IA
Natural Sciences Tripos Part IA
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
PRACTICAL BOOKLET
Name............................. College..........................
Michaelmas Term 2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
2014-15
Course A Practical Booklet
Before attending each session, you will be required to do a short pre-lab exercise
which will involve answering a series of questions online based on reading through
the relevant script in advance of attending the class. Apart from the introductory
practical, AP0, these exercises will be assessed and count for a small amount towards
your IA Materials final grade. If you do not complete the exercise in advance of the
class and attend the class in person, you will not receive any credit.
This booklet contains scripts for all practicals for Course A (Atomic Structure of
Materials) of the Part IA Materials Science course – it is important that you bring this
booklet with you to every session.
Practical Page
Title
AP0
2
Amorphous Materials: Elastomers – Glasses or Rubbers?
AP1
7
Crystalline Materials & Close-Packing of Spheres
AP2
15
Lattice Planes, Miller Indices and X-ray Diffractometry
AP3
28
Phase Identification and Quantification for Cubic Crystal
AP4
37
Introduction to Optical Diffraction
Practicals are held in the Teaching Labs at the Department of Materials Science &
Metallurgy on the West Cambridge site. These are most easily accessed from the
Coton cycle path, entering the building from the south entrance, and turning right
when in the reception area. You will need your University Card to access the labs –
please bring this with you to every session.
http://map.cam.ac.uk/Department+of+Materials+Science+and+Metallurgy
Please make every effort to attend your allotted session. If, in extenuating
circumstances, this is not possible you should attend an alternative session of the same
practical and report to the Head of Class of the session that you attend. Practicals
begin at 2pm sharp and end at 5pm.
The first practical, AP0, will commence from Thursday 9th October 2014, and will
include an induction tour and safety briefing, followed by the practical. Note that the
online exercise for AP0 is not assessed, but should still be completed in advance of
practical to enable you to benefit fully from the session. You will be notified by email about when and how to complete the online exercise.
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AmorphousMaterials:Elastomers–GlassesorRubbers?
Aims and Objectives of this Practical



To introduce you to amorphous materials by using properties of elastomers as
an example. In the next practical, you’ll look at crystalline materials;
Understand how the elastic properties of elastomers change with temperature;
Carry out an experiment to study the glass transition and determine Tg.
Materials science is fundamentally concerned with understanding how microscopic
structure relates to macroscopic behaviour in a material. In this first practical, you will
investigate the change in the properties of a squash ball with temperature and the
microstructural explanation for this behavioural change.
Squash balls are made from mainly natural rubber (polyisoprene) and can display
either rubbery (bouncy) or glassy (hard/brittle) behaviour in response to an impact,
depending on the temperature and chemical structure. You may have been used to
thinking about the terms ‘glass’ and ‘rubber’ as referring to particular types of
material, but actually they relate to the microscopic structure of the system and its
physical state. In particular, at a certain temperature known as the glass transition
temperature, Tg – see Background section below, an abrupt change in the behaviour of
the squash ball is observed. In this practical, you will study this by measuring the
bounce height of a squash ball as it warms through the glass transition point from the
temperature of liquid nitrogen.
IMPORTANT HEALTH AND SAFETY INFORMATION
During this practical, you will be handling objects that have been cooled in liquid
nitrogen to 77 K (–196 ºC). Although the low thermal conductivity of polymer
samples will protect you to a certain extent from chill burns, you must observe all
safety precautions involving the use of liquid nitrogen, and be careful not to touch
objects below ice freezing temperature (0 ºC) with exposed skin. Use gloves and
goggles provided when handling liquid nitrogen. Report spillages to the Head of
Class.
1
Background - measuring bounce height of squash ball warming through Tg
Each group is provided with a squash ball. The main component of these is
polyisoprene (–[CH2-CH=C(CH3)]n–), originally extracted from the Para rubber tree,
Hevea brasiliensis, although this is heavily treated with a range of additives to give it
the right strength, consistency, colour and elasticity.
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Most importantly, the rubber is partially ‘cross-linked’ by addition of sulphur in a
process known as vulcanisation. Strong covalent linkages are formed between some
isoprene units on adjacent chains, shown schematically below.
Not all the isoprene monomer units are cross-linked, as then the ball would be very
brittle, like a thermosetting polymer (e.g. Bakelite). The degree of cross-linking is a
very important factor in determining elastic properties of the squash ball and also its
glass transition temperature, Tg (see below). A small amount of cross-linking will be
sufficient to prevent viscous flow of the polyisoprene, but further cross-linking will
limit its ability to deform and recover through bond rotations.
Most polymers are can exist in either a glassy or rubbery state, depending on the
chemical structure and temperature. In a cross-linked rubber, viscous flow is not
possible, but shape-change via local bond rotations gives rise to rubbery behaviour.
The time necessary for the polymer to change in shape is called the relaxation time,
and is temperature-dependent. For high temperatures, the relaxation time is very short,
and so shape changes can readily occur and rubbery behaviour is observed. For low
temperatures, there is not enough time for shape changes to occur by bond rotation,
with only bond deformation (i.e. flexing) possible, and the material behaves as a
glass. At the glass transition temperature, bond rotations can occur, but only slowly.
Away from Tg, the response to an applied stress is rapid, but around Tg the response is
slower, with much of the energy lost as heat rather than stored as elastic potential
energy.
Since it is obvious that the behaviour of a squash ball is rubbery at room temperature,
it is clear that Tg for polyisoprene must lie some way below this. In fact, Tg for pure
uncrosslinked polyisoprene is around 200 K (–73 ºC), rising to around 300 K (27 ºC)
when 15% of the isoprene monomers have been cross-linked. Note that Tg increases
with increasing cross-link density as shape changes by bond rotation become more
difficult to achieve. In this practical, therefore, we will consider that Tg is only
affected by the degree of cross-linking. To those who play squash, the change in
elastic behaviour of the squash ball with temperature, as the ball warms up during
play, will be well known. This heating effect comes from the dissipation of energy by
repeated uncoiling and coiling of polymer chains as the squash ball is deformed
during play.
In the vicinity of the glass transition, as stated above, the effect of the dissipation is so
great that any kinetic energy of the ball is quickly transformed. Far below the glass
transition, as no bond rotations are allowed, the energy dissipation will be low. Hence,
it is expected that a minimum in the bounce height should be observed in the vicinity
of Tg.
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In the following experiment, we will try to identify this minimum, and therefore
estimate Tg for the squash balls provided.
2
Measurement of bounce height at room temperature
You are given a squash ball, a metal plate (to control the rebound), and a large
diameter plastic tube (to confine the ball) to which a measuring scale has been
affixed. Arrange the plastic tube on the bench so that the bounce height of a ball
inside the tube can be clearly read off the scale, using the top of the ball as the
reference point.
You may find it helpful to get a rough idea about how high the ball bounces on the
first bounce, and then use subsequent bounces to refine your estimate. Use the stickybacked plastic strip to fix a position on the measuring scale, and then adjust after each
bounce. Try to estimate the error in your readings (for example: 0.3 cm). This is the
precision of the measurement, as opposed to the accuracy of the result, and should
reflect your confidence in having achieved the stated precision.
 Can you devise any procedures to minimize the error in your measurements?
You may also find it helpful to work in threes, with one person dropping the ball, and
the other two observing and recording the bounce height.
3
Measurement of bounce height at ice water temperature
Iced water will be available for communal use on each bench. Place the squash ball
into the iced water and leave for 5 minutes to fully cool.
Remove the ball from iced water, dry its surface quickly using a paper towel, and
measure the bounce height as soon as possible using the same methodology as earlier.
In this case you should aim to make your measurement using only as few bounces as
possible.
 Why is this? What happens to the bounce height as the ball warms up?
4
Measurement of bounce height at liquid N2 temperatures
Liquid nitrogen will be dispensed into small polystyrene cups on the bench tops.
The quantities provided are not sufficient to be dangerous when handled properly,
but if a spillage occurs, move quickly away from the area and inform the Head of
Class. Do NOT put anything other than squash balls into liquid N2 or attempt to
take the liquid N2 out of the containers.
Place the squash ball carefully into the cup, using the tongs provided, being careful
not to splash liquid N2. Wear goggles to protect your eyes. Do not use the tongs to
keep the ball under the liquid N2 surface.
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Once the ball has fully cooled and nitrogen has stopped boiling off, remove the ball
carefully with tongs, and, using protective gloves, swiftly transfer the ball to a heatresistant tile. Start a stopwatch as soon as the ball is removed from the liquid N2. Be
sure to only grip the squash ball lightly with the tongs, as if the ball is held too
tightly it is likely to shatter.
Being careful to avoid direct contact between exposed skin and squash ball, measure
the bounce height about four minutes after removal from N2 using the same
methodology as earlier. Bouncing the ball too soon after removal from N2 may result
in it shattering! Again, ensure the squash ball is only lightly gripped.
Make repeated measurements of the bounce height at 1-minute intervals for a period
of 20 minutes after the squash ball is removed from the liquid N2. Use no more than 2
or 3 bounces (taking 10 to 15 seconds) to determine each bounce height. You may
need to periodically clean the frozen condensation from the squash ball carefully
using a piece of paper towel. Be careful not to burn yourself during this process, and
wear gloves.
The temperature of an identical squash ball has been measured as a function of time
during the heating process, and is shown below. Since direct temperature
measurement tends to interfere with the measurement of bounce height, use the
calibration curve to transform your data of average bounce height versus time into
average bounce height versus temperature. In order to help you do this, a cubic
polynomial fit has been added to the calibration curve to help you transform from
time, t, (in minutes) to temperature, T (in degrees Celsius). This is not based on any
physical law, but is intended to allow easy conversion of time into temperature. The
fitted equation for transforming between these two quantities is:
T = 0.0447 t3 – 1.9697 t2 + 32.102 t – 199.69
0
-20
0
2
4
6
8
10
12
14
16
18
20
-40
-60
]
C
º[ -80
T
,
-100
re
tu
-120
ra
e
p
-140
m
e
T
-160
Heating curve
Poly. (Heating curve)
T = 0.0447 t3 - 1.9697 t2 + 32.102 t - 199.69
-180
-200
Time, t [min]
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Course A: Atomic Structure of Materials
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Carry out the transformation for your data, and plot bounce height against
temperature.
 Check your curve using reference points taken at ambient and ice water
temperatures. Do they agree with the data points from your continuous curve? Is
the form of the curve as you would expect?
 What do you notice about the behaviour of the bounce height with temperature?
Use this to estimate the glass transition temperature of the squash ball.
 Assuming that the increase in Tg relative to pure uncrosslinked polyisoprene is
linearly proportional to the degree of cross-linking, estimate the degree of crosslinking in the squash ball, and state clearly any complicating factors.
5
Measurement of bounce height at elevated temperatures
Measure the bounce height at room temperature.
Heated water is provided in the form of a water bath set to 65 °C. Measure the actual
temperature of the water using a thermocouple provided.
Place the squash ball into the heated water and leave for 5 minutes to fully heat.
Remove from the heated water and measure the bounce height as soon as possible.
Once the heated ball has cooled again to room temperature, measure its bounce height
one more time.

Is the bounce height the same? What are the implications does this have for the
glass transition in polymers: i.e., would you say it was a chemical or a physical
type of phase transition?
Conclusions and take-home points
Polyisoprene is an example of an amorphous material, with no long-range order.
Amorphous materials can display different properties dependent on the temperature.
The glass transition is an example of a physical, reversible change whereas crosslinking represents a permanent, chemical change. You have determined the glass
transition temperature, Tg, for polyisoprene by a simple experiment, and quantified
(hopefully minimized!) any errors intrinsic to the experimental setup. In the next
practical, you will look at crystalline structures, which do possess long-range order
and can be described by a crystal lattice.
Please remember to fill in the feedback form with any comments on this practical
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Course A: Atomic Structure of Materials
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CrystallineMaterials&Close‐PackingofSpheres
Aims and Objectives of this Practical






To help you visualize crystal structures by building physical models.
To make you familiar with using the software package CrystalMaker to view
3-dimensional crystal structures.
Understand how to describe crystalline structures using the concepts of
coordination numbers, coordination polyhedra and unit cells.
Understand what is meant by “close-packing”.
Be able to differentiate between cubic and hexagonal close-packed structures.
and know how they have different stacking sequences of close-packed planes.
Understand the description of the structures in terms of close-packed planes
and how this relates to the conventional unit cell for both ccp and hcp
This practical has multiple parts. The first part (Section 1) is an exercise in building a
physical model of close-packed crystals, which are used later in the practical
(Sections 3-5), and the second part (Section 2) uses a computer to represent crystalline
structures. Your Head of Class will tell you whether to begin at Section 1 or Section
2, and tell you when to switch to the other sections.
1
Model building
You are supplied with a number of plastic balls with which to construct planes of
close-packed “atoms” possessing long-range order. Additionally, both the size and
shape of the spaces between atoms in close-packed structures are important (formally,
these are called interstices) so you will also build small ball models to illustrate two
kinds of interstice which occur in close-packing.
The balls need to be placed on the hexagonal template provided in a specified
arrangement so that each ball is touching its neighbours. You should stick them
together as follows:
Dip the small spatula into the solvent (butan-2-one) to collect a small amount. Put it
just above the point of contact of two balls as in figure (a). The solvent should flow
down the balls and be held by surface tension between the two balls; a disc of solvent
about 2 mm in diameter should be sufficient to stick the spheres together, figure (b).
Only a small quantity of solvent is needed – too much will dissolve the balls.
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Extensive inhalation of the solvent would be harmful, so replace the lids on the
jars after use. If you splash solvent on your hands, wash it off.
When all the balls in a particular arrangement have been stuck together, the tray
containing the models should be placed in an oven for 10 minutes to evaporate the
solvent, and then quenched in water to solidify the bonds. The models may then be
gently lifted off the template on to a sheet of card.
You will use a single metal tray to make your models: this has a number engraved on
the top so that you can retrieve the correct one from the oven after drying. The
suggested arrangement for balls within the tray is illustrated below.
At either end of the hexagonal template, stick together a set of six white spheres in the
pattern shown in figure (c). Then position a blue sphere so that it rests in the recess at
the centre of the model as shown in (d) and glue it in place. These models will be
referred to as X models.
Now, on the same hexagonal template, place six black spheres and twelve blue
spheres in the arrangement shown in (e), and stick them together.
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Next place six white balls into the recesses indicated by R in (e) and stick them in
position to make the model shown in (f). Make sure that X and Y models are not
glued to each other.
In between the two X models, glue together a group of three black balls in a triangle
as (h) below.
(h)
Make another set of three white balls in a triangle, as (h) above.
Finally, make another triangular group from white balls. However, this time position a
single white ball so that it rests on the recess at the centre of one of the triangles,
making a tetrahedron (i.e. each ball has its centre at one of the four corners of a
tetrahedron). Glue it in place.
Once your tray has come out of the oven and your models have been quenched, take
the two triangular groups that you have made and carefully position one set on top of
the other as shown in (i) below. Glue them together, making an octahedron and leave
the model to harden.
(i)
The X + Y models that you have made represent parts of two adjacent close-packed
layers, as found in the crystal structures of many metals. Once your models have set,
you should be able to invert each X model and insert it gently into a Y model as
shown in (g).
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(g)
You will use the models in section 3, and you can then take them home: they should
be useful throughout the course.
2
Representation of crystal structures
A crystal is characterised by having translational symmetry, which implies long-range
positional order of the atoms, and can be described by a repeating array of points
called a lattice. Each of these points (a lattice point) is in an identical environment,
i.e. the lattice points are equivalent, even in non-primitive lattices (see below). The
constraint of translational symmetry places a severe restriction on the number of
lattices that can exist. In fact, there are only 14 possible Bravais lattices in 3D, each
composed of a crystal system and lattice type (defined below).
A repeating unit of the lattice, from which the entire crystal can be built, is called a
unit cell. This is often, but not always, the smallest repeating unit. The shape of cell
(more precisely, the rotational symmetry elements it contains) defines the crystal
system (e.g. cubic, hexagonal, orthorhombic). The unit cell can either be primitive,
with lattice points only at the corners and one lattice point per unit cell, or nonprimitive, with lattice points at the corners and other locations in the unit cell and
more than one lattice point per unit cell. For a cubic lattice, there are 3 possible lattice
types: primitive (P), face-centred cubic (F) and body-centred cubic (I). The unit cell
can be represented in 3D as a parallelepiped described by 3 vectors a, b and c. In a
cubic unit cell, these vectors are orthogonal to each other and equal in length.
The structure of a crystal is fully described by a combination of the lattice (i.e. crystal
system and lattice type) and the motif. The motif is a list of the atoms associated with
each lattice point, with each atom’s coordinates expressed as a fraction of the unit cell
length with the lattice point at the origin. Since each lattice point is equivalent, the
arrangement of atoms about each one must be the same.
Another way of thinking about crystal structures is to use the coordination number,
which is the number of nearest-neighbour atoms around a given atom. In a crystal,
these nearest-neighbour atoms often form a regular polyhedron. For example, 4 atoms
surrounding another atom may define the corners of a tetrahedron, or 6 atoms may
define an octahedron, with the reference atom at the centre. An example is the
octahedral coordination of Na+ by Cl, or of Cl by Na+, in NaCl (see below). Hence,
in addition the description of structures by space-filling models (hard spheres) and by
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interatomic bonds (“a ball-and-spoke model”), they can also be described in terms of
coordination polyhedra linked at their corners or edges.
In this part of the practical, you will be using CrystalMaker, a graphics programme
helpful for viewing and manipulating models of crystals on the computer, to help you
visualise these structures in 3D. You may also find it helpful to look at the physical
“ball-and-stick” crystal models on display. If you are stuck, the CrystalMaker Users’
Guide can be found under the Help menu.
There is a large library of crystal structures available to view in CrystalMaker;
however, you should start by working through the following sequence.
-Fe
In the folder labelled “Crystal Structures Library” on the desktop, find the file “Iron –
gamma (fcc)” under the directory Minerals\Non-Silicates\Native Elements.
Open this and a window showing atoms in one unit cell of the structure should appear.
 What is the shape of the unit cell, outlined in red (you can rotate the image
using the mouse)? How many lattice points are there in one unit cell?
From the “Transform” menu, select “Set Range”. Put 4 in each of the “maximum”
boxes for the x, y and z-axes and click “Apply”. This will show what the 64 unit cells
look like.
 Do all the Fe atoms have the same coordination? What is the coordination
number?
NaCl (Halite)
Open this file (under the directory Minerals\Non-Silicates\Halides) and a single unit
cell of NaCl will be displayed (green = Cl, yellow = Na+). From the “Model” menu,
select “Space filling”. The unit cell will now be displayed with Na+ and Cl ions with
their correct relative sizes.
Alternate between the “Ball and Stick” and “Space Filling” models (in the “Model”
menu.
 Describe the coordination of Na+ by Cl and Cl by Na+.
 Which lattice does NaCl adopt? (give both crystal system and lattice type)
From the “Transform” menu, choose “Set Range” and put 2 into each of the
“maximum” boxes for the x, y and z axes. Convert the model to “Polyhedral” from the
“Model” menu.
 What is the basic polyhedral unit of the NaCl structure? How are these
polyhedra connected to each other?
Hexagonal Close Packing
Open the file “Hexagonal Close Packing - Layers” found under Basic Structure
Stypes\A.
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 What lattice does this structure have? (give both crystal system and lattice type)
What is the motif?
 Describe the coordination. How many lattice points are in the unit cell?
3
Hexagonal close-packing
Collect the models that you made earlier in Section 1, which should now have dried.
They should still have the smaller X models inserted into the larger Y models. Each
combined model now represents a pair of close-packed planes (assume all coloured
balls represent the same type of atom).
Identify the unit cell of a single two-dimensional close-packed layer and draw a 2-D
plan of four unit cells with a common corner. Draw touching circles to represent the
touching balls.
Mark on your plan the positions of holes into which the balls of the next layer could
fit.
 Are the two holes within each unit cell exactly equivalent?
 Could two balls be fitted in simultaneously?
 Are the two holes related by a lattice vector?
Hold one combined model up to the light; some of the gaps between spheres in the top
layer are above gaps in the lower layer, while others are above spheres in the lower
layer. Thus, in any one layer the gaps between the spheres are of two types. On your
drawing label the circles A, and the two types of recess B and C.
Place one combined model on the bench with the layer of white balls uppermost.
Superimpose the other combined model, in the same orientation, so that the blue
spheres of the top model lie directly over the blue spheres of the bottom model. The
sequence of layers is now ABAB and the model represents hexagonal close-packing.
Place your model on a light box and note that gaps run right through the stack. Thus
one set of sites (the C sites) is always unoccupied in hexagonal close-packing (hcp).
Compare your model with a ball-and-spoke model of hcp. Describe or sketch the
three-dimensional arrangement of nearest neighbours (ignoring the colours of the
balls) around a sphere in a white layer and around one in a blue-and-black layer.
 Are the arrangements different?
 Are all the atoms exactly equivalent, i.e. do they have exactly the same
environment?
 How many atoms are associated with each lattice point, i.e. how many atoms are
in the motif?
Find the three-dimensional unit cell of hcp and draw a large, accurate projection of
this structure on the close-packed plane showing four unit cells as above. Label the
axes in the close packed plane +x and +y, taking , the angle between them, as 120
Taking care to note that the axes are not orthogonal, write down the fractional
coordinates of the atoms in the unit cell. Mark on your plan the 3 close-packed
directions in the close packed plane, i.e. the lattice vectors joining the centres of
adjacent atoms in the close packed plane.
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Cubic close-packing
Remove the top two layers of the model used earlier. Take the X model from this
double sheet and place it on top of the two close-packed layers of the lower double
unit so that the six white spheres sit on the recesses above the gaps which run right
through the two lower layers. This pyramidal model has the stacking sequence
ABCA.
Invert the spare Y model and carefully add it to the combined model to show more of
the two upper close-packed layers. The two blue-and-black layers should now be
directly above and below each other. This model represents cubic close-packing (ccp).
Hold your model up to the light. Note there are no gaps running through the structure.
Describe or sketch the distribution of nearest neighbours about a sphere in each of the
layers of type A, B, and C. Again, ignore the colour of the spheres – they represent
chemically identical atoms. (A ball-and-spoke model is available for comparison.)
 Are the distributions different?
 Are all the spheres exactly equivalent?
 How many spheres are associated with each lattice point, i.e. how many spheres
are in the motif?
Identify the unit cell that has x and y in the blue-and-black close-packed layer and z
normal to the layers, and is thus similar to the one drawn in above. Draw a plan of
four of these unit cells projected onto the x-y plane.
 How many lattice points are there in the cell? Is this cell primitive?
The stacking sequence ABCA has, in fact, cubic symmetry. There are three other sets
of close-packed planes related by this symmetry. Locate these on your model. This is
most easily done by removing the top Y model added earlier, leaving the inverted X
model on top of the X + Y model.
In ccp the symmetry makes it possible to choose a unit cell geometrically simpler than
that used above. This conventional unit cell is a cube with atoms at the corners and at
the centres of the faces. It may be extracted from your model by removing the top Y
model and then lifting out the two X models together, keeping the same relative
orientation.
Place this cube with one of its faces in the smaller of the two square templates. The
edges of the face will lie at 45˚ to the edges of the template.
Choose the origin of the cubic unit cell on the blue sphere of the lower face of the
cube. Take the cubic reference axes x and y in the plane of the template and the z axis
vertical. Draw a plan of this cell projected onto the x-y plane, showing all the atoms.
Remove the top X model by lifting its blue ball to expose a close-packed plane. Make
a sketch of the arrangement of atoms in this close-packed plane and mark in the three
close-packed directions.
Replace the upper half of the cube. Locate the three other close-packed planes.
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5 Interstices in close-packed structures
Interstices in close-packed structures are sites where low concentrations of small
atoms such as H, C and N can be incorporated into crystal structures with minimal
distortion to the crystal structure. For example, cubic close-packed iron can take up to
1.7 wt% carbon into solid solution quite happily at 1130 ˚C. It is therefore important
to be able to appreciate what interstices are and their sizes relative to the atoms.
Examine the two models built in Section 1. They show the distribution of atoms
around so-called ‘octahedral’ and ‘tetrahedral’ interstices. These names are used since
the atoms around the two types of interstice are located at the corners of a regular
octahedron and tetrahedron.
Locate some of these interstices in your models of hexagonal and cubic closepacking. On the plan drawn earlier locate one interstice of each type, and give its
fractional coordinates.
Conclusions and take-home points
Physical models and computer software are useful tools for viewing and better
understanding crystal structures. We can describe crystalline structures in terms of the
coordination of one type of atom around another (coordination number) and the shape that
this coordination shell creates (coordination polyhedron). We can also visualise crystalline
structures in terms of repeating structures (unit cells) described by a lattice type, crystal
system and motif. The following facts about close-packed spheres are of great importance,
and will come up in the course time and again:
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

Stacking planes of close-packed layers ABAB… gives hcp, whereas the stacking
sequence ABCABC… gives ccp.
In both types, each atom has 12 neighbours, but the arrangements are different.
In ccp all atoms have identical environments, but this is not true in hcp where there
are two atoms in the motif.
For ccp a hexagonal unit cell of three atoms can be used, but the conventional unit
cell is a face-centred cube, with four atoms.
Both structures have tetrahedral and octahedral interstices. These are particularly
relevant for describing related structure types.
Please remember to fill in the feedback form with any comments on this practical
14
JAE M2014
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
LatticePlanes,MillerIndicesandX‐rayDiffractometry
** BEFORE ATTENDING THIS PRACTICAL **
To assist you with the assessed exercises, first work through the DoITPoMS TLP
entitled “Lattice Planes and Miller Indices”.
http://www.doitpoms.ac.uk/tlplib/index.php
Aims and Objectives of this Practical







Know what “Miller Indices” are and how they are used
Be confident in “indexing” sets of lattice planes in 2-D and 3-D
Be able to draw a set of lattice planes when given the Miller Indices
Be able to use CrystalMaker to construct lattice planes and directions
Use a mini X-ray diffractometer to observe diffraction from a LiF crystal
See Bragg’s law in action, observing that both incoming and outgoing X-ray
beams must be aligned correctly with respect to the crystal to observe a peak
Learn that X-rays are not simply reflected from the surface of a crystal but
can be diffracted from planes inside the crystal inclined to the surface
This practical consists of two halves. The first part (Sections 1-4) is a paper, pen and
computer-based exercise on Miller Indices, and the second part (Sections 5-9) is an
introduction to X-ray diffractometry. Your Head of Class will tell you whether to
begin at Section 1 or Section 5, and tell you when to switch to the other part.
1
Lattice planes in 2-dimensions
The indices are defined as (hkl) if the first plane away from the origin makes
intercepts on the +x, +y, and +z axes of a/h, b/k and c/l, respectively.
Rutile (TiO2) is tetragonal with a = b = 4.59 Å and c = 2.96 Å. A projection on (100),
i.e. looking down the x-axis onto the y–z plane, of the lattice is shown in Figure 1. The
intersections (traces) of portions of seven sets of lattice planes normal to the paper
have been drawn on this plan. What is the h index for all these planes? For each set of
planes, determine the indices k and l.
 How are (hkl) and ( hkl ) related?
A projection on (001) is shown in Figure 2. Use it to make a drawing similar to Figure
1 showing the traces of sets of lattice planes (110), (010), (210), ( 210 ) and ( 230 ).
A projection of a hexagonal crystal lattice (a = b, γ = 120°, the angle between x and y
axes) onto (001), i.e. looking down the z axis onto x-y plane is shown in Figure 3.
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JAE M2014
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
 Index the directions shown on the plan, and then draw the lattice vector [310]
from the origin to the edge of the plan. Add the trace of one of the (310) planes
passing the unit cell closest to the origin.
 What do you notice about the direction [310] relative to the trace of (310)
planes? Remember that the direction [hkl] is not necessarily perpendicular to the
(hkl) planes (this is only true in a cubic crystal).
2
Lattice planes in 3-dimensions
Figure 4 shows a view of a crystal lattice. Using the axes shown as your reference
axes, index the sets of lattice planes I to IV. Remember that in each case you may
choose any convenient lattice point as your origin.
Near the numeral IV, add to the diagram the traces of a few planes of the set ( 112 ).
Open the DoITPoMS TLP entitled “Lattice Planes & Miller Indices” on one of the
computers. Go to the page “Draw your own lattice planes” and satisfy yourself that it
provides the same answers for the 3-D planes you have drawn and indexed. Complete
the TLP questions (including the drag & drop game) if you haven’t already done so.
3
Lattice planes and directions
Assign some axes to the conventional unit cell of the
cubic close-packed structure (right).
 What are the Miller Indices of the shaded
planes?
 What are the lattice directions along which the
atoms in that plane touch? Use the Weiss Zone
Law to confirm that the directions you have
written down do indeed lie in that plane.
 For nickel, cubic close-packed a = 3.52 Å, calculate the separation of the shaded
planes using trigonometry. Check your answer by using the formula for
interplanar spacing for crystals with orthogonal axes from the Data Book. Now,
calculate the interplanar spacing of the (312) planes, and note that this answer
would be difficult to obtain purely by trigonometry.
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JAE M2014
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
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JAE M2014
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
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JAE M2014
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
2014-15
AP2
y
x
Figure 3
Figure 4
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JAE M2014
2014-15
4
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
CrystalMaker exercises
Open the file Iron - gamma (fcc).cmdf (Found in Crystal Structures Library\
Minerals\Non-Silicates\Native Elements). Open the Set Range window from the
Transform menu and set the range from 0 to 2 for x, y and z. It may be helpful to
display the axes (ModelShow Axes). You should now have 8 unit cells displayed.
Miller indices
Bring up the Lattice Plane menu (TransformLattice PlaneEdit) and display the
(111) plane. Using the Lattice Plane tool from the Tools palette, adjust the position of
the plane so it is not bisecting the atoms and then slice the model to show the closepacked plane (TransformSlice Model). The red (111) plane can be hidden
(TransformLattice PlaneHide). Change the model type to Space Filling and
compare it to the physical models you have made in AP1.
By resetting the model (from the Plot Range window) or by opening another Iron gamma (fcc).cmdf file, try cleaving along higher indexed planes, e.g. (431).
 What differences do you notice between the close-packed (111) and the higher
index planes (HINT: Think about how rough / smooth the planes are)?
Lattice vectors
In a close-packed structure, there are 12 close-packed directions (or six anti-parallel
pairs). The 6 in-plane directions are shown below:
Index the 12 close-packed directions for the cubic fcc/ccp unit cell and the 6
directions in a single close-packed plane for the hexagonal hcp unit cell. You may
find a 3D sketch helps to index the directions in the fcc cell (remember, close-packed
planes are of type {111} in fcc)
 Which rule could confirm that the directions you have found lie in a closepacked plane? How would you represent the 12 symmetry equivalent directions
using crystallographic notation?
Plot one of the directions you have indexed as a lattice vector in CrystalMaker. Select
one of the atoms in the close packed plane from the previous section using the Arrow
tool and define the direction from SelectionVectorAdd.
 Does it agree with your prediction?
20
JAE M2014
2014-15
5
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
Introduction to X-ray Diffractometry
In this part of the practical you will learn how to use a small, low-power X-ray
counter diffractometer to collect diffraction data from a LiF crystal. It will be
necessary to work in small groups of 2 to 3 students.
IMPORTANT HEALTH AND SAFETY INFORMATION
X-rays are potentially dangerous. In addition to a lead glass domed shield around the
X-ray tube, there is an interlocked plastic anti-scatter enclosure. Thus, the shutter can
only be opened allowing X-rays to exit the tube when the enclosure is correctly
positioned. You should be aware when the shutter is open and when it is closed, but
the interlocks are a backup to prevent any possible accident.
The LiF crystal may react with moisture, e.g. on your skin. Please wear gloves when
handling the crystals.
The portable diffractometer you will be using consists of a small solid-state high
voltage generator driving a copper anode X-ray tube that can be seen inside the lead
glass dome at the rear of the unit. X-rays generated at the anode emerge through a
thin glass port blown in the tube and then through a hole in the lead glass dome. After
passing through an interchangeable collimating slit, the beam passes over the central
turn-table on which a crystal can be mounted. X-rays diffracted by the crystal in a
particular direction pass through a receiving slit and are detected by a Geiger-Müller
(G.M.) tube positioned in a carriage moving concentrically with the crystal and
driven by the vernier wheel outside the transparent radiation protection cover.
The motion of the detector and crystal are coupled so that the angular velocity of the
former is twice that of the latter (known as a 2or 2 drive). The angular
setting (2) of the detector can be read off the scale around the perimeter of the top
plate of the unit while the setting () of the crystal is indicated on the scale
immediately around the central turntable.
Access to the working space is achieved by first checking the state of the X-rays. If
they are ON (red light, HT ON), turn (power) OFF, then carefully displace the
transparent cover sideways in the direction of the detector carriage and raise it. To
make a measurement, close the lid, then centre it (to engage the interlock) and turn on
the X-rays (Power ON and then X-rays ON).
Lights are top back and switches bottom front of instrument, see Figure 5.
Safety interlocks ensure that the high voltage generator can only be switched on when
the lid is both closed and centred. The high voltage generator is automatically
switched off if the lid is displaced from this position. However, this is a safety backup
only, and you always close the shutter manually before opening the lid.
The X-ray tube emits a relatively low level “white” radiation together with the strong
characteristic radiation Kand Kfrom the copper target; K is the stronger and has
the longer wavelength. You will use a filter to block out all but the Cu Kwavelength
X-rays.
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JAE M2014
2014-15
6
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
Operating Instructions
Figure 5: Schematic of the portable X-ray diffractometer.
Defining the outgoing beam. Carefully remove the G.M. tube from the carriage and
insert the 1 mm receiving slit vertically in the detector carriage in sprung slot 14, and
the 3 mm slit similarly in the rearmost slot 30.
Defining the direction of the incoming X-ray beam. Align the detector carriage
pointer with the 2 = 0° graduation. With no crystal in position check that the
collimator slit on the lead glass dome is vertical. You should be able to see the anode
clearly through all three aligned slits; this line of sight will define the direction of the
primary X-ray beam, 2 = 0°.
Zeroing both drives. When the detector is at 2 = 0° the crystal-holder (central
spindle) scale should read = 0° also, with the "reflecting" surface facing right, as
shown in Figure 5.
If the  scale does not read 0°, it is necessary to adjust it independently of the
detector carriage: loosen the knurled ring around the base of the crystal holder,
whereupon the setting can be adjusted as required. Tighten the knurled ring fingertight (without straining) so that the two-to-one linkage is re-engaged.
You might find it easier to stand over the set and look down to check that both the
crystal and detector are aligned at zero.
Orienting the crystal. Lithium fluoride is cubic. The lithium fluoride crystal supplied
is bounded by {100} cleavage planes. Insert the crystal into the holder at the centre of
the turntable with a large face against the vertical reference surface, and the longest
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JAE M2014
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
side vertical. If one of these large faces has a dull appearance due to slight abrasion,
mount the crystal so that this duller face is against the reference surface.
On looking through the collimating slits you should now see that the primary beam
direction lies in the surface of the crystal, as does the axis of rotation.
Note that, at this stage, when the detector carriage is moved, the normal to the crystal
face always bisects the angle between the primary beam direction and the centre-line
of the carriage, i.e. the crystal face is always symmetrically arranged with respect to
the incident and detected beam directions, i.e. = .
Installing the G.M. tube. Remove the 3 mm slit and replace the detector in sprung slot
24, with the cable emerging vertically upwards. The centre-line of the G.M. tube
should then be parallel to, but displaced from, that of the carriage. Place the antiscatter slit in slot 13. Place the Ni filter in slot 15. (Ensure that the cable is free of the
diffractometer cover. If you have any difficulty call a demonstrator.)
Setting the voltage. Check that the red switch on the top of the turntable is in the 30
kV position and that the time switch (front centre) is not at zero (suggest set for
~50mins) .
Checking the crystal holder. Move the detector around to the right (anti-clockwise)
through at least 15° of 2. Check that the crystal holder is arranged so that the X-rays
will be incident on the face of the crystal in contact with the vertical reference
surface. (If it appears that your unit has been set up so that the X-rays will be incident
on the other vertical surface of the crystal, consult a demonstrator at once.)
Powering up. Close and centre the radiation cover. Check that it cannot be re-opened,
simply by raising gently.
Switch on the mains supply: “POWER ON” switch. The white indicator lamp should
light up.
Turn on the high voltage generator to the X-ray tube by depressing the red
“X-RAYS ON” button on the front left of the unit. The red indicator lamp at the right
rear should light up. If it does not, check that the lid is correctly centred. Notice also
that the cathode of the X-ray tube is glowing.
7
Estimating the position of the 200 diffraction peak
LiF has the cubic NaCl structure; ball-and-spoke models of the NaCl structure are
available for inspection. The arrangement of the ions on a (001) plane is illustrated in
Figure 6 below.
 Estimate the cell parameter a of LiF from the ionic radii: Li+Å
FÅ.
Note  this approach provides a rough estimate. (Such information can often also be
obtained from databases.) In this practical you will obtain accurate values for the cell
parameters of the crystal that you have been given.
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JAE M2014
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Course A: Atomic Structure of Materials
2014-15
AP2
 Using this estimated value of a and the Bragg equation (see Data Book),
estimate 2200 and 2400 for CuKradiation ( = 1.54 Å).the cell parameter a
of LiF from the ionic radii: Li+ÅFÅ.
Li
F
Figure 6: Arrangement of ions on (001) plane for LiF.
8
Recording the 200 diffraction peak
With the diffractometer set according to the operating procedure outlined in the previous
section, the counter arm should be populated as follows:
slot 13: anti-scatter slide
slot 14: 1 mm slit
slot 15: Ni filter
slot 24: counter
By turning the counter though the appropriate angle (2 and hence the crystal by ,
locate the 200 reflection close to the 2 value estimated in section 3 above.
 Record the value of both 2and count rate (intensity) for this peak.
The crystal must now be oriented precisely. It has been assumed up to this point that a
zero reading on the crystal setting scale corresponds to the incident X-ray beam being
parallel to the (100) planes (i.e. the cleavage plane) of the LiF crystal. That assumption
is not necessarily precise: there may be a small zero-error in the crystal setting scale,
which can easily be eliminated by the following procedure:
Turn off the power and open the radiation cover. Note the reading for the crystal
setting (  1/2 2loosen the knurled ring at the base of the crystal holder to
disconnect the 2:1 gearing; tighten the knurled ring finger tight without straining, and
then loosen it by half a turn so that the 2:1 gearing is loosely engaged.
Lay the setting bar flat on the diffractometer table so that its two studs engage with
two holes in the centre plate. (It is often easiest to put the setting bar on the same side
as the detector – when they are on opposite sides, it is difficult to close the cover
properly.) Replace the radiation cover and switch on the X-rays.
Hold the counter arm at the 200 peak maximum determined above and adjust the
angle of the crystal by moving the setting bar slightly each way to achieve maximum
counts. Do this very slowly in order to allow the recorder time to respond. Move the
counter arm slightly either way to locate maximum response.
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JAE M2014
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP2
Repeat the cycle until adjustment of the crystal setting for maximum counts leads to
no change in the setting of the counter arm. The normal to (200) planes of the LiF
crystal will now bisect the angle between the incident and diffracted X-ray beams.
 Record this position of the inner ring in case you accidentally disturb it.
 Compare the intensity measured here with what you measured earlier, before
precise orientation. Is there any improvement?
Turn off the X-rays. Open the radiation cover, remove the setting bar and gently retighten the knurled ring. Close and centre the radiation cover.
Plot count rate (allowing the detector some 5 seconds to respond before each reading)
against 2every 0.5° of 2, from 1.5° below to 1.5° above the peak position found in
0, and so determine a value for 2200.
 Hence evaluate the cell parameter a for your LiF crystal.
Repeat the whole procedure in this section using the 400 reflection and the value of
2400 estimated in section 3, as this will lead to a more accurate measurement of the
cell parameter. If you want to see why this should be so, refer to the Appendix.
9
Recording the 420 diffraction peak
It is important to realize that Bragg “reflection” is not a surface reflection, and
therefore incident and diffracted beams are not necessarily symmetrical to the (100)
cleavage planes. To investigate this you are asked to find the diffracted beam arising
from the (420) planes.
Note  the (420) interplanar spacing has components both perpendicular and parallel
to the large surface face of the crystal, whereas the (400) interplanar spacing is
perpendicular to this surface. By combining these 2 measurements one can calculate
the cell parameter in both directions.
 Why might these values differ?
For a cubic crystal d(hkl) = a/(h2+k2+l2). Use this expression and the Bragg equation to
estimate 2420.
Evaluate the offset angle (100):(420) in order to calculate the angle through which the
crystal must be rotated (see Figure 7).
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JAE M2014
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Course A: Atomic Structure of Materials
2014-15
(a) For planes parallel to the sample
surface (1) Offset = 0
(a)
AP2
(b) & (c) For general planes (2)
Offset  0
(b)
2
( c)
2 0
n1 is the normal to planes parallel to the sample surface (1);
n2 is the normal to general planes (2)
Figure 7: Schematic demonstrating location of planes and angles between them.
Reset both the angles 2andto zero.
Loosen the knurled nut and use the setting bar to turn the crystal through the
(100):(420) offset angle calculated in above so that the (420) lattice plane is parallel
to the incident X-ray beam, as shown in Figure 7(b). After retightening the knurled
nut, moving the detector to 2will result in  moving to [1/2(2offset angle], see
Figure 7(c).
Replace the radiation cover and switch on the X-rays. Locate the 420 reflection.
Repeat the crystal alignment procedure in section 4. Then, by plotting counts against
2, obtain an accurate value of 2.
Calculate the value of a using this measured plane spacing.
 How does this result for cell parameter, calculated from 420 peak, compare with
those which you obtained in section 4 from the 200 and 400 peaks?
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Course A: Atomic Structure of Materials
AP2
Conclusions and take-home points
Planes are described using Miller Indices, a description based on how far away from the
origin the first plane intercepts each coordinate axis. The direction [321] is not
necessarily normal to the plane (321) – though it is for cubic crystals! The Weiss Zone
Law is the way to work out whether a direction [UVW] is contained within a plane (hkl) –
it works for any crystal system.
In order to measure diffraction peaks from a single crystal both the detector position,
2and the crystal position, must be carefully aligned. Measurements made at higher
2angles give more accurate cell parameters, due to the slower variation of sinat
higher diffraction angles. y tilting, or offsetting, the sample cell parameters can be
measured in different directions of the sample, e.g. both perpendicular to and parallel to
the surface of a thin film.
Please remember to fill in the feedback form with any comments on this practical
Appendix : Accurate determination of lattice parameter
d
d

 tan
Differentiate Bragg's law,   2dsin
For an error in , of 
d


d
tan
For the highest sensitivity in angle:
   tan.
d
 0, as   90
d
d
d
best as   90 ;
i.e. 2  180 
1.0

sin
0.8
0.6
 use the highest angle line to give
highest accuracy.
0.4
0.2
0.0
0
20
40
60
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JAE M2014
80
2014-15
MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP3
PhaseIdentificationandQuantificationforCubicCrystal
Aims and Objectives of this Practical





To show the relationship between “single crystal” and “powder” X-ray data
To learn how cell parameters and/or relative intensities can be used to
identify crystalline phases of a material and measure the amount of each
crystalline phase (or material) in a mixture (quantitative phase analysis)
To understand how the crystalline structure affects intensities by doing
structure factor calculations
To ‘index’ a cubic powder pattern, obtaining the lattice type and cell
parameter
Given possible trial structures with given lattice type, to determine which one
is best by comparing calculated intensities with the experimental values
Phase identification and quantification are important in materials science and have
many applications e.g. forensic work, drugs industry, corrosion, gas, oil & mining
exploration, catalysts, cement works and development of new materials & new
devices.
Both peak positions (which are related to cell parameters) and relative intensities
(related to the type of atom and their positions in the unit cell) are important
measurements to identify different crystalline phases. In a mixture, the relative
intensities are used to quantify each phase.
This practical is in several parts, and consists of a practical exercise involving the Xray diffractometers you used in AP2 plus a paper-and-pen exercise. It is important that
you attempt the whole practical – your Head of Class will tell you which section to
begin with, and when it is time to move on. You can return to any section if you have
time at the end.
IMPORTANT HEALTH AND SAFETY INFORMATION
X-rays are potentially dangerous. In addition to a lead glass domed shield around the
X-ray tube, there is an interlocked plastic anti-scatter enclosure. Thus, the shutter can
only be opened allowing X-rays to exit the tube when the enclosure is correctly
positioned. You should be aware when the shutter is open and when it is closed, but
the interlocks are a backup to prevent any possible accident.
The LiF crystal may react with moisture, e.g. on your skin. Please wear gloves when
handling the crystals.
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Course A: Atomic Structure of Materials
2014-15
1
AP3
Cell parameters
You are provided with these 4 crystals: LiF, NaCl, KCl and RbCl.
 From Bragg’s Law, calculate the angle at which you would expect to find the
strongest peak, 200 or 002, for the crystals with CuKα radiation (λ = 1.54 Å):
(For a cubic structure the position and intensity of the 200 and 002 reflections
are identical and they are named by various different conventions)
LiF a = 4.03 Å
NaCl a = 5.64 Å
KCl a = 6.28 Å
RbCl a = 6.58 Å
 Why do you think that these cell parameters are different?
You are provided with the same mini X-ray sets (TELTRONS) as in AP2. Take 1
crystal from the choice of those marked: Red, Green and Yellow. They could be: LiF,
NaCl, KCl, or RbCl. Mount the crystal and, using the same methodology as described
in the previous practical, find and measure the 200 reflection to determine the unit
cell (use the 3 mm slit rather than the 1 mm).
 From your measurement of the unit cell, work out which crystal you have.
 Do you think that you can measure the difference between KCl and RbCl?
If you have time, you may like to measure another crystal. Please replace crystals in
the correct tubes after use, as some will absorb water.
2
Intensities and cell parameters – calculation
The powder diffraction data for the 4 crystals you have been investigating are given
both graphically and in a table. Sample A is a mixture of 2 crystalline phases (i.e. LiF,
NaCl, KCl or RbCl). In this practical you will use the peak positions and intensities to
a) identify and quantify the crystalline phases present in the experimental powder
diffraction data of Sample A and b) gain some understanding of why the relative
intensities vary for cubic structures.
Counts Intensity (counts)
20 LiF
10
0
400
NaCl
200
0
KCl
500
0
2000
RbCl
1000
0
3000
2000
1000
0
Sample A
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JAE M2014
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20
30
40
50
60
2θº
CuKα
Position [°2Theta] (Copper (Cu))
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MATERIALS SCIENCE
Course A: Atomic Structure of Materials
2014-15
LiF
Height /cts
18
24
12
3
3
d /Å
2.33
2.02
1.42
1.22
1.16
NaCl

2
38.67
44.95
65.45
78.68
82.93
KCl
(hkl) Height /cts d /Å 2

111
002
022
113
222
41
455
279
10
89
39
5
104
78
3.26
2.82
1.99
1.70
1.63
1.41
1.29
1.26
1.15
AP3
27.37
31.70
45.45
53.87
56.47
66.23
73.07
75.29
83.99
RbCl
(hkl) Height /cts d /Å 2
(hkl) Height /cts d /Å 2 (hkl)
111
002
022
113
222
004
133
024
224
111
002
022
113
222
004
133
024
224
115
333
044

7
1040
656
4
214
93
1
246
181
1
1
59
3.63
3.14
2.22
1.89
1.81
1.57
1.44
1.40
1.28
1.21
1.21
1.11
24.54
28.41
40.61
48.02
50.30
58.78
64.65
66.55
73.88
79.20
79.20
87.89
799
2797
1930
394
645
290
158
782
582
116
116
185
119
402
402
3.80
3.29
2.33
1.98
1.90
1.64
1.51
1.47
1.34
1.27
1.27
1.16
1.11
1.10
1.10
 Identify which two materials are in Sample A. Use the graphs and/or tables.
 Using the strongest peak, roughly estimate the ratio (weight fraction) of these
two materials.
 Looking at the intensity scales on the graphs or intensities in the tables above
(they are the same) can you obtain a calibration coefficient for the strongest
peaks of these two materials?
Now we will calculate the relative intensities (I) of the strongest peak (002)

I hkl  Fhkl  Fhkl
where Fhkl is the structure factor:
The atomic scattering factors (also known as form factors, fn) are given below:
2 /o
0
39
45
27
32
25
28
23
27
Li
3
1.8
1.7
-
Na
11
8.8
8.5
-
Atomic Scattering Factors, fn
K
Rb
F
19
37
9
6.5
5.8
15.2
14.6
32
31.2
-
Cl
17
13.3
12.3
13.7
13
14.1
13.2
(note: fn increase with increasing atomic number and decrease with increasing angle)
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23.40
27.09
38.68
45.70
47.86
55.85
61.38
63.15
70.00
74.95
74.95
82.96
87.69
89.26
89.26
111
002
022
113
222
004
133
024
224
115
333
044
135
006
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All the crystals have the same cubic structure (NaCl). The atomic positions (x, y, z) in
fractional co-ordinates (i.e. as a proportion of the unit cell dimensions) are:
Metal cation (M)
Anion (A)
0,0,0
1/2,1/2,0 1/2,0,1/2 0,1/2,1/2
1/2,1/2,1/2 0,0,1/2
0,1/2,0
1/2,0,0
 Using your intensities calculated for the structure factor and the relative peak
heights of the strong 002 peak, now see what value you get for the weight
fractions in sample A.
Looking back at the graphs and tables of the data for all 4 cubic structures, note how
both the peak positions and intensities vary. In particular, compare say KCl to LiF or
RbCl and note that the relative intensities vary, e.g. (111) and (002): h+k+l is even as
compared with h+k+l is odd.
 For two of these materials, calculate the structure factors of (111) and (002).
You may have already done (002) in section 2 above.
 What factors contribute to the variation in the relative peak intensities?
3
Intensities and cell parameters - computational
In this section you will use the same mixture data as in section 2 (Sample A) and the
commercial software X’pert Highscore Plus, with a small reference database, to
identify the compounds in the mixtures and obtain a semi-quantitative measure of
their relative amounts. The current commercial ICCD database contains over one
hundred thousand unique crystalline phases. Here, for simplicity, this database only
contains data on the 4 cubic phases you are studying. The procedure for using the
software is described below:
i. Open the program X’pert Highscore Plus (HS+).
ii. (The database should be set to Mary mini cubic database. If a problem check
CustomiseProgram SettingsReference PatternsMary mini cubic
database.)
iii. Open the mixture data from the FileOpen menu. Start with Sample A, which
is printed in this document and considered in section 2
iv. Now, to account for the background scattering, open the TreatmentDetermine
Background window. Check the background fitted looks okay and click Accept.
v. To automatically find the peaks present, go to TreatmentSearch Peaks and
search for the peaks. Check the assignment looks okay and then click OK.
vi. Now that the peaks have been found, the phases present can be analysed. Go to
AnalysisSearch and MatchExecute Search and Match and ensuring the
Restriction Set is at None, click Search.
vii. Set the additional graphics pane to enable easy peak comparisons at Set
ViewAdditional GraphicsSeparate Patterns.
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viii. In the lists pane, you should now have the pattern list at the top and the list of
possible candidates at the bottom.
ix. Select the patterns in the bottom pane to check the match with the data on the
graph. Drag those that fit up to the Pattern list pane.
x. You may find it easier to see if you change the colours of the lines, using the tab
in the Pattern list pane.
xi. Remove candidates that don’t fit from the bottom pane.
xii. Once the peaks are assigned, scroll to the right-hand side in the Pattern list
pane. The last box should give the SemiQuant(%) value. This is the semiquantitative estimate of the relative amounts of each phase present.
xiii. (It is worth noting that once all the peaks are assigned then there will be no
more arrows on the lines denoting the peaks above the main pattern.)
 How well do the results of this analysis agree with your work in section 2?
The semi-quantitative analysis is calculated using the I/Ic values for each phase. I/Ic
(RIR- reference intensity ratio) is the intensity of the strongest peak compared with
the strongest peak in corundum Al2O3. This ‘calibration’ accounts for differences in
the structure factors, multiplicity, absorption and other factors allowing relative
intensities to be converted to weight fractions. In this analysis, the intensity is
calculated from a single peak height rather than using the areas of all the peaks which
gives more reliable results. Thus this quick RIR method gives a ‘semi-quantitative
result’.
(xiv In pattern list, if you double click on the reference data, all the crystallographic
information for that phase should appear as in the data base)
 Now, if you have time, try the same approach on some of the other data files –
Sample B, C &/or D (just using the HS+ software not manual calculations)
4
Interpretation of a cubic diffraction pattern
X-ray diffraction is the main technique for determining the structure of materials, i.e.
the crystal system, unit cell, space group and positions of the atoms in the unit cell.
Usually a full set of single crystal data are required (~1000 spots measured) but some
simple cubic structures can be determined from powder data with relatively few peaks.
Here we follow this method for the compound gallium phosphide (GaP).
For any powder diffraction pattern the interplanar spacing d of each powder line can
be calculated from the measured Bragg angle . For cubic powder patterns, in this
case from gallium phosphide, GaP, it is easy to go further and assign Miller indices to
each powder line and then determine the lattice type and unit cell parameter. Because
GaP is cubic and has a simple unit cell it is possible to do even better, and determine
the crystal structure by comparing the observed relative intensities of different powder
lines with those predicted on the basis of candidate crystal structures.
The sample used to obtain the powder pattern was sulphur-doped GaP provided by the
ESPRC III-V Semiconductor Facility in the University of Sheffield. This material is
used as the conducting substrate on which a thin film of gallium-aluminium-arsenic32
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phosphide is deposited for use in opto-electronic devices. A level of sulphur doping of
< 0.1 at.% is used to tailor the electronic properties of the GaP substrate.
4.1 Determination of unit cell parameter and lattice type
You are provided with a powder diffractometer trace, obtained using CuK1
radiation. Values of 2 for the centre of each peak, obtained from the diffractometer
output, are also given.
 Label each powder line in the pattern from n = 1 to n = 9 with increasing
diffraction angle . Set up a table of values of n and 2.
In view of the fact that the crystal is cubic, and thus:
1
h2 k2 l2



2
d hkl
a2 a2 a2
and, using the Bragg equation:
 2 =4d 2 sin 2  and N hkl 
a2
2
d hkl
Nhkl is thus proportional to sin2.
Add columns to your table for, sin2, (sin2n)/(sin21), N and hkl, where 1 is the
Bragg angle of the line with lowest 2, and n is the Bragg angle of the nth line.
Find the simplest set of integers, N, consistent with the values of (sin2n)/(sin21).
Using the values of N obtained above, index the diffraction pattern, i.e. assign hkl
values to all the powder lines.
Use the pattern of the indices of the observed reflections to determine the lattice type.
Use your value of the diffraction angle for the line with largest 2 to determine the
value of a. (Use the value of  (CuK1) given in the Data Book.)
Estimate the accuracy of your result for a, if the uncertainty in measuring 2 for each
powder line is ~ ± 0.02°.
5
The structure of gallium phosphide
The structure of this relatively simple cubic substance can be determined by
considering the alternative atomic arrangements possible for the atoms contained
within the unit cell. We may then distinguish between the alternatives by calculating
the ratios of the intensities of certain reflections for each possible structure and
comparing these with the ratios that are observed experimentally.
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5.1 Identifying the candidate crystal structures
From the evidence of section 4 it is clear that GaP is face centred cubic. From
measurements of density, it is found that there are 4 formula units of GaP per unit
cell. How many GaP units are there per lattice point?
Deduce the multiplicities of the 420 and 331 powder lines, e.g. by writing down all
the equivalent {420} and {331} planes.
GaP may exist as either the sodium chloride (NaCl) structure type or the sphalerite
(ZnS) structure type. For each structure type, write down the coordinates of the atoms
associated with the lattice point at the origin, i.e. the motif.
Figure 1: NaCl (top) and ZnS (zinc blende) (bottom) structure types for GaP.
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MATERIALS SCIENCE
Course A: Atomic Structure of Materials
AP3
Calculating the relative diffraction intensities for the candidate structures
Write down expressions for the intensities of the permitted hkl reflections for each of
the candidate structures, using the coordinates of atoms associated with the lattice
point at the origin only.
Hence obtain expressions for the intensities of the 420 and 331 reflections and for the
ratio of intensities I(420) / I(331) for each of the two structures.
Work out the ratio of intensities for the NaCl and ZnS (zinc blende) type structures,
using the following values for the atomic scattering factors:
fGa = 18.0, fP = 7.5 for the 420 reflection.
fGa = 18.3, fP = 7.6 for the 331 reflection.

Why are the values for a given atom not the same for both reflections?
Compare your qualitative observation of I(420) / I(331) for GaP with your calculated
estimates of the ratio for the two structures and hence determine the crystal structure
of sulphur-doped gallium phosphide as used for a substrate in opto-electronic devices.
Conclusions and take-home points
The size of the unit cell and the lattice type can be determined from the peak positions
in the diffraction pattern. The positions of atoms within the unit cell can be determined
from the relative peak intensities in the diffraction pattern.
X-ray diffraction is a powerful technique used to measure both the position and
intensity of diffraction peaks. The peak positions are related to the size of the unit cell
by Bragg’s law. The peak intensities are related to both the type and the positions of
atoms in unit cell. Thus by comparison with reference data, crystalline phases can be
identified and, with a suitable calibration factor, the amounts can be quantified.
Please remember to fill in the feedback form with any comments on this practical
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Course A: Atomic Structure of Materials
AP3
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Course A: Atomic Structure of Materials
2014-15
AP4
IntroductiontoOpticalDiffraction
** BEFORE ATTENDING THIS PRACTICAL **
Work through the DoITPoMS TLP entitled “Diffraction and imaging”.
http://www.doitpoms.ac.uk/tlplib/index.php
Aims and Objectives of this Practical


To understand the basic principles behind optical diffraction.
To understand how a lens can be used to form an image from a diffraction
pattern
 To understand the difference between bright field and dark field imaging
 To estimate the data density of a CD and DVD.
1
Introduction to diffraction
Diffraction concerns the spreading of light as it passes through a narrow slit or past
the edge of a small object. Laser light provides a narrow, coherent beam of light
and hence is ideal for use in diffraction experiments. Laser light is diffracted by a
mask (e.g. a grating composed of narrow slits) and the resulting diffraction pattern
is observed on a screen some distance away. The diffraction pattern arises from the
interference of light diffracted through each of the slits in the grating.

s

Figure 1: diffraction from a section of a diffraction grating.
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AP4
The path difference between the diffracted rays in Figure 1 is s sin , where s is the
slit spacing. Hence the condition for constructive interference is:
n  s sin 
(1)
where  is the wavelength of the light and n is an integer. Usually, we deal with just
first order diffraction, where n = 1.
If we assume that the distance from grating to an observation screen, L, is large in
comparison to the diffraction spot spacing, x, then sin   tan   x / L and so, the
relationship between the slit spacing, s, and the diffraction spot spacing, x, is:
s
L
x
(2)
This is a reciprocal relationship, i.e. as the slit spacing, s, is reduced the spacing in the
diffraction pattern, x, increases, as shown in Figure 2 below, and we say that the
diffraction pattern is a representation of the mask in reciprocal space, or is the
reciprocal lattice of the mask.
Figure 2: taken from www.doitpoms.ac.uk/tlplib/diffraction/
The gratings we have been discussing vary in one dimension, and so give a 1D
diffraction pattern, but diffraction gratings are not restricted to one dimension. If two
diffraction gratings are perpendicularly superimposed they create a 2D array of
apertures (see Figure 3). The diffraction pattern corresponding to this is analogous to
the reciprocal lattice of the array, and the separate patterns are repeated to form a 2D
array. In this part of the practical you will investigate both 1D and 2D diffraction
gratings.
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Figure 3: taken from www.doitpoms.ac.uk/tlplib/diffraction/
IMPORTANT HEALTH AND SAFETY INFORMATION
Laser light is very bright. The lasers you will be using are relatively low power, and are
similar in intensity to looking directly at the sun. Never look directly down the laser
beam, and be careful to avoid reflections. Take care not to shine the laser at anyone else.
You are provided with a slide comprised of various diffractive and refractive gratings,
shown below, and a laser pointer. Taking care that the beam does not shine at, or
reflect at, anyone, observe the diffraction patterns arising from the various gratings.
Diffractive Structures
1
2
4
9
5
10
Refractive Structures
3
6
11
7
12
14
15
16
17
18
19
8
13
Figure 4: Slide with gratings
Examine the gratings under the optical microscopes, noting the features of 10-13.
 Using the built-in graticule, can you calculate the spacings of the lines in 12
and 13?
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AP4
Now, set up the laser pointer and the laser bench, as shown in Figure 5 below. The
diffraction gratings can be held on the mounts using the metallic tape. The laser
pointer can be switched on and off by using the clamp to depress the on/off button.
Pointer in clamp
Magnetic mount
Grating with
metallic tape
Screen with paper
Switch
depressed
Figure 5: laser bench setup
Laser bench
You will first investigate the 1D diffraction pattern arising from the gratings labelled
12 and 13. For these two gratings, calculate the slit spacing, s, from the diffraction
patterns using equation (2). The wavelength of the laser light is 660 nm.
 Does this correspond to the spacing you measured under the microscopes?
Now look at the 2D pattern produced by the gratings 10 and 11.
 Given the reciprocal relationship between the mask structure and the
diffraction pattern, can you work out the grating shape necessary for the
pattern and their relative sizes?
 Does this result agree with what you observed under the microscopes?
2
Abbe theory and image formation
By introducing a convex lens to the setup, an image of the grating can be resolved, as
shown in the Figure 6. Note that both the diffraction pattern and the image can be
resolved depending on the position of the screen. The diffraction pattern is seen when
the screen is placed in the back focal plane of the lens and the image of the slits is
formed when the screen distance, v, satisfies:
1 1 1
 
u v f
(3)
with u, v and f defined as below.
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AP4
Figure 6: Abbe theory of imaging using all diffracted spots
In this part of the practical, you will investigate this effect by forming an image of one
of the diffraction gratings.
Some of the diffraction patterns produced by the mask require very complex
diffraction structures and hence are difficult to observe using this simple setup, so it is
recommended you restrict your observation to gratings 9, 10, 11, 12 and 13.
To form an image of the diffraction gratings, the convex lens needs to be introduced
to the laser bench. The spacing on the gratings is very small, so the image needs to be
large to resolve the features of the grating. To achieve this, position the lens one focal
length away from the grating, so that the image distance becomes very large, by
equation (3). The focal length of these lenses is 12.7 cm. The mirrors on the mounts
can be used to extend the image distance, and increase the magnification of the image.
Set up the laser bench as described to form an image of the grating 12. Check that the
diffraction pattern appears in the back focal plane of the lens by putting a sheet of
paper there.
Look at the images formed from the gratings listed. It may be easier to view the image
with a hand lens.
 For the 1D and 2D gratings, can you observe the reciprocal relationship
between slit spacing and diffraction pattern spacing (c.f. equation (2))?
 Describe / sketch the appearance of the images. Do these correlate with what
you observed under the microscope?
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AP4
It is possible to highlight particular information in the object (diffraction grating) by
isolating certain diffraction spots to form the image. In microscopy, this is known as
bright field and dark field imaging. For bright field, the image is formed from light
transmitted directly through the object, so that the clear/transparent areas appear
brighter. For dark field, the image is formed from light diffracted by the object, so that
the surrounding regions appear dark. Bright field images can be formed by selecting
the central spot in a diffraction pattern (i.e. the straight-through beam) and dark field
images by selecting an off-centre spot.
To observe this effect, it is recommended that, using the same set up as before, the
laser is shone on the gap between gratings 12 and 13, so both are illuminated. You
should see an image of the two gratings, with different spacings, and the clear glass
slide between them. Now, place the variable aperture in the back focal plane of the
lens and isolate the central spot. You should observe just the transmitted light.

Describe the bright field image.
Move the aperture so that only one of the dark field spots goes through the aperture
and contributes to the image. The image is now formed just by scattered light. If you
are very careful in the alignments, you may be able to make out different images for
the 2 gratings, as each diffracts light to a different position in the back focal plane,
and can therefore be imaged independently.
 Describe the dark field image.
Now, open or remove the aperture, and observe the complete pattern again.
3
Diffraction from CDs and DVDs
As shown in Figure 7, the information recorded on CDs and DVDs is laid out in
regularly spaced tracks. The spacing of these tracks is of the order of the wavelength
of optical light (in order to achieve high data storage densities) and so the tracks can
act as a diffraction grating. The same reciprocal relationship as discussed above
applies here also between the track spacing and disk capacity (i.e. data density).
Figure 7: Scanning electron microscope images of the surface of either a CD or DVD, shown at
different scales. The results of your measurements will indicate which one.
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AP4
Position the laser pointer in the clamp such that the beam is directed down onto the
surface of the CD as shown in Figure 8. Keep the screen close to the CD and adjust
the incidence angle of the laser beam so that you get the first diffraction maxima on
the screen. By moving the CD around slightly, you should be able to observe a beam
diffracted out to the side as shown below. The pattern you are looking for will have a
large spacing and will be horizontal. There may also be a closely-spaced vertical
pattern which is from the different layers in the disc, not the tracks.
viewing screen
CD
laser
pointer
Figure 8: Experimental setup, with an indication of the conditions leading to the first order diffracted
beam
Measure the angle between the direct beam and the first diffracted beam and estimate
the spacing between the tracks on the CD, using equation (2). The wavelength of the
light is marked on the laser pointer. Repeat for the DVD.
 Is this the result you would expect, based on the relative storage capacities of a
CD and DVD?
 Does figure 7 show the surface of a CD or a DVD?
 What diffraction pattern spacing would you expect for a multi-layer DVD?
Now repeat the experiment using the green laser pointer, if available, and estimate the
wavelength of the green light.
Conclusions and take-home points
The principles underlying optical, X-ray and electron diffraction are very similar,
and are based on the Abbe theory of image formation. Information about the
periodicity of a diffraction grating can be obtained from the diffraction pattern and
an image of the grating can be obtained by focusing this using a lens. Bright and dark
field imaging can be used to examine different spatial features of an object. The
relative storage capacities of CDs and DVDs can be related qualitatively to their
diffraction pattern.
Please remember to fill in the feedback form with any comments on this practical
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