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XRD

Fortgeschrittenpraktikum

High Resolution X-Ray Diraction

Mrz 2009

Walter Schottky Institut Zentralinstitut der Technischen Universitt Mnchen fr physikalische Grundlagen der Halbleiterelektronik Am Coulombwall 3 85748 Garching

2

CONTENTS

Contents1 Crystals1.1 Ideal crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 1.1.2 1.2 Denition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Miller indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 3 4 6

Real crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Theory of X-ray diraction2.1 2.2 2.3 2.4 2.5 2.6

9

Kinematic theory of X-ray diraction . . . . . . . . . . . . . . . . . . . . . . . 11 Reciprocal lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 X-ray diraction at periodic structures . . . . . . . . . . . . . . . . . . . . . . 13 Bragg equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Ewald sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Structure factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Experimental setup 4 Characterization of real crystals by HRXRD4.1 4.2 4.3 4.4

17 19

2--Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 -Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Reciprocal space map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

-Scan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5 Exercises (you should work on before conducting the experiment) 6 Experimental procedure6.1 6.2 6.3 6.4 6.5

27 27

Mosaicity of dierent ZnO-samples . . . . . . . . . . . . . . . . . . . . . . . . 27 Rocking curve of the 101-reex . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Epitaxial relationship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Lattice constants of ZnO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Determination of the Mg-content x of a M gx Zn1x O sample . . . . . . . . . 31

7 Report A ZnO reexes B Sapphire reexs

32 34 35

3

C Measurement report for rocking-curves and 2--scans D Measurement report for reciprocal space map E X-ray signal of the sample holder Bibliography

36 38 39 39

1 CrystalsAt 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 years before by Wilhelm Conrad Rntgen. For this purpose, von Laue worked out a theory for X-ray diraction at three-dimensional crystals. With the observation of diraction patterns by irradiating solids with X-rays, his coworkers Walther Friedrich and Paul Knipping veried both 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 of High Resolution X-Ray Diraction (HRXRD). In general, this technique is not used to determine the crystal structure, but to investigate deviations from an ideal crystal which can be induced i. 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 kinematic theory of X-ray diraction. Finally, the experimental setup is presented and the characterization of real crystals by HRXRD is discussed.

1.1 Ideal crystals1.1.1 DenitionAn ideal crystal is an innite, periodic array of a structural element. The structural element which 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 diraction pattern. Picture taken at the U-Bahn station in Garching (Forschungszentrum).

4

1 CRYSTALS

b b b a a

(a)

(b)

(a+b)

(c)

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 the so-called space lattice. The arrangement of the atoms looks identical viewed from every point of 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 the unit 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 the primitive lattice vectors. The number of atoms that its associated basis contains is as small as 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 into account other symmetries of the crystal than the translational symmetry, like rotational or mirror symmetry, 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 indicesAny plane containing lattice points is called a lattice plane. In an ideal crystal there is always an innite number of parallel lattice planes. As we will see later, it is convenient to label a set of 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 coordinate axes in units of the lattice constants, i.e. in multiples of the lengths of the lattice vector ai .1 2. Take the reciprocal values h = a , k = 1 b

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

c

Oa1 a3 a2

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 plane intersects the axes at a = 2, b = 1 and c = 4, the set of parallel planes it belongs to is labeled by the Miller indices ( . Negative values are marked with a bar over the index. If 241) the lattice plane is parallel to an axis, the intersection point is and therefore the corresponding index is equal to zero. To indicate that the triple (hkl) labels all parallel lattice planes, it is enclosed in parentheses. To designate crystallographicly equivalent planes, the Miller indices are enclosed in curly braces {hkl}. For instance lattice planes limiting the unit cell in a cubic lattice, i. e. the (100), (010), (001), ( , (0 and (00 planes, can be subsumed by 100) 10) 1) writing {100}. Similarly, directions in the lattice are indicated by a triple of integers [uvw] enclosed in square brackets. 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 oxygen atoms form two interpenetrating, hexagonal close-packed sublattices which are displaced along the c-direction of their hexagonal unit cell (gure 3). In this structure the zinc (oxygen) atoms are tetrahedrally coordinated which means that they are situated at the center of a tetrahedron which is formed by oxygen (zinc) atoms. The atoms are usually grouped in bilayers, which consist of two adjacent (00.1) lattice planes (the signication of the dot in this notation is explained 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 lattice vectors a1 and a2 lying in the basal plane as well as c being perpendicular hereon. The basis of the Wurtzite lattice consists of four atoms. By using the above described methods to label directions and planes, in hexagonal crystals crystallographicly equivalent directions and planes can be designated by a dierent type of triple (hkl) or [hkl]. For example the crystallographicly

6

1 CRYSTALS

B A B

} bilayer

[0001] AFigure 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 (1 and (10.0) 1.0) (gure 5 (a)). Therefore it is common to use a coordinate system with four axes and a quadruple of integers for indexing. This takes into account that there are three equivalent symmetry axes a1 , a2 and a3 in the plane perpendicular to the principal axis along the c-direction. The indices (hkil) are then determined analogously to the already presented algorithm. Thereby the relation

i = (h + k)

(3)

is valid, because a1 , a2 and a3 are not linearly independent as a3 = (a1 + a2 ). For designating directions, one has to assure to pick that linear combination of lattice vectors that satises equation (3). For instance, the direction along the a2 vector is not labeled [0 100] but [1210] (gure 5 (b)). If one chooses to use only three indices in a hexagonal lattice, one places usually a point, that shall represent the omitted fourth index, between the second and third index [hk.l]. Equation (3) has then not to be obeyed. Using four indices for indexing has the advantage that crystallographicly equivalent directions and planes are obtained by a simple cyclic permutation of the indices.

1.2 Real crystalsIn real crystals the requirement to minimize the free energy F = U T S induces deviations from the ideal crystal structure. Already small concentrations of defects can drastically inuence the properties of crystals. For example the electrical conductivity of a semiconductor can be signicantly increased by the incorporation of a small amount of extrinsic impurities which is known as doping. Therefore it is important to investigate the nature and density of defects. For this purpose, HRXRD is a suitable method which has the advantages that it does not damage the samples and that no complex sample preparation is necessary. In this laboratory exercise we investigate heteroepitaxially grown thin lms. This means that the underlying substrate and the epitaxial layer are dierent materials with typically dierent lattice constants and perhaps even a dierent crystal symmetry. To quantify the dierence in

1.2 Real crystals

7

[1120] or [11.0] (0001) or (00.1)

c

[1210] or [01.0]

c[0110] or [12.0]

(1100) or (11.0)

(1120) or (11.0)

a3

a2(1010) or (10.0)

a3

a2

(a)

a1

(b)

a1

Figure 5: Indexing of directions and planes in a hexagonal lattice.lattice constant the so-called lattice mismatch a/a is dened as follows

asubstrate arelax a f ilm = a asubstrate

(4)

where asubstrate and arelax refer to the respective relaxed lattice constants (see gure 6) in the f ilm plane parallel to the interface. For a small lattice mismatch the epitaxial lm can grow pseudomorphical, i. e. it adapts the in-plane lattice constant of the substrate thereby accumulating biaxial strain (gure 6 (b)). When the lm thickness exceeds a certain critical value, the built-up strain energy is released by the formation of one-dimensional defects, called dislocations, perpendicular to the interface. In the vicinity of a dislocation the crystal lattice is distorted and strain is accumulated, that decays rather slowly as one moves away from the dislocation. By moving along a 360-loop around a dislocation, one does not arrive at one's starting point. The dierence between starting and end point 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 Burgers vector 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 in gure 8 in order to minimize the strain energy. Such an agglomeration of dislocations is known as grain boundary. This type of two-dimensional defect can be seen as a boundary between two monocrystalline regions of a solid, called crystallites, which are twisted by an angle with respect to one another. In a similar way as edge-type dislocations result in a twist, screw-type dislocations can give rise to a tilt of the crystallites. In this laboratory exercise we investigate ZnO samples heteroepitaxially grown on c-plane sapphire substrates. Sapphire can be seen to have a quasi-hexagonal lattice, that is to say the O-sublattice exhibits sixfold symmetry whereas the Al-sublattice has only a threefold symmetry axis. The term c-plane means that the sapphire is cut parallel to its (0001) plane. The growth direction of ZnO on this substrate surface is the [0001] direction. In order to minimize the lattice mismatch, the ZnO lattice is rotated by 30 with respect to the sapphire substrate

8

1 CRYSTALS

unit cellscfilm arelax film relax

afilm= asubstrate

cfilm

csubstrate

csubstrate

(a)afilm

asubstrate

(b)

asubstrate afilm= afilm crelax film = cfilmrelax

cfilm

csubstrate

csubstrate asubstrate

(c)

(d)

asubstrate

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 b core

(a)

(b)

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 relationship between 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 very beginning of the lm growth. This results in a columnar growth mode (gure 10) with most of the dislocations perpendicular to the interface whereas there are few dislocations parallel to the 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 we will see later, this causes a broadening of specic X-ray reexes. However, the lattice mismatch between ZnO- and Zn1x Mgx O-layers is relatively small which allows pseudomorphical growth to take place. But the uppermost layer can also be partially relaxed or completely relaxed depending on the layer thickness and the magnesium content x in the Zn1x Mgx O layer. For a quantitative analysis, the degree of relaxation r

r=

af ilm asubstrate arelax asubstrate f ilm

(5)

is dened, where af ilm denotes the actually measured lattice constant of a thin lm and arelax f ilm the totally relaxed lattice constant. r is equal to 1 for fully relaxed lms and equal to 0 for pseudomorphical growth.

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

10

2 THEORY OF X-RAY DIFFRACTION

O Zn

ZnO [ 1 00] 1 Al 2 O 3 [ 2 0] 1 1

ZnO [ 2 0] 1 1 Al 2 O 3 [01 0] 1

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.

c1

c2

(a)

substrate

(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

P r R-r R R D SFigure 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 for diraction at ideal crystals. Since all the samples investigated in this laboratory exercise exhibit a mosaic structure, we constrain the discussion to the kinematic theory of X-ray diraction.

2.1 Kinematic theory of X-ray diractionThe interaction between an electromagnetic wave and an atom can be described with an oscillator model. For large distances R between X-ray source and sample, the incident wave can be approximated by a plane wave whose electric eld Ein is given by (gure 11)

Ein = E0 ei(kin (R+r)0 t)

(6)

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

|kin | = |kout | = |k| =

2

(7)

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

Eout = f Ein

eikout |R r| |R r|

(8)

Herein f denotes the scattering amplitude which depends on the type of atom and the frequency of 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) can be rewritten

Eout = f E0 ei(kin (R+r)0 t) eikout R = fwith

eikout r R(9)

E0 i(kin kout )r e R

E0 = E0 ei(kin R+kout R ) ei0 t

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

Eout

E0 RV

n(r) ei(kin kout )r dV =

E0 RV

n(r) eiqr 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 Fourier transformation of the electron density. By performing diraction experiments one can not detect the amplitude but only the intensity I of the scattered wave, thus loosing its phase information2

|E0 |2 I R2V

n(r) e

iqr

dV

(12)

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

2.2 Reciprocal latticeUntil now we made no use of the periodicity of the crystal. Yet, for a crystal the electron density n(r) has to be invariant under translations which constitute linear combinations of lattice vectors T = u a1 + v a2 + w a3 with u, v , w integer and ai the fundamental lattice vectors. Hence, n(r + T) = n(r) (13) Periodic functions that satisfy equation (13) can be expanded into a Fourier series. In one dimension the Fourier expansion reads

n(x) =m

nm ei(m2/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 be easily veried. Likewise, in three dimensions the Fourier expansion is dened as

n(r) =G

nG eiGr

(15)

In order to satisfy equation (13)

eiGT = 1 TThis involves

!

(16)

G T = 2nA suitable basis to construct vectors

T

n integer

(17) (18)

G = hb1 + kb2 + lb3that fulll equation (17) is given by

h, k, l integer a1 a2 a1 (a2 a3 )

b1 = 2

a2 a3 a1 (a2 a3 )

b2 = 2

a3 a1 a1 (a2 a3 )

b3 = 2

(19)

The vectors bi and their linear combinations are referred to as reciprocal lattice vectors, because their dimension is m1 and their length is inversely proportional to the length of the corresponding real lattice vectors. They span the so-called reciprocal lattice. It follows, that with every crystal structure there are two lattices associated, namely its space lattice in real space and its reciprocal lattice in Fourier space. The reciprocal lattice exhibits the same symmetries as the real crystal lattice. As both lattices are directly linked with each other via equation (19), by determining properties of the reciprocal lattice one easily obtains the same properties of the real lattice. Also, by rotating or translating the real lattice the reciprocal lattice is rotated or translated. One readily veries ai bj = 2 ij (20) and consequently all reciprocal lattice vectors constructed according to (18) satisfy equation (17). There is an important relation between the lattice planes of the crystal lattice and the reciprocal lattice vectors [2]: The reciprocal lattice vector Ghkl = hb1 + kb2 + lb3 is perpendicular to the lattice planes with Miller indices (hkl) and the distance dhkl between two such adjacent planes is given by 2 dhkl = (21) |Ghkl | For a hexagonal lattice with lattice constants a and c this yields a dhkl = 4 2 a 2 2 2 3 (h + k + hk) + ( c ) l (22)

2.3 X-ray diraction at periodic structuresWe now insert the Fourier expansion of the electron density (15) into equation (12)2

|E0 |2 I R2

nGG V

ei(Gq)r dV

(23)

14

2 THEORY OF X-RAY DIFFRACTION

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

ei(Gq)r dV = V

lx 2

ei(G1 q1 )x dx

ly 2

ei(G2 q2 )y dy

lz 2

ei(G3 q3 )z dz sin lz 1 (G3 q3 ) 2 1 (G3 q3 ) 2(24)

lx 2

ly 2

lz 2

= =

sin lx 1 (G1 q1 ) 2 1 (G1 q1 ) 2 lx ly lz = V 0

sin ly 1 (G2 q2 ) 2 1 (G2 q2 ) 2for lx , ly , lz large

for q = G otherwise

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 along the direction of kout , that is to say the phase factors of the waves scattered at dierent lattice points of the crystal dier only by a factor e2in (n integer) along kout . If the phase factor has a slightly dierent value than e2in , the contribution from the all lattice points average to zero very eectively due to the high number of scatterers. Hence, for constructive interference relation (23) yields for the intensity of the observed X-ray reex

I

|E0 |2 |nGq |2 V 2 R2

(26)

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

2.4 Bragg equationWe now try to give an intuitively clearer interpretation of the diraction process. By inserting the diraction condition (25) into equation (21) and taking into account that G = n Ghkl (n integer) one nds 2n 2n dhkl = = |q| 2 sin 2 Herein 2 represents the angle between incident and scattered wave (gure 12(a)) and n the diraction order. By rearranging this result, the famous Bragg equation can be obtained

2 dhkl sin = n

(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 requirement that the path dierence for waves reected by adjacent lattice planes has to be an integer multiple of the wave length .

2.5 Ewald sphere

15

kout q

q(hkl)

2q

q q

qdhkl dhkl sinq

kin

(a)

(b)

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

(205)qkin

4 2

kout

0

2

4 q

Figure 13: Ewald sphere

2.5 Ewald sphereIn the last section we tried to get a deeper insight in the diraction process by interpreting it in a more intuitively way. We saw that the Bragg equation relates every X-ray reex that can be observed with a set of parallel lattice planes (hkl). The Laue condition, however, assigns every X-ray reex to a reciprocal lattice point which make it convenient to label the reexes with the indices 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 is displayed. According to equation (11) and gure 12 (a), the maximum length of the scattering vector qmax in the case of backscattering ( = 90 ) is given by

qmax = 2|k| =

4

(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 given wavelength are accessible for diraction experiments are situated within the hemisphere with radius equal to 4 . 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

dV

r

ra atom a lattice point R

Figure 14: Denition of r and . r points to the center of an atom of the unit cell, to a point within that atom.beam according to the Bragg interpretation. designates the angle between the crystal surface and the incident X-ray beam, the angle between scattering lattice planes and incident X-ray beam and = is the oset angle between crystal surface and scattering planes. The reexes on the [001] axis, also labeled as q , are called symmetrical because for them = . All the other reexes are called asymmetrical. The components of the scattering vector parallel q|| and perpendicular q to the crystal surface expressed as a function of and 2 read

q|| = q

2 (cos(2 ) cos()) 2 (sin(2 ) + sin()) =

(29) (30)

2.6 Structure factorIn gure 13, the 00L reexes with L odd are represented by open circles, because these reexes are forbidden, i. e. no scattered intensity is observed. In order to understand this, we now also have 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 the Fourier coecient nGq in equation (26). nGq is related to n(r) by

n Gq =

1 VucVuc

n(r) eiGq r dV

(31)

which can be easily veried by inserting the Fourier expansion (15) of n(r) into (31). The integration herein extends over the Volume of the whole unit cell Vuc . In case the nuclei of the atoms are not too light the principal contribution to the scattered X-ray intensity arises from the core electrons whereas the delocalized valence electrons can be neglected. Therefore n(r) can be expressed as the sum over the electron densities n () of the various atoms of the unit cell n(r) = n () (r r ) (32)

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

17

rewritten

n Gq =

1 Vuc

eiGq r V

n () eiGq dVf (Gq )

(33)

Now we integrate only over the volume V of a single atom, multiply the result with the appropriate phase factor and sum up over all the atoms of the unit cell. The atomic scattering factor f (Gq ) dened in equation (33) can be seen as the Fourier transformation of the atomic electron density. If the electrons were point charges at the atomic centers r , f (Gq ) = 1 independent of Gq . But since the electron density in general stretches over some ngstrms around r , f (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 =

f (Gq ) eiGq r

(34)

The position r = u a1 + v a2 + w a3 of the atoms in the unit cell can be expressed by triples (u v w ) with u , v , w < 1. By taking into account equation (20), the structure factor then reads SHKL = f (Gq ) e2i(hu +kv +lw ) (35)

Finally, we want to evaluate the structure factor for the ideal Wurtzite lattice. The oxygen atoms 1 1 2 in the unit cell are situated at (000), ( 3 2 2 ), the zinc atoms at (00 3 ) and ( 1 3 7 ) (gure 3). 3 8 3 8 Therefore we obtain

SHKL = fO 1 + e2i( 3 + 3 + 2 ) + fZn e2i 8 l + e2i( 3 + 3 + 8 l)We now examine the structure factors for the symmetrical 00L reexes

h

2k

l

3

h

2k

7

(36)

S00L = fO 1 + eil + fZn ei 4 l + ei 4 l = fO 1 + (1)l + fZn ei 4 l + (1)l ei 4 l = 2 (fO + ei 4 l fZn ) 03 3 3

3

7

(37)

l even l odd

This shows that no intensity is observed for reexes (00L) with L odd. One can get a more intuitive picture by viewing the Wurtzit lattice as two interpenetrating, hexagonal close-packed 2 lattices. Each of the sublattices has two atoms per unit cell at (000) and ( 1 3 1 ). Therefore 3 2 dhkl is eectively reduced by a factor of 2. For L odd, the waves reected at the centered lattice planes interfere destructively with the ones limiting the unit cell.

3 Experimental setupFigure 15(a) shows schematically the setup of the HRXRD diractometer. In this laboratory exercise we use a Phillips X'Pert MRD diractometer. A X-ray tube with a copper cathode generates the incident CuK -X-ray beam ( = 1, 540595(2) ). The angle between the incident X-ray beam and the surface of the crystal is denoted as (cp. gure 13). The scattered beam

18

3 EXPERIMENTAL SETUP

j monochromatic X-ray beam W crystal W 2q 2q y

w

(a)

detector

(b)

substrate with epitaxial layer

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

kin kout

W

W

t

q2q

urface crystal slattic e pla nes

glue

tcorr

Figure 16: In order to obtain the measured angle has to be corrected by corrcan be detected by a silicon X-ray detector mounted on a rocking cantilever that encloses an angle 2 with the incident beam. The crystal, that shall be examined, is glued on the sample holder of a so-called Euler cradle. Therewith it can be rotated around the three Euler angles , and (gure 15 (b)). Normally the angle between disc and incident beam has a slightly dierent value than , because of an inaccurate sample mounting. Therefore the measured angle has to be corrected by corr to obtain (gure 16). By varying and the position 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

dhkl = cot + dhkl

(38)

Equation (38) shows that for the accurate determination of the distance dhkl between two adjacent lattice planes it is important to use a highly monochromatic incident beam and a detector with a very good angular resolution. By means of a Bartels monochromator (gure 17) whose functional principle is based on the Bragg reection of the primary beam at four germanium crystals with (220) surfaces, the spectral width of the incident CuK -X-ray beam can be reduced to < 1, 5104 . In order to enhance the spectral resolution of the diractometer,

19

Bartels monochromatorvacuum+ -

crystal on sample holder W 2q detector crosswise arranged aperture plates point focus

Wehneltcylinder e-

water in

out

X-ray tube analyzer

Figure 17: Optical path of the X-ray beam for the used goniometer.1 diaphragms with aperture angles ranging from 32 to 4 or an analyzer with an aperture angle of 12 arcsec can be used. Especially for higher indexed reexes with low intensities, the use of the analyzer is not always possible and therefore an appropriate diaphragm has to be used instead. Nevertheless, higher indexed reexes are preferable for the determination of the lattice constants because for 90 the inuence of in equation (38) can be signicantly suppressed.

4 Characterization of real crystals by HRXRDBy means of HRXRD, deviations form the ideal crystal structure are investigated. For this purpose the position and the width of X-ray reexes are measured by registering the scattered X-ray intensity while rotating the sample around an Euler angle or changing the detector position and thus varying the length and the direction of the scattering vector. In gure 18 the scattering geometry for a symmetrical and assymertical reex is shown together with the scan directions of a 2-- and a -scan.

4.1 2--ScanBy executing a 2--Scan only the length of the scattering vector is varied by rotating the sample around the -axis (gure 15(b)) and by simultaneously rotating the detector on the rocking cantilever with twice the angular velocity. The direction at which the scattering vector is pointing therby remains unchanged. For symmetrical reexes = and therefore the scan direction is along q (gure 18(a)). Particulariy this scan along q is useful to check if there are other crystalline phases incorporated in the crystal, which is true if other than the expected reexes are observed. Moreover, the occurrence of forbidden reexes hints at structural disorder in 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 the incident X-ray beam coherently, is limited by the layer thickness. As a consequence the integral

+ -

UH

+ - UW

Uacc

20

4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

q

- Scankin

2 - Scanq0

a)

4

2

2

kout2

4

q

q2 - Scan (20.5) - Scanqkin

kout

b)

4

2

0

2

4

q

Figure 18: Ewald construction illustrating the scattering geometry in the case of a (a) symmetrical and an (b) asymmetrical reex. The gray arrows show the scan directions for a 2--Scan and a -Scan. The reexes within gray semicircles are in a conventional scattering geometry only accessible in transmission.

4.1 2--Scan

21

q

0

2

4

q

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 reciprocal lattice vector G.2

|E0 |2 I R2where we dened

nGG V

ei(Gq)r dV

=

|E0 |2 sin (qz lz ) 2 2 |nGq | lx ly qz R

2

(39)

(lz but are broadened along the q -direction (gure 19). The function sin(qzqz ) is plotted in )2 gure 20. The full width at half maximum (FWHM) of the main maximum is proportional to 1 lz . The measured peak width (2) in [rad] of a symmetrical reex can thus be correlated with the layer thickness [5]

1 2 (G3

q3 ) = qz . Therefore the measured X-ray peaks are not -like,2

lz =

0.9 (2) cos

(40)

Thereby we assumed that the broadening of the reex along q is only due to the nite coherence length and other broadening mechanisms as for example heterogeneous strain are neglected. Furthermore, besides the main peak, secondary maxima are observed when the numermax = . ator | sin(qz lz )| is maximum. The distance between such secondary maxima is qz lz From this it can be deduced for symmetrical reexes that

lz =

sin sin 2

(41)

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

22

4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

100

80

sin (lz qz)/(qz)

2

60

FWHM ~ 1/N40

2

20

0

- /2

0 qz

/2

Figure 20: Plot of the the function l1 . z

sin2 (lz qz ) qz 2

2 for lz = 10. Its peak value is lz , its FWHM is

4.2 -ScanFor a -scan the sample is rotated around the -axis (gure 15(b)), which entails a variation of the scattering vector on a circular path around the origin (gure 18). The plot of the scattered X-ray intensity as a function of is often called rocking curve. As discussed in section 1.2, the investigated ZnO lms exhibit a columnar growth mode. Therefore in the lateral plane perpendicular to the [0001]-direction the typical length for coherent scattering is limited by the crystallite size. This involves similar to the case of a nite layer thickness a broadening of the reexes along the direction of limited coherence length (gure 21(a)). For symmetrical reexes the -scan-direction is quasi parallel to this direction. If one assumes that the limited coherence size is the only reason for the broadening of a symmetrical rocking curve, by determining its FWHM a lower limit for the lateral crystallite size L can be estimated [5] 0.9 L = (42) sin However, a rocking curve of a symmetrical reex in general is not exclusively broadened by a nite lateral coherence length, but also by the tilt of the crystallites (gure 21(b)). This can 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 mosaic crystal is built up by many crystallites with dierent tilt angles with respect to the [0001] direction. Thus, according to the crystallites can be grouped in ensembles. By rotating the sample around the -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 is inuenced by a nite crystallite size as well as the tilt of the crystallites with respect to one another. Therefore the FWHM 002 of the 002 rocking curve is often used as a gure of

4.2 -Scan

23

q

q

0

(a)

2

4

q

0

(b)

2

4

q

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 crystallites to a broadening along circular paths around the origin.

dW

q

2q

(a)

substrate

(b)

substrate

lateral coherence length of the X-ray beam

lateral coherence length of 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

n

Winoffset t

kin

Win Wout Ym kout

kin

Woutm kout n

(a) conventional geometrie (top view)

(b) skewed geometry

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 = .merit to evaluate the degree of mosaicity of a thin epitaxial lm. The smaller 002 , the better the 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 the 002 rocking curve because it produces no variation of the vertical position of the (0001) lattice planes. Instead, it can be seen as inducing a tilt of the {10 lattice planes. In order to 10} investigate and quantify the twist of the crystallites, we thus have to determine the FWHM of the 100 rocking curve 100 . Yet, as the {10 planes of the investigated ZnO thin lms 10} are perpendicular to the substrate plane, the measurement of the 100 is not strait forward. One has to determine the FWHM of rocking curves H0L whose associated lattice planes {h0hl} enclose successively smaller angles with the {10 lattice planes. From this 100 10} can be extrapolated. For this purpose the measurements have to be conducted in a so-called pseudo-symmetrical, skewed geometry, because in the conventional geometry the reexes whose associated lattice planes enclose the smallest angles with the {10 lattice plane (reexes in 10} the gray semicircles in gure 18) are not accessible. In the conventional diraction geometry the oset angle of the asymmetrical lattice planes {h0hl} is compensated by an adjustment of the angle (section 6, see gure 23(a))). In the skewed geometry the sample is rotated by = around the -axis in order to move the {h0hl} lattice planes in a position perpendicular to the diraction plane that is dened by the direction of incident and scattered X-ray beam (gure 23(b))). This geometry is also called pseudo-symmetrical because similar as for symmetrical reexes = since is compensated by an adjustment of the Euler angle . For a proper extrapolation, it is important that the FWHMs of the reexes with a large angle can be determined. In order to extrapolate 100 , the measured h0l are plotted as a function of and then tted by

h0l =

(0 cos )2 + (/2 sin )2

(43)

4.3 Reciprocal space map

25

0 ,4

202 203 2041 4 1 2

201

101 102

/2

0 ,3

[]

hkl

104 1060 ,2

103

intensity [a. u.]

1 0 8 6 4 2 0 1 7 ,0

(10.1)0,35

D W

002 004

105

W1 7 ,5 1 8 ,0

[]1 8 ,5 1 9 ,0 1 9 ,5

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

1 ,0

1 ,2

1 ,4

1 ,6

Y

[rad]

Figure 24: Measured FWHMs of rocking curves as a function of the angle tted by equation (43). In the inset the 101 rocking curve is shown.where the tting parameter 0 corresponds to 100 . This is exemplarily shown in gure 24. As this procedure is rather time-consuming, it is common to settle for approximating 100 with the FWHM of the 101 rocking curve 101 which can be directly measured. The reex 201 would be better suited for this purpose, but it is often dicult to measure for thin lms because of its low intensity. Figure 24 shows that 101 = 0.35 and the extrapolated value 100 = 0.4 dier only by 12.5%. Therefore in order to study tendencies for samples grown under 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 dislocations whereas the twist is due to edge-type dislocation. Hence, by assuming that the broadening of the rocking curves originates only from tilt and twist, it is possible to calculate from the FWHMs 002 and 101 the dislocation densities by using the relations given by Dunn and Kogh [1]

screw = edge

2 002 4.35 |bscrew |2 2 101 = 4.35 |bedge |21 3

(44) (45)

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

[1120] for

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

26

4 CHARACTERIZATION OF REAL CRYSTALS BY HRXRD

0.743

0.742

c = 5,2024 0,0005 a = 3,2475 0,0005

[cps]2.200 3.286 5.921 8.557 11.19

0.741

13.83 16.46 19.10

q [r.l.u.]

0.740

lateral coherence length mosaicity

0.739

0.738 0.546 0.547 0.548 0.549 0.550

q || [r.l.u.]

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 2--scan mode is employed to change the length of the scattering vector by a small amount q , then again a -scan is performed and so on . . . The result of such a 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 size and due to the tilt of the crystallites. The broadening due to the nite layer thickness can be neglected. Since the intensity of the higher indexed reexes is low, it was not possible to use the analyzer. Therefore the two principal broadening mechanisms could not be resolved which is why the peak exhibits an ellipsoidal form. From the position of the principal axis it can be deduced that the broadening is due to both mechanisms. A reciprocal space map of an asymmetrical reex allows the determination of both the a and the c lattice constant. As discussed in section 3 the high indexed reexes are best suited for this because the contribution of the limited angular resolution in equation (38) decreases with . a and c can be calculated by determining the components q and q|| of the peak center.

a=

1 h q|| (rlu) 3 1 c= l q (rlu) 2

(46) (47)

In the above equations q and q|| have to be inserted in reciprocal lattice units (rlu), i. e. in units of 4 which is equivalent to normalizing the radius of the Ewald sphere to 1.

4.4 -ScanBy performing a -Scan, and are kept constant while the sample is rotated around the -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 -Scan of 360 , six peaks can be observed which is in accordance with the sixfold symmetry of the [0001]-axis. By conducting such a 360 --Scan for the substrate and the thin lm on top, the epitaxial relationship between substrate and thin lm can be established.

5 Exercises (you should work on before conducting the experiment)1. Calculate the angle 2 for the 002 reex of ZnO (cZnO = 5.20 ). 2. Starting with equation (21), derive equation (22). 3. Calculate d10.1 the distance of the (10 ) lattice planes. Calculate the oset angle 11 of these lattice planes with respect to the (0001) lattice planes (cZnO = 5.20 , aZnO = 3.25 ).

6 Experimental procedure6.1 Mosaicity of dierent ZnO-samplesIn the rst part of this laboratory exercise we record - and 2--scans of two dierent ZnO samples 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. In order to correct the small oset angle corr which is due to an imperfect sample mounting the following steps have to be carried out (see also the form for the measurement report in appendix C):

28

6 EXPERIMENTAL PROCEDURE1. Select the -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 2 for the 002-reex of ZnO given in appendix A and set the oset , as well as and to zero. Enter the x-, yand z-position of the sample on the mounting disc (these values a given to you by your supervisor). 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 in the middle between the two peaks. The sample disc is now rotated to the thus chosen new position. Note the new -position. 3. Now the oset corr due to an imperfect sample mounting can be corrected by a rotation around the -axis. Therefore, select the -scan mode, change the scan range to 2 , the step size to 0.02 , and the time per step to 0.8s and execute the scan without using an 1 aperture slit. If not otherwise specied, the step size should be always 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 modeon the peak center thereby compensating the oset corr . Note the new oset angle.

4. Select the 2--scan mode, set the range to 1 , change the step size accordingly, and 1 execute the scan with a 32 -aperture slit in front of the detector. Again, position the line on the peak center to correct for the deviation from the theoretical 2 value. Note the new 2 position.1 5. Repeat the -scan with a scan range of 1 , a time per step of 2s, and a 32 -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 2--scan of the 002-reex of sample A. Use the same aperture slit and scan parameters as in the last -scan except for the scan range and step size which you should set to 0.3 and 0.003 , respectively. Correct for the new 2--position, note the new 2 angle and save the scan. Finally, we turn to sample B. Repeat the above described steps for this sample and record a rocking curve of its 002-reex. The peak widths of this sample should be substantially smaller and the peak intensity substantially higher as compared to sample A. Therefore, adjust the scan range and step size for the - and 2--scans as indicated in appendix C. An integration time of 0.2s should be sucient for all scans. 1 After having executed the -scan, do a -scan and a 2--scan with a 32 -aperture slit to nd the peak and then redo the scans by using the analysator in front of the detector and by reducing the range and step size in order to be able to position the line in Move mode properly on 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 holder to the mounting behind the analysator. Then you have to change the Diracted beam path in the Control -window to Lower (gur 27). Record a -scan of the 002-reex with a scan range of 0.05 and a step width of 0.0002 as well as a 2--scan of the 002-reex with a scan range of 1 and a step width of 0.001 and save them. Furthermore, record a longer 2--scan without using an aperture plate or the analysator. Set 2 = 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 break or at the end of this laboratory exercise).

6.2 Rocking curve of the 101-reexWe now record a rocking curve of the 101-reex of sample A in skewed geometry. To calibrate the sample position, we carry out the scans according to the form for the measurement report in appendix C. 1. Select Phi as scan axis, set the scan range to 360 , the step size to 1 , and the time per step to 0.2s. Employ the values given in appendix A for the 101-reex. Important: As we record the scan in skewed geometry, you have to position the {10 lattice planes 11} perpendicular to the diraction plane by rotating the sample around the -axis! Therefore you have to set the -angle to the oset value given in appendix A. Set the oset and to zero and enter the x-, y-, and z-positions of sample A. Execute the scan without using an aperture slit. The scan should exhibit four to six peaks. Position the line in Move mode on one of the peaks, thereby selecting one set of the six cristallographicly equivalent lattice plane sets {10 . Note the new -position. 11} 2. Redo a -scan but this time with a smaller range of 3 in order to position the line properly on the peak center. 3. Now again we have to correct for an imperfect sample mounting. Since for an asymmetric reex is already xed, this is mainly achieved by adjusting and . Nevertheless, by

30

6 EXPERIMENTAL PROCEDUREchanging these two angles, also and 2 have to be readjusted. Begin by recording a -scan with the transverse aperture plate (range = 5 and time per step 0.8s you may have to increase the time per step adequately if the scan is too noisy). Adjust and note its new value. 4. Select -scan and set the scan range to 3 . Execute the scan without using an aperture slit, adjust and note the oset angle. 5. Record a 2--scan with a value.1 4 -aperture

slit (time per step 1.5s), adjust 2 and note its

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

6.3 Epitaxial relationshipRecord a -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 the sapphire 113-reex (see appendix B), but set the step size to 0.5 .

6.4 Lattice constants of ZnOIn order to determine the exact a- and clattice constants of ZnO, we record a reciprocal space map (rsm) of the asymmetric 205-reex of sample B. To adjust the sample and detector positions follow the steps indicated in the form for the measurement report of a rsm (appendix D). 1. Select the -scan mode, set the scan range to 360 , the step size to 1 , the time per step to 0.2s, enter the values of 2 and (no skewed geometry!) given in appendix A for the 205-reex, and adjust the x-, y-, and z-position for sample B. Execute the -scan without an aperture slit, select one peak by adjusting to its maximum and note its new value. 2. Record a -scan (scan range 2 , time per step 0.8s) without an aperture slit, adjust to the peak maximum and note its value. 3. Select the 2--mode, set the scan range to 1 , and the time per step to 2s. Execute the scan with a 1 -aperture slit, adjust 2 very thoroughly to the peak center and note 4 its value as well as the absolute width of the peak. 4. Execute a -scan without an aperture slit (time per step = 2s), adjust thoroughly to the peak maximum, and note its value. 5. Record a -scan using the transverse aperture slit as well as a scan range of 8 and a time per step of 3s. Adjust accurately and note its new value. 6. Now, you have to determine the oset corr due to an imperfect sample mounting by which the recorded reciprocal space map will be subsequently corrected in order to increase the accuracy of the determined lattice constants. Since is already xed, corr will be

6.5 Determination of the Mg-content x of a M gx Zn1x O sample

31

corrected by and . For this purpose, adjust these two angles to the peak maximum of the 002-reex. Therefore select the -scan mode, enter the 2-value of the 002-reex 1 and set to zero while remains unchanged. Execute the scan with a 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 to the peak maximum and note its new value. 7. Execute a -scan with a transverse aperture slit (scan range = 8 , time per step = 0.2 s), adjust to the peak maximum, and note its value. 8. Select a 2--scan, set scan range to 0.3 , and time per step to 0.2, and execute the 1 scan with a 32 -aperture slit. Adjust 2 thoroughly to the peak maximum and note its new value. 9. Record a -scan with a scan range of 0.05 , a step size of 0.0002 , and a time per step 1 of 0.2s with a 32 -aperture slit. Adjust thoroughly to the peak maximum and note its value. 10. Execute a -scan with the transversal aperture slit (scan range 6 , time per step 0.2s). Adjust thoroughly to the peak maximum and note its new value. 11. By repeatedly conducting a -scan (scan range 6 , time per step 3s, 2 and as determined in step 3 and 4) of the 205-reex and a -scan of the 002-reex (scan range 6 , time per step = 0.2s, 2 and as determined in step 8 and 9) with the transverse aperture slit, and are adjusted. This iteration ends when the dierence between two successive - or -scans is smaller than 0.03 .1 12. Record a -scan of the 205-reex (scan range = 1 , time per step = 2s) with the 4 aperture slit in order to determine its absolute width and note it. Determine the peak center, note its value and save this scan.

13. Execute a -scan of the 002-reex (scan range = 0.05 , step size = 0.0002 , time per 1 scan = 0.2s) with the 32 -aperture slit. Adjust thoroughly to the peak maximum, note its value, and save the scan. The reciprocal space map of the 205-reex will be later corrected by this oset angle. The actual recording of the reciprocal space map will take about 10 hours and will therefore be conducted during the night. Your supervisor will write a program in order to do this.

6.5 Determination of the Mg-content x of a M gx Zn1x O sampleIn the last part of this laboratory exercise we investigate the variation of the c-lattice constant for a Zn1x Mgx O sample with a Mg-content x. For this purpose, measure the 2 angle of the 006-reex of this sample. As the exact value of the c-lattice parameter of the Zn1x Mgx O thin lm is unknown, you rst have to use the 006-reex of the sapphire substrate in order to determine corr . 1. Select the -scan mode, set the scan range to 360 , the step size to 1 , the time per step to 0.2s, enter the value of 2 for the 006-reex of sapphire given in appendix B, and adjust the x-, y-, and z-position for sample C. Execute the -scan without an aperture slit and proceed as in 6.1 to determine the new value of , which you should note down.

32

7 REPORT1 2. After this, determine corr by executing a -scan by using the 32 -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 Zn1x Mgx O thin lm. Enter the theoretical value of 2 for ZnO given in appendix A and execute a 2--scan with a scan range of 5 , 1 a step size of 0.05 , a time per step of 0.8s, and with 32 -aperture slit. Determine the value of 2 and note it down. 4. Record a -scan with a scan range of 0.1 , a step size of 0.001 , and a time per step of 0.8s.1 5. Finally, execute a 2--scan by using the 32 -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 of 2 is crucial for a precise determination of the Mg content x!

7 Report1. Plot the long 2--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 ASCIIformat by using the program xrd2asc which is given to you by your supervisor. Copy that program into your data folder and execute it. You then have to enter the name of the le in which you have saved the 2--scan name_of_your_datafile.d00 and then the program converts the .d00-le into a le in ASCII-format with the lename name_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). Compare the obtained values! Conclusion? 4. Estimate the screw-type dislocation densities of samples A and B and their lateral crystallite 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-reex respectively. 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 convert the saved le of the reciprocal space map into ASCII-format by using the program xrd2asc. You have to use the DOS-shell, go to your data folder and execute the command xrd2asc /q name_of_your_datafile.a00. Subsequently, you have to enter the oset angle of the 002-reex (Neues Oset fr Omega [36.4...] : ) The program then corrects the peak position of the 205-reex for this oset and creates the new le name_of_your_datafile.ASC. In case you use OriginT M 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 identify for the broadening of its 205-reex? Calculate the lattice mismatch by assuming that the ZnO lm on the sapphire substrate is fully relaxed and by taking into account the formation of a coincidence lattice. 8. Determine the c-lattice constant of sample C. Why was the 006-reex instead of the 002-reex used for this purpose? Sadofev et al. [3] demonstrated the validity of Vegard's rule and found for the variation of the c-lattice constant with the Mg-content x for Zn1x Mgx O:

cZn1x M gx O = cZnO 0.17 x

(48)

Use this relation with the above determined value for cZnO to calculate the Mg-content of 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 [] = c [] = [] =h 0 0 0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 3 3 1 1 1 1 1 Reex k l 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 1 2 3 1 2 3 4 5 2 [] 17.0297 34.4509 52.7441 72.6357 95.5191 125.3449 36.2815 47.5786 62.9131 81.4651 104.2535 136.7633 69.1442 77.0279 89.6997 107.5591 134.1320 113.2130 121.7149 138.1529 59.5988 68.0060 81.0673 98.7229 123.1122

3.2475 5.2024 1.540598 Oset [] 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 61.6043 42.7657 31.6579 24.8181 20.3024 17.1345 74.8744 61.6043 50.9614 42.7657 36.4984 79.7849 70.1808 61.6043 72.6660 58.0263 46.8828 38.6942 32.6513

35

B Sapphire reexs

a [] = c [] = [] =Reex h k l 0 0 0 0 0 1 1 1 1 1 1 2 2 3 4 4 1 1 1 1 1 1 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 6 9 12 15 2 5 8 4 11 14 10 13 12 8 11 0 3 6 9 12 15 0 3 6 9 12 2 [] 20.4936 41.6822 64.5069 90.7230 125.6066 25.5819 41.0361 61.3117 35.1563 85.4565 116.6276 89.0155 117.9066 129.9219 124.6453 165.7714 37.7872 43.3642 57.5110 77.2502 102.8457 142.3658 80.7253 84.3829 95.2779 114.1062 148.3697

4.7577 12.9907 1.540598 Oset [] 0.0000 0.0000 0.0000 0.0000 0.0000 57.6112 32.2344 21.5098 38.2457 15.9936 12.6915 32.2344 25.8760 38.2457 57.6112 48.9043 90.0000 61.2174 42.3070 31.2481 24.4692 20.0046 90.0000 74.6408 61.2174 50.5102 42.3070

36

C MEASUREMENT REPORT FOR ROCKING-CURVES AND 2--SCANS

C Measurement report for rocking-curves and 2--scansSample AScan-mode Range/Step size/Int. time Result Aperture

002-Reex 2- 2-

360 2 1 1 0.3

/ 1 / 0.02 / 0.01 / 0.01 / 0.003

/0.2 /0.8 /0.8 /2.0 /2.0

: : 2: : 2:

1 32 1 32 1 32

101-Reex 2-

360 3 5 3 3 2

/ / / / / /

1 0.03 0.05 0.03 0.03 0.02

/0.2 /0.8 /0.8 /0.8 /1.5 /2.0

: : : : 2: :

transverse 1 4

Sample BScan-mode Range/Step size/Int. time Result Aperture

002-Reex 2- 2- 2-

360 1 1 0.05 0.5 0.03 1

/ 1 / 0.005 / 0.005 / 0.0002 / 0.001 / 0.0003 / 0.001

/0.2 /0.2 /0.2 /0.2 /0.2 /0.2 /0.2

: : 2: : 2: : 2:

1 32 1 32

analysator analysator analysator analysator

37

Sample CScan-mode Range/Step size/Int. time Result Aperture

006-Reex (Al2 O3 )

360 / 1 /0.2 / 0.002 /0.2 2

: :

1 32

004-Reex (ZnO)2- 2-

3 / 0.03 /0.8 0.1 / 0.001 /0.8 0.3 / 0.003 /3.0

2: : 2:

1 32 1 32 1 32

38

D MEASUREMENT REPORT FOR RECIPROCAL SPACE MAP

D Measurement report for reciprocal space mapSample BScan-mode Range/Step size/Int. time Result Aperture

205-Reex 2-

360 2 1 1 8

/ / / / /

1 0.02 0.01 0.01 0.08

/0.2 /0.8 /2.0 /0.8 /3.0

: : 2: : :

transverse

1 4

002-Reex 2- (205) (002) (205) (002) (205) (002) (205) (002) (205) (002) (205) (002) (205) (002) (205) (002) (205) (002) (205) (002)

1 8 0.3 0.05 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

/ 0.001 / 0.08 / 0.003 /0.0002 / 0.06 / / / / / / / / / / / / / / / / / / / / 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06 0.06

/0.2 /0.2 /0.2 /0.2 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2 /3.0 /0.2

: : 2: : : : : : : : : : : : : : : : : : : : : : :

transverse1 32 1 32

1 32

transverse

: : : : : : : : : : : : : : : : : : : :

transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse transverse1 4 1 32

(205) (002)

1 / 0.01 /2.0 0.05 /0.0002 /0.2

: :

39

E X-ray signal of the sample holder

in te n s ity [a r b . u n its ]0

2 0

4 0

6 0

2 []

8 0

1 0 0

1 2 0

1 4 0

1 6 0

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

References[1] C. G. Dunn and E. F. Koch. Acta Metall., 5:584, 1957. [2] K. Kopitzki. Einfhrung in die Festkrperphysik. Teubner Studienbcher, 3rd edition, 1993. [3] S. Sadofev, S. Blumstengel, J. Cui, J. Puls, S. Rogaschewski, P. Schfer, Y. G. Sadofyev, and F. Henneberger. Growth of high-quality ZnMgO epilayers and ZnO/ZnMgO quantum well structures 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.