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PROFESSOR: OK, OK,
Settle down.
Settle down.
It's 11:05.
It's time to learn.
All right.
Last day we were looking at the
emissions spectrum of the
target in the x-ray tube.
The target is the anode, and
it's being a bombarded by
electrons that have been
accelerated across a potential
difference of some tens of
thousands of volts, and this
was what we saw as the output,
intensity versus wavelength.
We see this whale shaped curve
which we have attributed to
bremsstrahlung, which is
the breaking radiation.
And then if the voltage
is high enough, high
enough to do what?
High enough to eject inner shell
electrons, we will get
the cascade, and the cascade
will give rise to specific
values of wavelength associated
with transitions
within the target atom.
There's too much talking.
I want it absolutely silent or
else somebody is going to
leave. That's the only
way it works here.
I talk.
You listen.
And if you have a problem with
that, there's a door here.
There's a door there.
And there's a door
at the back.
So let's get it straight
right now.
It's the only way
it works here.
Now what happens is that these
characteristic lines are
calculable in part.
The K alpha line, the L alpha
line we saw from Moseley's
law, which is up here
on the slide.
and the leading edge of the
whale shape curve, the
bremsstrahlung is calculable
by the Duane-Hunt Law.
The rest of this is
not calculable.
So there it is.
What I want to do today is
harness this radiation.
The reason we were studying
x-rays in first place was that
we said they had a length scale
that was comparable to
that of atomic dimensions.
And so now we want to close
the circle here and probe
atomic arrangement by
x-ray diffraction.
And it goes by this
three-letter
initialization XRD.
And in order to use x-ray
diffraction to probe atomic
structure, we're going to
make another model.
And this model is going
to be simple.
And it's going to be inaccurate,
but it's going to
be good enough to explain
the phenomena that
we're going to study.
So the first thing I want to
do in constructing a model
that will allow to use x-ray
diffraction, is to model the
atoms as mirrors.
That's the first assumption
that we make.
We model the atoms as mirrors.
And when we model the atoms as
mirrors, this means that we
will invoke the laws of
specular reflection.
So if this table top is a plane
of atoms, and I have a
beam coming down at an angle,
the laws of specular
reflection say that the angle
of incidence will equal the
angle of reflection.
So that's all we're doing by
modeling the atoms as mirrors.
Laws of speculative
reflection apply.
So that simply means that the
angle of incidence, theta
incidence equals theta
reflection.
And you'll see how that comes
into play in a minute.
And then the second thing we're
going to do in order to
get intensities is to use the
concept of constructive and
destructive interference.
OK.
So we apply interference
criteria.
And I will explain what
that means in a
second with a diagram.
But in dense text, it simply
means that in phase rays--
it sounds like this rain in
Spain falls mainly in the
plain-- in-phase rays amplify.
And out-of-phase rays dampen.
So at some point, you'll study
this in physics, but we need
it right now, so I'll give you a
little bit of foreshadowing,
and then you'll be ahead
of the game when you
meet this in physics.
So what I'm going to do is I'm
going to show you two rays,
and they both have the
same wavelength.
And I want to show
you in phase.
And so I want to illustrate this
concept here, in-phase
rays amplify.
So let's give one ray here.
I'm going to depict it as
a wave. All right, so
that's ray number 1.
And beneath it, is
ray number 2.
It's got the same wavelength.
So that means crest-to-crest
distance is the same, and if
it's in phase, in phase means
that crest lines up with
crest, trough lines
up with trough.
So I'm going to make another
one seemingly identical.
It's supposed to be identical
to the capacity of
my ability to draw.
And so what this means is
that these will amplify.
So 1 plus 2, 1 plus 2 will
give me something
that looks like this.
I'm going to line it up again.
Only now the amplitude is double
the amplitude of a
single ray.
That's what I'm trying
to depict here.
OK.
So this is amplification.
Now over here, I want to
do the same thing.
Only I'm going to have
two rays, same
wavelength, but out of phase.
OK, so in phase is here, out
of phase is to the right.
So we'll have ray number
3, which does this.
And ray number 4, I'm going to
line up with the crest of ray
number 3, the trough of ray
number 4, but it's going to be
the same wavelength.
So the distance between
successive crests is the same.
Only they're going to be
offset by exactly half
wavelength.
So the crest here of wave
3 lines up with the
trough of wave 4.
The trough of wave 3 lines up
with the crest of wave 4, and
so on, and so on, and so on.
And if I drew this thing
accurately, it would be
obvious to you.
But since I can't draw very
well, then we're going
to take 3 plus 4.
And the sum of the crest
and the trough is 0.
The sum of the trough
and the crest is 0.
And so the sum here is 0.
All right this is total
destruction.
This is cancellation.
So in the extreme, you can have
destructive interference
that dampens to the point
of obliteration.
OK, so that's the two extremes,
full amplification
and complete cancellation.
So now, I want to take this
concept of planes as mirrors
and rules of interference
criteria, and now what I want
to do is illuminate.
And I want to use the
full board here.
So here's what we're
going to do.
We're going to take in
phase, coherent--
that's what coherent means.
Coherent means that the
radiation is in phase and
monochromatic.
What's that mean?
Monochromatic means one color,
which means single wavelength.
So I have something of a single
wavelength in phase.
Coherent, monochromatic,
incident radiation, so I've
got a plurality of rays.
I have this thing flooded
by a beam.
So what I'm going to
do, is I'm going to
draw some atoms here.
And I'll put a second
batch on top.
And now I'm going to model
these as mirrors.
So I'm going to have the
incident beam come in like so.
The incident beam comes in like
so at an angle theta.
So this is theta incident.
This is the angle
of incidence.
And according to our model, the
beam is reflected at the
same angle.
So this is theta, reflection
of theta.
Reflection equals
theta incidence.
So that's the first
thing that we do.
And then the second thing we
do, is we use the laws of
interference.
So I'm going to take a second
beam, and it's going to go
down to this atom here.
So I'm going to bring
it down like so.
And it's going to come
in at the same angle.
And it's going to leave
at the same angle.
And if you'll excuse
the drawing, these
lines should be parallel.
They should be coming in
parallel, and they should be
leaving parallel where these
angles theta i and
theta r are the same.
Now how do we invoke the
interference criteria?
Well what I'm going to do
is put a marker here.
I'll put a marker.
And I'll call the first one ray
number 1, and the second
one ray number 2.
And what I'm going to show is if
I start from ray number 1--
maybe it's time for
colored chalk--
so if I start from ray number 1,
to show that it's in phase
with ray number 2, I'm going to
draw this little waveform,
and they're both lined up.
So you can see crest lines up
with crest. Trough lines up
with trough, and they both
have the same wavelength.
So that makes the point
coherent, monochromatic.
But now things get interesting,
because you can
see that ray number 1 travels
a shorter distance than ray
number 2, and we can put on some
coordinates here and mark
the geometry, so we can say that
the point at which the
ray number 2 has to start
traveling a longer distance,
I'll drop a normal down,
and I'm going to
call this point A.
This is point A.
The bottom here is point B.
And then up here where it
catches up with ray number 1,
the reflected ray number
1, I'm going to
call this point C.
So you can you can see that
ray number 2 has to travel
this extra distance,
AB plus BC.
And now, if I want to get out
to here where I'm now in the
reflected zone, and I want ray
number 1 reflected and ray
number 2 reflected to continue
to be in phase, there's a
geometric constraint on
the dimension of AB
plus BC, isn't there?
That length, AB plus BC must
be a whole number of
wavelengths.
It must be an integer number
of wavelengths of whatever
this thing is, otherwise,
these two will
not being in phase.
So that's the interference
criterion.
And n can equal 1, 2, 3, and so
on, so this is called the
order of reflection.
And then we can go through
the trigonometry.
I'm not going to
do it in class.
It's just a waste of time.
It'll bore everybody to tears.
But if you want to do that at
home at some point when you've
got nothing to do, you
can go through.
You've got all the
numbers here.
You know the value of lambda.
You know the distance AB,
and you know what
this distance is.
This distance between successive
planes is your
dhkl, isn't it?
So if you go through all of this
analysis, you will show
that n lambda is equal to 2
times the d of the hkl spacing
times the sign of theta.
And this is called
Bragg's Law.
And Bragg's Law governs the
reflection of incident
radiation by a crystal.
Now you'll notice I didn't
put an apostrophe here.
Some people put the apostrophe
here, because the
man's name was Bragg.
Some people put the apostrophe
here because, in fact, it was
father and son.
And they both together won
the Nobel Prize in 1915.
Now there have been people who
lived to see, as Nobel Prize
winners, one of their children
win the Nobel Prize. but this
is the only time in history a
father and son team together
won the Nobel Prize
for the same work.
And there are undoubtedly people
sitting in this room
who think the fact that the
father and son could work
together in physics for an
extended period of time alone
is deserving of the
Nobel Prize.
Now in 3.091, I'm going to keep
it simple, always choose
first order reflection, always
n equals 1 in Bragg's Law.
So therefore, we will write
Bragg's Law as lambda equals
2d sin theta.
And to make the point, the d
is specific to a particular
set of planes.
So it's a d spacing
of the hkl planes.
And it's the theta associated
with the correct reflection of
the hkl planes.
Now how does destructive
interference come into play?
Destructive interference comes
into play should there be a
situation where I have
a third ray.
I'm going to bring a third
ray in also at the
same angle of incidence.
And that ray is halfway between
these two rays.
And what situation
could that be?
That could be a situation in
which the crystal structure
has in a plane out from the
board, an atom that sits
halfway between these
other two atoms.
And if that happens,
you can see.
We can go through
the derivation.
But there's going to be a path
length difference that's
exactly 1/2 wavelength
different.
That's bad.
It's not just there's going to
be some measure of reduction.
It's going to be total
cancellation.
And so by going through the set
of crystal structures and
recognizing that when you have
1/2 wavelength difference, you
get destructive interference.
And therefore, you
will see no line.
The incident radiation will come
in at this angle, and at
the reflection angle for that
particular plane, there will
be nothing detected.
So you can say that you have a
combination of selection rules
that involve the integral of
both the rules of interference
and interaction with the
crystal structure.
So let's generalize this and
say if we take interference
criteria plus the crystal
structure, that is to say, the
instant relationship of atom
positions for a particular
specimen, the combination of
that will give rise to the set
of expected reflections.
So only when you satisfy the
Bragg criterion do you get
reflection.
So you move the specimen.
And only at that special angle
will you get the reflections,
will you get constructive
interference.
So this, in fact, is
a fingerprint.
This is a fingerprint
of the crystal.
In this case, we're not getting
the chemical identity.
We're getting the structural
identity.
So we can determine
if something is
BCC, FCC, and so on.
So for example in BCC, that's
exactly the case that I just
illustrated.
So in BCC we have atoms
at the eight corners.
And we have an atom
dead center.
There's a central atom.
And the central atom lies in
0 0 2 plane, doesn't it?
The base of this is 0 0 1.
And the top is an 0 0 1.
But the central atom
is an 0 0 2.
And 0 0 2 is halfway between
successive 0 0 1.
So can you see that light
that comes in?
And it's going to be
constructively reinforced.
Off of 0 0 1 is going to be
destructively reinforced off
of 0 0 2, and the result is
that in BCC, no 0 0 1
reflection observed.
The same thing happens in face
centered cubic, right?
What's the face of face centered
cubic look like?
It looks like this,
doesn't it?
So there's an 0 0 1.
Here's an 0 0 1.
What about this space
centered atom?
Where is it?
That's an 0 0 2.
It's halfway between 0 0 1.
So at the angle where you
would've expected, if you use
the Bragg Law and calculate the
angle at which for your
particular crystal, because
you know the d spacing.
You fix the wavelength
of your x-radiation
coming out of the generator.
At the angle where you would
expect to see reflection off
of 0 0 1, you will see nothing,
because 0 0 2 is
canceling 0 0 1.
So you don't have to worry
about all this.
This has all been tabulated
for you.
Somebody has gone through
and done all of this.
Well, this is just
making the point.
See, there's a simple cubic.
All the planes are reflecting.
There's body-centered cubic.
That's a/2.
There's face-centered
cubic a/2.
0 0 2 is going to
cancel 0 0 1.
You're not going to see
anything there.
So people have gone through.
And they've made this set
of rules for reflection.
So simple cubic, you
get reflection
from all the planes.
Body-centered cubic, you just
get these planes here.
And it turns out that there's
a simple rule that compactly
represents which planes are
going to reflect in
body-centered cubic.
And in BCC, the hkl, it's the
planes for which h plus k plus
l is an even number.
h plus k plus l is even.
And 0 was counted as even.
So for example, 0 0 1 h
plus k plus l is 1.
It doesn't reflect.
0 0 2, 0 plus 0 plus 2 is 2.
It does reflect.
And so on.
You just go down
the whole line.
And so these are in ascending
order of h squared plus k
squared plus l squared, because
that's a nice way of
deciding how to add them up.
And now in FCC the
selection rule is
a little bit different.
In FCC you get reflection from
planes when you write the HKL
such that h, k, and l must be
either all even numbers, or
all odd numbers, or some people
like to say unmixed.
Some crystallographers say
unmixed, meaning you can't
have a combination of even
numbers and odd number.
So for example, 0 0 1 won't
work because 0 is a zero.
That's even.
1 is odd.
But 0 0 2, h plus k plus l all
even or all odd, there you go
all even or all odd unmixed.
So this will work.
And then so on.
So you can go through and
see which ones work.
All right.
So the next one here
is in the sequence.
This sums to 1.
And what would sum to 2?
It would be 0 1 1.
And 0 1 1 doesn't work either
because zero is even.
And 1 is odd.
The next one of the
sequence is 1 1 1.
These are all odd.
So that one reflects.
So in FCC, you don't
see 0 0 1, 0 1 1.
But 1 1 1 does reflect.
And then 0 0 2, which is the
next one, because 0 squared
plus 0 squared plus
2 squared is 4.
This is even even.
So that one works.
So there's the sequence.
And so now what we can do is use
this technique in order to
make measurements.
But before we do so, I want to
show you the experimental
measurement, one way.
There's several ways of
conducting the measurement.
And so the first way I'm going
to show you is called
diffractometry.
And you can do this over
in Building 13.
If you get yourself a UROP, you
might be assigned to make
some measurements.
So diffractometry is a form
of x-ray diffraction.
It's one of the techniques.
And the way it works is you fix
the wavelength and vary
the angle of incidence,
you don't
rotate the x-ray generator.
You rotate specimen and present
continually varying
angles to the thing.
So this shows the technique
in operation.
So you have a specimen
sitting here.
The specimen can either be
a thin film, or in other
instances, we have a finely
ground powder.
So each of the powder grains
presents a different angle.
And coming out of the
collimator is the
monochromatic.
We want a single wavelength.
So this collimator means it's
monochromatic and coherent.
So that beam comes and
strikes the specimen.
And for historical reasons,
instead of calling this the
angle of incidence, and this the
angle of reflection, they
call the angle that goes to
the detector 2 theta.
In other words, the projection
of the beam, this is theta
incident equals theta r.
So this is really 2
theta incident, or
it's 2 theta reflected.
It's the same thing.
So you'll see x-ray data
usually reported
in units of 2 theta.
And here's the detector.
And then all you do is you
rotate the specimen.
And by rotating the specimen
with a detector, you're able
to get the entire set
of reflections.
And whenever you move through
an angle that satisfies the
Bragg Law, you get a
peak in intensity.
And here's what the output
would look like.
So you're plotting intensity
as a function of 2 theta.
So somewhere along here at
around, it looks like about 38
degrees, you have satisfied the
Bragg angle, and you get
constructive interference.
When you move off of 38
degrees, destructive
interference reigns supreme,
and you get almost no
reflection.
Then you get to the right
angle for 2 0 0, you get
reflection.
Well, all you get
is these lines.
You don't get these numbers.
These numbers you don't
get for free.
So here's the experiment
that we're going to do.
We're going to run our
x-ray generator
with a copper target.
Why do I tell you the
copper target?
Because you're going to
fix the wavelength.
And you fix the wavelength
by choosing the target.
So we've got copper target in
our x-ray generator, OK?
So remember, the sample
is not the target.
The sample is what is
being irradiated.
This is being bombed by the
electrons in the tube, copper
target in x-ray tube, and that
means that the lambda of the
radiation is lambda copper.
And I'm going to use lambda
copper K alpha because I know
it's wavelength to five
significant figures.
It's 1.5418 angstroms.
And here's the data set.
I looked this up in
the literature.
Here's the data set from
the experiment.
And that's all you get,
a set of 2 thetas.
So here's my challenge to you.
I'm telling you you've got an
unknown sample that's cubic,
and there's your data set.
The task for you is determine
two parts to the question.
And we're going to
do it together.
First part, determine the
crystal structure.
So it's either FCC, BCC,
or simple cubic.
And the second part is
determine the lattice
constant, the value of
the lattice constant.
So you can get quantitative
measurements.
So I'm going to show
you how to do it.
So how are we going to do it?
We're going to use
this technique.
This is my self-help
book for you.
And here's the key.
Here's how I came
up with this.
There's a way to unravel this.
And the way you unravel this
is to take these two
relationships.
You have lambda equals
2d sin theta is 1.
And you also know that d--
in fact, I'm going to keep
writing hkl subscripts here--
you know the dhkl
is equal to a.
That's this lattice constant.
Be sure this is lattice
constant.
This is the cube edge
measurement lattice constant
divided by the square root
of h squared plus k
squared plus l squared.
So what I do is I
combine the two.
If you combine these two,
you can end up with this
relationship.
If you combine the two, you
get lambda squared over 4a
squared equals sin squared theta
over h squared plus k
squared plus l squared.
And this is the key.
Why is it the key?
The value of lambda
is set by you.
You chose the target.
The value of a is set
by the sample.
That's the lattice constant.
So the ratio of two constants
is a constant.
Agreed?
So this is a constant.
So if the left side of the
equation is a constant, the
right side of the equation
must be a constant.
But h, k, and l vary.
And theta varies.
But the ratio of the variation,
when mapped into
sin square theta over h squared
plus k squared plus l
squared, this must
be a constant.
And that's going to be the
way I work through
this delightful problem.
So let's see.
Start with 2 theta values, and
generate a set of sin squared
theta values.
That's what Sadoway says first.
So I took 2 theta.
And all I did was make 1 theta,
and then took the sin
of it, and then took
the square the sin.
So this is the data set
transmogrified into sin
squared theta.
All right, what's the
next thing he says?
Normalize by dividing through
with the first value.
So I'm just going to take
this whole series and
divide it by 0.143.
And now I end up, instead
of 0.143, 0.191.
I have 1, 1.34, and so on.
That's what I mean
by normalize.
Normalize to the first entry.
OK, What's next thing he says?
Clear fractions.
Clear fractions from the
normalized column.
So I'm going to multiply this
by a common number.
Because 1.34, remember this
is experimental data.
It's noisy.
1.34 looks like 4/3.
Doesn't it?
It's fuzzy logic.
And 2.67 looks like 2
and 2/3, 3 and 2/3.
That's roughly 4.
5 and 1/3.
The 3 looks like
a magic number.
So if a multiply this column by
3, I get 3, 4, 8, 11, 12.
What does it say next?
Speculate on the h, k, l values,
that if expressed as h
squared plus k squared plus l
squared would generate the
sequence of the clear
fractions column.
And then that's going to take
me to the selection rule.
So I say, well how do I get 3?
It's 1 plus 1 plus 1.
4 is 2 squared plus 0 squared
plus zero squared.
8 is 2 squared plus 2 squared
plus 0 squared.
So I'm generating this thing.
And now it's pretty
obvious, right?
Because now I've got
to use these.
And what do I see here?
Well, 1 1 1 are all odd.
2 0 0, all even.
2 2 0, all even.
Gee, this looks like it conforms
to the selection
rules for Bragg reflection
from an FCC crystal
And then to make sure that
I'm on to something,
what I do is this.
Compute for each theta the value
of sin square theta over
h squared plus k squared plus
l squared on the basis of
those assumed values.
What I'm doing is I'm saying if
my hunch is right, whatever
I choose for the theta and the
assumed value h squared plus k
squared plus l squared, that
should be a constant that
doesn't change.
So let's test it.
And sure enough, look.
0.0477, 0.0478, so it looks
like I'm on to something.
And furthermore, I know
what this is.
That 0.0477 is this.
This is 0.0477.
And I know my lambda.
That's 1.5418.
So now I can calculate my a.
So now I've determined
that it's FCC.
Plus if I go ahead and I
calculate the a value, I get
3.53 angstroms. And if it's FCC
and it's 3.53 angstroms, I
bet it's nickel.
So that's how you index
this stuff.
So there we are.
There are the selection rules.
Now here's a little trick.
Let me show you.
Because when I go to index this,
the first thing I do is
I go for low hanging fruit.
If you start looking for whether
it's simple cubic,
face-centered cubic, or
body-centered cubic, let's
look at the sum.
So I've got simple cubic,
body-centered cubic, and
face-centered cubic.
So the first plane on simple
cubic, it's going to give me
everything.
It's going to give me
1, 2, 3, 4, 5, 6.
And FCC, what does it give us?
If you start looking down
there, it gives you
3, 4, 8, 11, 12.
Well, this is so different.
3, 4, 8, 11 is so different
from 1, 2, 3, 4.
What does BCC give you?
It gives you 2, 4,
6, 8, 10, 12.
Can you see you have
a problem here?
It's pretty easy to
pull out FCC.
But look at these two.
Since you don't know, you're
just normalizing.
You can't tell.
If I give you a number sequence
that is in the order
2, 4, 6, 8, 10, how do you know
that that couldn't be 1,
2, 3, 4, 5?
There's a hook here, though.
Look at the sequence of h
plus h squared plus k
squared plus l squared.
Do you notice that there's no
combination of h squared plus
k squared plus l squared
that gives you 7?
You get 6.
And then the next one is 8.
But there is a sequence
that gives you 14.
So if you have a seventh line,
if the seventh line to the
first line, the ratio of h
squared plus k squared plus l
squared for angle number
7 to h squared plus k
squared plus l squared.
For angle number 1, if
it's equal to 7--
and point of fact it's
not 7:1, it's 14:2--
and if it's equal to 8 for
the seventh angle--
then it must be simple cubic.
And now you've sorted
it all out.
So you're an expert now.
Now you can do it.
OK, so you're going to get some
practice on homework.
So you can index crystals.
You can determine a
crystal structure.
All right, a couple of other
things we can do.
There is a second technique.
And it's called Laue diffraction
after von Laue,
who got the Nobel
Prize in 1914.
He beat the Braggs by one
year for this technique.
So he shines.
In this case for Laue, it's a
slightly different technique.
What Laue does, is he
uses light x-rays.
That means variable lambda.
So how would you get a fixed
value of lambda?
Well, you would pick off one of
those lines, like maybe the
K alpha line because it's
a nice, singular line.
And repress the bremsstrahlung
and so on.
But if I want a variable lambda,
where do I go for
variable lambda?
I repress the lines.
And I go for the
bremsstrahlung.
So now I've got lines varying
all across the x
region of the spectrum.
So I use variable lambda.
And I fixed the angle of
incidence, whereas with
diffractometry, I use a single
value of lambda and vary.
So here's the cartoon which
shows what's going on.
I'll do it one more time.
So this is called
camera obscura,
which is darkened room.
That's all this means
is dark room.
So let's say I've got the
specimen on the back wall of
this thing, and what I've got
is white x-rays coming in.
So the white x-ray enters
through the face.
I've got the plate here, and
I've got the specimen sitting
somewhere in between.
And what happens is I
get a spot pattern.
I get a spot pattern.
And the spot pattern is
imitative of the symmetry of
atomic arrangement.
So I have to say a little bit
about symmetry so we know what
that means.
So let's take a look
at symmetry.
So I'm going to talk a little
bit about rotational
symmetries.
So let's look at that.
So first one I'm going to look
at is an 0 0 1 plane.
So if I just take a projection
of an 0 0 1 plane,
it looks like this.
Doesn't it?
It's just the cube edge,
or the cube bottom.
So it's got a and
a as the edge.
And now, the rotational
axis, it's about a
normal rotational axis.
A normal rotational axis.
So what I'm going to ask is how
many degrees do I have to
go through before I get
this same image back.
You can see that you go 90
degrees, it'll come back.
If I stop at 45 degrees, it's
going to look like this.
You're going to say I know
he moved the specimen.
If I go 90 degrees, you
can't tell it apart.
So we define fold.
The fold is equal to
360 divided by the
basic angle of rotation.
So we would say that this plane
here exhibits 4-fold
symmetry, OK?
Let's do a second one.
The second one is 0 1 1.
So if I look at 0 1 1,
0 1 1 at this plane.
0 1 1 goes across
the diagonal.
So I'm going to take the
diagonal here and
plot it like so.
So it's going to have
an edge of a.
And it's going to have the
diagonal, which is
root 2 times a.
And if I put a rotational axis
in the center of that, clearly
if I go only 90 degrees, I'm
going to tell that the thing
is rotated.
I have to go 180 degrees before
I can't tell that
there's been any disturbance.
So this one here has
2-fold symmetry.
We'll do one more.
Because there's only three
major ones that we
have to deal with.
And that is going to
be the 1 1 1 plane.
And the 1 1 1 plane.
And the 1 1 1 plane, I think
I've got an image of it here.
You can see the 1 1 1 plane
drawn in this cube.
So what's the face of
1 1 1 look like?
The face of 1 1 1
looks like this.
That's face of 1 1 1.
And if you go through this
analysis, this is root 3a
because it's a diagonal of
a difference persuasion.
So this is 3-fold symmetry.
This has 3-fold rotational
symmetry.
And that means if I take
a crystal such
as this, for example.
I've told you before, this is
my silicon wafer, and this
silicon wafer has been cut
from a single crystal.
So I'm looking at the edge
of an atomic plane.
The question is, which
one is it?
The crystal didn't come with
a label on the side.
I don't know what the crystal
growth axis was.
I started with a single crystal,
and I dipped it into
molten silicon.
And just like rock mountain
candy, I drop the heating
coils and cause the
liquid to solidify
around the seed crystal.
And I grow this salami that's
about 8 inches in diameter,
about 2 meters long.
So now I go and I cut these
things with the diamond wheel.
So I'm cutting them normal to
the growth axis, but I still
don't know what plane
I'm looking at.
So if I use the Laue technique
and put the crystal like so,
bombard it with white
x-rays, and I
look at the spot pattern.
The spot pattern is going
to give me one of those
symmetries.
And if I get a 3-fold
symmetry, I know I'm
looking at 1 1 1.
I mean, it's not going to
be some wacko plane.
It's going to be either face,
edge, or diagonal.
And on the basis of the symmetry
in the Laue pattern,
I can tell which plane
I'm looking at.
They use this stuff.
They use it to make the devices
that are in your cell
phone and your computer.
I'm not just telling
you a story.
All right.
So let's look at some others.
We'll go back to the
Escher prints.
Everything has symmetry.
OK, so what's this one?
See this?
What's the symmetry?
4-fold.
How about this one?
What do you think?
3-fold.
I even went into Photoshop
just to show you
how dedicated I am.
So I took this image.
And I told Photoshop to give
me 120 degrees rotation.
And I got this.
You can even see the fold
in the book there.
So I'm really doing it.
It's coming back.
It's 120 degrees rotation.
How about this one?
That's 3.
Now these are real
Laue patterns.
This is 4-fold.
This is obviously 3-fold.
This one is hard to see.
But it turns out it's 2-fold.
The one axis is just
a little bit longer
than the other axis.
What about this one?
That's our simple cubic puppy.
Yes, 1-fold.
Translational symmetry without
rotational symmetry.
You can make this by just
moving them sideways.
But you have to go all
the way around.
So that's 1-fold symmetry
translation.
Look at this one?
What's going on here?
Is that one of the
Bravais lattices?
This is called Penrose tile,
again rotational symmetry
without translational
symmetry.
You can take a patch of this
and move it rotationally.
But you can't take and
cover a wall with it.
So that's rotational without
translational.
See?
You can go around this way.
1, 2, 3, 4, 5, 6, 7, 8, 9,
10, and reproduce this.
But you can't use this as a unit
cell and keep it hopping
laterally and tile
a wall with it.
Here's another one, not
a Bravais lattice.
All right.
So I told you at the beginning
of this unit that there are
ordered solids and disordered
solids.
We've been focusing
on ordered solids.
They have a unit cell.
They're periodic.
And we call them crystals.
And next week we're going
to start looking
at disordered solid.
So they have no building block,
no long range order.
And we call them glasses.
For a long time, people thought
that solis have to
fall into one box or
the other box.
And then the best thing
that can happen in
science is not, oh, yeah.
That's exactly what
I expected.
It's, I wonder what
that means.
And there was one of these
moments in 1982.
In 1982, a man by the name of
Danny Shechtman from Technion
in Israel, was over here in the
United States working at
the National Institute of
Standards and Technology,
which used to be National Bureau
of Standards down in
Gaithersburg, Maryland.
And he was looking at a set of
aluminum-manganese alloys.
So aluminum is a metal.
Manganese is a metal.
You can make a solid solution
which we call an alloy.
And these are highly ordered.
And what he found in his Laue
measurements, were rotational
symmetries that are impossible
in a crystal.
He was getting 5-fold
symmetry.
You can't get 5-fold symmetry.
There is no set of planes
here that will give
you a 5-fold symmetry.
They lack translational
symmetry.
They were called aperiodic.
And the popular name for
them was quasicrystals.
So here's the Laue pattern of
one of Danny Shechtman's
aluminum-manganese specimens.
I think this is 25%
manganese and
aluminum if I'm not mistaken.
So let's count, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10.
You can't have 10 or 5.
Atoms don't work that way.
But he got it.
He got it, 5- fold,
72 degrees.
Now this is really,
really exciting.
This is a shot of
his lab book.
You're looking over
his shoulder.
These are the specimen
numbers.
And this is aluminum 25%
manganese, April 8, 1982.
So it's just another
day at work.
You could be doing this.
And he can't believe
what he's getting.
He's wondering what's
going on.
You can't contribute this to
a calibration error on the
instrument.
This thing is on unassailably
5-fold symmetry.
The joke is that this is not
far from the Pentagon.
Only within shouting distance
of the Pentagon would you
discover 5-fold symmetry.
So there it is.
So where have we come
with all of this?
We've come with the ability to
start with some very simple
ideas about x-ray generation.
And we've come to the point
where we can characterize
crystals, get quantitative
measurements of their lattice
constants, and by using Laue
techniques, investigate symmetry.
OK.
Let's adjourn.
We'll see you on Friday.
