Hi and welcome to the second lecture of the
transmission electron microscope. As you can
see todays lecture will be conducted in the
lab here with the microscope and the topic
of day is electron diffraction, and I already
prepared to samples that I put into the sample
holder and that we are going to study. One
of the samples is MoO3, that is a orthorhombic
crystal structured material. What that means
is that it looks like a simple cubic, but
the lengths of the cube sides is different,
so it is not a straight cube. In order to
know the lattice spacing of the crystal that
were are studying we need to calibrate the
system and usually you do that by putting
in a known sample that you know the lattice
spacing of. Today that sample will be aluminum,
so I also have the second sample here that
is a sputtered aluminum sample. By checking
that we can calibrate the system and we can
also demonstrate the effects of poly crystalline
materials and how the electron diffraction
pattern looks in that case. That I can demonstrate
also with this. So let's start by putting
in the sample. As you can see know I put you
to view the camera image of the microscope,
it's easier because it's dark in the room
now. The thing that you see now that is the
TEM-grid with the sputtered aluminum on top
of it. The bright areas that is a thin layer
that has been coated, we can magnify this
and look at just one of these transparent
squares. This is how the sputtered aluminum
sample looks like in the bright field image
mode of the TEM. You can see that it looks
like the hole surface is covered by small
grains, that's correct. When you sputter aluminum
you get a surface that looks like this. What
we have here is a multiple amount of small
aluminum crystals that are oriented in completed
random directions. We are going to do electron
diffraction on this sample now, and I just
switch over and then I discuss the image.
So this is electron diffraction mode, here
we can see the electron diffraction of the
sputtered aluminum. We can see the pattern
is several concentric rings. These rings are
created because we have tremendous amounts
of crystals that we are making diffraction
patterns out of. If you stack all these patterns
on top of each other in random directions
they will create these rings. The idea is
to use this calibration sample to calibrate
the length distances in the image, this because
we know the lattice spacing of the aluminum.
By measuring the diameter of these circles
we can calculate a proportional factor that
we later can use when we measure our real
sample: MoO3. I said this was the diffraction
pattern of the hole region that is illuminated
by the electron beam. We can select a smaller
region instead, then we use the selected area
aperture, lets try that! I go back to the
bright field image now, this is the bright
field image again and now I put in the selected
area aperture like so. You see it is a small
disc that selects out a region of the sample
and this region we will take the electron
diffraction from. Here you see the electron
diffraction pattern again, because we selected
a smaller region less crystals will contribute
to the pattern and that means that the circles
will be less apparent in the pattern. We can
try an even smaller aperture to increase this
effect. Now you see I try the smallest aperture,
the diameter of this disc is much smaller
than the last one, that means we will get
very few crystals, you can almost count them
in the image. Lets see how the pattern looks
like from this one, here you see the circles
are almost barely visible now because so few
crystals contribute. We are starting to approach
more and more towards a single crystal electron
diffraction pattern. This demonstrates the
effect of poly crystallinity and single crystals.
If the samples instead would have been amorphous
then we wouldn't seen any dots att all, just
a blurred region in the center of the image,
that's how you can determine if it is an amorphous
sample or if it is a crystalline sample. Let's
switch to the MoO3 crystal and see how that
pattern looks like. That will be from a single
crystal and here we can see that it will have
this orthorhombic pattern, and it will basically
look cubic.
Now we are back from the lab and we are done
with all the measurements of the electron
diffraction. We have diffraction patterns
from this aluminum, the calibration sample
and the MoO3. You remember I said that we
are going to measure the distances, the dimeters
of the rings in this sputtered aluminum pattern.
We are going to measure the four inner rings
of this pattern, if you measure the horizontal
diameter, then you will get for the inner
ring 220px and for the vertical direction
you get 221px, that is sort of the same but
as you can see the second inner ring has 253px
vs 259px in horizontal vs vertical diameter
and this is caused due to astigmatism of the
system of course the circles should be complete
round. What we know of the aluminum sample
is that its lattice spacing is 4.049Å, this
we can take advantage of and use that to calibrate
our system. There is a simple rule that's
governs this we write this as a constant K
thats equal the S parameter times the d parameter.
The S parameter that is the length that you
measure on screen in pixels, of course there
is also a third factor involved here and that's
the wavelength of the beam
that you shine on the sample with, but because
we use the same accelerating voltage in both
the calibration sample and in the MoO3 sample
then we don't need to take that into account.
The d parameter that is the lattice spacing
that correspond to the measured pixel lengths.
This means if we know what lattice plane that
causes these rings then we can calculate the
d value because we know the a-vector
from the aluminum, thus it's a known sample.
We calibrate the system by calculating this
K constant, for example for the inner ring
of the aluminum we get that K equals, and
we take the average between the vertical and
horizontal diameters and that is 200.5px times
the d-spacing, and for this inner ring that
is the 1-1-1 plane and it will be 2.338Å,
if you calculate all this it will give you
a K factor of 515.5pxÅ. We now do the same
calculation for all the four inner rings of
this pattern and then we take the average
of those to be the final K constant. I get
an average 
of 517.2pxÅ. 
Lets look on the MoO3 pattern,
you know that it have this square shaped pattern.
Because it is orthorhombic that means its
cube lengths in this orthogonal crystal will
be of different lengths, we don't have one
"a"-vector, we will have "a,b, and c"-vectors
that are all different. For MoO3 the "a"-
and "b"-vectors is sort of the same, it is
the "c"-vector that's different with a greater
length, its value is around 13Å or something,
but "a" and "b" is around 3.7Å and 3.9Å.
We measure two orthogonal directions in our
diffraction pattern of MoO3 and I take these
ones just to simplify it. I measure those
to me 129px for the "a" length and measure
136px for the "b" length. Now we are going
to calculate the lattice spacings and we use
our constant from before that was K=517.2pxÅ.
Then we divide that one with the pixel distances
on the screen, we got 129px for a value and
this gives 4.0Å and this is for the 1-0-0
plane so that means this will be the lattice
parameters right of, thus otherwise we would
have had to multiply thus with the square
root of the miller indexes, but here it was
one so we multiplied by one. That means that
the a-vector is 4.0Å. We do the same thing
for the b-vector, that was 517.2pxÅ divided
by 136px which gives us 3.8Å. What we know
now is that we have this cube of the lattice,
we now know two dimensions of it "a", and
"b", the bottom of the cube that is. We don't
know the c-vector or the height of the cube unit
cell and that will remain unknown for us,
and in order to measure that one we need to
tilt the crystal.
What I want you to do know is a complex problem
that is sort of the end of this chapter of
the course and this one is a little of a fun
problem I think, it is a completely different
topic compared to electron diffraction but
you know that the in the TEM you have very
thin samples, down to 50nm in thickness, that
is sort of standard. Let's say that you have
a sample that is 50nm in thickness and then
you shoot electrons at it. Let's say it is
200keV of energy that goes into the sample,
you now that the electrons interact with each
other when they passes through the sample
and so on, because of allt these things that
you can read in the book :-). What I want you
to do now is to calculate how many electrons
that comes from the beam that exist at the
same time in the sample. I mean you can see
the beam as a discrete array of particles
that travels through the sample, so what I
want you to do now is to calculate how many
of these electrons are at the same time inside
the sample. You now the speed of the electrons
due to the kinetic energy that they have from
the accelerating voltage so then you can know
the speed of them. You know the beam current,
we can say the beam current is 5 micro ampere.
So then you know how many electrons that travels
also. So how many electrons will be in this
same volume of the sample and because it is
a very thin sample I don't think you can expect
them to be som many but I also think that
you will be surprised of what the results
will be after that you have done this calculation.
Do that complex problem and then you post
the results of your calculation as the procedure
I have given below. See you on the next chapter.
