The aim of these worked examples is to help you with the course in general as well as with the question sheets and examples classes.
Question 1
At high temperatures, the low-loss dielectric zirconium titanate, ZrTiO4, has a random arrangement of zirconium and titanium ions on cation sites. Neutron powder diffraction data show that the structure is orthorhombic, with a = 4.804 Å, b = 5.483 Å and c = 5.031 Å. Ion positions are as follows: O : x, y, z; 1/2 – x, 1/2 – y, 1/2 + z; 1/2 + x, 1/2 – y, – z; – x, y, 1/2 – z; – x, – y, – z; 1/2 + x, 1/2 + y, 1/2 – z; 1/2 – x, 1/2 + y, z; x, – y, 1/2 + z. with x = 0.27, y = 0.10 and z = 0.07. Zr, Ti : 0, v, 1/4; 0, – v, 3/4; 1/2, 1/2 + v, 1/4; 1/2, 1/2 – v, 3/4. with v = 0.265.
(a) Draw an accurate plan of 2 × 2 unit cells of this structure projected on (100). Specify the Bravais lattice of this crystal structure.
(b) Determine the nearest neighbour oxygen anions of the oxygen anion at x, y, z. Hence or otherwise show that the oxygen anions form a distorted hexagonal close packed array on (100). Relative to this array of oxygen anions, where are the cations located?
(c) On cooling ZrTiO4 below 1200 °C, a phase transition occurs which is associated with partial ordering of the cations, but otherwise the crystal structure remains essentially unchanged. Stating your reasons, specify whether you think this phase transition is likely to be
(i) reconstructive or (ii) displacive. Solution
(a) A plan of the structure projected on (100) is shown overleaf. The Bravais lattice is P. (b) There are twelve nearest neighbour oxygen anions of the oxygen anion at
0.27, 0.10, 0.07. Six of these anions are at heights x ~ 0.25, three at a height ~ 0.75 and three at a height ~ – 0.25.
A ‘unit cell’ of the distorted hexagonal close packed array on (100) is shown shaded in the diagram overleaf of the crystal structure.
Cations are located in half the octahedral interstices of this close packed hexagonal array of oxygen ions, as in the example shown in the ‘unit cell’.
C6H11 – 2 – C6H11
Plan of the structure of ZrTiO4 projected on (100).
[001] [010] 0.73 0.27 0.77 0.23 0.77 0.23 0.73 0.27 Oxygen ion at x ~ 0.75 Metal ion at x = 0 Oxygen ion at x ~ 0.25 Metal ion at x = 0.5 x Hex y Hex
(c) The phase transition is reconstructive. In order to induce partial ordering of the cations, cations will have to change places, i.e. break bonds and form new ones. This contrasts with what happens in a displacive phase transition where bonds remain unbroken, but changes in local environments occur, e.g. in the normal → ferroelectric phase transition in BaTiO3.
Question 2
ZnO has the wurtzite structure, with u not exactly 1/
8 and dependent on temperature. Experimental
work on ZnO shows that at 300 K, the lattice parameters are a = 3.24992 Å and
(a) Draw an accurate plan of 2 × 2 unit cells of ZnO projected on (001), labelling the positions of the Zn atoms and the O atoms. Mark on the (210) and (120) sets of planes on this diagram. (b) Write down the indices of all the planes related by symmetry to the (100) planes in ZnO. (c) What the distance in Å between the oxygen ion at 1, 0, 0 and the zinc ion at
2/3, 1/3, u ?
(d) What is the interplanar spacing in Å of (i) the (110) planes and (ii) the (110) planes? (e) What is the volume in Å3 of the unit cell of ZnO?
(f) Does ZnO exhibit piezoelectricity?
Solution (a) y
x
Oxygen
Zinc
(210) planes (120) planes – u u u u 1 2+ u 1 2 1 2 1 2 1 2 1 2+ u 1 2+ u 1 2+ u 1 2+ u 1 2+ u 1 2+ u 1 2+ u 1 2+ u (b) (100), (100), (010), (0 10), (110), (110).(c) 1.97453 Å (note that the dimensions of the unit cell have been given to five decimal places, so the Zn-O distance can also be specified to five decimal places).
(d) (i) 2.81451 Å; (ii) 1.62496 Å. (e) 47.62429 Å3.
(f) Yes – the structure is not centrosymmetric, so it will exhibit piezoelectricity. (Piezoelectricity can arise in all non-centrosymmetric crystals unless they have a point group of 432 - see J.F. Nye, Physical Properties of Crystals, O.U.P., 1985, pages 123-124. Zinc oxide has a non- centrosymmetric hexagonal crystal structure.)
C6H11 – 4 – C6H11
Question 3
At room temperature, the lattice parameter of silicon is 0.54306 nm and the lattice parameter of NiSi2 is 0.5406 nm. NiSi2 has the CaF2 structure. When a layer of NiSi2 is deposited on the (111)
plane of single crystal silicon, it can take one of two orientations with respect to the silicon. In orientation A all planes and directions are coincident in the two crystals. In orientation B the two crystals have the (111) plane and [111] direction in common, but the [110] direction in the silicon crystal corresponds to a [114] direction in the NiSi2 layer.
Draw an accurate stereogram centred on (001) including the [111], [110] and [114] directions of silicon. By constructing a suitable small circle show that if the stereogram is rotated by 180° about [111] the new position of the [110] pole coincides with the original position of the [114] pole. Assuming that the NiSi2 layer nucleates by the formation of a close-packed monolayer of Ni atoms
on the Si (111) plane, suggest reasons for the occurrence of the two growth orientations A and B.
Solution
A suitable stereogram is shown below.
[110] : [111] = cos-1 {2/ 6} = [111] : [114] = 35.26°. Furthermore, [110], [111] and [114] are
coplanar, since by inspection they all lie in the (110) plane.
010 100 [110] 001 [111] [114]
Accurate stereogram centred on (001) showing selected poles and a small circle of angular radius 35.26˚ drawn about [111]
From the calculations and the construction of a small circle of angular radius 35.26° about [111], it is evident that a rotation of 180° about the vector [111] brings the vector [110] into the position where [114] used to be, and visa-versa.
The stacking sequence of the (111) planes in silicon can be described in the form ...ABCABCABC..., since silicon has the cubic F Bravais lattice.
However, A, B and C in this case do not represent close packed planes, but rather pairs of non- close packed planes, because of the motif of Si – an atom at 0,0,0 and an atom at 1/
4, 1/4, 1/4 (see
Data Book).
The stacking sequence of (111) planes of NiSi2 can likewise be described in the form
..ABCABCABC.., since it too has a cubic F Bravais lattice, but again because of the motif, A, B and C do not represent close packed planes, but rather three sets of planes (2 planes each with Si atoms in between planes containing only Ni atoms).
Suppose the Si (111) planes terminate in a C layer. Then we can place either an A or a B layer of the NiSi2 stacking sequence on top of this Si C layer (if we assume that a NiSi2 C would be
unfavourable as it would be directly on top of C). The four possible sequences are therefore:
(i) ..ABCABCABCABCABCABCABCABCABC.. (ii) ..ABCABCABCABCBCABCABCABCABCA.. (iii) ..ABCABCABCABCACBACBACBACBACB.. (iv) ..ABCABCABCABCBACBACBACBACBAC..
Of these four sequences, (i) and (ii) reproduce the same sequence in the NiSi2 as in the Si, whereas
the NiSi2 sequences in (iii) and (iv) are in effect mirror images of the ..ABC.. stacking sequence in
the Si.
Thus, two distinct growth orientations arise depending on the exact stacking sequence chosen by the NiSi2 layer. These are twinned with respect to one another by a twinning operation of 180° about
the (111) plane normal. Directions in one twin orientation are thus reflected in the (111) plane to their positions in the other orientation, a mirror operation which is equivalent to a 180° rotation about the [111] direction followed by an inversion (2 = m).
Question 4
GaAs has the sphalerite structure given in the Data Book.
Draw sketch stereograms centred on 001 of the environments of (a) Ga atoms around As atoms,
and (b) As atoms around Ga atoms in sphalerite. Specify the nature of the tetrad axis parallel to
[001] in GaAs and (ii) the orientation of the mirror planes parallel to [001]. What is the point group of GaAs?
If it were possible to grow single crystals of GaAs what would be the shapes of the {100}, {110} and {111} forms?
C6H11 – 6 – C6H11
Solution
The environments of (a) Ga atoms around As atoms, and (b) As atoms around Ga atoms in
sphalerite are both tetrahedral, with the tetrahedra for (a) and (b) inverted with respect to one
another.
Suitable sketch stereograms centred on (001), with the x–axis down the page and the y–axis to the
right and with respect to sphalerite are drawn below:
As atoms around Ga Ga atoms around As
The environments of As atoms around Ga atoms are identical for all gallium atoms (since all the lattice points are able to coincide with the gallium atoms).
For the same reason, the environments of Ga atoms around As atoms are necessarily identical for all arsenic atoms.
The tetrad axis parallel to [001] is an inverse tetrad. (110) and (110) are the mirror planes parallel to [001].
The point group of GaAs is 43m.
Shapes of the various forms: {100}: cube;
{110}: rhombic dodecahedron (a regular figure with twelve rhombus faces);
{111}: since there are 2 symmetrically distinct sets of planes of the form {111}, the shape is a tetrahedron. There are two possible tetrahedra for the {111} form inverted with respect to one another.
Question 5
Rutile is tetragonal with a = 0.459 nm and c = 0.296 nm. Further details about the crystal structure
of rutile are given in the Data Book. Draw a plan of this structure on (001). Hence or otherwise confirm the Bravais lattice. Determine (i) the nature of the screw tetrad along [001] in the structure and (ii) the nature of the screw axes along <100> and (iii) the glide planes parallel to {100}. Confirm that {110} planes are mirror planes and that there are diad axes parallel to <110> directions. What would be the most complete description for the space group of rutile?
Solution [100] [010] Ti4+ O2-
3 A
KEY 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 mirror planes [110] 1 4 1 4 1 4 1 4 1 2C6H11 – 8 – C6H11 Note that it not necessary to know the details of the lattice parameters to answer this question – the fact that the structure is tetragonal is sufficient.
A suitable plan to scale of 2 × 2 unit cells is shown on the previous page, with some (but not all!) symmetry elements present.
(i) There are 42 screw axes along [001] intersecting the x-y plane at 1/2, 0, 0 and equivalent positions.
(ii) There are 21 screw axes along <100>, such as the one along [100] intersecting the y-z plane at
0, 1/4, 1/4.
(iii) The glide planes parallel to {100} are diagonal glide planes (denoted by n) intersecting the
[100] direction at x = 1/4 and 3/4.
The most complete description for the space group of rutile would be P 42/m 21/n 2/m. The
conventional description is P42/mnm, space group no. 136.
Question 6
Cadmium has the h.c.p. crystal structure with a = 2.98 Å, c = 5.62 Å. It slips on {0001} <1120>.
Figure 1 is a stereographic projection with the z axis at the centre mark and the +y axis as shown.
Using a Wulff net, mark on the directions <0001>, <1010>, <1120> and <1123> which lie in the
northern hemisphere.
A single crystal of cadmium grown in the form of a long cylinder of length 50 mm with a diameter of 3 mm is stressed parallel to its length. The axis T of the cylinder is marked on the stereographic
projection.
(i) Using a Wulff net, measure the angle φ between T and [0001] and the acute angles λ1, λ2 and
λ3 between T and the possible slip directions.
(ii) Suppose that the crystal cylinder is subjected to an increasing tensile stress parallel to the axis
T. From a consideration of the Schmid factor for each of the three possible slip systems,
decide which will be the first one to operate. Taking the critically resolved shear stress for cadmium as 0.15 MPa, calculate the tensile stress and the actual load when slip begins. (g =
9.81 ms–2)
(iii) Draw on your stereographic projection the way in which the tensile axis changes direction as the crystal slips, i.e. draw the path of its pole.
(iv) From the relevant equations given in the Data Book, calculate the new values of φ and λ when the crystal has extended to 75 mm. Plot on the stereogram the new position of the axis of the cylinder.
+ y z
T
C6H11 – 10 – C6H11
Solution
Directions on the primitive are straightforward to plot. The stereogram below shows the directions <0001>, <1010>, <1120> and <1123> directions which plot as poles in the northern hemisphere.
The angle [1123] : [0001] is given by θ = tan–1 (a/c) = 27.9°. This enables the <1123> directions to
be plotted.
(i) From the stereogram, φ = 52°, λ1 = 54°, λ2 = 41° and λ3 = 80°, where λ1 = T : [2110], λ2 = T : [1120] and λ3 = T : [1210].
(ii) The resolved shear stress τ = σ cos φ cos λ. At the onset of slip, τ = τcrit = constant, and so slip occurs first on the system with highest Schmid factor, i.e. the highest value of cos φ cos λ. Since φ is the same for all three slip systems, it follows that slip occurs first on the system for which cos λ is greatest, i.e. for the slip system [1120](0001) for which λ2 = T : [1120] = 41°. τcrit = 0.15 MPa, so therefore from the formula τ = σ cos φ cos λ, it follows that when φ = 52°
Now σ = F/A where F is the applied force and A the area over which the force acts. Thus the
mass m required to cause slip is given by
m = σ A g =
0.32 × 106×π 1.5 × 10–3 2
9.81 kg = 0.230 kg = 230 gm
(iii) The path of the tensile axis as a function of increasing crystal slip is drawn on the stereogram as the thick black line between T and [1120].
(iv) The relevant equations are
l1 l0 = cos φ0 cos φ1 = sin λ0 sin λ1
where the ‘0’ subscript denote the original values of φ and λ and the length l, and where the
‘1’ subscript represents their values after slipping has occurred.
Here, l1/l0 = 1.5 and so it follows that φ1= 66° and λ1= 26°. The new position of the axis of the cylinder after the crystal has been extended to 75 mm is denoted by TT on the stereogram.
Question 7
In higher dimensional projectional geometry used in quasicrystallography, two orthogonal low dimensional spaces V (of n dimensions) and W (of m dimensions)are defined in a suitable higher dimensional space H of m + n dimensions in which the unit vectors spanning the space are
orthonormal (i.e. of unit length and orthogonal to one another). Tilings can then be generated in V
space by projecting the hypercube of H centred at the origin onto W to form the corresponding projected shape R in m-dimensional space and then selecting only those points on the integer
lattice in H which project into R to form a vertex of the tiling in V space by its projection onto V. The principle of this can be demonstrated graphically in 2D. V and W are then perpendicular lines. The 2D hypercube centred at the origin is a square with vertices at (1/2, 1/2), (– 1/2, 1/2),
(1/2, – 1/2) and (– 1/2, – 1/2).
Suitable templates then have the 2D integer lattice marked on with a suitable origin and a square centred at the origin. Using templates with a 10 mm square mesh on which dots are placed to represent the 2D integer lattice, examine the periodicity of the pattern generated in V if:
(i) W is the rational direction [230] (so that V is the direction [320]); (ii) W is the irrational direction [1τ0], where τ = 5 + 1
C6H11 – 12 – C6H11
Solution
(i) If W is the rational direction [230] (so that V is the direction [320]), then the repeat sequence along V is LLSLS, where ‘L’ is a long vector and ‘S’ a short vector (see page 13). This is a periodic sequence.
(ii) If W is the irrational direction [1τ0], where τ = 5 + 1
2 , then the sequence along V is not
periodic (see page 14).
Thus, if W is rational, the sequence generated along V is periodic, whereas if W is irrational, the sequence generated is not, although there are ‘patches’ arbitrarily large in size which can be identified in the sequence, e.g. in the example on page 14, SLSLL occurs, but not periodically. In fact for this particular choice of W, there cannot be SS or LLL in the sequence along V.
The sequence along V in (ii) is very interesting – in reciprocal space there are peaks of intensity from this sequence, but the peaks are spaced aperiodically. A small computer program can be
W V L S S L L S L S L x y L S L S L L S L S L L S L S L L S L S L L S L
C6H11 – 14 – C6H11 W V L S x y S L L S L S L L S L L S L S L L S S L L L S L L S L L L S S L L