460 Final, 2014
Note: while “Yes” can sometimes be a reasonable answer, you of course need to provide
more than just this.
Section 1: Basic Facts (Explain/Justify with 1-2 paragraph answers and/or a short
diagram, 4 pts each)
1. Explain briefly the principle of aligning the condenser aperture.
Note: the question says “principle”. Best is a diagram showing the shift of the
illumination when the condenser aperture is not appropriately centered.
2. Why are there no intensity oscillations near the middle of the thickness for an inclined
dislocation in a thick sample?
This is due to adsorption on one of the two Bloch waves – assuming 2-beam condition.
Some explanation of the difference and that the contrast is due to interference of the two
Bloch waves is needed.
3. Why is the column approximation better than expected on a zone axis?
This is due to channeling. Some explanation of how the electrons become confined to
atomic-like states and most of the scattering is associated with deeply bound 1s states.
4. Why is EDS not so good for light elements?
Two contributors: a) adsorption by the window etc of the detector, and b) more Auger
processes for light elements. N.B., the probability of exciting a core-electron is less for
higher energies (not more). It is also appropriate to mention resolution limits and
background which matter more for lower energies.
5. What is the angle between (009) and [370]?
(009) = 9C ; [370] = 3a + 7b where A,B,C are the reciprocal lattice vectors, and a,b,c are
the real-space lattice vectors. Since C.a=C.b=0 the dot product is zero.
It is ninety degrees. Remember that (009) is a direction in reciprocal space, [370] in real
space so the dot-product is simple and zero no matter what the axes are. It is WRONG to
call (009) a direction normal to a plane – that is a simple concept used in introductory
classes. Also, it is wrong to take the complimentary of the angle.
Section 2: Logic (10 pts)
6. Explain briefly whether the statement below is right or wrong, and why:
“Theory indicates that the contrast in the image of a uranium atom on a thin carbon film
is 10 times greater in dark field images than in bright field images. In practice, however,
a noisy background contributes considerably to the difficulties in visualizing such an
atom.”
From R. M. Henkelman and F. P. Ottensmeyer, Proc Nat. Acad. Sci. USA, Vol. 68, No.
12, pp. 3000-3004, December 1971 (note the year!)
The statement is right, admittedly only part of the paper, and some explanation is needed.
A thin carbon film is amorphous, and will give speckle in dark field due to local regions
that diffraction into where the aperture is. (The question does say dark-field, not ADF or
HAADF.) Due to the lack of order the scattering is approximately incoherent so scales as
the carbon thickness and density – and carbon is light. (Remember that scattering factors
scale roughly as Z.) Uranium is heavy, so a single uranium atom will scatter a lot. The
contrast will scale something like ZU /(n*ZCarbon) perhaps squared, with n the number of
carbon atoms along the beam direction. ZU=92, C=6, so this is relatively large if n <10
(for instance).
In bright field the carbon signal is one minus the total dark field, also for the uranium.
Since the scattering is not large it will be much harder to see (you can estimate this).
Note, to detect a single atom you do not have to “see” it as a ~0.2nm object, all you need
is a signal that differentiates it from whatever background there is. Remember that you do
not “see” the dislocations in BF/DF, rather you see their strain field. However, for the
strain to be present there has to be a dislocation.
N.B., detection of single atoms was done in the 70’s. Clever experiments are better than
expensive microscopes.
N.N.B., grading was mainly on my opinion of how much you used your TEM/TED
knowledge in an unfamiliar context, rather than looking for you to replicate my answer.
Indeed you could score high even if you disagreed with the statement, so long as you
gave good reasons.
Section 3: Diffraction (10 pts)
7. For a bcc sample with a lattice parameter of 3.5 Angstroms:
a) Sketch the diffraction pattern along the [111] zone axis, labelling the ZOLZ spots.
Important think to remember, bcc h+k+l is even only.
Two smallest vectors normal to [111] are (11̅0) and (101̅) which are at 60 degrees to
each other (dot product is 1, then divide by 2 for ½). Fortunately both are even so any
sum of them is even so you get a hexagonal mesh. (3)
b) What is the spacing of the smallest diffraction spot in nm-1?
Spacing is √2/0.35 = 4.04 nm-1 (2)
c) What is the excitation error for (3̅12) assuming the beam is exactly down the zone
axis? (5)
First, the distance to the spot is √9 + 1 + 4/0.35 nm-1 = 10.69 nm-1 = D
From a simple Ewald sphere construction,
𝑘 2 = 𝐷2 + (𝑘 − 𝑠𝑧 )2
You can do this accurately or in a small
angle approximation using k=400 nm-1
sz = 0.143 nm-1
Section 4: Research Strategy - 20 points
8. Some AlMg alloys used for the decks of ships are known to become more susceptible
to corrosion at grain boundaries with time. You have two 10x1x1 cm pieces of the same
alloy, one which has been on a ship for 1 year, another for 10. Provide a reasonable
strategy to work out what has changed using transmission electron microscopy.
Remember that time is money.
First you have to make the sample. Crushing won’t work as these are ductile metals.
Obviously cut a small piece out somehow, being careful not to ignite it! Since Al & Mg
are not that different there is a chance of electrochemical thinning working, otherwise
either conventional ion-beam or FIB, although at first you won’t know what to FIB cut so
not my favorite.
Note: there is a difference between looking for where the corrosion is and looking for any
global changes in the microstructure. For the former you would want to look near the
surface but you would also need to know more about how they have been treated – which
side (if any) was exposed?
Next, put in microscope and look. Precipitates? Featureless solution or what?
Based upon this you want to analyze more carefully the g.b.’s, and you need to think
about different strategies for different cases:
a) Precipitates at g.b. – composition/structure
b) Nothing at g.b. – local chemistry
c) Something different – so be careful!
For a) use EDX/EELS plus diffraction if precipitates are large. The phase is probably a
known one.
For b) use EDX/EELS across the boundary at relatively high resolution – increase the
resolution until you see something.
When you have some information you can start to consider what next. This becomes
tricky because you should only use a specific technique if it is going to provide you
something relevant. Just using a technique “because it is there” is not clever.
N.B., the enhanced corrosion is associated with precipitates at the g.b.s.
Section 5: Micrograph - 20 points
9. For the attached micrograph, explaining/justifying briefly your answers
1) BF, DF, Diffraction? (3)
2) What are the features marked A, B, C and D? (3 each)
3) Estimate the size of the scale bar in the bottom left. (5)
1) BF – you can see well many different crystal regions.
2) A – thickness fringes from the darker grain running across (~ 2:00) – it has a strong
facet.
B – A bit tricky, either a stacking fault, a twin or a low-angle boundary, all essentially
normal to the beam direction. It cannot be a large angle boundary as the bend contours
are quasi-continuous across it. But, since they are not completely continuous it is not
simply a stacking fault.
I would go with either a twin or a low-angle boundary.
C – Bend contour.
D – Inclined dislocation (classic snake)
3) The right answer is 0.5 microns = 500nm. You can get close to this by noting that the
snake oscillations and the thickness fringes in A have to be something like 20nm.