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High Resolution X-ray Diffraction

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Fortgeschrittenpraktikum

High Resolution X-Ray Diraction

März 2009

Walter Schottky InstitutZentralinstitut der Technischen Universität Münchenfür physikalische Grundlagen der Halbleiterelektronik

Am Coulombwall 385748 Garching

2 CONTENTS

Contents

1 Crystals 3

1.1 Ideal crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Miller indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Real crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Theory of X-ray diraction 9

2.1 Kinematic theory of X-ray diraction . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 X-ray diraction at periodic structures . . . . . . . . . . . . . . . . . . . . . . 13

2.4 Bragg equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5 Ewald sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.6 Structure factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Experimental setup 17

4 Characterization of real crystals by HRXRD 19

4.1 2θ-Ω-Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.2 Ω-Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3 Reciprocal space map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.4 ϕ-Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5 Exercises (you should work on before conducting the experiment) 27

6 Experimental procedure 27

6.1 Mosaicity of dierent ZnO-samples . . . . . . . . . . . . . . . . . . . . . . . . 27

6.2 Rocking curve of the 101-reex . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.3 Epitaxial relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6.4 Lattice constants of ZnO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6.5 Determination of the Mg-content x of a MgxZn1−xO sample . . . . . . . . . 31

7 Report 32

A ZnO reexes 34

B Sapphire reexs 35

3

C Measurement report for rocking-curves and 2θ-Ω-scans 36

D Measurement report for reciprocal space map 38

E X-ray signal of the sample holder 39

Bibliography 39

1 Crystals

At the end of the 20th century the micro structure of solids was still under debate. Max von

Laue had the idea to clarify this issue by using X-rays which were discovered some 20 yearsbefore by Wilhelm Conrad Röntgen. For this purpose, von Laue worked out a theory forX-ray diraction at three-dimensional crystals. With the observation of diraction patterns byirradiating solids with X-rays, his coworkers Walther Friedrich and Paul Knipping veriedboth the periodic space structure of most solids and the wave character of X-rays.In this laboratory exercise we will address the structural characterization of solids by means ofHigh Resolution X-Ray Diraction (HRXRD). In general, this technique is not used to determinethe crystal structure, but to investigate deviations from an ideal crystal which can be inducedi. e. by defects, mosaicity or strain.We rst give an overview on the basic concepts of how to describe an ideal crystal. Subsequently,we treat irregularities and defects in real crystals. We then give an introduction in the kinematictheory of X-ray diraction. Finally, the experimental setup is presented and the characterizationof real crystals by HRXRD is discussed.

1.1 Ideal crystals

1.1.1 Denition

An ideal crystal is an innite, periodic array of a structural element. The structural elementwhich consists of an atom or a group of (identical or dierent) atoms is called basis. A crystal

Figure 1: Irradiation of a thin slab of a crystal with "white" X-rays produces a distinct dirac-

tion pattern. Picture taken at the U-Bahn station in Garching (Forschungszentrum).

4 1 CRYSTALS

(b)a

b

(c)½(a+b)

a

b

(a)a´

Figure 2: (a) Space lattice with two dierent choices of unit cells (b) basis (c) resulting

two-dimensional crystal.

can be built up by repeatedly placing the basis at well dened lattice sites which constitute theso-called space lattice. The arrangement of the atoms looks identical viewed from every pointof the space lattice. The lattice points can be reached by a translation

r = n1 a1 + n2 a2 + n3 a3 (1)

where all ni are integer. The lattice vectors ai have to be linearly independent. They span theunit cell, whose volume Vuc is given by the scalar triple product

Vuc = a1 · (a2 × a3) (2)

The unit cell with the smallest volume possible is called primitive and it is spanned by theprimitive lattice vectors. The number of atoms that its associated basis contains is as smallas possible. For a given basis there is an innite number of possible choices for the unit cell(gure 2). Yet the volume of all possible unit cells is always the same. In order to take intoaccount other symmetries of the crystal than the translational symmetry, like rotational or mirrorsymmetry, it is often reasonable to choose a non-primitive unit cell, or so called conventional

unit cell. However, this results in a more complex structure of the basis of the crystal.

1.1.2 Miller indices

Any plane containing lattice points is called a lattice plane. In an ideal crystal there is alwaysan innite number of parallel lattice planes. As we will see later, it is convenient to label a setof parallel lattice planes according to the following algorithm:

1. Determine the intersection points a, b and c of any of the lattice planes with the coordinateaxes in units of the lattice constants, i.e. in multiples of the lengths of the lattice vectorai.

2. Take the reciprocal values h′ = 1a , k

′ = 1b and l′ = 1

c .

3. Multiply these values with the smallest number m possible so that h = m ·h′, k = m · k′and l = m · l′ are integer.

1.1 Ideal crystals 5

a1

a3

c

a2

O

Zn

Figure 3: Hexagonal unit cell of the Wurtzite lattice. Its primitive unit cell is limited by the

lattice vectors a1, a2 and c as well as the the dotted lines and it contains four atoms.

The thus obtained triple (hkl) is known as Miller indices. For example, if one lattice planeintersects the axes at a = −2, b = 1 and c = 4, the set of parallel planes it belongs to islabeled by the Miller indices (241). Negative values are marked with a bar over the index. Ifthe lattice plane is parallel to an axis, the intersection point is∞ and therefore the correspondingindex is equal to zero. To indicate that the triple (hkl) labels all parallel lattice planes, it isenclosed in parentheses. To designate crystallographicly equivalent planes, the Miller indicesare enclosed in curly braces hkl. For instance lattice planes limiting the unit cell in a cubiclattice, i. e. the (100), (010), (001), (100), (010) and (001) planes, can be subsumed bywriting 100.Similarly, directions in the lattice are indicated by a triple of integers [uvw] enclosed in squarebrackets. u, v and w denote the smallest possible integer components of a vector R =ua + vb + wc pointing along the designated direction.In this laboratory exercise we will investigate thin zinc oxide (ZnO) and zinc magnesium oxide(ZnMgO) lms which have crystallized in a hexagonal Wurtzite lattice. The zinc and oxygenatoms form two interpenetrating, hexagonal close-packed sublattices which are displaced alongthe c-direction of their hexagonal unit cell (gure 3). In this structure the zinc (oxygen) atomsare tetrahedrally coordinated which means that they are situated at the center of a tetrahedronwhich is formed by oxygen (zinc) atoms. The atoms are usually grouped in bilayers, whichconsist of two adjacent (00.1) lattice planes (the signication of the dot in this notation isexplained below). The staking order of the bilayers is ABABAB. . . (gure 4).In gure 3 the hexagonal unit cell of the Wurtzite lattice is shown. It is spanned by the latticevectors a1 and a2 lying in the basal plane as well as c being perpendicular hereon. The basisof the Wurtzite lattice consists of four atoms. By using the above described methods to labeldirections and planes, in hexagonal crystals crystallographicly equivalent directions and planescan be designated by a dierent type of triple (hkl) or [hkl]. For example the crystallographicly

6 1 CRYSTALS

bilayer

[0001]A

B

A

B

Figure 4: Sideview of a wurtzite crystal showing that its bilayers are stacked in the sequence

ABABAB. . .

equivalent planes limiting the unit cell parallel to the c-direction can be labeled (11.0) and (10.0)(gure 5 (a)). Therefore it is common to use a coordinate system with four axes and a quadrupleof integers for indexing. This takes into account that there are three equivalent symmetry axesa1, a2 and a3 in the plane perpendicular to the principal axis along the c-direction. Theindices (hkil) are then determined analogously to the already presented algorithm. Therebythe relation

i = −(h+ k) (3)

is valid, because a1, a2 and a3 are not linearly independent as a3 = −(a1 + a2). Fordesignating directions, one has to assure to pick that linear combination of lattice vectors thatsatises equation (3). For instance, the direction along the −a2 vector is not labeled [0100]but [1210] (gure 5 (b)). If one chooses to use only three indices in a hexagonal lattice, oneplaces usually a point, that shall represent the omitted fourth index, between the second andthird index [hk.l]. Equation (3) has then not to be obeyed.Using four indices for indexing has the advantage that crystallographicly equivalent directionsand planes are obtained by a simple cyclic permutation of the indices.

1.2 Real crystals

In real crystals the requirement to minimize the free energy F = U − TS induces deviationsfrom the ideal crystal structure. Already small concentrations of defects can drastically inuencethe properties of crystals. For example the electrical conductivity of a semiconductor can besignicantly increased by the incorporation of a small amount of extrinsic impurities which isknown as doping. Therefore it is important to investigate the nature and density of defects. Forthis purpose, HRXRD is a suitable method which has the advantages that it does not damagethe samples and that no complex sample preparation is necessary.In this laboratory exercise we investigate heteroepitaxially grown thin lms. This means thatthe underlying substrate and the epitaxial layer are dierent materials with typically dierentlattice constants and perhaps even a dierent crystal symmetry. To quantify the dierence in

1.2 Real crystals 7

a3a3a3a3

a1

a2

c

(1100)or

(11.0)

(1010)or

(10.0)

(0001)or

(00.1)

a3a3

a1

a2

c

(1100)or

(11.0)

(1010)or

(10.0)

(0001)or

(00.1)

a3

a1

a2

c

(1120)or

(11.0)

[1120]or

[11.0]

[0110]or

[12.0]

[1210]or

[01.0]

(a) (b)

Figure 5: Indexing of directions and planes in a hexagonal lattice.

lattice constant the so-called lattice mismatch ∆a/a is dened as follows

∆aa

=asubstrate − arelaxfilm

asubstrate(4)

where asubstrate and arelaxfilm refer to the respective relaxed lattice constants (see gure 6) in theplane parallel to the interface. For a small lattice mismatch the epitaxial lm can grow pseu-domorphical, i. e. it adapts the in-plane lattice constant of the substrate thereby accumulatingbiaxial strain (gure 6 (b)).When the lm thickness exceeds a certain critical value, the built-up strain energy is released bythe formation of one-dimensional defects, called dislocations, perpendicular to the interface. Inthe vicinity of a dislocation the crystal lattice is distorted and strain is accumulated, that decaysrather slowly as one moves away from the dislocation. By moving along a 360°-loop around adislocation, one does not arrive at one's starting point. The dierence between starting and endpoint denes the Burgers vector b. In general, two types of dislocations can be distinguished,namely screw- and edge-type dislocations (gure 7). For screw-type dislocations the Burgersvector b is oriented along the dislocation, whereas for edge-type dislocation it is perpendicular.Dislocations have the tendency to arrange themselves in the stablest conguration possible.Therefore, for example edge-type dislocations prefer to group themselves together as shown ingure 8 in order to minimize the strain energy. Such an agglomeration of dislocations is knownas grain boundary. This type of two-dimensional defect can be seen as a boundary betweentwo monocrystalline regions of a solid, called crystallites, which are twisted by an angle δ withrespect to one another. In a similar way as edge-type dislocations result in a twist, screw-typedislocations can give rise to a tilt of the crystallites.In this laboratory exercise we investigate ZnO samples heteroepitaxially grown on c-plane sap-phire substrates. Sapphire can be seen to have a quasi-hexagonal lattice, that is to say theO-sublattice exhibits sixfold symmetry whereas the Al-sublattice has only a threefold symme-try axis. The term c-plane means that the sapphire is cut parallel to its (0001) plane. Thegrowth direction of ZnO on this substrate surface is the [0001] direction. In order to minimizethe lattice mismatch, the ZnO lattice is rotated by 30° with respect to the sapphire substrate

8 1 CRYSTALS

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=

(d)

Figure 6: (a) Relaxed unit cells of lm and substrate. Schematic representation of (b) a

pseudomophical (c) a partially relaxed and (d) a fully relaxed heteroepitaxial lm.

b

core

b

(b)(a)

Figure 7: (a) screw-type dislocation for which the Burgers vector b is parallel to the dislocation.

(b) edge-type dislocation with Burgers vector b perpendicular to the dislocation.

9

d

b

D

Figure 8: Grain boundary.

and thus forms a coincidence lattice therewith. Figure 9 illustrates the epitaxial relationshipbetween the ZnO lm and the sapphire substrate.But even then the lattice mismatch ∆a/a = -18.4% is quite substantial. As a consequence,the critical layer thickness for pseudomorphical growth is less then one monolayer of ZnO.Therefore the formation of dislocations, through which strain is released, starts with the verybeginning of the lm growth. This results in a columnar growth mode (gure 10) with mostof the dislocations perpendicular to the interface whereas there are few dislocations parallel tothe interfacial plane. Hence, the vertical crystallite size is only limited by the layer thickness.The individual columnar crystallites are twisted and tilted with respect to one another. As wewill see later, this causes a broadening of specic X-ray reexes.However, the lattice mismatch between ZnO- and Zn1−xMgxO-layers is relatively small whichallows pseudomorphical growth to take place. But the uppermost layer can also be partiallyrelaxed or completely relaxed depending on the layer thickness and the magnesium content xin the Zn1−xMgxO layer. For a quantitative analysis, the degree of relaxation r

r =afilm − asubstratearelaxfilm − asubstrate

(5)

is dened, where afilm denotes the actually measured lattice constant of a thin lm and arelaxfilm

the totally relaxed lattice constant. r is equal to 1 for fully relaxed lms and equal to 0 forpseudomorphical growth.

2 Theory of X-ray diraction

In this section we discuss the theory of X-ray diraction. Though the dynamic theory in whichthe Maxwell equations are solved for a medium with a periodic and complex dielectric functionis more accurate, it is perfectly sucient to treat diraction at real crystals with a mosaic

10 2 THEORY OF X-RAY DIFFRACTION

ZnO [1100]Al2O 3 [1210]

ZnO [1210]Al2O 3 [0110]

OZn

Figure 9: (a) Formation of a coincidence lattice of the ZnO lm and the sapphire substrate by

a 30° rotation. For a clearer view only the sapphire O-sublattice and the ZnO Zn-sublattice is

depicted. The arrows indicate the distances which are relevant for the calculation of the lattice

mismatch.

c1c2

substrate(a) (b)

Figure 10: (a) Schematic side view of an epitaxial thin lm showing the tilt of the crystallites.

(b) Top view illustrating the twist.

2.1 Kinematic theory of X-ray diraction 11

S

r

R´-r

R

P

D

Figure 11: Schematic representation of the scattering process. D marks the position of the

detector for which R′ r is valid.

structure within the less complex kinematic theory. The dynamic theory has to be only used fordiraction at ideal crystals. Since all the samples investigated in this laboratory exercise exhibita mosaic structure, we constrain the discussion to the kinematic theory of X-ray diraction.

2.1 Kinematic theory of X-ray diraction

The interaction between an electromagnetic wave and an atom can be described with an os-cillator model. For large distances R between X-ray source and sample, the incident wave canbe approximated by a plane wave whose electric eld Ein is given by (gure 11)

Ein = E0 · ei(kin·(R+r)−ω0t) (6)

It induces harmonic oscillations of the shell electrons of an atom at point P and thus theemission of a spherical electromagnetic wave (Hertz dipole). This process is known as Thomsonscattering. Given the fact that this is an elastic process, the scattered wave exhibits the samefrequency and the same norm of the wave vector as the incident wave

|kin| = |kout| = |k| =2πλ

(7)

where λ denotes the wave length of the X-rays. The amplitude of the scattered wave reads

Eout = f Eineikout·|R

′−r|

|R′ − r|(8)

Herein f denotes the scattering amplitude which depends on the type of atom and the frequencyof the incident wave. At large distances from the scattering atom (R′ r) again the scattered

12 2 THEORY OF X-RAY DIFFRACTION

wave can be approximated by a plane wave and thus by inserting equation (6) equation (8) canbe rewritten

Eout = f E0 · ei(kin·(R+r)−ω0t) eikout·R′ e−ikout·r

R′

= fE′0R′

ei(kin−kout)·r (9)

withE′0 = E0 · ei(kin·R+kout·R′)e−iω0t

The deduced expression (9) is valid for the scattering of a single atom. We now address thescattering of X-rays by an entire crystal. It is reasonable for the interaction between X-raysand solids to neglect multiple scatter processes (Born approximation). As a consequence, thescattered amplitude is proportional to the electron density n(r) of the crystal. Within theFrauenhofer approximation (R′ r), all wave vectors kout are parallel irrespective of theposition P of the scattering atom. The scattered amplitude is then obtained by integratingover the whole crystal volume V

Eout ∝E′0R′

∫V

n(r) ei(kin−kout)·r dV =E′0R′

∫V

n(r) e−iq·r dV (10)

where we have dened the scattering vector q

q = kout − kin (11)

Equation (10) shows that the amplitude of the scattered wave is proportional to the Fouriertransformation of the electron density. By performing diraction experiments one can not detectthe amplitude but only the intensity I of the scattered wave, thus loosing its phase information

I ∝ |E0|2

R′2

∣∣∣∣∣∣∫V

n(r) e−iq·r dV

∣∣∣∣∣∣2

(12)

Therefore the electron density can not be simply obtained by an inverse Fourier transformationof the obtained diraction pattern. However, equation (12) states the important result that theobserved intensity is proportional to the modulus of the Fourier transformation of the scatteringcrystal lattice.

2.2 Reciprocal lattice

Until now we made no use of the periodicity of the crystal. Yet, for a crystal the electrondensity n(r) has to be invariant under translations which constitute linear combinations oflattice vectors T = ua1 + v a2 + w a3 with u, v, w integer and ai the fundamental latticevectors. Hence,

n(r + T) = n(r) (13)

Periodic functions that satisfy equation (13) can be expanded into a Fourier series. In onedimension the Fourier expansion reads

n(x) =∑m

nm ei(m2π/a)x (14)

2.3 X-ray diraction at periodic structures 13

The validity of equation (13) for a displacement of an arbitrary lattice vector Tu = u a can beeasily veried. Likewise, in three dimensions the Fourier expansion is dened as

n(r) =∑G

nG eiG·r (15)

In order to satisfy equation (13)

eiG·T != 1 ∀ T (16)

This involvesG ·T = 2πn ∀ T ∧ n integer (17)

A suitable basis to construct vectors

G = hb1 + kb2 + lb3 h, k, l integer (18)

that fulll equation (17) is given by

b1 = 2πa2 × a3

a1 · (a2 × a3)b2 = 2π

a3 × a1

a1 · (a2 × a3)b3 = 2π

a1 × a2

a1 · (a2 × a3)(19)

The vectors bi and their linear combinations are referred to as reciprocal lattice vectors, becausetheir dimension ism−1 and their length is inversely proportional to the length of the correspond-ing real lattice vectors. They span the so-called reciprocal lattice. It follows, that with everycrystal structure there are two lattices associated, namely its space lattice in real space and itsreciprocal lattice in Fourier space. The reciprocal lattice exhibits the same symmetries as thereal crystal lattice. As both lattices are directly linked with each other via equation (19), bydetermining properties of the reciprocal lattice one easily obtains the same properties of thereal lattice. Also, by rotating or translating the real lattice the reciprocal lattice is rotated ortranslated.One readily veries

ai · bj = 2π δij (20)

and consequently all reciprocal lattice vectors constructed according to (18) satisfy equa-tion (17).There is an important relation between the lattice planes of the crystal lattice and the reciprocallattice vectors [2]: The reciprocal lattice vector Ghkl = hb1 + kb2 + lb3 is perpendicular tothe lattice planes with Miller indices (hkl) and the distance dhkl between two such adjacentplanes is given by

dhkl =2π|Ghkl|

(21)

For a hexagonal lattice with lattice constants a and c this yields

dhkl =a√

43(h2 + k2 + hk) + (ac )2 l2

(22)

2.3 X-ray diraction at periodic structures

We now insert the Fourier expansion of the electron density (15) into equation (12)

I ∝ |E0|2

R′2

∣∣∣∣∣∣∑G

nG

∫V

ei(G−q)·r dV

∣∣∣∣∣∣2

(23)

14 2 THEORY OF X-RAY DIFFRACTION

For a macroscopic crystal, whose side lengths lx, ly, and lz are typically 107 − 108 times thelattice constants, the integral in equation (23) shows a δ-like behaviour.

∫V

ei(G−q)·r dV =

lx2∫

−lx2

ei(G1−q1)·x dx

·

ly2∫

−ly2

ei(G2−q2)·y dy

·

lz2∫

−lz2

ei(G3−q3)·z dz

=

(sin(lx

12(G1 − q1)

)12(G1 − q1)

(sin(ly

12(G2 − q2)

)12(G2 − q2)

(sin(lz

12(G3 − q3)

)12(G3 − q3)

)

=lxlylz = V for q = G≈ 0 otherwise

for lx, ly, lz large (24)

This can be interpreted the way that only if the Laue condition

q = G (25)

is fullled, which states that only if the scattering vector is equal to a reciprocal lattice vector,a X-ray reex can be observed. Because then the scattered X-rays interfere constructively alongthe direction of kout, that is to say the phase factors of the waves scattered at dierent latticepoints of the crystal dier only by a factor e2πin (n integer) along kout. If the phase factorhas a slightly dierent value than e2πin, the contribution from the all lattice points average tozero very eectively due to the high number of scatterers. Hence, for constructive interferencerelation (23) yields for the intensity of the observed X-ray reex

I ∝ |E0|2

R′2|nGq |2 V 2 (26)

Equation (26) expresses the important result that for coherent X-ray scattering the scatteredintensity is proportional to V 2 and thus to N2, where N denotes the number of lattice points.

2.4 Bragg equation

We now try to give an intuitively clearer interpretation of the diraction process. By insertingthe diraction condition (25) into equation (21) and taking into account that G = nGhkl (ninteger) one nds

dhkl =2πn|q|

=2πn

2 sin θ 2πλ

Herein 2θ represents the angle between incident and scattered wave (gure 12(a)) and n thediraction order. By rearranging this result, the famous Bragg equation can be obtained

2 dhkl sin θ = nλ (27)

The Bragg equation (27) has a simple interpretation (gure 12(b)): The lattice planes (hkl)partially reect the incident wave. The diraction condition then amounts to the requirementthat the path dierence for waves reected by adjacent lattice planes has to be an integermultiple of the wave length λ.

2.5 Ewald sphere 15

kin

kout

q

(hkl)q 2q

(a) (b)

dhkl

d sinqhkl

q q

q

Figure 12: (a) The so-called scattering triangle. (b) Illustration of the Bragg equation.

q∥

q ⊥

−4

−2

0 2

4

q

(205)

kin

kout

Figure 13: Ewald sphere

2.5 Ewald sphere

In the last section we tried to get a deeper insight in the diraction process by interpreting it ina more intuitively way. We saw that the Bragg equation relates every X-ray reex that can beobserved with a set of parallel lattice planes (hkl). The Laue condition, however, assigns everyX-ray reex to a reciprocal lattice point which make it convenient to label the reexes with theindices of their corresponding reciprocal lattice points HKL. This is illustrated in gure 13.In the upper part, the (H0L) plane of reciprocal space of a crystal with Wurtzite structure isdisplayed. According to equation (11) and gure 12 (a), the maximum length of the scatteringvector qmax in the case of backscattering (θ = 90) is given by

qmax = 2|k| = 4πλ

(28)

Whenever the scattering vector is equal to a reciprocal lattice vector, a X-ray reex is observed,as shown exemplarily for the (205) reex. As a consequence, all reexes which for a givenwavelength λ are accessible for diraction experiments are situated within the hemisphere withradius equal to 4π

λ .In the lower part of gure 13, the lattice planes are depicted which scatter the incident X-ray

16 2 THEORY OF X-RAY DIFFRACTION

lattice point R

atom a

r

dV

ra

Figure 14: Denition of rα and ρ. rα points to the center of an atom α of the unit cell, ρ to

a point within that atom.

beam according to the Bragg interpretation. Ω designates the angle between the crystal surfaceand the incident X-ray beam, θ the angle between scattering lattice planes and incident X-raybeam and τ = Ω − θ is the oset angle between crystal surface and scattering planes. Thereexes on the [001] axis, also labeled as q⊥, are called symmetrical because for them Ω = θ.All the other reexes are called asymmetrical. The components of the scattering vector parallelq|| and perpendicular q⊥ to the crystal surface expressed as a function of Ω and 2θ read

q|| =2πλ

(cos(2θ − Ω)− cos(Ω)) (29)

q⊥ =2πλ

(sin(2θ − Ω) + sin(Ω)) (30)

2.6 Structure factor

In gure 13, the 00L reexes with L odd are represented by open circles, because these reexesare forbidden, i. e. no scattered intensity is observed. In order to understand this, we now alsohave to include the basis of the crystal lattice into our considerations.More generally, to be able to predict the intensity for dierent reexes, we have to evaluate theFourier coecient nGq in equation (26). nGq is related to n(r) by

nGq =1Vuc

∫Vuc

n(r) e−iGq·r dV (31)

which can be easily veried by inserting the Fourier expansion (15) of n(r) into (31). Theintegration herein extends over the Volume of the whole unit cell Vuc. In case the nuclei of theatoms are not too light the principal contribution to the scattered X-ray intensity arises fromthe core electrons whereas the delocalized valence electrons can be neglected. Therefore n(r)can be expressed as the sum over the electron densities nα(ρ) of the various atoms of the unitcell

n(r) =∑α

nα(ρ) δ(r− rα) (32)

where ρ denotes the distance form the center of a given atom α and rα the position of anatom α with respect to the origin of the unit cell (gure 14). With this equation (31) can be

17

rewritten

nGq =1Vuc

∑α

e−iGqrα

∫Vα

nα(ρ) e−iGq·ρ dV

︸ ︷︷ ︸fα(Gq)

(33)

Now we integrate only over the volume Vα of a single atom, multiply the result with theappropriate phase factor and sum up over all the atoms of the unit cell. The atomic scatteringfactor fα(Gq) dened in equation (33) can be seen as the Fourier transformation of the atomicelectron density. If the electrons were point charges at the atomic centers rα, fα(Gq) = 1independent of Gq. But since the electron density in general stretches over some Ångstrømsaround rα, fα(Gq) is not constant but decreases for higher indexed reexes.The Fourier coecient nGq is also known as structure factor Shkl (Gq = hb1 + kb2 + lb3).

SHKL =∑α

fα(Gq) e−iGq·rα (34)

The position rα = uαa1 + vαa2 + wαa3 of the atoms in the unit cell can be expressed bytriples (uαvαwα) with uα, vα, wα < 1. By taking into account equation (20), the structurefactor then reads

SHKL =∑α

fα(Gq) e−2πi(huα+kvα+lwα) (35)

Finally, we want to evaluate the structure factor for the ideal Wurtzite lattice. The oxygen atomsin the unit cell are situated at (000), (1

323

12), the zinc atoms at (003

8) and (13

23

78) (gure 3).

Therefore we obtain

SHKL = fO

(1 + e−2πi(h3 + 2k

3+ l

2))

+ fZn

(e−2πi 3

8l + e−2πi(h3 + 2k

3+ 7

8l))

(36)

We now examine the structure factors for the symmetrical 00L reexes

S00L = fO

(1 + e−πil

)+ fZn

(e−πi

34l + e−πi

74l)

(37)

= fO

(1 + (−1)l

)+ fZn

(e−πi

34l + (−1)le−πi

34l)

=

2 (fO + e−πi34lfZn) l even

0 l odd

This shows that no intensity is observed for reexes (00L) with L odd. One can get a moreintuitive picture by viewing the Wurtzit lattice as two interpenetrating, hexagonal close-packedlattices. Each of the sublattices has two atoms per unit cell at (000) and (1

323

12). Therefore

dhkl is eectively reduced by a factor of 2. For L odd, the waves reected at the centeredlattice planes interfere destructively with the ones limiting the unit cell.

3 Experimental setup

Figure 15(a) shows schematically the setup of the HRXRD diractometer. In this laboratoryexercise we use a Phillips X'Pert MRD diractometer. A X-ray tube with a copper cathodegenerates the incident CuKα-X-ray beam (λ = 1, 540595(2) Å). The angle between the incidentX-ray beam and the surface of the crystal is denoted as Ω (cp. gure 13). The scattered beam

18 3 EXPERIMENTAL SETUP

detector

crystalW

2q

W 2q

w

j

substrate withepitaxial layer(b)

mono-chromatic

X-ray beamy

(a)

Figure 15: (a) HRXRD diractometer. (b) Illustration of the Euler angles and the angles Ωand 2θ

crystal surface

lattice planes

gluetcorr

W´ tq

2q

kin

kout

W

Figure 16: In order to obtain Ω the measured angle Ω′ has to be corrected by τcorr

can be detected by a silicon X-ray detector mounted on a rocking cantilever that enclosesan angle 2θ with the incident beam. The crystal, that shall be examined, is glued on thesample holder of a so-called Euler cradle. Therewith it can be rotated around the three Eulerangles ω, ϕ and ψ (gure 15 (b)). Normally the angle Ω′ between disc and incident beam hasa slightly dierent value than Ω, because of an inaccurate sample mounting. Therefore themeasured angle Ω′ has to be corrected by τcorr to obtain Ω (gure 16). By varying Ω and theposition of the detector, the scattering vector can be adjusted according to equations (29, 30).Dierentiation of equation (27) and subsequent division of the result by the same equation (27)yields the dierential Bragg equation

∆λλ

= cot θ ·∆θ +∆dhkldhkl

(38)

Equation (38) shows that for the accurate determination of the distance dhkl between twoadjacent lattice planes it is important to use a highly monochromatic incident beam and adetector with a very good angular resolution. By means of a Bartels monochromator (gure17) whose functional principle is based on the Bragg reection of the primary beam at fourgermanium crystals with (220) surfaces, the spectral width of the incident CuKα-X-ray beam canbe reduced to ∆λ

λ < 1, 5·10−4. In order to enhance the spectral resolution of the diractometer,

19

+-

+-

+-

-e

water in out

UH

Uacc

UW

pointfocus

vacuumWehnelt-cylinder

Bartels monochromator

X-ray tube

crosswise arrangedaperture plates

crystal onsample holder

analyzer

detector

2q

Figure 17: Optical path of the X-ray beam for the used goniometer.

diaphragms with aperture angles ∆θ ranging from 132

to 4 or an analyzer with an aperture

angle of 12 arcsec can be used. Especially for higher indexed reexes with low intensities, theuse of the analyzer is not always possible and therefore an appropriate diaphragm has to beused instead. Nevertheless, higher indexed reexes are preferable for the determination of thelattice constants because for θ → 90 the inuence of ∆θ in equation (38) can be signicantlysuppressed.

4 Characterization of real crystals by HRXRD

By means of HRXRD, deviations form the ideal crystal structure are investigated. For thispurpose the position and the width of X-ray reexes are measured by registering the scatteredX-ray intensity while rotating the sample around an Euler angle or changing the detector positionand thus varying the length and the direction of the scattering vector. In gure 18 the scatteringgeometry for a symmetrical and assymertical reex is shown together with the scan directionsof a 2θ-Ω- and a Ω-scan.

4.1 2θ-Ω-Scan

By executing a 2θ-Ω-Scan only the length of the scattering vector is varied by rotating thesample around the ω-axis (gure 15(b)) and by simultaneously rotating the detector on therocking cantilever with twice the angular velocity. The direction at which the scattering vectoris pointing therby remains unchanged. For symmetrical reexes Ω = θ and therefore the scandirection is along q⊥ (gure 18(a)). Particulariy this scan along q⊥ is useful to check if thereare other crystalline phases incorporated in the crystal, which is true if other than the expectedreexes are observed. Moreover, the occurrence of forbidden reexes hints at structural disorderin the examined crystal.In this laboratory exercise we investigate thin lms grown on a substrate. For columnar growth(gure 10) the vertical coherence length, i. e. the length of the crystallites that scatter theincident X-ray beam coherently, is limited by the layer thickness. As a consequence the integral

20 4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

q∥

q ⊥

2θθ

Ω - Scan

−4

−2

0 2

4

2θ−Ω - Scan

a)

b) q∥

q ⊥

−4

−2

0 2

4

q

(20.5)

q

2θ−Ω - Scan

Ω - Scan

kin kout

kin

kout

Figure 18: Ewald construction illustrating the scattering geometry in the case of a (a) symmet-

rical and an (b) asymmetrical reex. The gray arrows show the scan directions for a 2θ-Ω-Scanand a Ω-Scan. The reexes within gray semicircles are in a conventional scattering geometry

only accessible in transmission.

4.1 2θ-Ω-Scan 21

0 q∥

q ⊥

4

2

Figure 19: Broadening of the X-ray reexes for a nite vertical coherence length. It is the

same for all reexes and solely determined by the lm thickness lz.

in equation (23) is also non-zero for a scattering vector q slightly dierent from a reciprocallattice vector G.

I ∝ |E0|2

R′2

∣∣∣∣∣∣∑G

nG

∫V

ei(G−q)·r dV

∣∣∣∣∣∣2

=|E0|2

R′2|nGq |2 lxly ·

∣∣∣∣sin (∆qz lz)∆qz

∣∣∣∣2 (39)

where we dened 12(G3 − q3) = ∆qz. Therefore the measured X-ray peaks are not δ-like,

but are broadened along the q⊥-direction (gure 19). The function sin2(lz∆qz)(∆qz)2

is plotted ingure 20. The full width at half maximum (FWHM) of the main maximum is proportional to1lz. The measured peak width ∆(2θ) in [rad] of a symmetrical reex can thus be correlated

with the layer thickness [5]

lz =0.9λ

∆(2θ) · cos θ(40)

Thereby we assumed that the broadening of the reex along q⊥ is only due to the nite co-herence length and other broadening mechanisms as for example heterogeneous strain are ne-glected. Furthermore, besides the main peak, secondary maxima are observed when the numer-ator | sin(∆qz lz)| is maximum. The distance between such secondary maxima is ∆qmaxz = π

lz.

From this it can be deduced for symmetrical reexes that

lz =λ sin θ

∆θ · sin 2θ(41)

Herein ∆θ denotes half of the measured distance between secondary maxima in a 2θ-Ω-scan.

22 4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

0

20

40

60

80

100

Δqz

0 π/2

sin2 (l z Δ

q z)/(Δ

q z)2

- π/2

FWHM ~ 1/N

Figure 20: Plot of the the function sin2(lz ∆qz)

∆qz2for lz = 10. Its peak value is l2z , its FWHM is

∝ 1lz.

4.2 Ω-Scan

For a Ω-scan the sample is rotated around the ω-axis (gure 15(b)), which entails a variation ofthe scattering vector on a circular path around the origin (gure 18). The plot of the scatteredX-ray intensity as a function of Ω is often called rocking curve.As discussed in section 1.2, the investigated ZnO lms exhibit a columnar growth mode. There-fore in the lateral plane perpendicular to the [0001]-direction the typical length for coherentscattering is limited by the crystallite size. This involves similar to the case of a nite layer thick-ness a broadening of the reexes along the direction of limited coherence length (gure 21(a)).For symmetrical reexes the Ω-scan-direction is quasi parallel to this direction. If one assumesthat the limited coherence size is the only reason for the broadening of a symmetrical rockingcurve, by determining its FWHM ∆Ω a lower limit for the lateral crystallite size L‖ can beestimated [5]

L‖ =0.9 · λ

∆Ω · sin θ(42)

However, a rocking curve of a symmetrical reex in general is not exclusively broadened bya nite lateral coherence length, but also by the tilt of the crystallites (gure 21(b)). Thiscan be understood by examining gure 22: For an ideal single crystal the Bragg equation (27)denes exactly the allowed angle of incidence for which a reex can be observed. But a mosaiccrystal is built up by many crystallites with dierent tilt angles δΩ with respect to the [0001]direction. Thus, according to δΩ the crystallites can be grouped in ensembles. By rotating thesample around the ω-axis, dierent ensembles are selected for which the Bragg equation (27)is fullled.In conclusion, we have seen that the width of the rocking curve of a symmetrical reex isinuenced by a nite crystallite size as well as the tilt of the crystallites with respect to oneanother. Therefore the FWHM ∆Ω002 of the 002 rocking curve is often used as a gure of

4.2 Ω-Scan 23

(b)

q ⊥

q∥2

00 q∥(a)4

4

q ⊥

2

Figure 21: Broadening of the X-ray reexes for a mosaic crystal. The nite coherence length

in the growth plane gives rise to a broadening of the reexes along q||, the tilt of the crystallitesto a broadening along circular paths around the origin.

substrate

lateral coherence of the X-ray beam

length

substrate

dW

(b)(a)

q 2q

lateral coherence lengthof the X-ray beam

Figure 22: X-ray diraction for a (a) ideal single crystal and (b) a mosaic crystal where the

tilt of the crystallites with respect to one another broadens the rocking curve.

24 4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

Y

kin

kout

m

n

offset t

conventional geometrie(top view)

Win

Wout

Win

Wout

(a) (b) skewed geometry

kout

kin

n

m

Figure 23: Illustration of the (a) conventional and (b) skewed geometry. For the conventional

geometry the vector n normal to the sample surface and the vector m normal to the lattice

plane lie in the diraction plane, for the skewed geometry they enclose a angle ψ = τ .

merit to evaluate the degree of mosaicity of a thin epitaxial lm. The smaller ∆Ω002, the betterthe individual crystallites are aligned and the larger their size.On the other hand, the twist of the crystallites in a mosaic crystal has no inuence on the002 rocking curve because it produces no variation of the vertical position of the (0001) latticeplanes. Instead, it can be seen as inducing a tilt of the 1010 lattice planes. In order toinvestigate and quantify the twist of the crystallites, we thus have to determine the FWHM ofthe 100 rocking curve ∆Ω100. Yet, as the 1010 planes of the investigated ZnO thin lmsare perpendicular to the substrate plane, the measurement of the ∆Ω100 is not strait forward.One has to determine the FWHM of rocking curves ∆ΩH0L whose associated lattice planesh0hl enclose successively smaller angles with the 1010 lattice planes. From this ∆Ω100

can be extrapolated. For this purpose the measurements have to be conducted in a so-calledpseudo-symmetrical, skewed geometry, because in the conventional geometry the reexes whoseassociated lattice planes enclose the smallest angles with the 1010 lattice plane (reexes inthe gray semicircles in gure 18) are not accessible. In the conventional diraction geometry theoset angle τ of the asymmetrical lattice planes h0hl is compensated by an adjustment of theangle Ω (section 6, see gure 23(a))). In the skewed geometry the sample is rotated by ψ = τaround the ψ-axis in order to move the h0hl lattice planes in a position perpendicular to thediraction plane that is dened by the direction of incident and scattered X-ray beam (gure23(b))). This geometry is also called pseudo-symmetrical because similar as for symmetricalreexes Ω = θ since τ is compensated by an adjustment of the Euler angle ψ.For a proper extrapolation, it is important that the FWHMs of the reexes with a large angleψ can be determined. In order to extrapolate ∆Ω100, the measured ∆Ωh0l are plotted as afunction of ψ and then tted by

∆Ωh0l =√

(∆0 cosψ)2 + (∆π/2 sinψ)2 (43)

4.3 Reciprocal space map 25

- 0 , 2 0 , 0 0 , 2 0 , 4 0 , 6 0 , 8 1 , 0 1 , 2 1 , 4 1 , 60 , 1

0 , 2

0 , 3

0 , 4

1 7 , 0 1 7 , 5 1 8 , 0 1 8 , 5 1 9 , 0 1 9 , 502468

1 01 21 4

Y

DW

π

W

Figure 24: Measured FWHMs of rocking curves as a function of the angle ψ tted by equation

(43). In the inset the 101 rocking curve is shown.

where the tting parameter ∆0 corresponds to ∆Ω100. This is exemplarily shown in gure 24.As this procedure is rather time-consuming, it is common to settle for approximating ∆Ω100

with the FWHM of the 101 rocking curve ∆Ω101 which can be directly measured. The reex201 would be better suited for this purpose, but it is often dicult to measure for thin lmsbecause of its low intensity. Figure 24 shows that ∆Ω101 = 0.35 and the extrapolated value∆Ω100 = 0.4 dier only by 12.5%. Therefore in order to study tendencies for samples grownunder dierent growth conditions, the above approximation is quite reasonable.In section 1.2 we have seen that the tilt of the crystallites is due to screw-type dislocationswhereas the twist is due to edge-type dislocation. Hence, by assuming that the broadeningof the rocking curves originates only from tilt and twist, it is possible to calculate from theFWHMs ∆Ω002 and ∆Ω101 the dislocation densities by using the relations given by Dunn andKogh [1]

ρscrew =∆Ω2

002

4.35 · |bscrew|2(44)

ρedge =∆Ω2

101

4.35 · |bedge|2(45)

with the Burgers vector bscrew = [0001] for screw-type dislocations and bedge = 13 [1120] for

edge-type dislocations.

4.3 Reciprocal space map

The scan mode mapping a two-dimensional region of reciprocal space is known as reciprocalspace map. Such a scan can be carried out by combining the 2θ-Ω-scan mode with the Ω-scan mode in the following way: rst for a given length of the scattering angle a Ω-scan is

26 4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

0.546 0.547 0.548 0.549 0.5500.738

0.739

0.740

0.741

0.742

0.743

q ⊥ [r

.l.u.

]

q || [r.l.u.]

2.200

3.286

5.921

8.557

11.19

13.83

16.46

19.10

c = 5,2024 ± 0,0005 Åa = 3,2475 ± 0,0005 Å

mosaicity

lateralcoherence length

[cps]

Figure 25: Reciprocal space map of the 205 reex of a ZnO thin lm. In the inset the herefrom

calculated lattice constants are given.

performed, then the 2θ-Ω-scan mode is employed to change the length of the scattering vectorby a small amount δq, then again a Ω-scan is performed and so on . . . The result of sucha scan conducted at the position of the 205 reex of a ZnO thin lm is shown in gure 25.The arrows indicate the directions of the peak broadening due to a nite lateral crystallite sizeand due to the tilt of the crystallites. The broadening due to the nite layer thickness can beneglected. Since the intensity of the higher indexed reexes is low, it was not possible to usethe analyzer. Therefore the two principal broadening mechanisms could not be resolved whichis why the peak exhibits an ellipsoidal form. From the position of the principal axis it can bededuced that the broadening is due to both mechanisms.A reciprocal space map of an asymmetrical reex allows the determination of both the a andthe c lattice constant. As discussed in section 3 the high indexed reexes are best suited forthis because the contribution of the limited angular resolution in equation (38) decreases withθ. a and c can be calculated by determining the components q⊥ and q|| of the peak center.

a =1

q||(rlu)λ√3h (46)

c =1

q⊥(rlu)λ

2l (47)

In the above equations q⊥ and q|| have to be inserted in reciprocal lattice units (rlu), i. e. inunits of 4π

λ which is equivalent to normalizing the radius of the Ewald sphere to 1.

4.4 ϕ-Scan

By performing a ϕ-Scan, Ω and θ are kept constant while the sample is rotated around theϕ-axis. For our samples this corresponds to a rotation around the [0001]-axis. Hence, for

27

Figure 26: Computer interface by which the Euler cradle can be controlled.

asymmetric reexes HKL, by performing a ϕ-Scan of 360, six peaks can be observed which isin accordance with the sixfold symmetry of the [0001]-axis. By conducting such a 360-ϕ-Scanfor the substrate and the thin lm on top, the epitaxial relationship between substrate and thinlm can be established.

5 Exercises (you should work on before conducting the experiment)

1. Calculate the angle 2θ for the 002 reex of ZnO (cZnO = 5.20 Å).

2. Starting with equation (21), derive equation (22).

3. Calculate d10.1 the distance of the (1011) lattice planes. Calculate the oset angle τof these lattice planes with respect to the (0001) lattice planes (cZnO = 5.20 Å, aZnO= 3.25 Å).

6 Experimental procedure

6.1 Mosaicity of dierent ZnO-samples

In the rst part of this laboratory exercise we record Ω- and 2θ-Ω-scans of two dierent ZnOsamples in order to evaluate and compare their structural quality. We start with sample A.Figure 26 shows the computer interface by which the HRXRD diractometer is controlled. Inorder to correct the small oset angle τcorr which is due to an imperfect sample mountingthe following steps have to be carried out (see also the form for the measurement report inappendix C):

28 6 EXPERIMENTAL PROCEDURE

1. Select the ϕ-scan mode (scan axis) and set the scan range to 360, the step width to 1,and the time per step to 0.2s. Enter the theoretical value of 2θ for the 002-reex of ZnOgiven in appendix A and set the oset τ , as well as ϕ and ψ to zero. Enter the x-, y-and z-position of the sample on the mounting disc (these values a given to you by yoursupervisor). Execute the scan without using an aperture slit.

2. Normally, one should obtain two rather broad peaks. Right-click into the scan window,and select Move mode. Press the left mouse button and position the appearing line inthe middle between the two peaks. The sample disc is now rotated to the thus chosennew ϕ position. Note the new ϕ-position.

3. Now the oset τcorr due to an imperfect sample mounting can be corrected by a rotationaround the Ω-axis. Therefore, select the Ω-scan mode, change the scan range to 2, thestep size to 0.02, and the time per step to 0.8s and execute the scan without using anaperture slit. If not otherwise specied, the step size should be always 1

100 of the

scan range and the time per step should be chosen in a way that the total scan

time is approximately 1min. After executing the scan, position the line in Move mode

on the peak center thereby compensating the oset τcorr. Note the new oset angle.

4. Select the 2θ-Ω-scan mode, set the range to 1, change the step size accordingly, andexecute the scan with a 1

32-aperture slit in front of the detector. Again, position the line

on the peak center to correct for the deviation from the theoretical 2θ value. Note thenew 2θ position.

5. Repeat the Ω-scan with a scan range of 1, a time per step of 2s, and a 132-aperture

slit, correct the oset angle and note its new value. Save this scan (task bar: File →Save as, the maximum number of digits allowed is 8).

6. Record a 2θ-Ω-scan of the 002-reex of sample A. Use the same aperture slit and scanparameters as in the last Ω-scan except for the scan range and step size which you shouldset to 0.3 and 0.003, respectively. Correct for the new 2θ-Ω-position, note the new 2θangle and save the scan.

Finally, we turn to sample B. Repeat the above described steps for this sample and record arocking curve of its 002-reex. The peak widths of this sample should be substantially smallerand the peak intensity substantially higher as compared to sample A. Therefore, adjust the scanrange and step size for the Ω- and 2θ-Ω-scans as indicated in appendix C. An integration timeof 0.2s should be sucient for all scans.After having executed the ϕ-scan, do a Ω-scan and a 2θ-Ω-scan with a 1

32-aperture slit to

nd the peak and then redo the scans by using the analysator in front of the detector and byreducing the range and step size in order to be able to position the line in Move mode properlyon the peak center. Use the analysator for all further scans.For this, you have to move the detector from the mounting behind the aperture plate holderto the mounting behind the analysator. Then you have to change the Diracted beam path inthe Control -window to Lower (gur 27).Record a Ω-scan of the 002-reex with a scan range of 0.05 and a step width of 0.0002 aswell as a 2θ-Ω-scan of the 002-reex with a scan range of 1 and a step width of 0.001 andsave them. Furthermore, record a longer 2θ-Ω-scan without using an aperture plate or the

analysator. Set 2θ = 83.5, the scan range to 163, the step size to 0.2, and the time per

6.2 Rocking curve of the 101-reex 29

Figure 27: To use the analysator, the Diracted beam path has to be changed from Upper to

Lower.

step to 4s. Save the scan (as it will take about 50min you should do it during your lunch breakor at the end of this laboratory exercise).

6.2 Rocking curve of the 101-reex

We now record a rocking curve of the 101-reex of sample A in skewed geometry. To calibratethe sample position, we carry out the scans according to the form for the measurement reportin appendix C.

1. Select Phi as scan axis, set the scan range to 360, the step size to 1, and the time perstep to 0.2s. Employ the values given in appendix A for the 101-reex. Important: Aswe record the scan in skewed geometry, you have to position the 1011 lattice planesperpendicular to the diraction plane by rotating the sample around the ψ-axis! Thereforeyou have to set the ψ-angle to the oset value given in appendix A. Set the oset andϕ to zero and enter the x-, y-, and z-positions of sample A. Execute the scan withoutusing an aperture slit. The scan should exhibit four to six peaks. Position the line inMove mode on one of the peaks, thereby selecting one set of the six cristallographiclyequivalent lattice plane sets 1011. Note the new ϕ-position.

2. Redo a ϕ-scan but this time with a smaller range of 3 in order to position the lineproperly on the peak center.

3. Now again we have to correct for an imperfect sample mounting. Since for an asymmetricreex ϕ is already xed, this is mainly achieved by adjusting τ and ψ. Nevertheless, by

30 6 EXPERIMENTAL PROCEDURE

changing these two angles, also ϕ and 2θ have to be readjusted. Begin by recording aψ-scan with the transverse aperture plate (range = 5 and time per step 0.8s you mayhave to increase the time per step adequately if the scan is too noisy). Adjust ψ andnote its new value.

4. Select Ω-scan and set the scan range to 3. Execute the scan without using an apertureslit, adjust and note the oset angle.

5. Record a 2θ-Ω-scan with a 14-aperture slit (time per step 1.5s), adjust 2θ and note its

value.

6. Finally, record the rocking curve of the 101-reex without using an aperture slit with atime per step of 2s, and save it.

6.3 Epitaxial relationship

Record a ϕ-scan of the ZnO 112-reex of sample A in skewed geometry (scan range = 360,step size = 1, time per step = 0.2s) without an aperture slit and save it. Do the same for thesapphire 113-reex (see appendix B), but set the step size to 0.5.

6.4 Lattice constants of ZnO

In order to determine the exact a- and c−lattice constants of ZnO, we record a reciprocalspace map (rsm) of the asymmetric 205-reex of sample B. To adjust the sample and de-tector positions follow the steps indicated in the form for the measurement report of a rsm(appendix D).

1. Select the ϕ-scan mode, set the scan range to 360, the step size to 1, the time perstep to 0.2s, enter the values of 2θ and τ (no skewed geometry!) given in appendix Afor the 205-reex, and adjust the x-, y-, and z-position for sample B. Execute the ϕ-scanwithout an aperture slit, select one peak by adjusting ϕ to its maximum and note its newvalue.

2. Record a Ω-scan (scan range 2, time per step 0.8s) without an aperture slit, adjust τ tothe peak maximum and note its value.

3. Select the 2θ-Ω-mode, set the scan range to 1, and the time per step to 2s. Executethe scan with a 1

4-aperture slit, adjust 2θ very thoroughly to the peak center and note

its value as well as the absolute width of the peak.

4. Execute a Ω-scan without an aperture slit (time per step = 2s), adjust τ thoroughly tothe peak maximum, and note its value.

5. Record a ϕ-scan using the transverse aperture slit as well as a scan range of 8 and atime per step of 3s. Adjust ϕ accurately and note its new value.

6. Now, you have to determine the oset τcorr due to an imperfect sample mounting bywhich the recorded reciprocal space map will be subsequently corrected in order to increasethe accuracy of the determined lattice constants. Since ϕ is already xed, τcorr will be

6.5 Determination of the Mg-content x of a MgxZn1−xO sample 31

corrected by τ and ψ. For this purpose, adjust these two angles to the peak maximum ofthe 002-reex. Therefore select the Ω-scan mode, enter the 2θ-value of the 002-reexand set τ to zero while ϕ remains unchanged. Execute the scan with a 1

32-aperture slit

and a scan range of 1, a step width of 0.001, and a time per step of 0.2 s. Adjust τ tothe peak maximum and note its new value.

7. Execute a ψ-scan with a transverse aperture slit (scan range = 8, time per step = 0.2s), adjust ψ to the peak maximum, and note its value.

8. Select a 2θ-Ω-scan, set scan range to 0.3, and time per step to 0.2, and execute thescan with a 1

32

-aperture slit. Adjust 2θ thoroughly to the peak maximum and note its

new value.

9. Record a Ω-scan with a scan range of 0.05, a step size of 0.0002, and a time per stepof 0.2s with a 1

32-aperture slit. Adjust τ thoroughly to the peak maximum and note its

value.

10. Execute a ψ-scan with the transversal aperture slit (scan range 6, time per step 0.2s).Adjust ψ thoroughly to the peak maximum and note its new value.

11. By repeatedly conducting a ϕ-scan (scan range 6, time per step 3s, 2θ and τ as de-termined in step 3 and 4) of the 205-reex and a ψ-scan of the 002-reex (scan range6, time per step = 0.2s, 2θ and τ as determined in step 8 and 9) with the transverseaperture slit, ϕ and ψ are adjusted. This iteration ends when the dierence between twosuccessive ϕ- or ψ-scans is smaller than 0.03.

12. Record a Ω-scan of the 205-reex (scan range = 1, time per step = 2s) with the 14-

aperture slit in order to determine its absolute width and note it. Determine the peakcenter, note its value and save this scan.

13. Execute a Ω-scan of the 002-reex (scan range = 0.05, step size = 0.0002, time perscan = 0.2s) with the 1

32-aperture slit. Adjust τ thoroughly to the peak maximum, note

its value, and save the scan. The reciprocal space map of the 205-reex will be latercorrected by this oset angle.

The actual recording of the reciprocal space map will take about 10 hours and will therefore beconducted during the night. Your supervisor will write a program in order to do this.

6.5 Determination of the Mg-content x of a MgxZn1−xO sample

In the last part of this laboratory exercise we investigate the variation of the c-lattice constantfor a Zn1−xMgxO sample with a Mg-content x.For this purpose, measure the 2θ angle of the 006-reex of this sample. As the exact valueof the c-lattice parameter of the Zn1−xMgxO thin lm is unknown, you rst have to use the006-reex of the sapphire substrate in order to determine τcorr.

1. Select the ϕ-scan mode, set the scan range to 360, the step size to 1, the time perstep to 0.2s, enter the value of 2θ for the 006-reex of sapphire given in appendix B, andadjust the x-, y-, and z-position for sample C. Execute the ϕ-scan without an apertureslit and proceed as in 6.1 to determine the new value of ϕ, which you should note down.

32 7 REPORT

2. After this, determine τcorr by executing a Ω-scan by using the 132-aperture slit and with

a scan range of 2, a step size of 0.004, and a time per step of 0.2s.

3. Now we turn to the 006-reex of the Zn1−xMgxO thin lm. Enter the theoretical valueof 2θ for ZnO given in appendix A and execute a 2θ-Ω-scan with a scan range of 5,a step size of 0.05, a time per step of 0.8s, and with 1

32-aperture slit. Determine the

value of 2θ and note it down.

4. Record a Ω-scan with a scan range of 0.1, a step size of 0.001, and a time per step of0.8s.

5. Finally, execute a 2θ-Ω-scan by using the 132-aperture slit with a scan range of 0.3, a

step size of 0.003, and a time per step of 3s and save it. An accurate measurement of2θ is crucial for a precise determination of the Mg content x!

7 Report

1. Plot the long 2θ-Ω-scan of sample B in a logarithmic scale. Which peaks can you identify?Therby you have to take into account the X-ray signal of the sample holder (appendix E).In order to read out your data, you rst have to convert the saved datale to ASCII-format by using the program xrd2asc which is given to you by your supervisor. Copythat program into your data folder and execute it. You then have to enter the nameof the le in which you have saved the 2θ-Ω-scan name_of_your_datafile.d00 andthen the program converts the .d00-le into a le in ASCII-format with the lenamename_of_your_datafile.ASC.

2. Estimate the layer thickness of sample A by using equation (40).

3. Determine the layer thickness of sample B by using equations (40) and (41). Comparethe obtained values! Conclusion?

4. Estimate the screw-type dislocation densities of samples A and B and their lateral crys-tallite size. Compare the structural quality of both samples.

5. Estimate the edge-type dislocation density of sample A.

6. Illustrate graphically which lattice planes are associated with the 112-and the 113-reexrespectively. Verify the epitaxial relation of ZnO on a c-plane sapphire substrate.

7. Determine the peak center of the 205-reex. In order to do that, you rst have to con-vert the saved le of the reciprocal space map into ASCII-format by using the programxrd2asc. You have to use the DOS-shell, go to your data folder and execute the com-mand xrd2asc /q name_of_your_datafile.a00. Subsequently, you have to enter theoset angle τ of the 002-reex (Neues Oset für Omega [36.4...] : ) The programthen corrects the peak position of the 205-reex for this oset and creates the new lename_of_your_datafile.ASC. In case you use OriginTM proceed as follows:

Import name_of_your_datafile.ASC

33

Set the third column to Z and select it. Select Edit→ Convert to matrix→ Random

XYZ

Select Gridding Method: Correlation and press enter.

Select the matrix window.

Select Plot → Contour Plot → Color Fill

Determine the a- and c-lattice constants of sample B. Which mechanisms can you identifyfor the broadening of its 205-reex? Calculate the lattice mismatch by assuming thatthe ZnO lm on the sapphire substrate is fully relaxed and by taking into account theformation of a coincidence lattice.

8. Determine the c-lattice constant of sample C. Why was the 006-reex instead of the002-reex used for this purpose?Sadofev et al. [3] demonstrated the validity of Vegard's rule and found for the variationof the c-lattice constant with the Mg-content x for Zn1−xMgxO:

cZn1−xMgxO = cZnO − 0.17Å · x (48)

Use this relation with the above determined value for cZnO to calculate the Mg-contentof sample C. Estimate the error of the obtained result by assuming that the equation (48)is accurate.

34 A ZNO REFLEXES

A ZnO reexes

a [Å] = 3.2475c [Å] = 5.2024λ [Å] = 1.540598

Reexh k l 2θ [°] Oset τ [°]

0 0 1 17.0297 0.00000 0 2 34.4509 0.00000 0 3 52.7441 0.00000 0 4 72.6357 0.00000 0 5 95.5191 0.00000 0 6 125.3449 0.0000

1 0 1 36.2815 61.60431 0 2 47.5786 42.76571 0 3 62.9131 31.65791 0 4 81.4651 24.81811 0 5 104.2535 20.30241 0 6 136.7633 17.1345

2 0 1 69.1442 74.87442 0 2 77.0279 61.60432 0 3 89.6997 50.96142 0 4 107.5591 42.76572 0 5 134.1320 36.4984

3 0 1 113.2130 79.78493 0 2 121.7149 70.18083 0 3 138.1529 61.6043

1 1 1 59.5988 72.66601 1 2 68.0060 58.02631 1 3 81.0673 46.88281 1 4 98.7229 38.69421 1 5 123.1122 32.6513

35

B Sapphire reexs

a [Å] = 4.7577c [Å] = 12.9907λ [Å] = 1.540598

Reexh k l 2θ [°] Oset τ [°]

0 0 3 20.4936 0.00000 0 6 41.6822 0.00000 0 9 64.5069 0.00000 0 12 90.7230 0.00000 0 15 125.6066 0.0000

1 0 2 25.5819 57.61121 0 5 41.0361 32.23441 0 8 61.3117 21.50981 0 4 35.1563 38.24571 0 11 85.4565 15.99361 0 14 116.6276 12.6915

2 0 10 89.0155 32.23442 0 13 117.9066 25.8760

3 0 12 129.9219 38.2457

4 0 8 124.6453 57.61124 0 11 165.7714 48.9043

1 1 0 37.7872 90.00001 1 3 43.3642 61.21741 1 6 57.5110 42.30701 1 9 77.2502 31.24811 1 12 102.8457 24.46921 1 15 142.3658 20.0046

2 2 0 80.7253 90.00002 2 3 84.3829 74.64082 2 6 95.2779 61.21742 2 9 114.1062 50.51022 2 12 148.3697 42.3070

36 C MEASUREMENT REPORT FOR ROCKING-CURVES AND 2θ-Ω-SCANS

C Measurement report for rocking-curves and 2θ-Ω-scans

Sample A

Scan-mode Range/Step size/Int. time Result Aperture

002-Reexϕ 360 / 1 /0.2 ϕ : Ω 2 / 0.02 /0.8 τ : 2θ-Ω 1 / 0.01 /0.8 2θ: 1

32

Ω 1 / 0.01 /2.0 τ : 132

2θ-Ω 0.3 / 0.003 /2.0 2θ: 132

101-Reexϕ 360 / 1 /0.2 ϕ : ϕ 3 / 0.03 /0.8 ϕ : ψ 5 / 0.05 /0.8 ψ : transverseΩ 3 / 0.03 /0.8 τ : 2θ-Ω 3 / 0.03 /1.5 2θ: 1

4

Ω 2 / 0.02 /2.0 τ :

Sample B

Scan-mode Range/Step size/Int. time Result Aperture

002-Reexϕ 360 / 1 /0.2 ϕ : Ω 1 / 0.005 /0.2 τ : 1

32

2θ-Ω 1 / 0.005 /0.2 2θ: 132

Ω 0.05 / 0.0002 /0.2 τ : analysator2θ-Ω 0.5 / 0.001 /0.2 2θ: analysator

Ω 0.03 / 0.0003 /0.2 τ : analysator2θ-Ω 1 / 0.001 /0.2 2θ: analysator

37

Sample C

Scan-mode Range/Step size/Int. time Result Aperture

006-Reex (Al2O3)ϕ 360 / 1 /0.2 ϕ : Ω 2 / 0.002 /0.2 τ : 1

32

004-Reex (ZnO)2θ-Ω 3 / 0.03 /0.8 2θ: 1

32

Ω 0.1 / 0.001 /0.8 τ : 132

2θ-Ω 0.3 / 0.003 /3.0 2θ: 132

38 D MEASUREMENT REPORT FOR RECIPROCAL SPACE MAP

D Measurement report for reciprocal space map

Sample B

Scan-mode Range/Step size/Int. time Result Aperture

205-Reexϕ 360 / 1 /0.2 ϕ : Ω 2 / 0.02 /0.8 τ :

2θ-Ω 1 / 0.01 /2.0 2θ: 14

Ω 1 / 0.01 /0.8 τ : −−−ϕ 8 / 0.08 /3.0 ϕ : transverse

002-ReexΩ 1 / 0.001 /0.2 τ : 1

32

ψ 8 / 0.08 /0.2 ψ : transverse

2θ-Ω 0.3 / 0.003 /0.2 2θ: 132

Ω 0.05 /0.0002 /0.2 τ : 132

ψ 6 / 0.06 /0.2 ψ : transverse

ϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverseϕ (205) 6 / 0.06 /3.0 ϕ : ∆: transverseψ (002) 6 / 0.06 /0.2 ψ : ∆: transverse

Ω (205) 1 / 0.01 /2.0 τ : 14

Ω (002) 0.05 /0.0002 /0.2 τ : 132

39

E X-ray signal of the sample holder

0 2 0 4 0 6 0 8 0 1 0 0 1 2 0 1 4 0 1 6 0

int

ensity

[arb.

units]

θ

Figure 28: X-ray intensity scattered by the sample holder vs. 2θ.

References

[1] C. G. Dunn and E. F. Koch. Acta Metall., 5:584, 1957.

[2] K. Kopitzki. Einführung in die Festkörperphysik. Teubner Studienbücher, 3rd edition, 1993.

[3] S. Sadofev, S. Blumstengel, J. Cui, J. Puls, S. Rogaschewski, P. Schäfer, Y. G. Sadofyev, andF. Henneberger. Growth of high-quality ZnMgO epilayers and ZnO/ZnMgO quantum wellstructures by radical-source molecular-beam epitaxy on sapphire. Applied Physics Letters,87:091903, 2005.

[4] H. Vogel. Gerthsen Physik. Springer-Verlag, 20th edition, 1999.

[5] B. E. Warren. X-ray Diraction. Dover Publications, Inc., New York, 1st edition, 1990.