice because you think there's a systematic absence.
But if you look at it much, much more closely, you discover that in fact, it's not perfectly cancelling.
Okay. And we're happy. All right. So.
Right. We can do the same thing with FCC lattices.
Consider the case of FCC. FCC lattices.
So for FCC we can think of FCC equals equals a basic, simple cubic times a basis simple cubic.
Time's a basis. Incidentally, I should I should have mentioned this.
If we were looking at nuclear scattering lengths,
then there's absolutely no reason you should ever expect any two would be the same independent of their atomic number.
Because, I mean, you could have a strange coincidence that the scattering length of hydrogen happens
to happen to match the scattering length of iron or something crazy like that.
But just because they are close to each other in the periodic table doesn't mean that they have close nuclear scattering links completely unrelated.
So it's sometimes good if you think there's a near cancellation to switch from x rays to neutrons,
because a near cancellation with x rays won't be a near cancellation with neutrons generally.
All right, back to FCC. The FCC. If we can think of it as simple cubic with a basis where the basis equals basis equals 0000 and one half,
one half zero and one half, zero one half and zero one half one half.
Actually, I think I have a picture of that as well. Remember this from from a couple of lectures ago.
The FCC lattice is for inter penetrating simple qubits.
Okay, so we can write the structure factor.
The structure factor is going to be the sum over all these four four lattice points within the conventional unit cell of E to the two pi,
and I'll abbreviate this hk l dotted into you the w point alpha, which we can then write as one plus e to the i pi h plus.
So one is from 000. Then we'll have h plus k plus.
Either the i pi h plus l plus the i pi k plus l,
which is then equal to one plus minus one to the h plus k plus minus one to the plus L plus minus one to the K plus l.
And I claim that this vanishes unless h cancel h can't l all even or all odd.
And to convince you of that, if at all.
Even then, all those sets of exponents are even so it's one plus one plus one plus one.
If h can't l are all odd. Then again two odds added together always give you an even.
So it's again one plus one plus one plus one. But if you have, for example, one odd and two evens, let's think about that case.
Let's suppose H is odd and canal or even then h plus k is odd, h plus l is odd, but k plus l is.
Even so we'll have one minus one, minus one plus one and vanishes.
And you can convince yourself that unless h can either all even or all odd, you'll get that will vanish.
So this is a selection role rule for for FCC.
So where do these where do these selection rules actually come from?
If you remember back to the last lecture or the lecture before we discussed families of lattice planes and how how
it is that certain sets of Miller indices do not correspond to families of lattice planes for BCS and FCC lattices.
So remember, for example, with 100 family of lattice planes are reciprocal, lattice vector corresponds to this these sets of planes for simple cubic.
If I tried to use that 1004 BC, it would not be a family of last planes because you would miss the the guy in the centre of the conventional use.
So you have to go up to 200 in order to get that second the extra plane in the middle,
a smaller plane spacing in order to catch that additional additional last point in the middle of the conventional unit.
So OC two, does this sound familiar? Vaguely familiar.
I hope so. Now, this is why 200 satisfies the basic selection rule,
whereas 100 dozen for BC the selection rule is that h plus k plus l should be even that is true for 200.
It is not true for 100. That's the origin of the selection rule.
The selection rule is asking when is this set of no indices actually a reciprocal lattice vector for this particular lattice?
That's where it's coming from.
You can see another case with one one, one, one, one one makes these diagonal planes here cutting across the unit cell here.
But it would miss the back point if it were thinking about PCC.
We'd be missing the point in the centre of the unit cell.
If you went to 2 to 2, you have twice the density of of lattice plane of planes here and you would catch the,
the lattice point in the centre of the unit.
So 111 does not satisfy the BC selection rule that h plus capable cell should be even whereas 2 to 2 does.
Okay, so that's how you tell when you have a real reset.
When the indices actually represent a reciprocal lattice vector,
they represent a reciprocal lattice vector when they satisfy the selection role and
they do not represent a reciprocal lattice vector when they do not represent it,
satisfy the selection rule. Okay. Makes sense. Yeah. Good.
All right. One thing that's really cool about these about these selection rules is that, in fact, when you can we can describe any crystal,
we can describe a crystal as a lattice, any one of these 14 lattice types times a basis.
This is one of the things we went over when we did crystal structure, for example, sodium chloride is an FCC lattice,
but then times a basis which tells you there's a sodium is 000 and a chlorine at one half, one half, one half.
The selection rules remain the same independent of the basis selection rules.
Rules are independent, independent in depth of basis.
So there's a mnemonic for how it is that people remember.
This is frequently said that S is S for the lattice times s for the basis.
So if s for the lattice vanishes, then the whole thing vanishes.
Okay, so let's see if we can put some, you know, some equations behind this.
What I mean by s for the lattice times s for the basis. All right.
So let's recall that S is going to be a sum over atoms, alpha and units l of f sub alpha in the i g r alpha.
And then we can write any atom alpha in terms of a lattice point and a basis vector x.
So this is lattice point, they are flat point.
An X is the basis factor basis.
So that its position are alpha is the position of the lattice vector R for that atom and the basis vector X for that atom.
So for example, x for sodium here is zero.
It's sitting right on the lattice where X for chlorine is one half, one half,
one half, which tells you that you take to find the position of a chlorine,
you go to a lattice point and then you just place it by a vector X for chlorine, which is one half, one half, one half.
So I'm just writing the. Coordinates of the atom in terms of its lattice coordinate and the basis displacement.
Okay. All right. So then we can write s in terms of this as some over R in lattice, some over X in basis.
Of F the form factor is only going to depend on X because it's only cares what type of atom you're talking about, whether it's a sodium or a chlorine.
So F only depends on X here times each the i g and then the position are smaller is bigger plus bigger plus b x.
Okay, then we can factor this.
Write it as equal to sum over r in lattice times e to the i g dot r times sum over x and basis in basis f sub x in the i g dot x.
Okay, so this is what we just calculated above.
This is ask for the lattice and this here is s for the bases.
So it's really pretty much strictly true that the structure factor for any crystal is a structure factor for the lattice alone times,
the structure factor this factor here for the basis.
So if the structure factor for the lattice vanishes, in other words,
if you're not satisfying the selection rule for that, for the lattice type, then the whole thing is going to vanish.
So it doesn't matter how complicated your basis is.
If it's FCC, in order to get scattering, you have to satisfy the FCC scattering rules that h can all have to be all even or all odd.
Okay. Now there's a way to interpret this statement that that you have to satisfy the selection
rule for the lattice alone in order to get scattering from a complicated crystal.
So let's think about this picture here with sodium chloride for a second. Imagine for a second that we ignore the chlorine.
Just get rid of the chlorine. And then what we have left is sodium is on an FCC lattice.
In order to get X-ray scattering or neutron scattering from those sodium, you have to satisfy the selection rule.
Otherwise you get no scattering from it. Now let's get rid of the sodium and put the chlorine.
Then the chlorians are also forming a complete lattice and in order to get scattering from the chlorians, you have to satisfy the selection rule.
Now, if you put them back, both back in by superposition in order, it's just going to be the sum of the two.
Scattering is a scattering from the sodium and the salt and the scattering from the chlorians.
So if you're not satisfying the selection rule for one of them, you're not you're not going to get any scattering at all.
So it's basically just saying that you add up the scattering from each of the pieces separately.
Okay. All right. So in the next lecture, what we're going to do is we're going to yeah, we're going to actually look at some real X-ray data,
and we're going to analyse the X-ray data and deduce what is in some real materials.
But one of the things that we're expected to know that sort of has to be discussed is how you make X-rays in the first place, how to make make X-rays.
So there's two ways to make X-rays, which we may only be able to get to the simplest of them today.
But the simple way, the cheap way, simple and cheap is to use an X-ray tube.
X-ray tube. This is a method discovered by Röntgen way, back in around 1919, 1900, 1901.
The idea is you take a great big voltage here and you put it across cathode on this side,
just a piece of metal on this side and a big target on this side.
Target on this side. And by having this huge voltage, you spit electrons off the cathode and the electrons hit the target and off comes X-rays.
So what's really happening here is that the electrons are fast electrons.
Fast electrons kick out electrons from core orbitals kick out electrons from core.
Orbitals from core. And the x rays you see,
the x ray lines you actually see are mainly from other electrons falling down
from the high orbitals down to the look to the core orbitals to refill them.
So see emission from transitions, from electron transitions.
So an electron in a high up orbital drops down to fill one of these empty spaces in the in
the core orbitals and you will see a nice x ray line coming and nice x ray coming off.
Now the energy of that of that transition is basically about the rig burg times the atomic number squared.
Remember, when you have a hydrogen atom, the energy of the are the I can say to the hydrogen atom, I'll go as the nuclear charge squared.
So, you know, the red berg is 13 electron volts, but the atomic number of your atom,
if you're using something like copper, something like 40, so you can have a huge, huge energy.
This can be tens of kilowatts of kbps.
And that will correspond very nicely to what we want to correspond to what we want, which is lambda approximately one angstrom.
Now, there's a couple of really nice things about this technique.
One thing that's really nice about this technique is since it's an atomic transition, it's very, very sharp.
In other words, there is a very sharp line, a very, very distinct transition between two distinct energy levels.
You know, very clear energy level up here, a very clear energy level down there.
And so you get a very, very well-defined, well-defined, defined lambda out of this type of technique.
And this is basically what what Röntgen was doing, you know, 100, 115 years ago.
Now, in modern in the modern era, I may or may not have have a picture of this.
Let me see if I have it. Uh, yeah.
Here in the modern era, there's another way to make X-rays, which is sort of favoured as a better technique, is only these X-ray tubes.
They're they're incredibly cheap.
You can you know, this is what you have in the the dentist's office when they X-ray your mouth is one of these things or,
you know, X-ray broken bones. It's these things. And there's probably dozens them around this building.
You can build one yourself if you wanted to, but frequently you need a much more high tech X-ray source.
And the modern high tech X-ray source is known as the Synchrotron Prana.
And it's in this picture. It's one of these big buildings. This is just down the road.
This is the X-ray synchrotron source at Diamond Lab, just just up the road by about 10 minutes by car.
These things are about ten to the ten times brighter. Ten to the ten.
Yeah, 10 to 10 times brighter. More photons than X-ray tubes.
Than X-ray tubes. And that's actually quite useful to have if for a lot of modern experiments which
examine either extremely small samples or you need to collect data extremely quickly.
I mean, at the end of the day, you're always counting photons. You're sort of counting how many photons are scattering in a particular direction.
If you have more photons, you always get better data.
If you're if it's particularly if you're looking at a very small sample where it's hard to focus those electrons in on the small sample.
The general idea of the synchrotron is you take this great big ring,
which you can see in this picture, this synchrotron source, you spin electrons around this ring.
So the electrons are going around this ring at very high velocities a bit gives so giga electron volts and then in order to get X-rays out of it,
you take your electrons and you send them through a bunch of magnets.
So north, south, south, north, north, south, this is known as an undulating or a Wigler sometimes.
And the idea is that the electrons going through this,
they're going through very fast and you make them accelerate through and as you know, accelerate accelerating electrons emits radiation.
And in this case, they will emit directed X-ray radiation, which is extremely colonnaded, extremely well defined frequency.
And if you're not happy with the sort of a technique, this is sort of important to know.
If you're not happy with how well-defined the frequency is, if you want to specify a frequency more precisely, specify a wavelength more precisely.
What you do is you refract off a known crystal different tract off the known crystal,
known crystal at a particular angle, and that will pick out a particular wavelength that you're interested in using.
Now, whenever you're doing this, you're throwing away lots and lots and lots of the photons that you have.
But if you have ten to the ten times the number of photons that you need to begin with,
you can throw away a ton of them and you can refract multiple times to really specify
that wavelength extremely precisely and do extremely high precision measurements.
I think that's all I have to say about this. I'll see you tomorrow. Wow.
Wow. Okay.