1 2 X-Ray Diffraction Crystallography3 4 Yoshio Waseda Eiichiro Matsubara Kozo Shinoda X-Ray Diffraction Crystallography...

X-Ray Diffraction Crystallography

Yoshio Waseda Kozo Shinoda

Eiichiro Matsubara

X-Ray Diffraction Crystallography Introduction, Examples and Solved Problems

With 159 Figures

123

Professor Dr. Yoshio Waseda Professor Kozo Shinoda Tohoku University, Institute of Multidisciplinary Research for Advanced Materials Katahira 2-1-1, 980-8577 Sendai, Aoba-ku, Japan E-mail: [emailprotected]; [emailprotected]

Professor Dr. Eiichiro Matsubara Kyoto University, Graduate School of Engineering Department of Materials Science and Engineering Yoshida Honmachi, 606-8501 Kyoto, Sakyo-ku, Japan E-mail: [emailprotected]

Supplementary problems with solutions are accessible to qualified instructors at springer.com on this book’s product page. Instructors may click on the link additional information and register to obtain their restricted access.

ISBN 978-3-642-16634-1 e-ISBN 978-3-642-16635-8 DOI 10.1007/978-3-642-16635-8 Springer Heidelberg Dordrecht London New York Library of Congress Control Number: 2011923528 c Springer-Verlag Berlin Heidelberg 2011 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: eStudio Calamar Steinen Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Preface

X-ray diffraction crystallography for powder samples is well-established and widely used in the field of materials characterization to obtain information on the atomic scale structure of various substances in a variety of states. Of course, there have been numerous advances in this field, since the discovery of X-ray diffraction from crystals in 1912 by Max von Laue and in 1913 by W.L. Bragg and W.H. Bragg. The origin of crystallography is traced to the study for the external appearance of natural minerals and a large amount of data have been systematized by applying geometry and group theory. Then, crystallography becomes a valuable method for the general consideration of how crystals can be built from small units, corresponding to the infinite repetition of identical structural units in space. Many excellent and exhaustive books on X-ray diffraction and crystallography are available, but the undergraduate students and young researchers and engineers who wish to become acquainted with this subject frequently find them overwhelming. They find it difficult to identify and understand the essential points in the limited time available to them, particularly on how to estimate useful structural information from the X-ray diffraction data. Since X-ray powder diffraction is one of the most common and leading methods in materials research, mastery of the subject is essential. In order to learn the fundamentals of X-ray diffraction crystallography well and to be able to cope with the subject appropriately, a certain number of “exercises” involving calculation of specific properties from measurements are strongly recommended. This is particularly true for beginners of X-ray diffraction crystallography. Recent general purpose X-ray diffraction equipments have a lot of inbuilt automation for structural analysis. When a sample is set in the machine and the preset button is pressed, results are automatically generated some of which are misleading. A good understanding of fundamentals helps one to recognize misleading output. During the preparation of this book, we have tried to keep in mind the students who come across X-ray diffraction crystallography for powder samples at the first time. The primary objective is to offer a textbook to students with almost no basic knowledge of X-rays and a guidebook for young scientists and engineers engaged in full-scale materials development with emphasis on practical problem solving. For the convenience of readers, some essential points with basic equations

v

vi

Preface

are summarized in each chapter, together with some relevant physical constants and atomic scattering factors of elements listed in appendices. Since practice perfects the acquisition of skills in X-ray diffraction crystallography, 100 supplementary problems are also added with simple solutions. We hope that the students will try to solve these supplementary problems by themselves to deepen their understanding and competence of X-ray crystallography without serious difficulty. Since the field of X-ray structural analysis of materials is quite wide, not all possible applications are covered. The subject matter in this book is restricted to fundamental knowledge of X-ray diffraction crystallography for powder samples only. The readers can refer to specialized books for other applications. The production of high-quality multi-layered thin films with sufficient reliability is an essential requirement for device fabrication in micro-electronics. An iron-containing layered oxy-pnictide LaO1x Fx FeAs has received much attention because it exhibits superconductivity below 43 K as reported recently by Dr. Hideo Hosono in Japan. The interesting properties of such new synthetic functional materials are linked to their periodic and interfacial structures at a microscopic level, although the origin of such peculiar features has not been fully understood yet. Nevertheless, our understanding of most of the important properties of new functional materials relies heavily upon their atomic scale structure. The beneficial utilization of all materials should be pursued very actively to contribute to the most important technological and social developments of the twenty-first century harmonized with nature. Driven by environmental concerns, the interest in the recovery or recycling of valuable metallic elements from wastes such as discarded electronic devices will grow significantly over the next decade. The atomic scale structure of various materials in a variety of states is essential from both the basic science and the applied engineering points of view. Our goal is to take the most efficient approach for describing the link between the atomic scale structure and properties of any substance of interest. The content of this book has been developed through lectures given to undergraduate or junior-level graduate students in their first half (Master’s program) of the doctoral course of the graduate school of engineering at both Tohoku and Kyoto universities. If this book is used as a reference to supplement lectures in the field of structural analysis of materials or as a guide for a researcher or engineer engaged in structural analysis to confirm his or her degree of understanding and to compensate for deficiency in self-instruction, it is an exceptional joy for us. Many people have helped both directly or indirectly in preparing this book. The authors are deeply indebted to Professors Masahiro Kitada for his valuable advice on the original manuscript. Many thanks are due to Professor K.T. Jacob (Indian Institute of Science, Bangalore), Professor N.J. Themelis (Columbia University), Professor Osamu Terasaki (Stockholm University) and Dr.Daniel Gr¨uner and Dr. Karin S¨oderberg (Stockholm University) and Dr. Sam Stevens (University of Manchester) who read the manuscript and made many helpful suggestions. The authors would like to thank Ms. Noriko Eguchi, Ms. Miwa Sasaki and Mr. Yoshimasa Ito for their assistance in preparing figures and tables as well as the electronic TeX typeset of this book. The authors are also indebted to many sources

Preface

vii

of material in this article. The encouragement of Dr. Claus Ascheron of SpringerVerlag, Mr. Satoru Uchida and Manabu Uchida of Uchida-Rokakuho Publishing Ltd should also be acknowledged. Sendai, Japan January 2011

Yoshio Waseda Eiichiro Matsubara Kozo Shinoda

Note: A solution manual for 100 supplementary problems is available to instructors who have adopted this book for regular classroom use or tutorial seminar use. To obtain a copy of the solution manual, a request may be delivered on your departmental letterhead to the publisher (or authors), specifying the purpose of use as an organization (not personal).

Contents

1

Fundamental Properties of X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1.1 Nature of X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1.2 Production of X-rays.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1.3 Absorption of X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 1.4 Solved Problems (12 Examples) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .

1 1 3 5 6

2

Geometry of Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 2.1 Lattice and Crystal Systems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 2.2 Lattice Planes and Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 2.3 Planes of a Zone and Interplanar Spacing . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 2.4 Stereographic Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 2.5 Solved Problems (21 Examples) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .

21 21 26 30 31 35

3

Scattering and Diffraction .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 3.1 Scattering by a Single Electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 3.2 Scattering by a Single Atom.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 3.3 Diffraction from Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 3.4 Scattering by a Unit Cell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 3.5 Solved Problems (13 Examples) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .

67 67 69 73 76 80

4

Diffraction from Polycrystalline Samples and Determination of Crystal Structure .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .107 4.1 X-ray Diffractometer Essentials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .107 4.2 Estimation of X-ray Diffraction Intensity from a Polycrystalline Sample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .108 4.2.1 Structure Factor .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .109 4.2.2 Polarization Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .109 4.2.3 Multiplicity Factor .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .110 4.2.4 Lorentz Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .110 4.2.5 Absorption Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .111

ix

x

Contents

4.3 4.4 4.5 4.6 4.7

4.8

4.2.6 Temperature Factor .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .112 4.2.7 General Formula of the Intensity of Diffracted X-rays for Powder Crystalline Samples . . . . . . . . . . . . . . . . .. . . . . . .113 Crystal Structure Determination: Cubic Systems . . . . . . . . . . . . . . .. . . . . . .114 Crystal Structure Determination: Tetragonal and Hexagonal Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .116 Identification of an Unknown Sample by X-ray Diffraction (Hanawalt Method) .. . . . . . . . . . . . . . . . . . . . . . .. . . . . . .117 Determination of Lattice Parameter of a Polycrystalline Sample . . . . .120 Quantitative Analysis of Powder Mixtures and Determination of Crystalline Size and Lattice Strain. . . . . . .. . . . . . .121 4.7.1 Quantitative Determination of a Crystalline Substance in a Mixture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .121 4.7.2 Measurement of the Size of Crystal Grains and Heterogeneous Distortion . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .123 Solved Problems (18 Examples) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .127

5

Reciprocal Lattice and Integrated Intensities of Crystals . . . . . . . . .. . . . . . .169 5.1 Mathematical Definition of Reciprocal Lattice . . . . . . . . . . . . . . . . . .. . . . . . .169 5.2 Intensity from Scattering by Electrons and Atoms . . . . . . . . . . . . . .. . . . . . .171 5.3 Intensity from Scattering by a Small Crystal . . . . . . . . . . . . . . . . . . . .. . . . . . .174 5.4 Integrated Intensity from Small Single Crystals. . . . . . . . . . . . . . . . .. . . . . . .175 5.5 Integrated Intensity from Mosaic Crystals and Polycrystalline Samples.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .177 5.6 Solved Problems (18 Examples) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .179

6

Symmetry Analysis for Crystals and the Use of the International Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .219 6.1 Symmetry Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .219 6.2 International Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .224 6.3 Solved Problems (8 Examples).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .228

7

Supplementary Problems (100 Exercises) . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .253

8

Solutions to Supplementary Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .273

A

Appendix . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .289 A.1 Fundamental Units and Some Physical Constants . . . . . . . . . . . . . .. . . . . . .289 A.2 Atomic Weight, Density, Debye Temperature and Mass Absorption Coefficients (cm2 =g) for Elements.. . . . . . . . . . . . . . . . .. . . . . . .291 A.3 Atomic Scattering Factors as a Function of sin =.. . . . . . . . . . . .. . . . . . .295 A.4 Quadratic Forms of Miller Indices for Cubic and Hexagonal Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .298 A.5 Volume and Interplanar Angles of a Unit Cell . . . . . . . . . . . . . . . . . .. . . . . . .299 A.6 Numerical Values for Calculation of the Temperature Factor . .. . . . . . .300

Contents

xi

A.7 Fundamentals of Least-Squares Analysis.. . . . . . . . . . . . . . . . . . . . . . .. . . . . . .301 A.8 Prefixes to Unit and Greek Alphabet.. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .302 A.9 Crystal Structures of Some Elements and Compounds . . . . . . . . .. . . . . . .303 Index . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .305

Chapter 1

Fundamental Properties of X-rays

1.1 Nature of X-rays X-rays with energies ranging from about 100 eV to 10 MeV are classified as electromagnetic waves, which are only different from the radio waves, light, and gamma rays in wavelength and energy. X-rays show wave nature with wavelength ranging from about 10 to 103 nm. According to the quantum theory, the electromagnetic wave can be treated as particles called photons or light quanta. The essential characteristics of photons such as energy, momentum, etc., are summarized as follows. The propagation velocity c of electromagnetic wave (velocity of photon) with frequency and wavelength is given by the relation. .ms1 /

c D

(1.1)

The velocity of light in the vacuum is a universal constant given as c D 299792458 m=s (2:998 108 m=s). Each photon has an energy E, which is proportional to its frequency, E D h D

hc

.J/

(1.2)

where h is the Planck constant (6:6260 1034 J s). With E expressed in keV, and in nm, the following relation is obtained: E.keV/ D

1:240 .nm/

(1.3)

The momentum p is given by mv, the product of the mass m, and its velocity v. The de Broglie relation for material wave relates wavelength to momentum. D

h h D p mv

(1.4)

1

2

1 Fundamental Properties of X-rays

The velocity of light can be reduced when traveling through a material medium, but it does not become zero. Therefore, a photon is never at rest and so has no rest mass me . However, it can be calculated using Einstein’s mass-energy equivalence relation E D mc 2 . EDr

me 2 v 2 c 1 c

(1.5)

It is worth noting that (1.5) is a relation derived from Lorentz transformation in the case where the photon velocity can be equally set either from a stationary coordinate or from a coordinate moving at velocity of v (Lorentz transformation is given in detail in other books on electromagnetism: for example, P. Cornille, Advanced Electromagnetism and Vacuum Physics, World Scientific Publishing, Singapore, (2003)). The increase in mass of a photon with velocity may be estimated in the following equation using the rest mass me : mD r

me v 2 1 c

(1.6)

For example, an electron increases its mass when the accelerating voltage exceeds 100 kV, so that the common formula of 12 mv2 for kinetic energy cannot be used. In such case, the velocity of electron should be treated relativistically as follows: me 2 2 v 2 c me c 1 c s 2 me c 2 vDc 1 E C me c 2

E D mc 2 me c 2 D r

(1.7)

(1.8)

The value of me is obtained, in the past, by using the relationship of m D h=.c/ from precision scattering experiments, such as Compton scattering and me D 9:109 1031 kg is usually employed as electron rest mass. This also means that an electron behaves as a particle with the mass of 9:109 1031 kg, and it corresponds to the energy of E D mc 2 D 8:187 1014 J or 0:5109 106 eV in eV. There is also a relationship between mass, energy, and momentum.

E c

2

p 2 D .me c/2

(1.9)

It is useful to compare the properties of electrons and photons. On the one hand, the photon is an electromagnetic wave, which moves at the velocity of light sometimes called light quantum with momentum and energy and its energy depends upon

1.2 Production of X-rays

3

the frequency . The photon can also be treated as particle. On the other hand, the electron has “mass” and “charge.” It is one of the elementary particles that is a constituent of all substances. The electron has both particle and wave nature such as photon. For example, when a metallic filament is heated, the electron inside it is supplied with energy to jump out of the filament atom. Because of the negative charge of the electron, (e D 1:602 1019 C), it moves toward the anode in an electric field and its direction of propagation can be changed by a magnetic field.

1.2 Production of X-rays When a high voltage with several tens of kV is applied between two electrodes, the high-speed electrons with sufficient kinetic energy, drawn out from the cathode, collide with the anode (metallic target). The electrons rapidly slow down and lose kinetic energy. Since the slowing down patterns (method of loosing kinetic energy) vary with electrons, continuous X-rays with various wavelengths are generated. When an electron loses all its energy in a single collision, the generated X-ray has the maximum energy (or the shortest wavelength D SWL ). The value of the shortest wavelength limit can be estimated from the accelerating voltage V between electrodes. eV hmax c hc SWL D D max eV

(1.10) (1.11)

The total X-ray intensity released in a fixed time interval is equivalent to the area under the curve in Fig. 1.1. It is related to the atomic number of the anode target Z and the tube current i : Icont D AiZV 2

(1.12)

where A is a constant. For obtaining high intensity of white X-rays, (1.12) suggests that it is better to use tungsten or gold with atomic number Z at the target, increase accelerating voltage V , and draw larger current i as it corresponds to the number of electrons that collide with the target in unit time. It may be noted that most of the kinetic energy of the electrons striking the anode (target metal) is converted into heat and less than 1% is transformed into X-rays. If the electron has sufficient kinetic energy to eject an inner-shell electron, for example, a K shell electron, the atom will become excited with a hole in the electron shell. When such hole is filled by an outer shell electron, the atom regains its stable state. This process also includes production of an X-ray photon with energy equal to the difference in the electron energy levels. As the energy released in this process is a value specific to the target metal and related electron shell, it is called characteristic X-ray. A linear relation between the square root of frequency of the characteristic X-ray and the atomic number Z of the target material is given by Moseley’s law.

4

1 Fundamental Properties of X-rays

Fig. 1.1 Schematic representation of the X-ray spectrum

p

D BM .Z M /

(1.13)

Here, BM and M are constants. This Moseley’s law can also be given in terms of wavelength of emitted characteristic X-ray: 1 D R.Z SM /2

1 1 2 2 n1 n2

(1.14)

Here, R is the Rydberg constant (1:0973107 m1 ), SM is a screening constant, and usually zero for K˛ line and one for Kˇ line. Furthermore, n1 and n2 represent the principal quantum number of the inner shell and outer shell, respectively, involved in the generation of characteristic X-rays. For example, n1 D 1 for K shell, n2 D 2 for L shell, and n3 D 3 for M shell. As characteristic X-rays are generated when the applied voltage exceeds the so-called excitation voltage, corresponding to the potential required to eject an electron from the K shell (e.g., Cu: 8.86 keV, Mo: 20.0 keV), the following approximate relation is available between the intensity of K˛ radiation, IK , and the tube current, i , the applied voltage V , and the excitation voltage VK : IK D BS i.V VK /1:67

(1.15)

Here, BS is a constant and the value of BS D 4:25108 is usually employed. As it is clear from (1.15), larger the intensity of characteristic X-rays, the larger the applied voltage and current. It can be seen from (1.14), characteristic radiation is emitted as a photoelectron when the electron of a specific shell (the innermost shell of electrons, the K shell) is released from the atom, when the electrons are pictured as orbiting

1.3 Absorption of X-rays

5

the nucleus in specific shells. Therefore, this phenomenon occurs with a specific energy (wavelength) and is called “photoelectric absorption.” The energy, Eej , of the photoelectron emitted may be described in the following form as a difference of the binding energy (EB ) for electrons of the corresponding shell with which the photoelectron belongs and the energy of incidence X-rays (h): Eej D h EB

(1.16)

The recoil of atom is necessarily produced in the photoelectric absorption process, but its energy variation is known to be negligibly small (see Question 1.6). Equation (1.16) is based on such condition. Moreover, the value of binding energy (EB ) is also called absorption edge of the related shell.

1.3 Absorption of X-rays X-rays which enter a sample are scattered by electrons around the nucleus of atoms in the sample. The scattering usually occurs in various different directions other than the direction of the incident X-rays, even if photoelectric absorption does not occur. As a result, the reduction in intensity of X-rays which penetrate the substance is necessarily detected. When X-rays with intensity I0 penetrate a uniform substance, the intensity I after transmission through distance x is given by. I D I0 ex

(1.17)

Here, the proportional factor is called linear absorption coefficient, which is dependent on the wavelength of X-rays, the physical state (gas, liquid, and solid) or density of the substance, and its unit is usually inverse of distance. However, since the linear absorption coefficient is proportional to density ,.=/ becomes unique value of the substance, independent upon the state of the substance. The quantity of .=/ is called the mass absorption coefficient and the specific values for characteristic X-rays frequently-used are compiled (see Appendix A.2). Equation (1.17) can be re-written as (1.18) in terms of the mass absorption coefficient. I D I0 e

x

(1.18)

Mass absorption coefficient of the sample of interest containing two or more elements can be estimated from (1.19) using the bulk density, , and weight ratio of wj for each element j. X D w1 C w2 CD wj 1 2 j jD1

(1.19)

6

1 Fundamental Properties of X-rays

Fig. 1.2 Wavelength dependences of mass absorption coefficient of X-ray using the La as an example

Absorption of X-rays becomes small as transmittivity increases with increasing energy (wavelength becomes shorter). However, if the incident X-ray energy comes close to a specific value (or wavelength) as shown in Fig. 1.2, the photoelectric absorption takes place by ejecting an electron in K-shell and then discontinuous variation in absorption is found. Such specific energy (wavelength) is called absorption edge. It may be added that monotonic variation in energy (wavelength) dependence is again detected when the incident X-ray energy is away from the absorption edge.

1.4 Solved Problems (12 Examples)

Question 1.1 Calculate the energy released per carbon atom when 1 g of carbon is totally converted to energy. Answer 1.1 Energy E is expressed by Einstein’s relation of E D mc 2 where m is mass and c is the speed of light. If this relationship is utilized, considering SI unit that expresses mass in kg, E D 1 103 .2:998 1010 /2 D 8:99 1013

J

The atomic weight per mole (molar mass) for carbon is 12.011 g from reference table (for example, Appendix A.2). Thus, the number of atoms included in 1 g carbon is calculated as .1=12:011/ 0:6022 1024 D 5:01 1022 because the numbers of atoms are included in one mole of carbon is the Avogadro’s number

1.4 Solved Problems

7

.0:6022 1024/. Therefore, the energy release per carbon atom can be estimated as: .8:99 1013 / D 1:79 109 .5:01 1022 /

J

Question 1.2 Calculate (1) strength of the electric field E, (2), force on the electron F , (3) acceleration of electron ˛, when a voltage of 10 kV is applied between two electrodes separated by an interval of 10 mm.

Answer 1.2 The work, W , if electric charge Q (coulomb, C) moves under voltage V is expressed by W D VQ. When an electron is accelerated under 1 V of difference in potential, the energy obtained by the electron is called 1 eV. Since the elementary charge e is 1:602 1019 (C), 1eV D 1:602 1019 1 D 1:602 1019

(C)(V) (J)

Electric field E can be expressed with E D V =d , where the distance, d , between electrodes and the applied voltage being V . The force F on the electron with elementary charge e is given by; F D eE

(N)

Here, the unit of F is Newton. Acceleration ˛ of electrons is given by the following equation in which m is the mass of the electron: ˛D .1/ E D

eE m

104 .V/ 10 .kV/ D 2 D 106 10 .mm/ 10 .m/

.m=s2 / .V=m/

.2/ F D 1:602 1019 106 D 1:602 1013 .3/ ˛ D

1:602 1013 D 1:76 1017 9:109 1031

.N/

.m=s2 /

Question 1.3 X-rays are generated by making the electrically charged particles (electrons) with sufficient kinetic energy in vacuum collide with cathode, as widely used in the experiment of an X-ray tube. The resultant X-rays can be divided into two parts: continuous X-rays (also called white X-rays) and characteristic X-rays. The wavelength distribution and intensity of continuous X-rays are usually depending upon the applied voltage. A clear limit is recognized on the short wavelength side.

8

1 Fundamental Properties of X-rays

(1) Estimate the speed of electron before collision when applied voltage is 30,000 V and compare it with the speed of light in vacuum. (2) In addition, obtain the relation of the shortest wavelength limit SWL of X-rays generated with the applied voltage V , when an electron loses all energy in a single collision.

Answer 1.3 Electrons are drawn out from cathode by applying the high voltage of tens of thousands of V between two metallic electrodes installed in the X-ray tube in vacuum. The electrons collide with anode at high speed. The velocity of electrons is given by, 2eV mv2 ! v2 D eV D 2 m where e is the electric charge of the electron, V the applied voltage across the electrodes, m the mass of the electron, and v the speed of the electron before the collision. When values of rest mass me D 9:110 1031 .kg/ as mass of electron, elementary electron charge e D 1:602 1019 .C/ and V D 3 104 .V/ are used for calculating the speed of electron v. v2 D

2 1:602 1019 3 104 D 1:055 1016 ; 9:110 1031

v D 1:002 108 m=s

Therefore, the speed of electron just before impact is about one-third of the speed of light in vacuum .2:998 108 m=s/. Some electrons release all their energy in a single collision. However, some other electrons behave differently. The electrons slow down gradually due to successive collisions. In this case, the energy of electron (eV) which is released partially and the corresponding X-rays (photon) generated have less energy compared with the energy (hmax ) of the X-rays generated when electrons are stopped with one collision. This is a factor which shows the maximum strength moves toward the shorter wavelength sides, as X-rays of various wavelengths generate, and higher the intensity of the applied voltage, higher the strength of the wavelength of X-rays (see Fig. 1). Every photon has the energy h, where h is the Planck constant and the frequency. The relationship of eV D hmax can be used, when electrons are stopped in one impact and all energy is released at once. Moreover, frequency () and wavelength () are described by a relation of D c=, where c is the speed of light. Therefore, the relation between the wavelength SWL in m and the applied voltage V may be given as follows: SWL D c=max D hc=eV D

.12:40 107 / .6:626 1034 / .2:998 108 / D .1:602 1019 /V V

This relation can be applied to more general cases, such as the production of electromagnetic waves by rapidly decelerating any electrically charged particle including

1.4 Solved Problems

9

electron of sufficient kinetic energy, and it is independent of the anode material. When wavelength is expressed in nm, voltage in kV, and the relationship becomes V D 1:240. 10

100kV

Intensity [a.u.]

8 80kV

6

4

60kV

2

40kV 20kV

0 0.01

0.02

0.05

0.1

0.2

0.4

Wavelength [nm]

Fig. 1 Schematic diagram for X-ray spectrum as a function of applied voltage

Question 1.4 K˛1 radiation of Fe is the characteristic X-rays emitted when one of the electrons in L shell falls into the vacancy produced by knocking an electron out of the K-shell, and its wavelength is 0.1936 nm. Obtain the energy difference related to this process for X-ray emission.

Answer 1.4 Consider the process in which an L shell electron moves to the vacancy created in the K shell of the target (Fe) by collision with highly accelerated electrons from a filament. The wavelength of the photon released in this process is given by , (with frequency ). We also use Planck’s constant h of .6:626 1034 Js/ and the velocity of light c of .2:998 108 ms1 /. Energy per photon is given by, E D h D

hc

Using Avogadro’s number NA , one can obtain the energy difference E related to the X-ray release process per mole of Fe. E D

0:6022 1024 6:626 1034 2:998 108 NA hc D 0:1936 109 11:9626 107 D 6:1979 108 J=mole D 0:1936

10

1 Fundamental Properties of X-rays

Reference: The electrons released from a filament have sufficient kinetic energy and collide with the Fe target. Therefore, an electron of K-shell is readily ejected. This gives the state of FeC ion left in an excited state with a hole in the K-shell. When this hole is filled by an electron from an outer shell (L-shell), an X-ray photon is emitted and its energy is equal to the difference in the two electron energy levels. This variation responds to the following electron arrangement of FeC . Before release After release

K1 L8 M14 N2 K2 L7 M14 N2

Question 1.5 Explain atomic density and electron density.

Answer 1.5 The atomic density Na of a substance for one-component system is given by the following equation, involving atomic weight M , Avogadro’s number NA , and the density . Na D

NA : M

(1)

In the SI system, Na .m3 /, NA D 0:6022 1024 .mol1 /, .kg=m3 /, and M .kg=mol/, respectively. The electron density Ne of a substance consisting of single element is given by, Ne D

NA Z M

(2)

Each atom involves Z electrons (usually Z is equal to the atomic number) and the unit of Ne is also .m3 /. The quantity Na D NA =M in (1) or Ne D .NA Z/=M in (2), respectively, gives the number of atoms or that of electrons per unit mass (kg), when excluding density, . They are frequently called “atomic density” or “electron density.” However, it should be kept in mind that the number per m3 (per unit volume) is completely different from the number per 1 kg (per unit mass). For example, the following values of atomic number and electron number per unit mass (D1kg) are obtained for aluminum with the molar mass of 26.98 g and the atomic number of 13: 0:6022 1024 D 2:232 1025 .kg1 / 26:98 103 0:6022 1024 Ne D 13 D 2:9 1026 .kg1 / 26:98 103

Na D

Since the density of aluminum is 2:70 Mg=m3 D 2:70 103 kg=m3 from reference table (Appendix A.2), we can estimate the corresponding values per unit volume as Na D 6:026 1028 .m3 / and Ne D 7:83 1029 .m3 /, respectively.

1.4 Solved Problems

11

Reference: Avogadro’s number provides the number of atom (or molecule) included in one mole of substance. Since the atomic weight is usually expressed by the number of grams per mole, the factor of 103 is required for using Avogadro’s number in the SI unit system. Question 1.6 The energy of a photoelectron, Eej , emitted as the result of photoelectron absorption process may be given in the following with the binding energy EB of the electron in the corresponding shell: Eej D h EB Here, h is the energy of incident X-rays, and this relationship has been obtained with an assumption that the energy accompanying the recoil of atom, which necessarily occurs in photoelectron absorption, is negligible. Calculate the energy accompanying the recoil of atom in the following condition for Pb. The photoelectron absorption process of K shell for Pb was made by irradiating X-rays with the energy of 100 keV against a Pb plate and assuming that the momentum of the incident X-rays was shared equally by Pb atom and photoelectron. In addition, the molar mass (atomic weight) of Pb is 207.2 g and the atomic mass unit is 1amu D 1:66054 1027 kg D 931:5 103 keV.

Answer 1.6 The energy of the incident X-rays is given as 100 keV, so that its momentum can be described as being 100 keV=c, using the speed of light c. Since the atom and photoelectron shared the momentum equally, the recoil energy of atom will be 50 keV=c. Schematic diagram of this process is illustrated in Fig. 1.

Fig. 1 Schematic diagram for the photo electron absorption process assuming that the momentum of the incident X-rays was shared equally by atom and photoelectron. Energy of X-ray radiation is 100 keV

On the other hand, one should consider for the atom that 1amu D 931:5103 keV is used in the same way as the energy which is the equivalent energy amount of the rest mass for electron, me . The molar mass of 207.2 g for Pb is equivalent to

12

1 Fundamental Properties of X-rays

207.2 amu, so that the mass of 1 mole of Pb is equivalent to the energy of 207:2 931:5 103 D 193006:8 103 keV=c. When the speed of recoil atom is v and the molar mass of Pb is MA , its energy can be expressed by 12 MA v2 . According to the given assumption and the momentum described as p D MA v, the energy of the recoil atom, ErA , may be obtained as follows: ErA D

1 p2 .50/2 M A v2 D D 0:0065 103 D 2 2MA 2 .193006:8 103 /

.keV/

The recoil energy of atom in the photoelectron absorption process shows just a very small value as mentioned here using the result of Pb as an example, although the recoil of the atom never fails to take place. Reference:

1:66054 1027 .2:99792 108 /2 D 9:315 108 .eV/ 1:60218 1019 On the other hand, the energy of an electron with rest mass me D 9:109 1031 .kg/ can be obtained in the following with the relationship of 1 .eV/ D 1:602 1019 .J/: Energy of 1 amu D

E D me c 2 D

9:109 1031 .2:998 108 /2 D 0:5109 106 1:602 1019

.eV/

Question 1.7 Explain the Rydberg constant in Moseley’s law with respect to the wavelength of characteristic X-rays, and obtain its value.

Answer 1.7 Moseley’s law can be written as, 1 D R.Z SM /2

1 1 2 2 n1 n2

(1)

The wavelength of the X-ray photon ./ corresponds to the shifting of an electron from the shell of the quantum number n2 to the shell of the quantum number of n1 . Here, Z is the atomic number and SM is a screening constant. Using the elementary electron charge of e, the energy of electron characterized by the circular movement around the nucleus charge Ze in each shell (orbital) may be given, for example, with respect to an electron of quantum number n1 shell in the following form: 2 2 me 4 Z 2 En D (2) h2 n21 Here, h is a Planck constant and m represents the mass of electron. The energy of this photon is given by,

1.4 Solved Problems

13

h D En2 En1 D E D

2 2 me 4 2 Z h2

1 1 2 2 n1 n2

The following equation will also be obtained, if the relationship of E D h D is employed while using the velocity of photon, c: 2 2 me 4 2 1 D Z ch3

1 1 2 2 n1 n2

(3) hc

(4)

If the value of electron mass is assumed to be rest mass of electron and a comparison of (1) with (4) is made, the Rydberg constant R can be estimated. It may be noted that the term of .Z SM /2 in (1) could be empirically obtained from the measurements on various characteristic X-rays as reported by H.G.J. Moseley in 1913. RD

2 .3:142/2 .9:109 1028 / .4:803 1010 /4 2 2 me 4 D 3 ch .2:998 1010 / .6:626 1027 /3 D 109:743 103 .cm1 / D 1:097 107 .m1 /

(5)

The experimental value of R can be obtained from the ionization energy (13.6 eV) of hydrogen (H). The corresponding wave number (frequency) is 109737:31 cm1, in good agreement with the value obtained from (5). In addition, since Moseley’s law and the experimental results are all described by using the cgs unit system (gauss system), 4:803 1010 esu has been used for the elemental electron charge e. Conversion into the SI unit system is given by (SI unit velocity of light 101 ) (e.g., 5th edition of the Iwanami Physics-and-Chemistry Dictionary p. 1526 (1985)). That is to say, the amount of elementary electron charge e can be expressed as: 1:602 1019 Coulomb 2:998 1010 cm=s 101 D 4:803 1010 esu The Rydberg constant is more strictly defined by the following equation: 2 2 e 4 ch3

(6)

1 1 1 D C m mp

(7)

RD

Here, m is electron mass and mP is nucleus (proton) mass.The detected difference is quite small, but the value of mP depends on the element. Then, it can be seen from the relation of (6) and (7) that a slightly different value of R is obtained for each element. However, if a comparison is made with a hydrogen atom, there is a difference of about 1,800 times between the electron mass me D 9:109 1031 kg and the proton mass which is mP D 1:67 1027 kg. Therefore, the relationship of (6) is usually treated as D m, because mP is very large in comparison with me .

14

1 Fundamental Properties of X-rays

Reference: The definition of the Rydberg constant in the SI unit is given in the form where the factor of .1=4 0 / is included by using the dielectric constant 0 .8:854 1012 F=m/ in vacuum for correlating with nucleus charge Ze . 2 2 e 4 RD ch3

1 4 0

2

D

me 4 8 02 ch3

D

9:109 1031 .1:602 1019 /4 8 .8:854 1012 /2 .2:998 108 / .6:626 1034 /3

D

9:109 .1:602/4 10107 D 1:097 107 .m1 / 8 .8:854/2 .2:998/ .6:626/3 10118

Question 1.8 When the X-ray diffraction experiment is made for a plate sample in the transmission mode, it is readily expected that absorption becomes large and diffraction intensity becomes weak as the sample thickness increases. Obtain the thickness of a plate sample which makes the diffraction intensity maximum and calculate the value of aluminum for the Cu-K˛ radiation. t

dx I0 Surface side

x

t-x

I Back side

Fig. A Geometry for a case where X-ray penetrates a plate sample

Answer 1.8 X-ray diffraction experiment in the transmission mode includes both absorption and scattering of X-rays. Let us consider the case where the sample thickness is t, the linear absorption coefficient , the scattering coefficient S , and the intensity of incident X-rays I0 as referred to Fig. A. Since the intensity of the incident X-rays reaching the thin layer dx which is at distance of x from the sample surface is given by I0 ex , the scattering intensity dIx0 from the thin layer dx (with scattering coefficient S ) is given by the following equation: (1) dIx0 D SI0 ex dx The scattering intensity dIx passes through the distance of .t x/ in the sample and the absorption during this passage is expressed by the form of e.t x/ . Therefore, the scattering intensity of the thin layer dx after passing through the sample may be given in the following form: dIx D .SI0 ex dx/e.t x/ D SI0 et dx

(2)

1.4 Solved Problems

15

The scattering intensity of the overall sample will be equal to the result obtained by integrating the intensity of the thin layer dx with respect to the sample thickness from zero to t. Z t SI0 et dx D SI0 t et (3) I D 0

The maximum value of I is given under the condition of dI =dt D 0. dI D SI0 .et tet / D 0; dt

t D 1

!

tD

1

(4)

We can find the values of mass absorption coefficient .=/ and density ./ of aluminum for Cu-K˛ radiation in the reference table (e.g., Appendix A.2). The results are .=/ D 49:6 cm2=g and D 2:70 g=cm3, respectively. The linear absorption coefficient of aluminum is calculated in the following: D

D 49:6 2:70 D 133:92 .cm1/

Therefore, the desired sample thickness t can be estimated as follows: 1 tD D

1 133:92

D 7:47 103 .cm/ D 74:7 . m/

Question 1.9 There is a substance of linear absorption coefficient . (1) Obtain a simple relation to give the sample thickness x required to reduce the amount of transmitted X-ray intensity by half. (2) Calculate also the corresponding thickness of Fe-17 mass % Cr alloy .density D 7:76 106 g=m3 / for Mo-K˛ radiation, using the relation obtained in (1).

Answer 1.9 Let us consider the intensity of the incident X-rays as I0 and that of the transmitted X-rays as I . Then, I D I0 ex If the condition of I D

I0 2

(1)

is imposed, taken into account, one obtains, I0 D Iex 2 1 D ex 2

(2) (3)

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1 Fundamental Properties of X-rays

When the logarithm of both sides is taken, we obtain log 1 log 2 D x log e. The result is log 2 D x, as they are log1 D 0, and loge D 1. Here, natural logarithm is used and the required relation is given as follows: xD

0:693 log2 '

(4)

The values of mass absorption coefficients of Fe and Cr for the Mo-K˛ radiation are 37:6 and 29:9 cm2 =g obtained from Appendix A.2, respectively. The concentration of Cr is given by 17 mass %, so that the weight ratio of two alloy components can be set as wFe D 0:83 and wCr D 0:17. Then, the mass absorption coefficient of the alloy is expressed in the following: D wFe C wCr Alloy Fe Cr D 0:83 .37:6/ C 0:17 .29:9/ D 36:3 .cm2=g/ Next, note that the unit of the density of the Fe–Cr alloy is expressed in cgs, 7:76 106 g=m3 D 7:76 g=cm3. We obtain, Alloy D 36:3 7:76 .cm1 / D 281:7 .cm1 / xD

0:693 D 0:0025 cm D 0:025 mm D 25 m 281:7

Question 1.10 Calculate the mass absorption coefficient of lithium niobate .LiNbO3 / for Cu-K˛ radiation.

Answer 1.10 The atomic weight of Li, Nb, and oxygen (O) and their mass absorption coefficients for Cu-K˛ radiation are obtained from Appendix A.2, as follows: Atomic weight Mass-absorption coefficient (g)

= .cm2 =g/

Li

6.941

0.5

Nb

92.906

145

O

15.999

11.5

The molar weight(molar mass) M per 1 mole of LiNbO3 is given in the following: M D 6:941 C 92:906 C .15:999 3/ D 147:844 .g/ The weight ratio wj of three components of Li, Nb, and O is to be obtained.

1.4 Solved Problems

wLi D

17

6:941 D 0:047; 147:844

wNb D

92:906 D 0:628; 147:844

wO D

47:997 D 0:325 147:844

Then, the mass absorption coefficient of lithium niobate can be obtained as follows: X D wj D 0:047 0:5 C 0:628 145 C 0:325 11:5 LiNbO3 j D 94:8 .cm2 =g/

Question 1.11 A thin plate of pure iron is suitable for a filter for Co-K˛ radiation, but it is also known to easily oxidize in air. For excluding such difficulty,we frequently utilize crystalline hematite powder (Fe2 O3 :density 5:24106 g=m3 ). Obtain the thickness of a filter consisting of hematite powder which reduces the intensity of Co-Kˇ radiation to 1/500 of the K˛ radiation case. Given condition is as follows. The intensity ratio between Co-K˛ and Co-Kˇ is found to be given by 5:1 without a filter. The packing density of powder sample is known usually about 70% of the bulk crystal.

Answer 1.11 The atomic weight of Fe and oxygen (O) and their mass absorption coefficients for Co-K˛ and Co-Kˇ radiations are obtained from Appendix A.2, as follows: Atomic / for Co-K˛ / for Co-Kˇ weight (g)

(cm2 =g)

(cm2 =g)

Fe

55.845

57.2

342

O

15.999

18.0

13.3

The weight ratio of Fe and O in hematite crystal is estimated in the following: MFe2 O3 D 55:845 2 C 15:999 3 D 159:687 55:845 2 D 0:699; wO D 0:301 wFe D 159:687 The mass absorption coefficients of hematite crystals for Co-K˛ and Co-Kˇ radiations are, respectively, to be calculated. ˛ D 0:699 57:2 C 0:301 18:0 D 45:4 .cm2 =g/ Fe2 O3 ˇ D 0:699 342 C 0:301 13:3 D 243:1 .cm2=g/ Fe2 O3

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1 Fundamental Properties of X-rays

It is noteworthy that the density of hematite in the filter presently prepared is equivalent to 70% of the value of bulk crystal by considering the packing density, so that we have to use the density value of f D 5:24 0:70 D 3:67 g=cm3 Therefore, the value of the linear absorption coefficient of hematite powder packed into the filter for Co-K˛ and Co-Kˇ radiations will be, respectively, as follows: ˛ ˛ D f D 45:4 3:67 D 166:6 .cm1 / Fe2 O3 ˇ ˇ D f D 24:1 3:67 D 892:2 .cm1 / Fe2 O3 The intensity ratio of Co-K˛ and Co-Kˇ radiations before and after passing through the filter consisting of hematite powder may be described in the following equation: I ˇ eˇ t ICoKˇ D 0˛ t ICoK˛ I0 e ˛ From the given condition, the ratio between I0˛ and I0ˇ is 5:1 without filter, and it should be 500:1 after passing through the filter. They are expressed as follows: 1 eˇ t 1 D 500 5 e˛ t

!

1 D e.˛ ˇ /t 100

Take the logarithm of both sides and obtain the thickness by using the values of ˛ and ˇ . .˛ ˇ /t D log 100

.* log e D 1;

log 1 D 0/

.166:6 892:2/t D 4:605 t D 0:0063 .cm1/ D 63 . m/

Question 1.12 For discussing the influence of X-rays on the human body etc., it would be convenient if the effect of a substance consisting of multielements, such as water (H2 O) and air (N2 , O2 , others), can be described by information of each constituent element (H, O, N, and others) with an appropriate factor. For this purpose, the value of effective element number ZN is often used and it is given by the following equation: ZN D

q

2:94

a1 Z12:94 C a2 Z22:94 C

where a1 ; a2 : : : is the electron component ratio which corresponds to the rate of the number of electrons belonging to each element with the atomic number

1.4 Solved Problems

19

Z1 ; Z2 ; : : : to the total number of electrons of a substance. Find the effective atomic number of water and air. Here, the air composition is given by 75.5% of nitrogen, 23.2% of oxygen, and 1.3% of argon in weight ratio.

Answer 1.12 Water (H2 O) consists of two hydrogen atoms and one oxygen atom, whereas the number of electrons are one for hydrogen and eight for oxygen. The values of atomic weight per mole (molar mass) of hydrogen and oxygen (molar mass) are 1.008 and 15.999 g, respectively. Each electron density per unit mass is given as follows: For hydrogen For oxygen

0:6022 1024 1 D 0:597 1024 .g1 / 1:008 0:6022 1024 NeO D 8 D 0:301 1024 .g1 / 15:999 NeH D

In water (H2 O), the weight ratio can be approximated by 2=18 for hydrogen and 16=18 for oxygen, respectively. Then, the number of electrons in hydrogen and oxygen contained in 1 g water are 0:597 1024 .2=18/ D 0:0663 1024 and 0:301 1024 .16=18/ D 0:2676 1024 ,respectively, so that the number of electrons contained in 1 g water is estimated to be .0:0663 C 0:2676/ 1024 D 0:3339 1024. Therefore, the electron component ratio of water is found as follows: 0:0663 D 0:199 0:3339 0:2262 aO D D 0:801 0:3339 p 2:94 ZN D 0:199 12:94 C 0:801 82:94 p p 2:94 2:94 D 0:199 C 362:007 D 362:206 D 7:42

aH D

1

1

Here, we use the relationship of ZN D X y ! lnZN D y1 lnX ! ZN D e y lnX On the other hand, the molar masses of nitrogen, oxygen, and argon are 14.01, 15.999, and 39.948 g, respectively. Since 75.5% of nitrogen (7 electrons), 23.2% of oxygen (8 electrons), and 1.3% of argon (18 electrons) in weight ratio are contained in 1 g of air, each electron numbers is estimated in the following: For nitrogen For oxygen For argon

0:6022 1024 0:755 7 D 0:2272 1024 14:01 0:6022 1024 0:232 8 D 0:0699 1024 NeO D 15:999 0:6022 1024 0:013 18 D 0:0035 1024 NeAr D 39:948 NeN D

20

1 Fundamental Properties of X-rays

Therefore, the value of .0:2272 C 0:0699 C 0:0035/ 1024 D 0:3006 1024 is corresponding to the number of electrons in 1 g of air. The rate to the total number of electrons of each element is as follows: 0:2272 D 0:756 0:3006 0:0699 aO D D 0:232 0:3006 0:0035 aAr D D 0:012 0:3006 aN D

Accordingly, the effective atomic number of air is estimated in the following: p 0:756 72:94 C 0:232 82:94 C 0:012 182:94 p p 2:94 2:94 D 230:73 C 104:85 C 58:84 D 394:42 D 7:64

ZN D

2:94

Chapter 2

Geometry of Crystals

2.1 Lattice and Crystal Systems The origin of crystallography can be traced to the study for the external appearance of natural minerals, such as quartz, fluorite, pyrite, and corundum, which are regular in shape and clearly exhibit a good deal of symmetry. A large amount of data for such minerals have been systematized by applying geometry and group theory. “Crystallography” involves the general consideration of how crystals can be built from small units. This corresponds to the infinite repetition of identical structural units (frequently referred to as a unit cell) in space. In other words, the structure of all crystals can be described by a lattice, with a group of atoms allocated to every lattice point. Crystals can be classified into 32 point groups on the basis of eight kinds of symmetry elements. There are seven crystal systems for classification, which consist of 14 kinds of Bravais lattices. For convenience, these relations are illustrated in Fig. 2.1. Furthermore, if it is extended to include space groups, by adding point groups, Bravais lattices, screw axis, and glide reflection axis, there will be 230 classifications in total. In other words, all crystals “belong to one of 230 space groups,” the details available in other books on crystallography (see for example International Tables for X-ray Crystallography published by the International Union of Crystallography). Let us consider the three-dimensional arrangement of points called a point lattice, as shown in Fig. 2.2.When the atomic position or configuration in crystal is described by a lattice point, any point indicates exactly the same environment (Didentical surroundings) as any other point in the lattice. This means the lattice point can be reproduced by repeating a small unit. This small repeating unit is referred to as a unit cell (or sometime called unit lattice) where all sequences can be given by three vectors a, b, and c (or those lengths a, b and c) and the interaxial angles between them, ˛, ˇ, and . The relationship between a, b, c and ˛, ˇ, is illustrated in Fig. 2.3, and these lengths and angles are called the lattice parameters or lattice constants of the unit cell. As shown in Fig. 2.2, there is more than one way to choose a unit cell, so that it is better to select a unit cell in the direction where three axes have the highest

21

22

2 Geometry of Crystals

Fig. 2.1 Symmetry elements in crystals and their relationships for classification

Fig. 2.2 Different ways for selecting a unit cell in a point lattice

symmetry. On the other hand, all three lengths a, b, and c in case of Fig. 2.3 have different values, and all three angles ˛, ˇ, and are also found to be different from each other. This case called as “triclinic system” shows an axis with the altissimo symmetry of only a onefold axis (no symmetry) or 1N rotatory inversion (or rotoinversion) axis. For convenience, some essential points on “how atoms are arranged in a substance” are given below.

2.1 Lattice and Crystal Systems

23

Fig. 2.3 Example of a unit cell

Four macroscopic symmetry operations or symmetry elements are well known: “reflection,” “rotation,” “inversion,” and “rotatory inversion.” For example, several planes of symmetry in a cube are readily noticed. In this symmetry of reflection, if a body shows symmetry with respect to a certain plane passing through it, reflection of either half of the body in the plane as in a mirror makes a body coincident with the other half. A body shows n-fold rotation symmetry around an axis, when a rotation by .360=n/ı or .2=n/ brings it into self-coincidence. One can easily understand that a cube has a fourfold rotation axis normal to each face, a threefold axis along each body diagonal, and twofold axis joining the centers of opposite edges. In general, there are one, two, three, four and six fold axes for a rotation axis, but a onefold axis corresponds to no symmetry at all. A body having an inversion center can bring itself into coincidence, when every point in the body is inverted or reflected at the inversion center. The corresponding points of the body are at equal distances from the center on a line drawn through the center. A cube is known to have such a center at the intersection of its body diagonals. In general, there is either one, two, three, four or sixfold axes for a rotatory-inversion axis. It is noted for an n-fold, rotatory-inversion axis exists when a body comes into coincidence with itself by coupling the rotation by .360=n/ı around the axis followed by inversion operation in a center lying on the axis. By putting lattice points at the corner of these crystal systems for finding a certain minimum set of symmetry elements, seven kinds of crystal systems are obtained as shown in Table 2.1. That is, only seven different kinds of cells are necessary to cover all possible point lattices or all crystals can be classified into one of the seven crystal systems. Nevertheless, there are other ways for fulfilling the condition that each point has identical surroundings. In this regard, Auguste Bravais (physicist in France) found that there are 14 possible point lattices and no more and we use Bravais lattices as shown in Fig. 2.4. Since the unit cell including two or more lattice points is chosen in the Bravais lattice for convenience, some of the Bravais lattices can be expressed by other simple lattices. For example, the face-centered cubic lattice is also described by a trigonal (rhombohedral) lattice which contains only one lattice point (see Question 2.5). The symbols P , F , I , etc. in Fig. 2.4 or Table 2.1 are given on the basis of the following rule. When a unit cell has only one lattice point, it is called a primitive (or simple) cell, and usually represented by P . In addition, although the trigonal (rhombohedral) crystal system can also be classified into primitive, we use R as the symbol. Other symbols are nonprimitive cells and more than one lattice point per cell is included. It may be suggested that any cell containing lattice points only at

24

2 Geometry of Crystals

Table 2.1 Summary of seven crystal systems and Bravais lattices System Axial lengths and angles Bravais lattice Cubic Three equal axes at right angles Simple a D b D c, ˛ D ˇ D D 90ı Body-centered Face-centered

Lattice symbol P I F

Tetragonal

Three axes at right angles, two equals a D b ¤ c, ˛ D ˇ D D 90ı

Simple Body-centered

P I

Orthorhombic

Three unequal axes at right angles a ¤ b ¤ c, ˛ D ˇ D D 90ı

Simple Body-centered Base-centered Face-centered

P I C F

Trigonal

Three equal axes, equally inclined a D b D c, ˛ D ˇ D ¤ 90ı

Simple

R

Hexagonal

Two equal coplanar axes at, 120ı third axis at right angles a D b ¤ c, ˛ D ˇ D 90ı , D 120ı

Simple

P

Monoclinic

Three unequal axes, one pair not at right angles a ¤ b ¤ c,

Simple Base-centered

P C

Simple

P

˛ ¤ D 90ı ¤ ˇ Triclinic Three unequal axes, unequally inclined and none at right angles a ¤ b ¤ c, ˛ ¤ ˇ ¤ ¤ 90ı Also called rhombohedral.

the corners is primitive, whereas one containing additional points in the interior or on a cell face is nonprimitive. Symbols I and F refer to body-centered and facecentered cells, respectively. The symbols A, B, and C represent base-centered cells where the lattice point is given at the center on one pair of opposite faces A, B, or C . Here, the face of C , for example, is the face defined by b-axis and a-axis. There are various substances and the atomic arrangements in these substances reveal a variety of crystal structures characterized by a certain periodicity. Of course, all structures cannot be covered here. However, many of elements in the periodic table are metals and about 70% of them have relatively simple crystal structure with high symmetry, such as the body-centered cubic (bcc), face-centered cubic (fcc), and hexagonal close-packed (hcp) lattices. Typical features of these three crystal structures are summarized in Fig. 2.5. Crystals can be broadly classified into three categories from the point of view of bonding: “metallic,” “ionic,” and “covalent.” In metallic crystals, a large number of electrons (conduction or valence electrons) are free to move the inside of the system, without belonging to specific atoms but shared by the whole system. This bonding arising from a conduction electron is not very strong. For example, the interatomic distances of alkali metals are relatively large, because the kinetic energy of conduction electrons is relatively low at the large interatomic distances. This leads to weak binding and simple structure. On the other hand, ionic crystals consist of positive and negative ions, and the ionic bond results from the electrostatic interaction

2.1 Lattice and Crystal Systems

25

Fig. 2.4 The fourteen Bravais lattices

of oppositely charged ions in the solid state. Typical examples are metal–halogen compounds, and two typical structures found for ionic crystals are sodium chloride and cesium chloride structures. In sodium chloride, NaC and Cl ions are arranged to form the structure as shown in Fig. 2.6. In ionic crystals, ionic arrangements that minimize electrostatic repulsion and maximize electrostatic attraction are preferred. In many cases, the negative ions (anions) of large size are densely arranged so as to avoid their direct contact, and the positive ions (cations) of small size occupy the positions equivalent to the vacant space produced by anions. Therefore, the correlation is recognized between crystal structure and the size ratio, for example, the ratio of ionic radii D rc =ra, where rc and ra are the radii of cation and anion, respectively. When the value of rc =ra is 0.225, one can find the tetrahedral arrangement with the coordination number of 4, and the octahedral arrangement with the coordination number of 6 in the rc =ra D 0:414 case. Thus, the value of rc =ra has a critical value for ionic configurations. For actual ionic crystals, the arrangement is quite likely to avoid direct contact of the same electric-charged ions mainly arising from energetic constraints. Therefore,

26

2 Geometry of Crystals

Fig. 2.5 Typical crystal structures of metallic elements Fig. 2.6 Crystal structure of sodium chloride

many combinations of rc =ra found in an actual ionic crystal show a little bit larger values than the critical values based on spherical models. The covalent bond, having strong directional properties, is the classical electron pair, and silicon and germanium are included in this category. These crystals have the diamond structure with atoms bonded to four nearest neighbors at tetrahedral angles.

2.2 Lattice Planes and Directions The key points for describing crystal planes and directions are discussed below. In order to show a lattice plane, Miller indices are usually employed. Miller indices are defined as the reciprocals of the fractional intercepts which the plane makes

2.2 Lattice Planes and Directions

27

with the crystallographic axes. For example, if a plane is described by the Miller indices of (h k l), the plane makes fractional intercepts of 1= h, 1=k, and 1= l with the axes a, b, and c, respectively. This reciprocal symbolism enables us to give the Miller indices being zero, when a plane is parallel to an axis. For example, the center position of a body-centered cubic lattice is expressed by 12 21 21 and the position of surface-centered lattice as 12 21 0; 12 0 21 ; 0 12 21 . Some generalized rules for presentation are as follows: (1) The distance from the origin to the intersection of the desired plane with each crystal axis is determined from the basis of unit length, such as a lattice parameter. As shown in Fig. 2.7, the a-axis intersects at the unit length of 1= h. (2) The reciprocals of three numbers are taken and let the minimum integer ratio (h k l) be the index of the corresponding plane. (3) If the desired plane is parallel to a certain axis, the distance from the origin in the axis to the intersection becomes infinite. In that case, the index is expressed by zero. For example, (h 0 0) represents a plane parallel to b-axis and c-axis. (4) Although a set of planes parallel to it can be found for every plane, Miller indices usually refer to that plane in the set which is nearest to the origin. (5) When a plane intercepts at the negative side in any axis, such negative value is N represented by writing a bar over the Miller indices, for example, (hN kN l). (6) There are sets of equivalent lattice planes related by symmetry, for example, the N N (001), and (001).They N planes of a cube, (100), (010), (100), (010), are called “planes of a form” and the expression of f001g is used. The number of the equivalent lattice planes in one plane of a form corresponds to the multiplicity factor and they are given for seven crystal systems in Table 2.2.

Fig. 2.7 Example of Miller indices for plane

28

2 Geometry of Crystals

Table 2.2 Multiplicity factors for crystalline powder samples Cubic hkl hkk hk0 hh0 hhh h00 48 24 24 12 8 6 Hexagonal hk l hh l h0 l hk 0 hh 0 h0 0 00 l 24 12 12 12 6 6 2 N Trigonal Referred to hkl kkh hkk hk0 kN hh hhh hh0 h00 rhombohedral axes 12 12 6 12 6 2 6 6 Referred to hk l hh l h0 l hk 0 hh 0 0h 0 00 l hexagonal axes 12 12 6 12 6 6 2 Tetragonal hkl hhl hh0 hk0 h0l h00 00l 16 8 4 8 7 4 2 Orthorhombic hkl hk0 h00 0k0 00l h0l 0kl 8 4 2 2 2 4 4 Monoclinic hkl hk0 0kl h0l h00 0k0 00l (Orthogonal axis: b) 4 4 4 2 2 2 2 Triclinic hkl hk0 0kl h0l h00 0k0 00l 2 2 2 2 2 2 2 In some crystals, planes having these indices comprise of two forms with the same spacing but different structure factor. In such case, the multiplicity factor for each form is half the value given here.

On the other hand, the direction of crystal lattice is given by any coordinates u v w on a line passing through the origin. Note that the indices are not necessarily integer, because this line will also pass through the point of 2u 2v 2w, etc. Nevertheless, the direction is described using the method based on the Miller indices. For example, translation of the origin is carried out to a certain point, and if the set of u v w is found the minimum integer, when the shift is made by moving the point by ua in the direction of a-axis, vb in the direction of b-axis, and wc in the direction of c-axis, the indices of the direction of the line is expressed as [u v w] in a square bracket. Negative indices are written with a bar over the number, for example, [Nu vN w]. N The equivalent direction related by symmetry is called “directions of a form” and described by hu v w i, similar to the plane case. As already described, the direction indices are not necessarily integer. Nevertheless, since all of [ 21 21 1, [112], [224], etc. indicate the same direction, they are usually described by [112]. For convenience, some examples of the plane indices and direction indices are shown in Fig. 2.8. With respect to the hexagonal system, a slightly different method for plane indexing is employed: the so-called Miller–Bravais indices refer to plane indices with four axes such as (h k i l), instead of Miller indices. The unit cell of a hexagonal lattice is given by two equal and coplanar vectors of a1 and a2 with 120ı to one another, and a third axis c at right angle, as shown in Fig. 2.9. In addition, the third axis of a3 , lying on the basal plane of the hexagonal prism, is symmetrically related to a1 and a2 and then it is often used with the other two. The complete hexagonal lattice is obtained by repeated translations of the points at the unit cell corners by

2.2 Lattice Planes and Directions

29

Fig. 2.8 Example of some indices for planes and directions in cubic system

Fig. 2.9 Example of the indices for unit cell, planes, and directions in hexagonal system

the vectors a1 , a2 , and c. It is noted for the Miller–Bravais indices that the relation of i D .h C k/ between h and k is always satisfied. This is because the value of i depends on the h and k values, since the intercepts of a plane on a1 and a2 determine its intercept on a3 (see Fig. 2.9).

30

2 Geometry of Crystals

The indices of directions in the hexagonal system, the notation using four indices [u v t w], is used, and in this case there is a relation t D .u C v/. The indices of some planes and directions in the hexagonal system are illustrated in Fig. 2.9. More details such as interaction of three indices and four indices in the hexagonal crystal lattice are given in Question 2.11.

2.3 Planes of a Zone and Interplanar Spacing As shown earlier, there are sets of equivalent planes by symmetry in any crystal lattice and they are called planes of a form. Atoms in crystals can be arranged not only on a lattice plane but also on a group of straight lines which are mutually parallel. This straight line is called “zone axis” and all the planes parallel to the direction of this line are called “planes of a zone.” Such planes have quite different indices and spacings, but their parallelism to a line is satisfied. For example, the plane of a zone which belongs to the zone axis [001] in a cubic system is shown in Fig. 2.10. If a plane belongs to a zone whose indices are (h k l) and the indices of zone axis are [u v w], the following relation is satisfied: hu C kv C lw D 0:

(2.1)

Let us consider any two planes to be planes of a zone, when they are both parallel to their line of intersection. If these two planes are denoted by (h1 k1 l1 ) and (h2 k2 l2 ), the indices of their zone axis [u v w] are given by the following relations: u D k1 l2 k2 l1 ;

v D l1 h2 l2 h1 ;

w D h1 k2 h2 k1 :

(2.2)

The value of the interplanar spacing d is a function of both the plane indices (h k l) and the lattice parameters (a, b, c, ˛, ˇ, and ). The relationship between the plane indices (h k l) and the interplanar spacing d depends on crystal systems.

Fig. 2.10 Example of the planes belonging to the zone axis [001] in cubic system

2.4 Stereographic Projection

31

For example, the interplanar spacing d of the plane of (h k l) for the cubic and tetragonal systems is given in the following equations: d D p

a

h2 C k 2 C l 2 a d D 2 h C k 2 C l 2 .a2 =c 2 /

(cubic) (tetragonal)

(2.3) (2.4)

It may be worth mentioning that lower indices the plane has, larger the value of interplanar spacing becomes, and the density of the lattice points in the corresponding plane also becomes large.

2.4 Stereographic Projection To display the angular relationships between planes and directions in a crystal distributed over three dimensions, various methods are employed. For this purpose, the stereographic projection based on spherical projection is very common in crystallography, because this projection method enables us to permit graphical solution of angular problems between planes. In spherical projection, the direction of a plane when placing the crystal at the center of the sphere is represented by a point that the straight line drawn in the direction that passes through the center of the sphere intersects the surface of the sphere. The sphere is called a reference sphere or a projection sphere. The direction of any plane can be represented by the inclination of the normal to that plane. Then, all the planes in a crystal can be described by a set of plane normals radiating from one point within the crystal. If a reference sphere is placed about this point, the plane normals intersect the surface of the sphere in a set of points called poles. The pole position on the sphere represents the direction of the corresponding plane. The plane can also be represented by the trace (line) the extended plane makes in the sphere surface. The spherical projection can accurately represent the symmetry of the angular relationships between planes and directions as well as zone, but the use of sphere is not always convenient, because the measurement of angles on a flat sheet is more convenient in comparison with measurements on the surface of a sphere. For this purpose, the stereographic projection is widely used. The method is similar to that used by geographers who want to transfer a world map from a terrestrial globe to a sheet of an atlas. Particularly, the equiangular stereographic projection is preferred in crystallography, because it preserves angular relationships faithfully, although area is distorted. Stereographic projection is one of the perspective projection methods. As shown in Fig. 2.11, the projection plane is normal to the line NS that connects two poles, N (north pole) and S (south pole), and at the midpoint of the diameter AB of projection sphere a crystal C is placed. A light source is set at S (south pole), whereas the observer views the projection from N (north pole) just opposite the light source.

32

2 Geometry of Crystals

Fig. 2.11 Fundamentals of stereographic projection Fig. 2.12 The intersection of lines drawn between the poles of planes in the northern hemisphere and the south poles is recorded on the equatorial plane in the stereographic projection

If a certain plane of the crystal has its pole at P, the stereographic projection of P can be obtained as P0 , by drawing the line NP, and it will intersect the projection plane. Alternatively stated, if a pole P is located in the southern hemisphere, its stereographic projection P0 is made from the arctic (north pole) N being the point of perspective and P0 corresponds to the intersection of a straight line NP with a projection plane. If a pole Q is in the northern hemisphere, consider the intersection Q0 with the straight line SQ which makes the Antarctic (south pole) S the point of perspective. It may be added that the stereographic projection of the pole Q is the shadow cast denoted by Q0 on the projection plane when a light source is placed at S. As shown in Fig. 2.12, a line for each of the poles in the northern hemisphere is projected to the south pole and its intersection with equatorial plane of the equator can be marked with a point. By this method, all poles can be depicted inside an equatorial circle (basic circle). In this case, it is required to distinguish the projecting point with N being the point

2.4 Stereographic Projection

33

of perspective and the projecting point with S being the point of perspective. Such issue is easily resolved by using different symbols, for example, for the former and ı for the latter. Great circles on the reference sphere project as circular arcs, whereas small circles project as circles, but their projected center does not coincide with their projection center. One can also select any arbitrary plane perpendicular to NS besides the equatorial plane as a projection plane. In this case, only the diameter of the basic circle changes but the relative positions of projections are unchanged. The net graphics obtained by projecting meridian circles and latitude circles at every 1ı or 2ı on the equatorial plane is called polar net. Polar net is used for obtaining the projecting point on the equatorial plane with respect to a point on a projection sphere. When considering a terrestrial globe, the longitudinal lines correspond to great circles, whereas the latitude lines are small circles, except the equator. The net graphics obtained by projecting the meridian circles and latitude circles on one meridian circle is called Wulff net. In this case, the longitude lines are drawn by the stereographic projection of the great circles connecting the north and south poles of the net at interval of 2ı , and they are displayed with thick arc (line) at every 10ı . The latitude lines on the Wulff net are obtained by the stereographic projection of the small circle extending from side to side at intervals of 2ı , corresponding to the intersection of a projection sphere with a plane perpendicular to NS axis. The Wulff net is quite convenient for estimating the angle between two planes. In the analysis using the Wulff net, a tracing paper containing the stereographic projection is usually placed on a copy of the Wulff net and the centers are made coincident and fixed by a tack. The stereographic projection is made on a tracing paper with the basic circle of the same diameter as that of the Wulff net. The center of the stereographic projection always coincide with the Wulff net center. Although the uncertainty of the angle determined from the Wulff net analysis is about 1ı , it is sufficient in most cases. It should be remembered that the angle between two poles is taken to average the angle between two normals, n and it is not the dihedral angle ( ) between the corresponding two planes P1 and P2 . Since the poles of the planes lie on great circle (the zone circle), these two angles are simply related as n D 180ı , as shown in Fig. 2.13. Some essential points of the stereographic projection are given below.

Fig. 2.13 Relationship between the angle of intersection of normal and the dihedral angle formed by two planes P1 and P2

34

2 Geometry of Crystals

Fig. 2.14 Methods for obtaining (a) the pole of great circles, (b) the pole of small circles, and (c) the angle between the two lattice planes A and B using the stereographic projection

In the stereographic projection, circles on the reference (projection) sphere project as circles or two circular arcs, if they do not pass through the points, N and S, being the point of perspective. Whereas, circles project as straight lines through the center of the projection, if they pass through the points, being the perspective point. In other words, projected great circles always pass the intersection between the basic circle and the straight line passing through a center of the basic circle. This is called “theorem of corresponding circle to circle.” On the other hand, the angle given by two stereographically projected great circles (it also includes the straight line case) is equal to the spherical angle of two great circles on the reference sphere. This is called “theorem of conformal mapping.” To obtain the pole of a great circle on the projection plane, with respect to a great circle ACB in Fig. 2.11 (see also Fig. 2.14a), the diameter FOG which is perpendicular to the diameter AOB is put on the equator of the Wulff net, and P is taken from C at 90ı . On the other hand, to obtain the pole P of a small circle on the projection plane (see Fig. 2.14b), the diameter AOBO1 CD is taken with the center of a small circle as O1 ; then, this is put on the equator of the Wulff net and set point P dividing equally the angle between BC into two parts. To know the angle between the lattice planes A and B, put on the projection on the Wulff net at first, and as shown in Fig. 2.14c, it is rotated and it is made for A and B so as to get on one meridian circle. Next, if we measure the value of angle on the meridian, it is equivalent to the angle of interest. Rotation of the stereographic projection is readily made by using the polar net and the Wulff net. In addition, the so-called standard projection is very useful for discussing problems of crystal orientation, because it gives the relative orientation of all the important planes in the crystal at a single look. Such projections are obtained by choosing some important lattice planes of low indices as the projection plane, such as the lattice planes of (110), (100), (111), or (0001), which are frequently encountered. In this process, the projection of main poles is made by placing a crystal so as to coincide the directions of [100], [110], [111], or [0001] with the north–south NS axis given in Fig. 2.11. Some standard projections, such as cubic crystals on (001) and on (011) are available in textbooks or handbooks for X-ray analysis. When this is utilized, one can obtain information about the relative relationships of main lattice planes in a crystal. It is also extremely convenient for dealing with problems of crystal orientation. Some selected examples are given in Questions 2.19–2.22.

2.5 Solved Problems

35

2.5 Solved Problems (21 Examples)

Question 2.1 Illustrate (100), (110), (111), and (112) planes in cubic lattice N and direction indices of [010], [111], [100], and [120].

Answer 2.1 About given Miller indices, planes are shown in Fig. 1 and directions are illustrated in Fig. 2.

Fig. 1 Examples of the plane indices in cubic lattice

Fig. 2 Examples of the direction indices in cubic lattice

In addition, there are sets of equivalent lattice planes related by symmetry, for examN N (001), and (001). N They are called ple, the faces of a cube, (100), (010), (100), (010), planes of a form and the expression of f001g is used. Similarly, with respect to the direction, it is shown for [110], [101], [011], etc. as h110i. Question 2.2 Answer the following questions about body-centered cubic (bcc) structure with the lattice parameter “a.” (1) Obtain the volume of void, supposing the case where the spherical atoms of radius rA are arranged in each lattice point. Calculate also the porosity and packing fraction.

36

2 Geometry of Crystals

(2) The position of the maximum void in this body-centered cubic lattice is known to be corresponding to the tetrahedral site (1/2, 1/4, 0), and to equivalent position. Obtain the radius of maximum sphere that fits to this space.

Answer 2.2 (1) In body-centered cubic (bcc) structure, atoms contacting each other are seen on diagonals and then we obtain the following relationship: p 4 rA D a 3 p 3 rA D a 4 In a unit cell of bcc structure, there are two atoms: one atom at eight corners (8 1=8 D 1) and one atom at the center. Therefore, the volume VA occupied by atoms is described by 4 VA D 2 3

p !3 p 3 3 3 a D a : 4 8

The unit cell volume is expressed by a3 , so that the void volume VH is as follows: p ! 3 3 a3 : VH D a VA D 1 8 The porosity in the bcc lattice is given in the following:

VH a3

p !3 3 D 1 D 0:32: 8

Therefore, the packing fraction of bcc lattice is 0.68. (2) If the radius of the sphere which fits to void is rX , the following relationship is obtained by geometric conditions (see Fig. 1): .rA C rX /2 D

a 2 4

C

a 2 2

:

p

Next, using the relation of rA D 43 a in the bcc lattice, the radius of the maximum sphere which fits to void is given by following equations: !2 p a 2 a 2 5 2 3 a C rX D a ; C D 4 4 2 16

p rX D

5 4

p ! 3

a

2.5 Solved Problems

37

Reference: You will understand that the maximum radius which fits to void in the bcc lattice is about 30% of the radius of the constituent atom in the following result: p rX D rA

p 5 3 a 4 p 3 4 a

p 5 D p 1 D 0:29: 3

Fig. 1 Tetrahedral voids in the bcc lattice. (Filled circle) Metal atoms, (open circle) Tetrahedral void

Question 2.3 At 278 K, iron (Fe) is found to show bcc structure with a lattice parameter of 0.2866 nm. Obtain the density of iron from this information.

Answer 2.3 The bcc structure includes two atoms per unit cell. If Avogadro’s number is NA , one mole of iron includes NA =2 unit cells. Therefore, the volume V per mole of Fe (atomic volume) is given by V D

.0:2866 109 /3 : 0:6022 1024 =2

The atomic weight M (molar mass) per 1 mol of Fe is 55.845 g is obtained from Appendix A.2. Therefore, from the relationship of V D M=, we can estimate the density value of as follows: D

55:845 2 D 7:88 106 g=m3: .0:2866 109 /3 0:6022 1024

Reference: The experimental value of density for Fe is 7:87106 g=m3 . Since some defects such as vacancy and dislocation are usually included in an actual crystal, there are some differences between the density estimated from the X-ray structure data and the experimental value.

38

2 Geometry of Crystals

Question 2.4 Beryllium (Be) mineral is expressed by a chemical formula (3BeO Al2 O3 6SiO2 ), and it is revealed that the structure is hexagonal with the lattice parameters a D 0:9215 nm and c D 0:9169 nm, and density 2:68 106 g=m3 . Obtain the numbers of molecules contained in a unit cell.

Answer 2.4 At first, we obtain the molecular weight of beryllium mineral using a chemical formula from the molecular weight per 1 mol of the individual oxide component: SiO2 D 60:08 g BeO D 25:01 g; Al2 O3 D 101:96 g; 3BeO C Al2 O3 C 6SiO2 D 537:47 g=mol If this molecular weight is divided by Avogadro’s number, one obtain the value equivalent to the weight of one beryllium mineral molecule.

Fig. 1 Geometric feature found in hexagonal system

Next, we estimate the volume of a unit cell for beryllium mineral from the given values of lattice parameters. As readily seen in Fig. 1, in a unit cell of hexagonal q 2 system, the value of c is given by twice the height of 3 a for the regular tetrahedron of length a of one qside,and the area of the parallelogram which corresponds 3 2 .Therefore, the volume V of a unit cell of hexagonal to the base is given by 4a system is given in the following equation (see also Appendix A.6): p 3 2 V D a c D 0:866a2 c 2 D 0:866 .0:9215 109 /2 .0:9169 109 / D 0:6743 1027 Œm3 : The product of the volume of a unit cell and the density corresponds to the weight of one beryllium mineral molecule, so that if this value is compared with the value

2.5 Solved Problems

39

calculated from molecular weight and Avogadro’s number, the number of molecules in a unit cell will be obtained: 0:6743 1027 2:68 106 D 2:02: 537:47 0:60221024

Thus, the number of molecules in a unit cell is estimated to be two. Question 2.5 (1) Illustrate that a trigonal cell (it is also called rhombohedral cell) is recognized in the face-centered cubic (fcc) cell. (2) The fcc structure is known in the close-packed arrangement of spheres with the identical size and its layer stacking sequence of ABCABC type. Illustrate that the layer stacking sequence of ABAB type found in the hexagonal close-packed (hcp) structure is also detected.

Answer 2.5 If six atoms in the six cell faces of the fcc lattice and two atoms of both sides of a diagonal line in a cubic are tied, a trigonal (rhombohedral) cell is built up as shown in Fig. 1. When the lattice parameterpof fcc lattice is set as “a0 ,” the lattice parameter of trigonal lattice is given by a0 = 2. The lattice parameters of the trigonal system are p a D b D c D a0 = 2;

˛ D ˇ D D 60ı :

Next, if you look at the (111) plane centering on the [111] direction of the fcc lattice, a part of the (0002) plane of hexagonal close-packed (hcp)-type stacking will be recognized, as seen in Fig. 2. In this case, when two layers corresponding to the (111) plane are considered as A and B layers, the atomic position of the corner of a cubic lattice in the diagonal line corresponds to the C layer.

Fig. 1 The trigonal (rhombohedral) lattice recognized in fcc lattice

40

2 Geometry of Crystals

Fig. 2 Stacking of atoms in fcc lattice and hcp lattice

Question 2.6 The atomic weight per 1 mol of copper (Cu) with face-centered cubic (fcc) structure and the density at 298 K are 63.54 g and 8:89 106 g=m3 , respectively. Estimate the nearest-neighbor distance of Cu atoms.

Answer 2.6 In the fcc lattice, four atoms are known to be included in a unit cell. When Avogadro’s number is denoted by NA , 1 mol Cu (63.54 g) includes NA =4 unit cells. If the lattice parameter is set as “a,” the volume of 1 mol Cu (Dthe atomic volume) V can be expressed as V D a3 NA =4. On the other hand, we obtain the relationship of a3 NA =4 D M=; using the atomic weight M and density , the lattice parameter can be estimated as follows: a3 D

4 63:54 0:6022 1024 8:89 106

a D 3:621 1010 m:

The nearest-neighbor distance r of Cu atoms can be calculated since Cu atoms are in contact along the diagonal line of a cell face in the fcc structure: p r D a= 2 D 2:560 1010 m D 0:2560 nm:

2.5 Solved Problems

41

Question 2.7 Answer the following questions about gold (Au), which has fcc structure with the lattice parameter a D 0:4070 nm. (1) Obtain the nearest-neighbor distance, the second nearest-neighbor distance, and their coordination numbers. (2) Obtain the values of density and packing fraction when the density of gold atoms are considered as hard spheres. (3) Obtain the maximum radii of the spheres which just fit the octahedral and tetrahedral voids produced in the fcc structure consisting of hard spheres.

Answer 2.7 (1) When you look at the characteristic feature of the fcc lattice, the distance between the atoms which occupy the corners of a unit cell is equal to the lattice parameter “a,” and the distance between the atom located at a center pof p position D a= 2. the cell face and the atom which occupies a corner position is a 2=2 p Thus, the nearest-neighbor distance is estimated to be r1 D p a= 2 and the second nearest-neighbor distance is r2 D a. It is seen that a= 2 < a. a 0:4070 D 0:2878 nm r1 D p D 1:4142 2

r2 D a D 0:4070 nm:

Fig. 1 Geometric feature found in fcc lattice

With respect to the coordination number in the nearest-neighbor distance, let us consider the atom A occupying the corner position (see Fig. 1). At the distance of r1 from atom A, there are four atoms marked by B1 which occupy the center position of the cell face, considering the four equivalent planes for a unit cell around A. One can also find two planes cross at right angles to this face and consider the situation in both upper and lower sides. For example, there are four atoms marked by B2 in both sides and similarly we find four atoms corresponding to B3 . Thus, the total nearest neighbors are 3 4 D 12. You may also estimate the coordination numbers in nearest-neighbor region from the characteristic features of the fcc lattice, which has close-packed

42

2 Geometry of Crystals

Fig. 2 Stacking of atoms in fcc lattice

arrangement of spheres with the identical sizes and layer stacking sequence of ABCABC type. When considering the environments with respect to the sphere which is located in the layer B with the help of Fig. 2, there are three atoms in both layers of A and C and six atoms in layer of B, so that we can find of total of 12 atoms in the nearest-neighbor distance. For the second nearest-neighbor case, let us consider the environments around the A atom occupying the corner position in Fig. 1. There are four atoms in the plane and two atoms at both sides. Therefore, the coordination number of the second nearest-neighbor atoms is 6. (2) The volume V of a unit cell for fcc structure is given by a3 . On the other hand, one atom located at a corner is shared by eight cells and atom in the face center is shared by two cells. Then the number of atoms “n” which belongs to a unit cell is n D 18 8 C 12 6 D 4. Since the atomic weight of 1 mol. Au is 196.97 g from Appendix A.2, the mass m of one atom of Au can be computed using Avogadro’s number as follows: mD

196:97 D 327:0 1024 Œg: 0:6022 1024

Then, we obtain the density value of : D

4m 4 .327:0 1024 / 1:308 1021 D D D 19:41 106 g=m3 : 3 9 3 a .0:4070 10 / 0:0674 1027

(Reference: Measured density value D 19:28 106 g=m3 .) In fcc structure, the atoms are in contact along the diagonal line of a cube face, so that p the lattice parameter a is related to the atomic radius r by the relation 4r D a 2: p a 2 a ! r D 3=2 : rD 4 2 The volume V 0 of four atomic spheres contained in a unit cell is obtained: 0

V D4

a 3 a3 4 4 3 16 a3 r D 4 3=2 D p D p : 3 3 3 2 16 2 3 2

2.5 Solved Problems

43

Since the volume V of a unit cell is given by a3 , the packing fraction is obtained from the relation of V 0 =V : a3 D p 3 2

a3 D p D 0:741: 3 2

(3) Let us consider the case where the hard sphere with the atomic radius r D a=23=2 is arranged in the fcc lattice. As shown in Fig. 3a, the void is found at the center of a unit cell as well as the midpoint of each edge-line. The void at the center of unit cell is surrounded by six spheres and constitutes the octahedral interstitial site. Since such void position forms the fcc lattice with the lattice parameter “a,” the octahedral void has four equivalent positions per unit cell. The tetrahedral void of the regular tetrahedron surrounded by four spheres is shown in Fig. 3b. Such tetrahedral voids form a simple cubic lattice with the lattice parameter a=2. This tetrahedral void has eight equivalent positions per unit cell.

Fig. 3 (a) Octahedral void and (b) tetrahedral void in fcc lattice

Next, we will estimate the maximum radius of the sphere which is just fit to the octahedral void, ro , and tetrahedral void, rt , in fact structure. With the help of the relationships found in Fig. 3a, geometry of the octahedral void is illustrated in Fig. 4 and we obtain the value of ro in the following way: p AC D 2.r C ro / D .AB2 C BC2 /1=2 D 2 2r p ro D r. 2 1/ D 0:414r With respect to the tetrahedral voids, found in Fig. 5 is helpful. p a relationshipp For triangle ABC, we find AC D a 2 and AB D 3r. The relationship of AB D BC is also recognized. p Considering triangle ABM (or CBM), the distance BM is given as BM D 2r=2. When the center of the tetrahedral void is denoted by O, the position of p O corresponds to the center of the distance BM, so that it is given by OM D 2r=2.

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2 Geometry of Crystals

Fig. 4 Geometry of the octahedral void in fcc lattice

Fig. 5 Geometry of the tetrahedral void in fcc lattice

On the other hand, we take triangle AOM that AM is a half of AC and r D a=23=2 and the angle AMO is 90 degrees, so that from the relationship of AO2 D AM2 C OM2 , we obtain p !2 a 2 2r AO D 3=2 C D 2 2 3 1 2 r D r2 D 1C 2 2 r 3 r: AO D 2 2

p !2 p 2 2r r 2 C p .* a D 2 2r/ p 2 2 2

Using the relationship of AO D r C rt , the desired value is obtained as follow: r r C rt D

3 r; 2

r rt D

! 3 C 1 r D 0:225r: 2

From these results, we obtain that a sphere of radius about 41% of main constituent atoms can fit into the octahedral void in fcc structure and a sphere with radius about 23% of main constituent atoms can fit into the tetrahedral void.

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Question 2.8 If spheres of equal size are used to fill space, there are two ways for arranging spheres; in square form and in hexagonal form. (1) Compute the percentage of void of these two cases for two-dimensional array of spheres. (2) Explain the packing fraction of three-dimensional array of spheres.

Answer 2.8 In two-dimensional array of spheres, each sphere is found to be in contact with four spheres in the square form. On the other hand, each sphere contacts with six spheres in the hexagonal form, as illustrated in Fig. 1. This implies that the coordination numbers of the nearest neighbors are 4 in the square form and 6 in the hexagonal form, respectively. (1) When the radius of a sphere is given by r, the area produced by one set of sphere array in the square form is expressed as 2r 2r D 4r 2 . Since the area occupied by a sphere is r 2 , the percentage of void area AV in the square form is as follows: 4r 2 r 2 D 1 D 0:215: AV D 4r 2 4

Fig. 1 Two-dimensional array of spheres in the square and hexagonal forms

In the hexagonal form, our notice focuses the area of an equilateral triangle with p one edge-line of 2r (see Fig. 1). Since thepheight of the equilateral triangle is 3r, the area of this triangle is given by 3r 2 . The sum of an (adjacent) interior angle of the equilateral triangle is 180ı . This suggests that the area occupied by a sphere is expressed with r 2 =2 , because it is equivalent to half of a sphere. Therefore, the percentage of void area AV in the hexagonal form is p 2 r2 3r 2 D1 p D 0:093: AV D p 2 3r 2 3

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2 Geometry of Crystals

(2) With respect to the packing fraction of three-dimensional array of the equal size spheres, the value in the hexagonal form is higher than the square case. This is readily found from the results of Question (1). Question 2.9 Iron (Fe) is present as -phase characterized by face-centered cubic (fcc) structure with the lattice parameter of 0.3647 nm at temperatures near 1,273 K. The Fe–C alloy containing 2.0 mass% of carbon (C) can form either the interstitial and substitutional solid solution. Calculate the value of density in these two cases and compare with the experimental value of 7:65 106 g=m3.

Answer 2.9 From Appendix A.2, the atomic weight of Fe and C are 55.845 g and 12.011 g, respectively. For the Fe–C alloy containing 2.0 mass% of carbon (C), the atomic percent of the constituent elements can be calculated. At first, we compute the molar values of each component such as 98:0=55:845 D 1:7549 for Fe and 2:0=12:011 D 0:1665 for C, and 1:7549 C 0:1665 D 1:9214 for alloy. Then, the desired values are given by the following: Atomic % of Fe .1:7549=1:9214/ 100 D 91:33 at %: Atomic % of C .0:1665=1:9214/ 100 D 8:67 at %: Density is mass per unit volume. Taking into consideration that four atoms per unit cell are included in fcc structure, the density can be calculated if Fe forms a solid solution with C of either interstitial or substitutional type. (1) Density in interstitial-type solid solution is given by (mass of Fe C mass of C)/ unit volume. Mass Volume Density

4 f55:845 C .8:67=91:33/ 12:011g D 227:94 g .0:3647/3 1027 0:6022 1024 D 29:21 106 m3 227:94=.29:21 106 / D 7:80 106 g=m3

(2) Density in substitutional-type solid solution may be calculated as (mass/unit volume) Mass Volumes Density

4 .0:9133 55:845 C 0:0867 12:011/ D 208:18 g .0:3647/3 1027 0:6022 1024 D 29:21 106 m3 208:18=.29:21 106 / D 7:13 106 g=m3

By comparing the two calculated density values with measured density of 7:65 106 g=m3 , it can be concluded that the Fe–C alloy forms an interstitial solid solution.

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Question 2.10 Copper (Cu) is known to form substitutional solid solution with nickel (Ni) and show face-centered cubic (fcc) structure. For Cu (lattice parameter: 0.3625 nm) which contains 0.001 mass% of Ni, calculate the distance between Ni atoms in this solid solution. Answer 2.10 From Appendix A.2, the atomic weights per mole for Cu and Ni are 63.55 and 53.69 g, respectively. First calculate the atomic percent of Cu and Ni in the alloy. The moles of each component are obtained as 0:001=58:69 D 0:000017 for Cu and 99:999=63:55 D 1:57355 for Ni, and 1:57355 C 0:000017 D 1:573567 for the alloy. Atomic % of Ni Atomic % of Cu

.0:000017=1:573567/ 100 D 0:0011 100 0:0011 D 99:9989

Considering the concentrations of the two components in atomic percent, if there are one million atoms of the alloy in total, 999,989 are Cu atoms and 11 are nickel atoms. This is approximately equivalent to there being one nickel atom as an impurity in 100,000 Cu atoms. Four atoms are contained in a unit cell of fcc structure, 100,000 Cu atoms form 25,000 unit cells. If Ni atoms are thought to replace the position of Cu atoms at random, one Ni atom will be homogeneously distributed in these 25,000 unit cells. In other words, Ni atoms are separated, at least, by multiples of the unit cell; in the present case, .25;000/1=3 D 29:24 unit cells. Therefore, the distance between the impurity Ni atoms contained in Cu is estimated to be 29.24 the lattice parameter .0:3615 nm/ D 10:57 nm. This result suggests that the impurity atoms are located in a relatively close region even for the cases where the impurity content is very small at the level of 0.001%. The impurity effect, for example at the level of 0.001%, cannot necessarily be ignored. Reference: If there are 100 atoms denoted by A, 25 unit cells are known to be formed in fcc structure. Let us consider the A–B binary alloy containing 25 at.% of B atoms with fcc structure. When B atoms replace the position of A atom at random, one atom per unit cell will be replaced with B atom in this alloy. On the other hand, if B atoms are not distributed randomly, but occupy some designated positions, an ordered structure (frequently called superlattice) will be formed. Question 2.11 A slightly different method than the usual Miller indices is employed for indexing planes in the hexagonal system. It is called the Miller– Bravais indices. Explain the essential points of the Miller–Bravais indices including some of its merits for indexing both planes and directions.

Answer 2.11 As shown in Fig. 1, a unit cell of a hexagonal lattice is provided by two equal and coplanar vectors a1 and a2 which are 120ı to one another and the

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2 Geometry of Crystals

third axis c perpendicular to vectors a1 and a2 . The complete hexagonal lattice is given by repeated translations of the points at the unit cell corners by the vectors a1 , a2 , and c. The plane indexed by the usual Miller indices is allowed for hexagonal system. However, it can sometimes cause confusion. For example, “Is the plane of N (100) completely equivalent to the plane of (110)?” Miller–Bravais indices refers to plane indices with four axes such as (h k i l), instead of Miller indices. Then, the use of Miller–Bravais indices enables us to provide one way by showing (100) ! N N N respectively. (1100) and (110) ! (1010), The key points of Miller–Bravais indices for hexagonal system are summarized below: (1) As shown in Fig. 1, we employ four axes, by adding the vector a3 to vectors a1 and a2 . Vector c remains perpendicular to vectors a1 and a2 . (2) Direction of the planes is decided in the same manner as for Miller indices, except for the use of four coordinate axes a1 , a2 , a3 , and c. (3) As a result, the plane indices refer to plane indices with four digit number such as (h k i l); they always have a relationship i D .h C k/. Third point suggests that the value of i depends on h and k values, so that by replacing the index i by a dot, a different plane symbol written as (hk l) is sometimes employed. However, this method is not strongly recommended because the advantage of Miller–Bravais indices (for example, similar indices to similar planes) is not always retained. For example, all six side planes of a hexagonal prism of Fig. 2 are considered crytallographically equivalent and such mutual relationship can be N (0110), N (1100), N N N N readily recognized from (1010), (1010), (0110), and (1100) in the Miller–Bravais notation. Such mutual relationship is not obtained directly from the N N N and (110). N abbreviated symbols ; (100), (010), (110), (100), (010)

Fig. 1 Method for describing the planes in hexagonal system

However, the Miller–Bravais indices for directions in hexagonal system is slightly complicated. The directions in hexagonal system are simply described by three basic vectors a1 , a2 , and c, then we use only three digits such as [U V W ] to represent a direction referred to the three axes. Some examples for direction indices are illustrated in Fig. 2 together with some planes. Just as the case for plane,

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Fig. 2 Typical planes in hexagonal system

we can use four indices [u v t w]. For convenience, some additional notes are given below. The use of four indices for directions in hexagonal system is based on four component vectors, parallel to a1 , a2 , a3 , and c. The third index is known to be equivalent to the sum of the first and second indices with change in sign, t D .u C v/. When [U V W ] are the direction indices referred to three axes, [u v t w] corresponds to the four axes case. The relationships between these two indices are: U Dui V Dvt W Dw

u D .2U V /=3 v D .2V U /=3 t D .u C v/ D .U C V /=3 wDW

N and [210] ! [1010]. N To facilitate underTherefore, we find that [100] ! [21N 10] standing, Fig. 3 shows a simple example of the straight line which passes along the origin on the bottom plane. The coordinates of the desired point in crystal lattice are generally given by distances of parallel translations with respect to each axes required to reach the points from the origin. For describing the directions in hexagonal system, we frequently make a detour so as to obtain the distance equal to the negative value of the sum of the distances x and y, ..x C y// moved by parallel translations with respect to a1 and a2 , respectively. It is noteworthy that this is equivalent to the distance moved by parallel translation with respect to a3 (see Fig. 3). The direction is given by the ratio of the distances moved in parallel to each axes. The slightly complicated procedure for the hexagonal system is based on the objective of providing similar indices to similar directions.

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2 Geometry of Crystals

Fig. 3 Examples of the directions in hexagonal system

Question 2.12 Calculate the bond angle, , of O–Si–O in silica (SiO2 ) assuming that Si and O atoms form regular tetrahedron with Si at the center.

Answer 2.12 Let us consider the case where the peak positions of a cube characterized by the length of a are connected to form regular tetrahedron as shown in Fig. 1.

A

q 2 O P

B

Fig. 1 Geometry in regular tetrahedron

The position of O in this figure corresponds to a center of both cube and regular tetrahedron. The positions denoted by A and B represent corners of regular tetrahedron and P gives the midpoint of regular tetrahedron which is characterized by the length of AB. The angle AOB is equivalent to the desired angle, , of the O–Si–O bond. Therefore, the following relationship is readily found in triangle AOP: p p p a 2 a a 2=2 AB D OP D tan D D 2 AP D 2 2 2 2 a=2 D 109:48ı D 54:74ı 2

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Question 2.13 Silica (SiO2 ) is known to form chain-like, sheet-like or network structures on the basis of tetrahedral unit of SiO4 4 . Figure A shows the atomic arrangement of ˇ-cristobalite at 583 K, which is the high temperature phase where Si atoms form diamond structure and each Si is surrounded by four oxygen atoms to form a tetrahedral unit. (1) Estimate the radius of SiO4 4 ion assuming the condition that the lattice parameter of ˇ-cristobalite is given by 0.716 nm and the radius of oxygen ion is 0.140 nm. (2) Estimate the radius ratio assuming the condition that Si4C ions make contact with O2 ions.

Fig. A Structure of ˇ-cristobalite

Answer 2.13 (1) If the distance between Si–Si is given by R, it is found equivalent to the distance between (000) and .1=4; 1=4; 1=4/, from geometric condition: .R/2 D

1 a 4

2

C

1 a 4

2

C

1 a 4

2

D3

a 2 4

:

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2 Geometry of Crystals

O2 ion combines with two Si4C ions and the condition of keeping the distance of R for two silicon ions, thus we obtain the following relationship: " #1=2 0:716 2 R D 2rSi C 2rO D 3 D 0:310 nm: 4 The radius of SiO4 4 D 0:155 nm: 2rSi D 0:310 2 .0:140/ 2rSi D 0:03

rSi D 0:015 nm

.Reference Si4C D 0:041 nm/

(2) Since the length of edge of SiO4 4 tetrahedron is 2rO (see Fig. 1) and the bond angles of a tetrahedron is given by 109.48ı (see Question 2.12), the following relationship is readily found:

109:48 sin 2

D

rO D 0:816 rO C rSi

rO D 0:816.rO C rSi / D

0:816 rSi .1 0:816/

rSi 1 0:816 D 0:225 D rO 0:816

Fig. 1 Geometry of SiO4 4 tetrahedron

Question 2.14 In ionic crystals, anions of the relatively larger size are densely arranged so as to avoid their direct contact, whereas cations of relatively smaller size occupy the positions equivalent to the vacant space produced by anions. For this reason, if the radii of cation and anion are described by rc and ra , respectively, some correlations are recognized between the coordination numbers and the size ratio of rc =ra. Estimate the specific values of rc =ra for

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cases that cations are surround by anions with the coordination numbers of 3, 4, 6, and 8.

Fig. A Four typical geometries for arraying equal spheres

Answer 2.14 (1) When the coordination number is 3, a cation is likely to occupy the position equivalent to the center of an equilateral triangle of ABC formed by three anions as shown in Fig. 1. The height and the center ofpgravity for the equilateral triangle with its one edge being p 2ra are given by 3ra and the height 2 2 , respectively. Then, the value of 3ra is exactly equivalent to ra C rc 3 3 p 2 3 ra ra C rc D 3

)

rc D

! p 2 3 1 ra 3

p rc 2 3 1 D 0:155 D ra 3

Fig. 1 Geometric relation for the 3-coordination case

(2) When the coordination number is 4, a cation is at the center of a regular tetrahedron, formed by four anions as shown in Fig. 2. Consider the triangle ABC corresponding to the plane where the vacant space existed in the regular tetrahep dron with its one edge being 2ra . We find the relationship, AB D BC D 3ra , so that ABC is an p isosceles triangle. With p respect to the triangle ABM, AM D ra and AB D 3ra . Hence BM D 2ra . If O is the center of the vacant space the distance of AO is exactly equivalent to ra C rc .

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2 Geometry of Crystals

Since the center O of the vacant space is also the center of regular tetrahep dron, O is equivalent to the midpoint of BM. Therefore, OM D 12 BM D 22 ra . Next, if triangle AOM is considered, we obtain the following relationships:

1 AM .AO/ D .ra C rc / D .OM/ C 2 2

2

r ra C rc D

3 ra 2

2

r )

rc D

2

D

! 3 1 ra 2

!2 r p !2 2 3 2 ra C .ra / D ra ; 2 2 rc D ra

r

3 1 D 0:2247: 2

Fig. 2 Geometric relation for the 4-coordination case

(3) When the coordination number is 6, the cation is located at the center of an octahedron formed by six anions. As shown in Fig. 3, the distance of AC or BD, corresponding to the diagonal line of the cross-sectional view of ABCD portion of octahedron, is exactly equal to twice the value of .ra Crc /. When considering that one edge of the cross-sectional view of the square ABCD is 2ra , we obtain the following result: p 2.ra C rc / D 2 2ra

)

p rc D . 2 1/ra

Fig. 3 Geometric relation for the 6-coordination case

p rc D . 2 1/ D 0:414: ra

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(4) For 8-coordination, it is difficult to visualize the cation in the vacant space of a polyhedron formed by anions. Nevertheless, if the appropriate value of rc =ra is given, one can obtain the ionic arrangements in which a cation is located at a center of body-centered cubic (bcc) lattice formed by eight anions. In this case, the diagonal line of the cross-sectional view of ABGF portion of bcc lattice with one edge of 2ra just corresponds to twice the value of .ra C rc / as readilypseen in Fig. 3. Here, the lengths of each edge of square ABGF are 2ra and 2 2ra , respectively. Therefore, 2.ra C rc / D

p

3 .2ra /

ra C rc D

p 3ra

)

p rc D . 3 1/ra

p rc D 3 1 D 0:732 ra

Fig. 4 Geometric relation for the 8-coordination case

Reference: .BF/2 D .BG/2 C .FG/2 p p Œ2.rc C ra /2 D .2 2ra /2 C .2ra /2 D .12ra/2 D .2 3ra /2 :

Question 2.15 Caesium chloride (CsCl) crystal has cubic structure in which CsC ions occupy the center position of a unit cell and its corner positions are occupied by Cl ions. The density of caesium chloride is 3:97 106 g=m3 . (1) If the ionic radii of CsC ion and Cl ion are 0.169 and 0.181 nm, respectively, compute the lattice parameter and compare with the value estimated from density. (2) There is a threshold value of the ratio of the radii of positive/negative ions for alkali halides having CsCl-type structure. The radius r C of a positive ion needs such as to just fit the space formed by eight negative ions of the radius of r without their direct contact. Calculate the minimum value of the ratio of r C =r .

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2 Geometry of Crystals

Answer 2.15 (1) If CsC and Cl ions are represented by black and white circles, respectively, geometry of a unit cell of thepCsCl is shown in Fig. 1. Let a be the lattice parameter. Thus, AB D a; AC D 3a. The CsC and Cl ions are arranged along the diagonal AC such that AC D 2.r C C 2r / D 2.0:169 C 0:181/ D aD

p 3a;

2.0:169 C 0:181/ p D 0:404 nm: 3

Fig. 1 Geometry of CsCl-type structure

The unit cell accommodates one CsC ion and one Cl ion, because the ion at corners is shared by eight unit cells .8 18 D 1/. The molecular weight of CsCl is 132:90 C 35:45 D 168:35 g (sum of atomic weights). Therefore, the mass m of the unit cell is mD

168:35 D 2:796 1022 g 0:6022 1024

Using the relation between the density () and the unit cell volume (a3 ) a3 D aD

2:796 1022 m D D 0:0704 1027 m3 ; 3:97 106

p 3 0:0704 1027 D 0:413 109 m D 0:413 nm:

This value of a D 0:413 nm is larger by 2% than the lattice parameter a D 0:404 nm calculated from their ionic radii. Thus, the lattice parameter of ionic crystals computed from measured density is realistic. The radii of ions such as CsC and Cl complied in handbooks are usually derived from mean values of interionic distance obtained from various ionic crystals. The lattice parameter of CsCl crystal determined by X-ray diffraction is a D 0:4123 nm (see Appendix A.9).

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(2) The following relationship may be obtained by referring to the condition of Fig. 1: p AC D 2.r C C r / D a 3; AB D 2r D a: When two relationships are coupled, the radius ratio obtained is equal to that estimated in Question 2.14 for the coordination number 8. p rC C r AC D D 3; AB r

p rC D 3 1 D 0:732: r

From this result, if the ratio of the ionic radii between cations and anions is smaller than 0.732, direct contact of anions will be allowed, and then this makes an ionic crystal unstable due to the strong repulsion between ions with the same electric charge. Question 2.16 Water (H2 O) crystallizes to form ice, when cooled below 273 K (0ı C) under 1 atmospheric pressure. Ice has hexagonal crystal structure with the lattice parameters of a D 0:453 nm and c D 0:741 nm. The density of ice at 1 atmospheric pressure at 273 K is 0:917 106 g=m3 . Estimate the number of water molecules (H2 O) contained in a unit cell of ice crystal.

Answer 2.16 In hexagonal crystal system, a unit cell is given by a prism with a vertical axis perpendicular to a rhombus-shaped base and the equal edges of which are at 60ı and 120ı with respect to each other (see Fig. 2.9). The length of the rhombus edge is designated by two equal vectors, a1 and a2 . Considering that a1 and a2 axes make an angle of 60ı , the volume V of a unit cell can be obtained as follows (refer to Appendix A.5): V D a2 sin60ı c D .0:453/2 .0:866/ .0:741/ D 0:132 1027 nm3 The molecular weight of water (m) obtained from Avogadro’s number (NA ) and density () assuming one molecule per unit cell is; m D NA V D 0:917 106 0:6022 1024 0:132 1027 D 72:9 g The molecular weight of water is 1:0082C15:999 D 18:015 g, using values of the atomic weights compiled in Appendix A.2. Then, we obtain 72:9=18:015 D 4:05. This suggests that four water molecules are included in a unit cell of ice crystal. As a result, each oxygen is quite likely to be surrounded by four oxygen in ice crystal (see Fig. 1). Note: The error in the measurement of lattice parameter and density is responsible for non-integer value 4.05. Water can crystallize in other structures such as trigonal lattices, tetragonal lattices etc., depending on temperature and pressure.

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2 Geometry of Crystals

Fig. 1 Arrangement of water molecules in ice crystal

Question 2.17 Many compounds with the formula ABX3 have the perovskite structure and a typical example is the natural mineral CaTiO3 . In the Perovskite structure, A atom occupy the center of a cubic unit cell and B and X atoms occupy the cell corners and face centers, respectively. There are two types of perovskite structure. Explain the essential points of two types including correlations of the unit cell. Also answer the following questions. (1) In barium titanate (BaTiO3 ), titanium occupy the center of a cubic unit cell, whereas barium and oxygen share the corner and the center of the cell faces, respectively. Assuming that titanium atoms share the vacant space formed by Ba-O lattice, what is the nature of the coordination and the resulting radius ratio constraint? (2) Why does titanium occupy the vacant space located in this position.

Answer 2.17 Two types of the perovskite structures are characterized by two different unit cells. For example, MoF3 and ReO3 belong to the so-called ReO3 structure, in which Mo atoms share all corner positions of a cubic unit cell, whereas F atoms occupy the midpoint of all edge-lines (see Fig. 1). Perovskite A-type structure can be obtained by adding A of ABX3 to the center of a cubic unit cell of the ReO3 base consisting of B and X. The structure can also be generated by adding X to the midpoint of all edge-lines of a cubic unit cell in which A and B form the CsCl type atomic arrangement. Such correlations are readily seen in Fig. 1. In the structure of intermetallic compounds such as Cu3 Au and Cu3 Pt, Au atoms occupy all corner positions of the fcc lattice, whereas Cu atoms share all center positions of cell faces of fcc lattice. Perovskite B-type structure is obtained by adding B to the center of a cubic unit cell of the Cu3 Au base consisting of A and X. Such correlations are illustrated in Fig. 2. It is also noted that the description of layer, as shown in Fig. 3, facilitates our understanding of the characteristic of the two perovskite structures.

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Fig. 1 Perovskite A-type structure

Fig. 2 Perovskite B-type structure

Since the atomic arrangement in perovskite structure is characterized by a formation where A and X are closely packed and B fits into its vacant space as shown in Figs. 1 and 2, it is desirable that the magnitude of A and X are equal and B is small. (1) The structure of barium titanate (BaTiO3 ) can be readily understood by referring to the B-type unit cell in Fig. 2. Ba2C ions occupy the position of (000) corresponding A-site, whereas Ti4C ions corresponding to theB site occupy 1 to 11 the position 2 2 2 . Three oxygens (O) keep the positions 0 21 21 ; 12 0 21 , and 1 1 2 2 0 of X. Here, B is found to be surrounded by six X. Incidentally, there are 12X distributed around A. Therefore, Ti4C ions share the center position of octahedron consisting of six oxygen atoms. There are four positions corresponding to such vacant space given by octahedron in an fcc type unit cell formed by A and X (refer to Question 2.7) and one of them (25%) is occupied by Ti4C ion.

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2 Geometry of Crystals

Fig. 3 Layer structure found in perovskite

(2) The octahedral position except for the octahedral void surrounded by six oxygen atoms in a unit cell is found at the midpoint of edge-lines in a unit cell and it will be surrounded by two Ba2C ions and four oxygen atoms. Since the structure becomes unstable when two kinds of positive ions such as Ba2C and Ti4C are approaching, it is thought that Ti4C ions preferentially fit into the octahedral void surrounded by six oxygens. The radii of Ba2C ion and O2 ion are 0:135 and 0:140 nm, respectively, so that barium titanate shows a slight deviation from an ideal perovskite structure. The dipole resulting from this lattice distortion is responsible for the ferroelectric properties. Question 2.18 Explain the procedure of stereographic projection for an octahedron by setting a plane of projection to the equator.

Answer 2.18 By setting the plane of projection to the equatorial plane, the equatorial plane becomes the basic circle of stereographic projection. Figure 1 shows the relation when a light source is placed at south pole and the projection is made to the equatorial plane are shown in Fig. 2. In the present case, the stereographic projection of a pole located on the upper half of the northern hemisphere is obtained by finding the intersection on the projection plane of the line connecting the target pole and south pole S. Conversely, one can get the stereographic projection of the pole located on the lower half of the southern hemisphere by finding the intersection on the projection plane of the line connecting the target pole and north pole N. To show the hemisphere in which the point is located, signs such as ˚ and ı, are used for differentiation.

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61 N 111 111

111

111

111 111 111

111

S

Fig. 1 Procedure of Stereographic projection of octahedron

Fig. 2 The results of projection to equatorial plane

Question 2.19 Explain two types of nets, such as the polar net and the Wulff net. In addition, explain the procedure that 40ı rotation of a pole P1 about any Q1 axis on a plane of projection.

Answer 2.19 (1) The polar net is a figure (refer to Fig. 1) of the meshes at every 2ı (or 1ı ) obtained by projecting all meridian circles and latitude circles on an equatorial plane. The pole net is useful for finding the projecting point to the equatorial plane about the point on a projection sphere with the given coordinates (; ) as shown in Fig. 2. (2) The Wulff net is a figure of the meshes at every 2ı (or 1ı ) obtained by projecting a sphere drawn with parallels of latitude and longitude on a plane parallel to the north–south axis of the sphere. For example, the longitude lines correspond

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2 Geometry of Crystals

Fig. 1 Polar net

Fig. 2 Example of the use of the polar net

Fig. 3 Projection of the grid net of globe for producing the Wulff net

to great circles connecting the north and south poles of the net and the latitude lines are small circles extending from side to side of the net. Figure 3 shows the stereographic projection of the grid net of sphere (the NS direction perpendicular to the N0 S0 one) for drawing the Wulff net. This drawing includes the positions of the angular coordinates and the corresponding pole distance . Wulff net (refer to Fig. 4) is used, for example, to estimate the angle between the lattice planes A and B, as shown in Fig. 5.

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Fig. 4 Wulff net

63

Fig. 5 Example of the use of the Wulff net

Fig. 6 Example of the rotating operation of a pole P1 by 40ı about an arbitrary axis Q1 on the projection plane

To rotate by 40ı around an axis perpendicular to a projection plane (corresponding to the case of rotating projection about the NS axis in Fig. 2.11), put the stereographic projection on the pole net and rotate projection at required angle (40ı ) about a central axis. On the other hand, to rotate the pole P1 by 40ı around the arbitrary axis Q1 on projection plane (40 degrees rotation around the arbitrary axis on the circle ADBE in Fig. 2.11), the following procedures are made using the Wulff net. Rotate so as to fit the north–south axis of the Wulff net with the axis Q1 . As a result, Q1 moves to Q2 , and P1 moves to P2 , respectively, as shown in Fig. 6. Next, P2 moves to P3 , by moving the pole P2 only by 40ı along the latitude line on

64

2 Geometry of Crystals

the Wulff net. Furthermore, rotate reversely the projection only by the same angle as the first operation, and return the axis Q3 to Q1 . As a result, P3 is set to P4 . Question 2.20 Find the geocentric angle between Sendai, Japan (38ı of north latitude and 141ı of east longitude) and Los Angeles, USA (33ı of north latitude and 120ı of west longitude) using the Wulff net.

Answer 2.20 Put the tracing paper on a copy of the Wulff net. Fix the centers by a tack as shown in Fig. 1. This makes the centers always coincident. Subsequently, the positions of Sendai () and Los Angeles (X) are marked in the tracing paper. Next, rotate the tracing paper so as to get the marked two points on the longitude line of great circle of Wulff net. In Fig. 1, two points marked by and X will be on the same longitude line of the great circle when rotating the tracing paper from position 1 to position . 2 We obtain the value of 76ı for the angle from the scale of Wulff net. Note: An uncertainty of 1ı (or 2ı ) is generally encountered when using the Wulff net of approximately 90 mm in diameter.

Fig. 1 Method for obtaining the angle of two positions with the Wulff net

Question 2.21 In the standard stereographic projection of cubic crystals, when setting a pole of (111) on the center of basic circle, find other poles of f111g, f100g, and f110g.

2.5 Solved Problems

65

Answer 2.21 The results are summarized in Fig. 1. Since the orientation of any plane in crystal can be represented by the inclination of the normal to that plane itself, the indices of a pole located at the position of 90ı relative to each zone (i.e., the set of planes) present the zone axis.

Fig. 1 Standard stereographic projection of cubic crystals on a (111) pole

Putting a crystal on the center of a reference sphere, the plane normal intersects the surface of the sphere in a set of points called poles. Each plane is represented by the intersection between normal and the reference sphere using the Miller indices. If the axis of a zone is given by the indices [u v w], and any plane belongs to that zone denoted by the indices (h k l), the well-known relation (hu C kv C lw D 0) called Weiss rule for the zone is satisfied. It may be noted that the Weiss rule is independent of the crystal system. If two planes of (h1 k1 l1 ) and (h2 k2 l2 ) belong to one zone axis of [u v w], the following relationships are obtained: h1 u C k1 v C l1 w D 0 and h2 u C k2 v C l2 w D 0; .ph1 C qh2 /u C .pk1 C qk2 /v C .pl1 C ql2 /w D 0; where p and q are arbitrary integers. In other words, if a zone axis [u v w] contains two planes (h1 k1 l1 ) and (h2 k2 l2 ), planes represented by p.h1 k1 l1 / C q.h2 k2 l2 / N also belong the same zone. For example, the planes (120) and (520) set as p D 1 N and q D 2 belong to a zone [001] D [002] including the planes of (100) and (110). Reference: If any two nonparallel planes of (h k l) and (h0 k 0 l 0 ) cross on one straight line and its direction is given by [u v w] as shown in Fig. 2, this direction

66

2 Geometry of Crystals

is called zone axis [u v w]. The indices of the zone axis [u v w] are related to the indices of the two planes and the vectors of a, b, and c which define a unit cell. ˇ ˇ1 0 1 0 0 1 0 ˇa b c ˇ u kl lk 0 ˇ ˇ @ v A D @ lh0 hl 0 A @ ua C vb C wc D ˇ h k l ˇ A ˇ ˇ ˇ h0 k 0 l 0 ˇ w hk 0 kh0

uvw

h´k´l´ hkl

Fig. 2 The line of intersection of two planes relevant to zone axis

Chapter 3

Scattering and Diffraction

An X-ray beam is an electromagnetic wave characterized by an electric field vibrating at constant frequency, perpendicular to the direction of movement. This variation of the electric field gives electrons (charged particles) a sinusoidal change with time at the same frequency. As a result of periodic acceleration and deceleration of the electron, a new electromagnetic wave, i.e., X-rays are generated. In this sense, X-rays are scattered by electrons. This phenomenon is called Thomson scattering. On the other hand, the physical phenomenon called “diffraction as a function of atomic position” is also found when an X-ray beam encounters a crystal whose atomic arrangement shows the long range periodicity. The intensity of diffracted X-rays depends on not only the atomic arrangement but also the atomic species. When considering diffraction of X-rays from a crystal, one needs information about “atomic scattering factors” which provide a measure of the scattering ability of Xrays per atom. Since the nucleus of an atom is relatively heavy compared with an X-ray photon, it does not scatter X-rays. The scattering ability of an atom depends only on electrons, their number, and distribution.

3.1 Scattering by a Single Electron In X-ray scattering by electrons, the scattered X-rays have the same frequency (wavelength) as the incident beam and is “coherent” with the incident X-rays. As shown in Fig. 3.1, if the incident X-ray beam traveling along the X -axis meets a single electron with mass m (kg) and charge e coulombs (C) located at the origin O, the intensity I of the scattered X-rays at position P in the X -Z plane at a distance r (m) from the origin may be expressed by the “Thomson equation”: 2 e 4 K 1 C cos2 2 0 2 sin ˛ D I0 2 I D I0 4 m2 r 2 r 2

(3.1)

where I0 is the intensity of incident X-rays, 0 D 4 107 .m kgC2 /, and ˛ is the angle between the scattering direction and that of the acceleration of electron. On the other hand, 2 in (3.1) is the angle between the line of OP connecting from 67

68

3 Scattering and Diffraction

Fig. 3.1 The relationship of the components of the electric vector of the incident radiation at a point O to the components of that of the scattered radiation at a point of observation P

the origin O to the point of measurement P, and the X -axis, the direction of the incident X-ray beam. The constant K D .2:8179 1015 /2 .m2 / is equivalent to the square of the electron radius re in classical electromagnetic theory. X-rays are scattered by an electron in all directions, but (3.1) clearly shows that the intensity of scattered X-rays decreases as the inverse square of the distance from the electron at origin, as well as a function of the scattering angle. In addition, the intensity of scattered X-rays is larger both in forward and backward directions compared to the direction at right angle to the incident X-ray beam. The term in the parenthesis in the last expression of (3.1) is called the “polarization factor.” For example, the ratio of (I =I0 ) at the position of 0.01 m away from an electron at origin is extremely small about 7:94 1026 . Nevertheless, the intensity of scattered X-rays can be amplified for detection without any difficulty, because the number of electrons contained in 1 mg of substance is of the order of 1020 1021 and such a large number of electrons interfere with each other. Finding the absolute value of the scattering intensity from one electron using (3.1) is a very difficult task either by measurement or by calculation. However, for most applications only the relative values of intensities of X-ray scattering and X-ray diffraction are required. It may be assumed that all terms in (3.1) except for the polarization factor may be considered as a constant. There is another way for X-ray scattering by an electron and it is characterized by a quite different mechanism from Thomson scattering. Compton scattering occurs when X-rays encounter a loosely bound or free electrons. It is relatively easy to understand Compton scattering by considering X-rays as particles (photons) which have energy h0 rather than as waves. As shown in Fig. 3.2, if a photon collides with a loosely bound electron and the collision between photon and electron is considered an elastic one, in a way similar to two billiard balls, the electron is knocked aside at an angle and the direction of the incident X-ray photon is altered by an angle 2 (it is called recoil phenomenon). In this collision process, a part of the energy h0 of the incident X-ray photon is converted to kinetic energy of the electron. As a result, the energy of the incident X-ray photon after collision becomes h which is smaller than h0 . For this reason, the wavelength after collision becomes slightly longer than the wavelength of the

3.2 Scattering by a Single Atom

69

Fig. 3.2 Collision of photon and an electron (Compton scattering)

incident X-ray beam 0 before impact and the relationship is given in unit of nm as follows: h .1 cos 2/ D 0:002426.1 cos 2/ (3.2) D 0 D mc The term of .h=mc/ D 0:002426 nm is called the “Compton wavelength.” According to (3.2), the increase in wavelength due to the Compton scattering depends only on the scattering angle 2 and values are estimated to be zero at 2 D 0 and 0.005 nm at 2 D 180ı , respectively. The phase of X-rays produced by Compton scattering is not the same as that of the incident X-ray beam, because of the change in wavelength. The Compton modified radiation cannot take part in diffraction because it has no specific phase relation with the incident X-ray beam and therefore cannot produce any interference effect. For this reason, Compton scattering is often called “incoherent scattering.”

3.2 Scattering by a Single Atom If an X-ray beam encounters an atom consisting of the nucleus and a certain number of electrons, each electron produces coherent scattering intensity given by (3.1), the so-called Thomson equation. Since the mass of the nucleus is much larger than that of electron, the X-ray beam cannot oscillate the nucleus to any appreciable extent. The acceleration and deceleration of the nucleus to emit X-ray are not functional. Therefore, when considering the net effect of X-ray scattering from an atom, one needs to take into consideration only scattering by electrons associated with the atom. This is also evident from (3.1) since the square of mass of scattering particle appears in the denominator. The scattering amplitude of a single atom with atomic number Z containing Z electrons is equal to Z times the scattering amplitude from one electron in the forward direction. Because in direction the scattering angle is zero .2 D 0/, the phases of X-rays scattered by all electrons in one atom are completely coincident, so that the amplitude of the scattered X-rays can be simply added. However, when the scattering angle has nonzero values, there is variation in the phase of X-rays

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3 Scattering and Diffraction

scattered from individual electrons in an atom. In other words, the X-rays scattered by electrons located at, for example, the point A and point B in space, have phase difference because of difference in optical path, when the scattering angle is not zero. As a result, the scattering amplitude of a single atom decreases with increase in . The scattering amplitude of an atom also depends on the wavelength of the incident X-rays. For example, at the same scattering angles 2, the shorter the wavelength is, the larger the phase difference becomes. This implies that the scattering amplitude becomes relatively small when the shorter wavelength is used. In order to calculate the scattering amplitude of X-rays for atoms containing more than two electrons, it is necessary to bear in mind that the charge of electrons are not focused at fixed points; rather it is distributed in space like a cloud. The electron density function .r/ as a function of distance r away from the nucleus at origin is useful for describing electron distribution. Let us consider that the wave vectors of the incident X-rays and the scattered X-rays are given by .s0 / and .s/, respectively. The scattered X-rays at the distance r will produce the optical path difference .s s0 / r in comparison with the scattered X-rays at origin. If the wavelength of the incident X-rays is given by , the scattering amplitude of X-rays irradiated in the direction of s can be described by 2 i .s s0 / r dV exp

(3.3)

The amplitude of the coherent scattering from a single electron may be obtained by integrating over the volumes occupied by electron with the help of the phases relevant to dV . Then, the scattering factor fe per electron in electron unit is obtained Z 2 i .s s0 / r dV (3.4) fe D exp Note that the value of fe corresponds to the ratio between the amplitude of coherent scattering by a single electron which has distribution and that of X-rays scattered by a single electron whose location is fixed at a point according to the classical view. Further, the wave vectors of s0 and s are related q D s s0

)

jqj D q D

2 sin

(3.5)

Here, the vector q is frequently called “scattering vector (or wave vector).” This is the vector required to turn the direction of incident X-rays to that of the scattered radiation by angle 2, thus, s D s0 C q. Since the electron density distribution of inner shell is quite likely to be approximated by spherical symmetry, the scattering factors for atom containing more than two electrons can be readily estimated. For example, if the electron density distribution around the nucleus set at origin is given by D .r/ as a function of distance r, the scattering factor fe can be obtained as

3.2 Scattering by a Single Atom

71

Z fe D

1 0

4 r 2 .r/

sin 2qr dr 2qr

(3.6)

The amplitude of the coherent scattering per atom including n electrons is computed by the sum of fe regarding all electrons using the following equation. f D

X j

fen D

XZ j

1 0

4 r 2 j .r/

sin 2qr dr 2qr

(3.7)

This f is usually called “X-ray atomic scattering factor” or simply “the atomic scattering factor.” Sometimes f is called “the form factor” because it depends on the distribution of electrons around the nucleus. The quantity f provides the efficiency or ability of the coherent scattering per atom and it is defined as a ratio of the amplitude of the wave scattered from one atom to that scattered from one electron under the same condition. Therefore, f D Z for any atom which scatters in the forward direction. Equation (3.7) implies that f is a function of (sin =) for any atom. The electron density distributions in atoms have been provided from the electron wave functions by using several techniques, such as Hatree–Fock and Fermi–Thomas–Dirac approximation and a number of theoretical calculations for the atomic scattering factors of elements were carried out as a function (sin =). Such results are compiled in the International Tables for X-ray Crystallography, Vol.C (Kluwer Academic Pub., London, UK, 1999). The atomic scattering factors as a function (sin =) are shown in Fig. 3.3 using the results for Al, Fe, and Ag as examples. The f value decreases as (sin =) increases. The variable of Q defined by 2q D 2

sin 2 sin D 4 DQ

Fig. 3.3 Atomic scattering factors of Al, Fe, and Ag

(3.8)

72

3 Scattering and Diffraction

The variable of Q is often used. It can be obtained from (2q) in (3.7) using (3.5). Equation (3.7) can be rewritten as XZ 1 X sin Qr dr (3.9) fen D 4 r 2 j .r/ f D Qr 0 j

j

It is useful to summarize some important points regarding scattering of X-rays from a single atom. When monochromatic X-ray beam encounters an atom, two scattering processes, coherent and incoherent, simultaneously occur in all directions. The intensity of incoherent (Compton) scattering for light elements (with small atomic number) is found to increase with decreasing atomic number Z. Furthermore, as the quantity of (sin =) increases intensity of incoherent (Compton) scattering increases. Thus, the intensities of unmodified (coherent) scattering and modified (Compton) scattering change in opposite ways with Z and (sin =). In addition, the sum of the coherent and incoherent scattering intensities is equal to the classical scattering intensity per electron. If ie represents the intensity of incoherent scattering in electron unit for the completely unpolarized incident X-ray beam, the following equation can be obtained: e4 e4 1 C cos2 2 1 C cos2 2 2 fe C I0 2 4 2 ie I0 2 4 2 m c R 2 m c R 2 1 C cos2 2 e4 (3.10) D I0 2 4 2 m c R 2 It is note worthy that (3.10) can be simply rewritten in the form, ie D 1 fe2 . As previously mentioned, the phase of radiation attributed to the Compton scattering effect has no fixed relation to that of the incident beam, because of the variation in wavelength after collision, suggesting that interference effect cannot be produced by the modified Compton radiation. Therefore, the incoherent scattering intensity i.M / per atom may be given by the simple sum of the incoherent scattering intensity of the respective electrons. i.M / D

X j

ien D Z

Z X

2 fen

(3.11)

j D1

Figure 3.4 shows the calculated results of f and i.M / for Li atom assuming that the electron density distributions for three electrons, two of K shell and one of L shell, are considered to be similar to that of the hydrogen atom. One clearly recognizes the reverse relationship in variation of the intensities of coherent scattering and Compton scattering with respect to (sin =). As readily obtained from (3.9), the atomic scattering factor f of Li atom, corresponding to the coherent scattering intensity is represented by fLi D 2feK C feL , whereas the incoherent scattering intensity i.M / in electron unit is given by the relationship of i.M / D 2 2 feL (see (3.11)). 3 2feK

3.3 Diffraction from Crystals

73

Fig. 3.4 Coherent and incoherent scattering intensities of Li atom calculated based on the assumption that the electron density distributions have spherical symmetry and the interference between electrons in the atom is ignored

3.3 Diffraction from Crystals Atoms being the constituent of a crystal generates X-rays with the same wavelength as that of the incident X-ray beam by oscillating electrons and the generated X-rays are likely to be the spherical waves centering on respective atoms. This situation is similar to that a wave rolling from one side to the piles lined up on the same line at equal intervals in a pond and propagating to the other side. That is, the diffraction phenomena of X-rays by crystals is attributed to certain phase relations between two or more waves, such as differences in phase produced from the differences in path length of waves and a change in amplitude related to the phase difference. In addition, the most important point to know is that two waves are completely in phase if the difference in path lengths is zero or an integer multiple of wavelength. Some additional details are given below. The phase of any two waves generally shows deviation of , corresponding to their path difference. Since the phase of two waves is completely coincident (in phase) if the value of is given by an integer multiple of wavelength , two waves will combine to form one synthesized wave just like the original one, and its amplitude will be double. The path difference depends on direction of X-ray with respect to the crystal. When the value of is =2, the two waves cancel each other and they are completely out of phase, resulting from the fact that the waves have the same magnitude but opposite amplitude at any point along its path. Situations between these two extreme cases ( D and =2) are also encountered depending on path difference. The main target of X-ray diffraction by crystals is to know the particular condition in which the scattered X-rays from atoms and the incident X-rays are completely in phase and reinforce each other to produce a detectable diffraction beam. In other words, we have to find the common relationship that the differences in path length between X-rays scattered from crystals and that of the incident X-rays

74

3 Scattering and Diffraction

Fig. 3.5 Schematic diagram of diffraction of X-rays by a crystal (Bragg condition)

is an integer multiple of wavelength . For this purpose, the most important and familiar method is given by Bragg law which incorporates Bragg angle. In order to facilitate the understanding of Bragg law, Fig. 3.5 is useful. It is also required to remember the following two geometric relationships: 1. The angle between the incident X-ray beam and the normal to the reflection plane is equal to that between the normal and the diffracted X-ray beam. The incident X-ray beam, the plane normal, and the diffracted X-ray beam are always coplanar. 2. The angle between the diffracted X-ray beam and the transmitted one is always 2, and this angle is called “the diffraction angle.” If the incident X-rays of wavelength () strike a crystal where all atoms are placed in a regular periodic array with interplanar spacing d 0 , diffraction beam of sufficient intensity is detected only when the “Bragg condition” or “Bragg law” is satisfied (3.12) 2d 0 sin D n where n is called the order of reflection and is equal to the number of wavelengths in the path difference between diffracted X-rays from adjacent crystal planes (see Fig. 3.5). For fixed values of both and d 0 , the diffraction occurs at several angles of incidence such as 1 , 2 , 3 ; : : : ; corresponding to n D 1; 2; 3; : : : : In the first-order reflection (n D 1), the path difference between two scattered X-rays denoted by 1’ and 2’ in Fig. 3.5 is one wavelength. The path difference between X-rays 1’ and 3’ is two wavelengths, etc. The diffracted X-rays from all atoms in all the planes are considered completely in phase so as to produce the diffracted X-ray beam with appreciable intensity in a particular direction which satisfies the Bragg law. Equation (3.12) may be rewritten as 2d sin D (3.13) where d D d 0 =n. This form of Bragg law is frequently used.

3.3 Diffraction from Crystals

75

Table 3.1 Information of plane spacing for seven crystal systems 1 d2 1 d2

D

Hexagonal

1 d2

D

Trigonal

1 d2 1 d2

D

1 d2 1 d2

D

Cubic Tetragonal

Orthorhombic Monoclinic Triclinic

D

D D

h2 Ck 2 Cl 2 a2 2 h2 Ck 2 C cl 2 a2

2 4 h2 ChkCk 2 C cl 2 3 a2 .h2 Ck 2 Cl 2 / sin2 ˛C2.hkCklChl/.cos 2 ˛cos ˛/ a2 .13 cos2 ˛C2 cos3 ˛/ h2 k2 l2 C C 2 2 2 a b 2 c 2 2 2 k sin ˇ 2hl cos ˇ 1 h C b 2 C cl 2 ac sin2 ˇ a2 1 .S11 h2 C S22 k 2 C S23 k 2 C 2S12 hk C V2

2S23 kl C 2S13 hl/

On the triclinic system, V is the volume of a unit cell and the coefficients are given below. S11 D b 2 c 2 sin2 ˛;

S12 D abc2 .cos ˛ cos ˇ cos /;

S22 D a c sin ˇ;

S23 D a2 bc.cos ˇ cos cos ˛/;

S33 D a2 b 2 sin2 ;

S13 D ab2 c.cos cos ˛ cos ˇ/

2 2

2

In general, the nth order reflection from a certain crystal plane (h k l) with the interplanar spacing of d could be considered the first-order reflection from a plane (nh nk nl). Since the (nh nk nl) plane is parallel to the (h k l) plane, reflection from (nh nk nl) plane is equivalent to the first order reflection from planes spaced at the distance (d D d 0 =n). From such a point of view, 2 is called the diffraction angle in many cases. The diffraction angle 2 of any set of planes (h k l) can be computed by combining (3.13) with the plane-spacing equations (see Table 3.1) which relate distance between providing the relationship among the distance of adjacent planes to Miller indices and lattice parameters for each crystal system. For example, if the crystal is cubic with the lattice parameter a, the interplanar spacing d and Miller indices (h k l) are given by .h2 C k 2 C l 2 / 1 D (3.14) d2 a2 By combining the Bragg law with (3.14), one obtains for experiments using the wavelength of the following equation. sin2 D

2 2 .h C k 2 C l 2 / 4a2

(3.15)

Equation (3.15) suggests that the diffraction angle, corresponding to diffraction directions, can be determined from the shape and size of the unit cell. This is an important point for structural analysis of substances. The converse is also very valuable, since one can possibly determine an unknown crystal structure by measuring the diffraction angles. In other words, the particular directions of diffracted X-ray beams given by the diffraction angles are related directly to the positions of atoms in the unit cell.

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3 Scattering and Diffraction

3.4 Scattering by a Unit Cell As shown in Chap. 2, a crystal is defined as a solid consisting of atoms arranged in a periodic pattern defined by the unit cell. Therefore, when considering the scattering intensity from crystals, it is important to get information of phase differences based on the relationships between the scattering intensity and the atomic positions in one unit cell. Only some essential points are given below. Readers who want to have a deeper knowledge about this field should consult textbooks on structural determination by X-ray crystallography. The phase difference of X-rays whose path difference is one whole wavelength is 2 radians (360ı). Let us consider an atom A is placed at the origin of (000) and atom B at actual coordinates (x y z) which can be expressed by fractional coordinates (u v w) where u D x=a, v D y=b, and w D z=c, where a, b, and c are the lattice parameters along each axis. Under these conditions, the phase difference between the X-rays scattered by atom B and that scattered by atom A at the origin is given by the following equation for the (h k l) reflection. D 2.hu C kv C lw/

(3.16)

This relationship is applicable to a unit cell of any shape. If atoms A and B are of different elements, there will be differences not only in phase but also in amplitude. By adding waves scattered by all atoms in a unit cell including the atom at the origin, the resultant wave can be obtained. Each wave can be described by a vector whose length is equal to the amplitude and this vector is inclined to the horizontalaxis at the phase angle. Then, the amplitude and phase of the resultant wave are obtained simply by adding two vectors using the parallelogram law. Such a geometrical solution may be simplified if we use complex numbers to represent the vectors. Complex exponential function can also be employed. The variation in electric field intensity E with time t is frequently expressed in the form E D A exp.2 i t/ D A cos.2 t/ C iA sin.2 t/

(3.17) (3.18)

where A is the amplitude and is frequency. Equation (3.17) is called a complex exponential function and this is also convenient for discussion of wave behavior. The key points are displayed in Figs. 3.6 and 3.7. A complex number p is represented in the form (a C ib), where a and b are real numbers, whereas i D 1 is imaginary. A complex number is usually represented in the complex plane and as shown in Fig. 3.6, the real number is plotted as abscissa and the imaginary number as ordinate. Amplitude of the wave is given by A, the length of the vector, and phase by , the angle between the vector and the horizontal axis. With reference to (3.18) taken together with Fig. 3.6, the wave is now treated by the complex number, A cos C i A sin . Note that these two terms correspond to the horizontal component OM and the vertical one ON of the vector.

3.4 Scattering by a Unit Cell

77

Fig. 3.6 Wave behavior represented by complex numbers sin(2πiυt) A

φ

t

λ

cos(2πiυt)

λ t

Fig. 3.7 Cosine component is obtained by projecting the wave vector on the real-axis and sine component on the imaginary-axis

In the complex plane, multiplication of a vector by i is equivalent to the rotation of the vector counterclockwise by 90ı . Multiplication twice by i , (i 2 D 1), rotates the vector through 180ı , equivalent to reversing the direction. In other words, when multiplying the horizontal vector 2 in Fig. 3.6 by i converts it into the vertical vector 2i along abscissa. Multiplication twice by i converts the horizontal vector 2 into the horizontal vector 2 along the ordinate, pointed in the opposite direction.

78

3 Scattering and Diffraction

Using the power-series expansions of eix , cos x and sin x, it can be shown that Aei D A cos C iA sin

(3.19)

The left hand side of (3.19) is called a complex exponential function. The wave intensity is proportional to the square of the amplitude, so that we have to obtain the square of the absolute value of the wave vector, (A2 ). When the wave is described by the complex form, the A2 value can be obtained by multiplying the complex function by its complex conjugate, (i being replaced by i). The complex conjugate of Aei is Aei , so that jAei j2 D Aei Aei D A2 . Some useful relations, such as eni D .1/n , eni D eni , where n is any integer, eix Ceix D 2 cos x and eix eix D 2i sin x are very useful. The amplitude of the scattered wave from each atom in a unit cell is given by the atomic scattering factor of the atom f , and the value of (sin =) relevant to diffraction. Since the phase of each wave is given by (3.16) for the (h k l) reflection using fractional coordinates (u v w) of the atom, any scattered wave may be expressed in the complex exponential form. (3.20) Aei D f e2i.huCkvClw/ Based on this relationship, the sum of the scattered waves from atoms in a unit cell can be computed using the generalized equation. Fhkl D

N X

fj e2 i.huj Ckvj Clwj /

(3.21)

j D1

where N is the total number of atoms involved in the unit cell; for example, it is 2 for bcc and 4 for fcc. F defined by (3.21) is called the structure factor or geometrical structure factor and one can obtain the value of F , only if the position and the type of atoms in the unit cell are given. For this reason, (3.21) as well as Bragg condition given by (3.13) are considered very important relationships for analyzing the structure of crystals by X-ray diffraction. The structure factor F is generally given by a complex number and represents both the amplitude and the phase of the scattered wave obtained from the summation over all atoms in the unit cell. In addition, the absolute value of F represents the amplitude of the resultant scattered wave by adding together waves scattered by individual atoms in the unit cell, based on that of the wave scattered by a single electron. Therefore, the intensity of the resultant wave scattered from all atoms in the unit cell in the direction which satisfies the Bragg law, is simply proportional to jF j2 . The value of jF j2 can be obtained by multiplying F of (3.21) by its complex conjugate F . Although there is no theoretical difficulty in calculation of the structure factor given by (3.21), the usefulness and application of (3.21) can be fully appreciated only by working out some actual cases. For example, in the case of the bodycentered cell containing two atoms of the same kind, (position of 000 and 12 ; 12 ; 12 ) in unit cell, the structure factor is given as follows:

3.4 Scattering by a Unit Cell

79

Table 3.2 Relationship between Bravais lattice and reflections Crystal type Bravais Lattice type Reflections possible Reflections necessarily present absent Simple Primitive Any h; k; l None Body-centered Body-centered h C k C l even h C k C l odd Face-centered Face-centered h; k and l unmixed h; k and l unmixed Diamond cubic Face-centered cubic As fcc, but if all even h; k and l mixed and if all and h C k C l ¤ 4N ; even and then absent h C k C l ¤ 4N Base-centered Base-centered h and k both even or h and k mixed both odd h C 2k D 3N with l even Hexagonal close- Hexagonal h C 2k D 3N ˙ 1 with h C 2k D 3N with l odd packed l odd h C 2k D 3N ˙ 1 with l even These relationships apply to a cell centered on the C face. If reflections are present only when h and k are unmixed, or when k and l are unmixed, then the cell is centered on the B or A face, respectively.

uvw D 000; h

k

111 222

l

F D f e2 i 0 C f e2 i . 2 C 2 C 2 / D f Œ1 C f e i.hCkCl/ When the number of .h C k C l/ is evenW F D 2f; F 2 D 4f 2 When the number of .h C k C l/ is oddW F D 0; F 2 D 0 The results show that the reflections will be observed for planes such as (110), (200), and (211) whose indices satisfy the condition (h C k C l) is even, but not for the planes (111), (210), (300), etc., because the waves cancel each other being out of phase. In calculation of the structure factor, information on crystal systems is not required. In the previous example, the given information is only the body-centered cell containing two atoms. No information about the shape of the unit cell, such as cubic, tetragonal, and orthorhombic (see Bravais lattices in Fig. 2.4) was used. The structure factor is completely independent of the shape and size of the unit cell. This also illustrates an important point; any body-centered cell provides the missing reflections when (h C k C l) is an odd number without reference to the specific crystal system. In other words, information about missing reflections provides a clue about the actual atomic arrangement in crystals. The relationship between Bravais lattice and diffraction behavior is summarized in Table 3.2. The two or more kinds of atoms are included in a unit cell, it is necessary to take into consideration the atomic scattering factor of each atom in calculations using (3.21). They will be further illustrated in Questions 3.12 and 3.13.

80

3 Scattering and Diffraction

3.5 Solved Problems (13 Examples)

Question 3.1 The differential cross-section of the coherent scattering intensity for a free electron is expressed by the following equation with respect to per unit plane angle (radian), if the classical electron radius is set to re r2 d e D e .1 C cos2 / 2 sin d 2

.m2 =rad/

(1) Estimate e which is called the Thomson classical scattering coefficient by the integration about angle. (2) Express the Thomson classical scattering coefficient in barn unit. (1barn D 1 1028 m2 ). (3) When X-rays penetrate an aluminum foil of thickness 1 mm, calculate the probability of the coherent scattering induced from free electrons. The density of aluminum is 2:70 106 g=m3 . Answer 3.1 Figure 1 shows the schematic diagram of the process by which coherent scattering is produced per unit solid angle (steradian) in direction denoted by angle , when photon strikes a free electron. The relationship of d˝ D dA=r 2 D 2 sin d is readily established about area .A/ denoted by a part of a coaxial cone which is specified by the angle between and C d on the surface of a sphere of radius r.

Fig. 1 Gemometric feature found in the cone of diffraction from a free electron

(1) Integration is made from zero to with respect to the angle Z

Z re2 2 2 .1 C cos /2 sin d D re .sin C cos2 sin /d

e D 2 0 0 4 8 cos3 2 2 4 D re cos C D re2 D re 3 3 3 3 0

3.5 Solved Problems

81

where the following relationship is used. t D cos Z

cos2 sin d D Z

Reference:

!

dt D sin d Z t2 dt D

cos3 t3 D 3 3

1 n1 In2 sinn xdx D .sinn1 x cos x/ C n n Z Z Il D sin xdx D cos x; I0 D dx D x

In D

(2) The value of the classical electron radius re is defined by the following equation using the permittivity of free space 0 D 107 =.4c 2 / in electromagnetism. re D

e2 e2 .1:602 1019 /2 D D D 2:8719 1015 40 c 2 me me 107 9:109 1031 107

.m/

where me and e are electron rest mass and electron charge respectively. It may be noted for the dielectric constant of vacuum 0 D 8:854 1012 .Fm1 / is also employed as electric capacity per unit length. The important point for e obtained by question (1) is that the coherent scattering intensity is proportional to re2 . Because of an extremely large mass of the nucleus relative to the electron case, about 1,800 times that of the electron mass, the net effect on the coherent scattering intensity by the nucleus is very small. This is the reason why the scattering caused only by electrons in the atom is considered. The value of e is estimated as follows, using the relation of 1b D 1 1028 m2 . 8 2 8 re D .2:8179 1015 /2 D 66:52 1030 .m2 / 3 3 66:52 1030 D D 0:6652 .b/ 1 1028

e D

Note: The cross-sectional area per unit area is usually given in barn, because the value becomes very small if SI unit is used. (3) The mass of aluminum per mole (molar mass) is given as 26.98 g in Appendix A.2, and the number of electrons equivalent to the atomic number 13. The number of electrons Ne contained in 1 m3 from density value is estimated in the following way. Ne D

0:6022 1024 13 2:70 103 D 0:783 1030 26:98 103

.m3 /

Converting the number of electrons of Nefilm for aluminum foil with its thickness of 1 mm into the one per unit area .m2 /, Ne D 0:783 1027 .m2 /. Then, the probability P of the coherent scattering intensity due to electrons in this case is

82

3 Scattering and Diffraction

found as follows: P D e Nefilm D 66:52 1030 0:783 1027 D 0:052 Question 3.2 From high-precision scattering experiments, 0.002426 nm is obtained as the wavelength of Compton scattering. Calculate the effective mass of photon me using the so-called Einstein relation E D mc 2 showing relationship between the mass of photon m and its energy E. Answer 3.2 X-ray is electromagnetic radiation (photon) of exactly the same nature as visible light except that values of wavelength or energy (D frequency ) are different. The propagation speed of photon c is considered equal to the speed of light in vacuum, 2:998 108 m=s. The following equation is obtained from the relationship c D h by setting Planck constant to h. hc E D h D To calculate the effective mass of photon me , Einstein relation can be used me c 2 D

hc

!

me D

h c

Using h D 6:626 1034 .J s/, c D 2:998 108 .m=s/, and 1 nm D 109 m. me D

6:626 1034 D 9:110 1031 2:998 108 0:002426 109

.kg/

Note: The value of 9:109 1031 kg is also used as electron rest mass. If acceleration voltage exceeds 100 kV, an increase (relativistic) in mass accompanying the variation of speed of photon should be taken into account by using Einstein relation. If the mass of photon and its velocity are set to m0 and v, respectively, an increase (relativistic) in mass may be estimated from (1.8) in Chap. 1 for the case that acceleration voltage of 200 kV is applied to an electron. The energy of an electron with its rest mass m0 D 9:109 1031 .kg/ is computed in the following, by coupling with 1.eV/ D 1:602 1019 .J/. E D m0 c 2 D

9:109 1031 .2:998 108 /2 D 0:5109 106 .eV/ 1:602 1019

Since the energy given to an electron is equivalent to 0:2 106 .eV/ in the case that applied voltage is 200 kV, and the speed v of photon may be estimated as follows. s vDc

1

0:5109 0:2 C 0:5109

2

Dc

p 1 0:5165 D 0:6953c

3.5 Solved Problems

83

Because of cv D 0:6953, the increase in mass of photon in this case can be obtained in the following equation: mD q

m0 m0 m0 v 2 D p1 .0:6953/2 D 0:7187 D 1:39m0 1 c

Question 3.3 When incident X-rays collide with a free electron, a part of the energy of incident X-rays (photons) is given to the electron as kinetic energy. Accordingly, the energy of the photon after collision is less than the energy before collision. That is, the wavelength of X-rays after collision becomes slightly longer than the wavelength of the incident X-rays and it is called the Compton shift. Answer the following questions related to such incoherent scattering of X-rays. (1) Obtain the Compton equation for the case where a photon with the energy of h0 (momentum of h0 =c) collides with an electron at rest. (2) Compute the increment of Compton shift in wavelength produced at a scattering angle of 30ı . Here, h is Planck constant, c the speed of light in a vacuum, and the frequency. Answer 3.3 (1) Momentum is usually given by the product mv of mass m and speed v of a desired particle. On the other hand, the momentum of a photon may be h expressed by h c or using some relationships that the kinetic energy of a photon is described by E D h, the Einstein relation of E D mc 2 and c D , where is the wavelength. The collision between a photon and a free electron is considered an elastic one, as shown in Fig. 1. If the photon having the momentum of hc 0 collides with the electron at rest, the electron is knocked aside and the direction of the photon is deviated through an angle 2. Since a part of energy of the incident photon is given to the electron, the momentum of photon h c after collision is quite likely to become small in comparison with hc 0 before collision.

Fig. 1 The elastic collision of photon and an electron (the Compton effect)

84

3 Scattering and Diffraction

It is thought that the law of conservation of energy is satisfied before and after collision. Thus, h0 C m0 c 2 D h C mc 2 2

(1)

2

mc D h.0 / C m0 c D A C B

(2)

Taking square of both sides of (2) and re-arranging; .mc 2 /2 D A2 C B2 C 2AB D Œh.0 / 2 C .m0 c 2 /2 C 2hm0 c 2 .0 / 2

2

2

2 2

(3) 2

D .h0 / C .h/ 2h 0 C .m0 c / C 2hm0 c .0 v/

(4)

The electron before collisions is at rest, v0 D 0. The momentum of the electron before collision is also zero. Denoting the velocity of electron after collision by v, and invoking the law of conservation of momentum, one obtains h h0 D C mv c c

!

mvc D h0 h

(5)

By applying the “law of cosines” .A2 D B2 C C2 2BC cos / to (5), .mvc/2 D .h0 /2 C .h/2 2h2 0 cos 2

(6)

By subtracting (6) from (4); .mc 2 /2 .mvc/2 D .m0 c 2 /2 C 2hm0 c 2 .0 / 2h2 0 .1 cos 2/ (7) The left hand side of (7) may be rewritten as 2 2

2

2 2

.mc / .mvc/ D .mc /

1

v 2 c

D .m0 c 2 /2

(8)

Here, we use the relationship with energy in case where an increase in mass of a photon arising from change of its velocity v can be estimated in the following way. mc 2 D q

m0 c 2 2 1 cv

)

v 2 D .m0 c 2 /2 .mc 2 /2 1 c

(9)

Therefore, (7) can be summarized as; 2hm0 c 2 .0 / D 2h2 0 .1 cos 2/ h0 c.0 / D .1 cos 2/ m0 c

(10) (11)

3.5 Solved Problems

85 c 0

and D c , we obtain

h .1 cos 2/ m0 c

(12)

On dividing both sides of (11) by 0 and using 0 D the Compton equation, as follows. D 0 D

(2) Equation (12) suggests that the Compton shift of depends only on the scattering angle and it varies from zero in the forward direction .2 D 0/ and to twice the term of h=.m0 c/ in the backward direction .2 D 180ı/. It is also noted that the incoherent scattering due to the Compton effect is not always welcome in analyzing the structure of substances, because it increases the background intensity. Substituting h D 6:626 1034 .J s/; m0 D me D 9:109 1031 .kg/; and c D 2:998 108 .m=s/, in (12) 60626 1034 h D D 0:2426 1011 .m/ me c 9:109 1031 2:998 108 D 0:0243 108 .cm/ Next, further substitution of 2 D 30ı .cos 30ı D 0:866/ in (12) gives;

D 0:2426 1011 .1 0:866/ D 0:0325 1011 .m/ D 0:0033 108 .cm/

Question 3.4 A stream of X-ray quanta with the energy of 200 keV strikes a free electron and the incoherent scattering is produced in a direction 180ı relative to the incident beam. The energies of the scattered photon and the recoil electron were found 111.925 and 87.815 keV, respectively. Verify if this incoherent scattering process satisfies the law of conservation of momentum. Answer 3.4 According to the Einstein relation of E D mc 2 , the energy is equivalent to the mass and it is also noted that an increase in mass of a quantum is related directly to the change of its speed. If rest mass of an electron 9:109 1031 kg is converted into energy, we obtain 0:5109 106 eV D 510:9 keV. Total energy E (it is equivalent to mass) of the electron after collision may be given, in keV, by the sum of the energy of the scattered photon and that of the recoil electron. E C m0 c 2 D 87:815 C 510:9 D 598:715 keV The speed of electron v can be computed using (1.8) in the following way: s vDc

1

m0 c 2 E C m0 c 2

2

s Dc

1

510:9 598:715

2

D 0:521 c

86

3 Scattering and Diffraction

The energy (equivalent to mass) of the recoil electron is quite likely to increase because of collision and its momentum p may be given by mv and, then we obtain, p D 598:715 0:521 c D 311:931 c Since the incoherent scattering is produced at a scattering angle of 180ı, corresponding to the extreme backward direction to the incident X-ray beam, the sum of momentum of the incident X-rays and the scattered photon should be compared with the momentum of the recoil electron. For this reason, we obtain the following equation using p D meff c D h=c. h0 h C D .200 C 111:925/c D 311:925 c c c Although a very small difference of 0.006c is found between the two values, it may be inferred that the law of conservation of momentum is almost satisfied. Question 3.5 X-rays with energy 51.1 keV produced incoherent scattering by impacting with an electron in the outershell of an atom in a sample. Answer to the following questions. (1) Estimate the angle (so-called recoil angle ) between the direction of the recoil electron and that of the incident X-ray beam, when the scattering angle 2 is 20ı . (2) Compute the energy of the scattered photon in this case. Answer 3.5 (1) In the incoherent scattering process, the recoil angle is known to correlate with the scattering angles 2 in the following form. For convenience, Fig. 1 shows the schematic diagram of two angles together with the incident X-ray photon beam. h 2 1 D 1C tan 2 tan m0 c 2 In other words, this equation suggests that we can obtain the scattering angle 2 if the recoil angle is known or vice versa. Similarly, if either the energy .h0 / of the incident X-ray photon beam or the energy h of the scattered photon is given, the energy (Er ) of the recoil electron can be computed. Since the energy of the incident X-ray photon beam is given as 51.1 keV, one obtains h 51:1 103 D D 0:1 m0 c 2 0:5109 106 Here, we use the information that rest mass of an electron 9:109 1031 kg is equivalent to energy of 0:5109 106 eV D 510:9 keV (see Question 3.4).

3.5 Solved Problems

87

Fig. 1 Schematic diagram for the incoherent scattering process where the incident X-ray photon beam, a scattered photon, and a recoil electron are correlated

On the other hand, from the relationship of E.keV/ 1:240= .nm/ for the wavelength 0 of the incidence X-ray beam, we obtain, 0 D

1:240 D 0:0243 .nm/ 51:1

The recoil angle can be computed from the value of the scattering angle 20ı , as follows: 1 20 D .1 C 0:1/ tan tan 2 tan D 5:1557 ! D 79:0ı The recoil electron is released in the direction of 79ı with respect to the direction of propagation of the incident X-ray beam. (2) Calculate the Compton shift in the case where the scattering angle is 20ı , using (3.2). Since cos 2 D 0:9397, one obtains D 0:2426 1011 .1 cos 2 / D 0:2426 1011 0:0603 D 0:0001 109 .m/

Therefore, the wavelength of the scattered photon is D 0 C D 0:0243 C 0:0001 D 0:0244 .nm/ The energy of this photon may be estimated from E D h, E D h D

1:240 D 50:8 .keV/ 0:024

The energy of the recoil electron, Er , can be computed from the difference in energy between the incident X-ray beam and the scattered photon. Er D 51:1 50:8 D 0:3 .keV/

88

3 Scattering and Diffraction

The wavelength of the incident X-ray photon beam with energy of 51.1 keV becomes longer by 0.0001 nm after scattering and the energy of the scattered photon decreases to 50.8 keV. The recoil electron which has the energy of 0.3 keV is released at an angle of 79ı from the direction of propagation of the incident X-ray beam. Question 3.6 Complex number .A cos C iA sin / is widely used as an analytic expression for the wave. Using the given diagram of the complex plane, discuss the wave vector with the amplitude and phase. Also confirm the product of a complex number and its complex conjugate is always constant and it is equivalent to the square of the amplitude.

Fig. A

Wave vector represented by the complex number

Answer 3.6 A real number can express only one quantity, but a complex number can represent two ingredients (e.g., amplitude and phase of a wave). This particular feature makes the use of vectors much more convenient. A complex number is the sum of a real and an imaginary number and it is usually described by solid dot in the complex plane as shown in the diagram. Here, real numbers are plotted as abscissa and imaginary numbers as ordinates and if the vector drawn from the origin to the solid dot indicates the complex number .A D x C iy/, where x and y are real numbers and i is an imaginary number. The length of vector from the origin denoted by jAj corresponds to the amplitude and the phase is given by the angle between the vector and the axis of real number. Multiplication of a vector by i makes it rotate counterclockwise by 90ı . Thus, multiplication by i converts a horizontal vector x to a vertical vector ix. From the power-series expansions of eix , cos x, and sin x, we obtain eix D cos x C i sin x. Then, the wave vector is described analytically by either side of the following equation with respect to the complex number A.

3.5 Solved Problems

89

Aei D A.cos C i sin / x D A cos ;

(1)

y D A sin

(2)

Further, Aei of the left hand side of (1) is called a complex exponential function. The complex conjugate of A D x C iy is A D x iy, or Aei for Aei , respectively. It is also usually described by A . If a wave is described in the complex form, its quantity can be obtained by multiplying the complex expression for the wave by its complex conjugate. As shown in the given diagram of the complex plane, A D x C iy and A D x iy show the so-called mirror symmetry and the following relationship is readily obtained. jAi j2 D Aei Aei D A2 .ei ei / D A2

(3)

This relationship can also be found as follows. A.cos C i sin / A.cos i sin / D A2 .cos2 C sin2 / D A2

2

2

2

A A D .x C iy/.x iy/ D x C y D A

(4) (5)

where the relationship of i 2 D 1 is utilized. The operation i in a complex number is equivalent to the square root of 1. For example, when the operation i is additionally applied to the imaginary number iy ! i.iy/ D i2 y D y. Namely, it becomes the real number with the opposite sign by rotating by =2 counterclockwise. The product of a complex number and its complex conjugate becomes the square of the amplitude of the original complex number (see (3)–(5)). This is quite a useful relationship which can be used for calculating the structure factor, because the intensity of a wave is proportional to the square of the amplitude.

Fig. 1 Addition (a) and multiplication (b) of complex numbers

90

3 Scattering and Diffraction

Note: The addition of two complex numbers corresponds to the sum of two vectors on a complex plane. The multiplication of two complex numbers is explained by the relationship of two vectors on a complex plane as follows. The rotation by 2 following the 1 rotation is equal to the rotation by .1 C 2 /. These relationships are illustrated in Fig. 1. Question 3.7 X-rays belong to the electromagnetic spectrum and its propagation speed is equal to the velocity of light (c). When X-ray wavelength is set to , the frequency is given by D c=. The cyclic variation in electric field intensity E may be expressed by the following equation, if time and phase are set to t and ı, respectively. E D A cos 2.t C ı/ Discuss the variation in electric field E and its intensity I in the following two cases of superiomposition of two waves. (1) Two waves with equal amplitude, but their phases are different. (2) Two waves with difference in both amplitude and phase. Answer 3.7 The variations in electric field intensity related to X-rays cannot be directly observed. The meaning of intensity of X-rays may be used in two ways. One is the energy which transits a unit area perpendicular to the direction of propagation of X-rays (wave). Another is the amount proportional to the square of the amplitude of wave obtained as a result of the interference effect of scattered X-rays. The latter is more frequently used in X-ray diffraction crystallography and we usually discuss its relative value. For this purpose, the expression of each wave as a complex exponential function is convenient and the cyclic variation in electric field intensity E can be described as follows: E D Ae2 i. t Cı/ D A cos 2.t C ı/ C iA sin 2.t C ı/

(1)

where A is the amplitude. (1) The summation of two waves with equal amplitude and different phase is given in the following equation. 0

E 0 D Ae2 i. t Cı/ C Ae2 i. t Cı / D Ae2 i. t Cı/f1 C e D Ef1 C e

2 i.ı 0 ı/g

2 i.ı 0 ı/

g

g

Since the intensity I is proportional to the square of E 0 , we obtain,

(2) (3) (4)

3.5 Solved Problems

91 0

I D jE 0 j2 D jEE j D A2 f1 C e2 i.ı ı/ g2 2

(5)

D A 2f1 C cos 2.ı ı/g

(6)

Here, x D 2.ı 0 ı/ and the following relationships of exponential and trigonometric functions are utilized. x

1Ce D2 ix

x

ei 2 C ei 2 2

! x

ei 2 D 2 cos

x ix e 2 2

(7)

1 ˛ D .1 C cos ˛/ 2 2 x x x x ix 2 .1 C e / D 2 cos ei 2 2 cos ei 2 2 2 x x D 2.1 C cos x/ei 2 ei 2 D 2f1 C cos 2.ı 0 ı/g cos2

(8) (9) (10)

According to (6), one obtains I D 4A2 if two waves are said to be in phase (D no phase difference) or constructive interference, but I D 0 if the phase difference is (corresponding to the out of phase case or destructive interference). (2) If two waves differ, not only in amplitude, but also in phase, the resultant E and I are given in the following equation. ED

X

Aj e2 i. t Cıj / D e2 i t

j

I D jEE j D @e2 i t

D

8 k C j > > > > > > > = > > > > > > > > ;

(5.2)

For example, it is shown in (5.1) that the reciprocal-lattice axis b1 is found to be perpendicular to the plane of the crystal-lattice vectors a2 and a3 , and its length is equal to the reciprocal of spacing of the (100) plane. The same relationship is confirmed with respect to other reciprocal-lattice vectors b2 and b3 . Namely, the reciprocal-lattice axes b2 and b3 are normal to the (010) and (001) planes of the crystal lattice, respectively, and their lengths are equal to the reciprocals of the spacing of these planes. In other words, the point at the end of the b1 vector is labeled (100), that at the end of the b2 vector is labeled (010), and that at the end of the b3 vector is labeled (001), and then an array of points each of which is labeled with its coordinates by the basic vectors. Similar relationships are confirmed for all the planes with an arbitrary Miller indices .h k l/ of the crystal lattice. Of course, h k l are three integers. Considering these results, a vector Hhkl perpendicular to the .h k l/ plane is given by the following equation with the reciprocal-lattice vectors, b1 , b2 , and b3 . Hhkl D hb1 C kb2 C lb3

(5.3)

The length of this reciprocal-lattice vector Hhkl is equal to the reciprocal of a spacing d of the .h k l/ plane, as follows. jHhkl j D

1 dhkl

(5.4)

Every crystal structure has two lattices, the real-space lattice in the dimensions of (length), and the reciprocal lattice in the dimensions of (1/length) and a diffraction pattern produced by a crystal is a map of the reciprocal lattice of the crystal. It may also be noted that the reciprocal lattice is a lattice in the Fourier space associated with the crystal. Each point of reciprocal lattice can represent the spacing as well as the direction of a related crystal plane. That is, the wave diffracted by the crystal planes with a certain periodicity of atoms in the real-space lattice appears as a diffracted spot in reciprocal space, as seen in the Laue photographs. Such spots also produce a certain regular sequence to form the reciprocal lattice. Fourier transform that is widely utilized in crystallography corresponds to an exchange operation from the real space to the inverse space or from the inverse space to the real space. According to (5.1), we have following relationships between a reciprocal-lattice vector bj and the real-space-lattice vector ak .

5.2 Intensity from Scattering by Electrons and Atoms

bj ak D ıjk I ıjk D 0.j ¤ k/; ıjk D 1.j D k/

171

(5.5)

The complete set of relationships is as follows. b1 a2 D b1 a3 D b2 a1 D b2 a3 D b3 a1 D b3 a2 D 0 b1 a1 D b2 a2 D b3 a3 D 1 The reciprocal-lattice vector K may be expressed by bj and integers kj as follows. K D k1 b1 C k2 b2 C k3 b3

(5.6)

Similarly, the real-lattice vector R may be given by aj and integers nj in the following form. (5.7) R D n1 a1 C n2 a2 C n3 a3 Since the scalar product of the reciprocal-lattice vector and the real-space-lattice vector can be expressed in the following equation, the product of K with R is found to be integers. K R D k1 n1 C k2 n2 C k3 n3 (5.8) As the scalar product of K R is an integer, the relationship of ei2KR D 1 is established. Therefore, it may safely be said that the reciprocal lattice, K, corresponds to one set of the wave vector 2K that satisfies the relationship of ei2KR D 1 with respect to all the real-space vectors R. When the real-space lattice is primitive (simple), the reciprocal lattice becomes primitive, and similarly, if the real-space lattice is classified into the complex (nonprimitive) lattice, the reciprocal lattice is also a complex one. Such general properties that take on a very simple relationship for any crystal whose unit cell is based on mutually perpendicular vectors (are checked by (5.1)). Some examples for explaining the relationship between the real-space lattice and the reciprocal lattice are illustrated in Fig. 5.1 using two cases of cubic and hexagonal crystals. Hhkl is found to be normal to the .hkl/ plane, and its length is equal to the reciprocal of the spacing dhkl . We recall that if the volume of Bravais lattices is set to V , then the volume of the reciprocal lattice is 1=V .

5.2 Intensity from Scattering by Electrons and Atoms X-rays are characterized by the wave–particle duality. An X-ray beam is known to be an electromagnetic wave characterized by an electric field that exerts its force on an electron. For this reason, an electron is continuously accelerated and decelerated by the field of X-rays. In other words, an electron as set into oscillation by an electric field of X-rays is continuously accelerating and decelerating during its motion so that the corresponding electron emits a new electromagnetic wave. In this case, an electron is said to scatter X-rays. The scattered beam has the exactly

172

5 Reciprocal Lattice and Integrated Intensities of Crystals

Fig. 5.1 The relationships between the reciprocal lattice and the real crystal lattice in two cases. A cubic crystal with lattice parameter ˚ and a a1 D 0:4 nm .4 A/ hexagonal crystal with lattice parameter a1 D 0:25 nm ˚ The axes of a3 and b3 .2:5 A/. in both cases are normal to the drawing

same wavelength and frequency as the incident X-ray beam. This is called coherent scattering or unmodified scattering. On the contrary, the interaction of the X-ray beam with electrons may also involve the exchange of energy and momentum. This is frequently represented by an elastic collision like that of two billiard balls. In this process, a loosely bound electron is knocked out by an incident X-ray photon (or quantum), and some of the energy of the incident X-ray photon is consumed by giving kinetic energy to the electron. Such scattering is called incoherent scattering, modified scattering, or Compton scattering. In addition, there is another way in which the X-ray beam interacts with an electron. When the incident X-ray photon has sufficient energy to knock out an inner shell electron to produce a photo electron that is emitted, the atom will be left in the excited state with a hole in the inner electron shell. This is called photoelectron effect by X-rays. Usually, the resultant hole is quickly filled by an electron located in the outer electron shell, and we obtain an X-ray photon with an energy equal to the difference in the relevant electron energy levels. This corresponds to the production process for characteristic radiation. Taking these interactions between X-rays and electron into account, some fundamental points for the scattering intensity from an atom and a crystal are summarized below.

5.2 Intensity from Scattering by Electrons and Atoms

173

When an unpolarized X-ray beam, such as that irradiated from an X-ray tube, encounters a single electron located at the origin, the scattering intensity Ie at the distance R from the origin is given by the following equation, which is frequently called Thomson equation. Ie D I0

e4 2 m c 4 R2

1 C cos2 2 2

(5.9)

where, I0 is intensity of the incident X-ray beam, e is an elementary charge, m is a rest mass of electron, c is the speed of light in vacuum, and 2 is the scattering angle. The last term within parentheses in (5.9) is called polarization factor, and an alternative expression of this factor is needed when using a crystal monochromator. The constant term .e 2 =mc 2 /2 of (5.9) is equivalent to the square of the classical electron radius re given as .2:8179 1015 /2 m2 in SI units. The amplitude, f , of the coherent scattering per atom including more than one electron is given by the sum of the amplitude fej per one electron (refer to Chap. 3), and it is as follows. XZ 1 X sin Qr dr (5.10) fej D 4 r 2 j .r/ f D Qr 0 j

j

The atomic scattering factor f is a function of Q or .sin =/ and is equivalent to the efficiency of scattering of a given atom in a given electron. Since the number of electrons in the atom is equal to the atomic number Z, we obtain R1 P 2 4 r j .r/dr D Z. Then, it is clear that f is close to the atomic number j 0 Z for any atom if the value of Q or .sin =/ becomes very small (scattering in the forward direction). The atomic scattering factors mentioned here satisfy the following two assumptions: (1) The distribution of electrons around the nucleus in atom can be well approximated by spherical symmetry. (2) The wavelength of the incident X-ray beam is much shorter than, or far from, that of an absorption edge of the scattering atom. There are not many cases, but apparent deviation from the perfect spherical symmetry is found in the electron distribution around the nucleus in a carbon atom with diamond structure, and this is one example where the assumption of (1) is not satisfied. The wavelength of the incident X-ray beam is close to the absorption edge of the scattering atom, the assumption of (2) is no longer accepted, and a correction of the atomic scattering factor in the following complex form is required. f D f0 C f 0 C if 00

(5.11)

where f 0 and f 00 are the real and imaginary components of the anomalous dispersion, and both components depend on the incident X-ray energy (wavelength). f0 in

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5 Reciprocal Lattice and Integrated Intensities of Crystals

(5.11) corresponds to the normal atomic scattering factor whose numerical values are given in Table (see Appendix A.3). On the contrary, f 0 and f 00 are related only to inner-shell electrons, such as K or L. Since the spatial distribution of inner electrons is considerably smaller than the magnitude of the X-ray wavelength, the dipole approximation Œexp.iQ r/ ; 1 is well accepted. Thus, the Q-dependence of the anomalous dispersion components can usually be ignored. Normal X-ray diffraction measurements are carried out using wavelengths away from the absorption edge region without correction of the anomalous dispersion effect. The main mechanism of X-ray diffraction by crystals is coherent scattering, but incoherent scattering also occurs. The incoherent scattering intensity i.M / per atom can be described by the simple sum of the incoherent scattering intensity of respective electrons as follows (refer to Chap. 3.2): i.M / D

X j

iej D Z

Z X

fej2

(5.12)

j D1

5.3 Intensity from Scattering by a Small Crystal When setting the origin of a certain unit cell in a crystal to a position O, a position vector of another unit cell from the origin of O may be expressed in the form m1 a1 C m2 a2 C m3 a3 . Here, a1 , a2 , and a3 are the basis vectors of the unit cell, whereas m1 , m2 , and m3 are integers and the coordinates. When setting the position vector of j -atom with respect to the origin in a unit cell to the vector rj , the positions of j -atoms in the crystal can be described by Rmj D m1 a1 C m2 a2 C m3 a3 C rj . Let us consider the case where the monochromatic X-ray beam with the wavelength and its intensity denoted by I0 encounters a very small single crystal. The incident and diffracted X-rays are usually represented by the wave vectors s0 and s, respectively, and Rmj , s0 , and s are, in general, not coplanar (js0 j and jsj D 1= where is wavelength). This crystal is assumed to be so small, relative to the distance between the X-ray source and the corresponding crystal, and therefore, the incident X-ray beam can be treated by the plane wave approximation. Suppose the plane waves are passing through the origin O and are scattered by j -atoms located at the position vectors Rmj and we wish to detect the scattered intensity I from a j -atom at P at a distance R from the origin O, I is given as follows. .s s0 / N1 a1 ˚ I D Ie FF sin2 .s s0 / a1 ˚ ˚ sin2 .s s0 / N2 a2 sin2 .s s0 / N3 a3 ˚ ˚ sin2 .s s0 / a2 sin2 .s s0 / a3 2 2 e Ie D I0 mc 2 R sin

2

˚

(5.13) (5.14)

5.4 Integrated Intensity from Small Single Crystals

175

Here, F is a structure factor. The scattering intensity by a single electron (Thomson’s scattering equation (5.9)) is also used here. In addition, since (5.13) is obtained under the condition that the incident X-ray beam is polarized vertically to the plane of the drawing, the polarization factor becomes unity. When the incident X-ray beam is not polarized at all, it is necessary to use the polarization factor described by .1 C cos2 2/=2. It is also noted that a function of .sin2 .N k/= sin2 k/ that appears in (5.13) is called “Laue function.” When k is integer multiple (n) of , both denominator and numerator of the Laue function are zero. This condition of k D n (n: integer) is equivalent to the so-called “diffraction by a crystal” whose lattice is characterized by the atomic array in a line at the regular interval, and then the phase difference of waves generated from such structure is given by integer multiples of the wavelength. It may also be added that if k ! n, the value of the limit of the Laue function is given by N 2 . Accordingly, when the scattering intensity is represented by the Laue function normalized by N 2 , a peak at the position of k D n is found to be sharper with increasing n. Since n is considerably large in a crystal lattice, a peak is usually approximated by the delta function at the position of k D n. On the contrary, let us consider that the value of a changes in the form of k D .ss0 /a, at constant n. If the absolute value, jaj becomes triple, the frequency of Laue function shows the reverse, as 1/3. That is, the length of the real-space lattice can be described as 1/(length) in the reciprocal space lattice, and their relationships are vice versa (see Question 5.8).

5.4 Integrated Intensity from Small Single Crystals Since the incident X-ray beam is usually not completely parallel and strictly a monochromatic radiation, this produces variation in its direction vector s0 . For example, the usual monochromatic beam is simply said to be one strong K˛ component, but it has a width of about 0.0004 nm. For this reason, we have to check the effect on diffraction phenomena by a crystal by considering some deviation from the ideal case. As already shown in Chap. 4.7 (2), the so-called destructive interference is not perfect so that we will detect a certain diffracted intensity in the close vicinity of the Bragg angle region near 2B . This is particularly true when the number of planes completely satisfying the Bragg condition is not sufficient, for example, by reducing the particle size of a crystalline powder sample, even one consisting of perfect single crystals. Accordingly a measured diffracted peak has some width in 2. In other words, if the crystal is perfect, we have to consider the small size alone a crystal imperfection. A single crystal is often found to have a periodic array of defects such as dislocation if its structure is examined in detail. Such a single crystal is said to have a kind of crystal imperfection that affects, more or less, the diffraction phenomena. This is a kind of substructure into which a single crystal is broken up as shown in the schematic diagram in Fig. 5.2, and this is called “mosaic structure.” In other words, a crystal with mosaic structure does not have its atomic arrangement with a

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5 Reciprocal Lattice and Integrated Intensities of Crystals

Fig. 5.2 Schematic diagram of mosaic structure of a real crystal and geometric arrangement for calculating the integrated intensity from a crystal plate with mosaic structure

completely regular crystal lattice extending from one side of the crystal to the other side. The crystal lattice is rather broken up into a number of tiny crystal blocks in the order of 100 nm in length, and each block is slightly disorientated from one another disturbing the coherency in their crystal planes. These tiny blocks are quite likely to be identical with subgrains, and the region between the blocks are considered the dislocation wall. In addition, the maximum angle of disorientation between tiny crystal blocks changes from a very small value to as much as 1ı . In this case, the incident X-ray beam changes its direction, little by little, at each mosaic crystal structure so as to form a relatively sharp diffraction peak with a certain range in width. As a result, the integrated intensity of the diffracted beam from a crystal with mosaic structure increases, relative to that computed theoretically for an ideally perfect single crystal. As shown in the schematic diagram in Fig. 5.3, a large number of atoms arranged with a perfectly three-dimensional periodic array to form a crystal scatter X-rays in relatively few directions because the structural periodicity causes the so-called destructive interference of the scattered X-rays in all directions except only those predicted by the Bragg law in which constructive interference takes place. It may be added that a single atom scatters the incident X-ray beam in all directions in space. Although the scattering intensity from a small single crystal is expressed by (5.13), we have to use the integrated intensity for comparing with the measured intensity data. Here, let us consider the scattering intensity in the angular region between a slightly smaller angle near 2B and a slightly larger angle when rotating the crystal with its angular speed of !. Here, 2B is the Bragg angle. The integrated intensity P from the .h k l/ plane may be given by the following equation. ! 2 Nuc Fhkl e4 I0 P D ! m2 c 4 va

1 C cos2 2 2 sin 2

! (5.15)

where va is the volume of a unit cell and Nuc is the number of unit cell included in a crystal. The factor given in the second parenthesis to the right-hand side of (5.15) is called the Lorentz polarization factor with respect to the measurement for a single crystal using an unpolarized incident X-ray beam. Equation (5.15) clearly suggests that the experimental results can provide the structure factor Fhkl , from which the unknown crystal structure is able to be determined if the integrated intensity P

5.5 Integrated Intensity from Mosaic Crystals and Polycrystalline Samples

177

a

b

Intensity

Crystal

2q [degree]

Monoatomic gas

2q [degree]

Fig. 5.3 Schematic diagrams for scattering by a crystal (a) and by a single atom (b) together with their scattering patterns for comparison. The vertical scales are not equal

from a .hkl/ plane and the intensity I0 of the incident X-ray beam can be quantitatively obtained. If the Fhkl value for a pure substance with simple structure can be estimated from the measured intensity data with sufficient reliability, the atomic scattering factor f may also be computed.

5.5 Integrated Intensity from Mosaic Crystals and Polycrystalline Samples The integrated intensity of a single crystal with mosaic structure can be represented as follows. Each mosaic block in the single crystal is considered a perfect crystal, but a slight difference in direction of adjoining mosaic blocks is recognized so as not to interfere with each other. This is called “ideal mosaic single crystal.” A real single crystal is usually described by the intermediate state between perfect single crystal and ideal single mosaic crystal. If a desired crystal is regarded as an ideal mosaic single crystal, the integrated intensity can be computed by assuming that (5.15) is allowed to be used in each mosaic block. Let us consider the intensity of the incident X-ray beam I0 and its cross-section A0 and that the average volume of mosaic blocks is set to ıv (see Fig. 5.2). If the intensity I0 of the incident X-ray beam reaches the mosaic block located in a portion

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5 Reciprocal Lattice and Integrated Intensities of Crystals

between z and z C dz deep from the crystal surface, its intensity will be given by the form of I0 exp.2z= sin /, where is the linear absorption coefficient for a substance of mosaic block. Considering the intensity P0 D I0 A0 of the incident X-ray beam, the integrated intensity from an ideal mosaic single crystal can be given by the following equation. ! 2 3 Fhkl e4 P0 P D ! m2 c 4 2v2a

1 C cos2 2 2 sin 2

! (5.16)

A crystalline powder sample is approximated by a very small poly-crystalline aggregate, and each grain is found to have a crystallographic orientation different from that of its neighbors. When considered as a whole, the orientations of all the grains are randomly distributed. Let us consider that a monochromatic X-ray beam with the wavelength hits a powder crystalline sample where a tiny single crystal in a small part of this sample shows its direction so as to satisfy the Bragg law with respect to the .h k l/ plane. In this case, the plane-normal vector Hhkl of the .h k l/ plane of a tiny single crystal coincides with the direction of the scattering vector .s s0 /, which is defined by the vector s0 of incident X-ray beam and the vector s of diffracted X-ray beam. It is also noteworthy that the concepts of the reciprocal lattice and the Ewald sphere as well as the limiting sphere are very convenient for understanding the relationships between the incident and diffracted beam vectors and crystal plane. The details of such information are available in Question 5.7. When the total number of small single crystals in a powder sample is set to Nsc and multiplicity factor phkl , which indicates the number of the equivalent .h k l/ plane with the same spacing and structure factor, the total number of the planenormal vector Hhkl in the powder sample is given by Nsc phkl . Here, we take into account the distribution of the scattering vector from a powder crystalline sample satisfying the Bragg law, if the incident angle is in the region between . C ˛/ and . C ˛ C d˛/. Then, the diffracted intensity from the .h k l/ plane can be obtained by integrating all directions and the cross-sectional area followed by multiplying the scattered intensity I of one tiny single crystal given by (5.13), the number of crystals dNsc in d˛ of a sample and the area element dA in the cross-section. The intensity P computed in this procedure is equivalent to the total diffraction intensity uniformly distributed on the Debye ring so as to form a cone of half apex angle 2, as is seen in Fig. 5.4. Accordingly, the intensity that can be measured using a diffractometer corresponds to the intensity per unit length of this Debye ring. That is, it is equivalent to the intensity P 0 , which is divided P by the circumference of the Debye ring .2R sin 2/. Therefore, when the unpolarized X-ray beam with its intensity of I0 is employed, the integrated intensity measured at a distance R from the crystalline powder sample is given by the following equation: P0 D

I0 16R

e4 m2 c 4

2 V 3 mhkl Fhkl v2a

1 C cos2 2 sin sin 2

(5.17)

5.6 Solved Problems

179

Fig. 5.4 The relationships between the reciprocal lattice and the Ewald sphere (formation of a cone of diffracted X-rays) in the Debye–Scherrer method

where V is the irradiated volume. It is worth mentioning that the last term to the right-hand side of (5.17) is the Lorentz polarization factor for a crystalline powder sample. Equation (5.17) shows that the integrated intensity for the .h k l/ plane of the crystalline powder sample is proportional to three factors: multiplicity factor phkl , structure factor Fhkl , and Lorentz polarization factor. The most common X-ray diffraction measurement for crystalline powder samples is known to be the method using a diffractometer. In this case, the conditions are adjusted so as to satisfy the sample being a flat plate with infinite thickness. In this experimental condition, the irradiated volume of (5.17) may be replaced by V D A0 =2.

5.6 Solved Problems (18 Examples)

Question 5.1 The reciprocal lattice corresponding to a unit cell described by the primitive crystal-lattice vectors a1 , a2 , and a3 has a unit cell defined by the vectors b1 , b2 , and b3 given by the following equations, when the volume of the crystal unit cell is set to V . b1 D

a2 a3 ; V

b2 D

a3 a1 ; V

b3 D

a1 a2 V

This corresponds to the definition of the reciprocal lattice as a function of the crystal lattice. Show the crystal lattice as a function of the reciprocal lattice.

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5 Reciprocal Lattice and Integrated Intensities of Crystals

Answer 5.1 Recall a few fundamentals of vector algebra. (1) Product of two vectors A and B (it is called cross product or vector product) .A B/ D .B A/ (2) Product of three vectors A, B, and C (it is called triple-scalar product) A .B C/ D B .C A/ D C .A B/ D A .C B/ D B .A C/ D C .B A/ When three vectors A, B, and C correspond to the sides of parallelepiped as shown in Fig. 1, .B C/ is an equivalent vector to the area of parallelogram that constitutes the base of parallelepiped. Note that the vector product of two vectors A and B written by A B is corresponding to a vector C normal to the plane of A and B. The magnitude of C is equal to the area of the parallelogram constituted of A and B, whereas the direction of C is along the direction of movement forward of the right-hand screw if rotated in such a way so as to bring A to B. Therefore, the direction is found to be perpendicular to the plane made by B and C. Consequently, the triple-scalar product of three vectors, A .B C/, may be described by the projection of the vector A onto .B C/ This is equivalent to the volume V of parallelepiped. Reference: The scalar product of two vectors A and B or the inner product (or dot product) is expressed as A B. It is also noted that A B D B A due to a scalar quantity.

Fig. 1 Graphical representation of A .B C/

Next, obtain the cross product of the reciprocal-lattice vectors b2 and b3 , i.e., (b2 b3 ) b2 b3 D

1 a3 a1 a1 a2 D 2 f.a3 a1 / .a1 a2 /g V V V

(1)

When u D .a3 a1 / is set and the known formula u.vw/ D v.uw/w.uv/ for vector products is simultaneously employed, the following equations may be readily obtained. u .a1 a2 / D a1 .u a2 / a2 .u a1 / D a1 f.a3 a1 / a2 g a2 f.a3 a1 / a1 g

(2)

5.6 Solved Problems

181

The second term of (2) contains the same vector a1 twice so that it vanishes as one of the characteristic features of vectors includes u .u w/ D 0. In addition, if the relationship of .a3 a1 / a2 D V is taken into account, (1) may be rearranged in a simplest form as follows. a1 (3) b2 b3 D V On the contrary, the volume V of reciprocal lattices is given by the following equation. .a2 a3 / a1 1 (4) D V D b1 .b2 b3 / D 2 V V Therefore, the following relationship can be obtained by combining (3) and (4). a1 D

b2 b3 ; V

a2 D

b3 b1 ; V

b3 D

b1 b2 V

(5)

When comparing the given equation in question with (5), it turns out that similarity between two formulas is well recognized, and only right-hand side and left-hand side are changed. Namely, the reverse of the reciprocal-lattice vectors is equal to the crystal-lattice vectors, and the reciprocal of the volume of the crystal lattice is equivalent to the volume of the reciprocal lattice. Question 5.2 Demonstrate that the reciprocal-lattice vector Hhkl D hb1 C kb2 C lb3 is perpendicular to the .hkl/ plane, regardless of crystal systems, and that the magnitude of this reciprocal-lattice vector is equal to the reciprocal of the spacing .dhkl /.

Answer 5.2 The relationship between the crystal-lattice vectors a1 , a2 , and a3 with respect to the .hkl/ plane and the reciprocal lattice Hhkl is shown in Fig. 1 from which the target is given in the following. If the scalar product (inner product) of two vectors included in the .hkl/ plane and the reciprocal-lattice vector is obtained, and its value is found to be zero and then the two kinds of vectors are said to show the perpendicular relationship. In Fig. 1, two non parallel vectors in the .hkl/ plane are found, ah1 ak2 and ak2 al3 . Hhkl

a

1

h

a a2 a2 1 D .hb1 C kb2 C lb3 / k k h a1 a2 D hb1 C 0 C 0 0 C kb2 C0 D11D0 h k (1)

Here, we use the characteristic properties of real and reciprocal-lattice vectors, such as bj ak D 1 if j D k and bj ak D 0 if j ¤ k. In other words, this is based on the useful relationship that since b3 , for example, is perpendicular to both a1 and a2 , the scalar product with either one of the crystal and reciprocal vectors is zero. The

182

5 Reciprocal Lattice and Integrated Intensities of Crystals

Fig. 1 The relation of vector and unit vector n of normal direction that is in (h k l) plane

same result is readily confirmed in the combination of Hhkl ah1 ak2 . Therefore, Hhkl is said to be perpendicular to the .h k l/ plane. If the unit vector on the normal that goes to the .h k l/ plane is set as n, the following equation is obtained. nD

Hhkl jHhkl j

(2)

The spacing dhkl corresponds to the projection to the direction of normal vector n that goes to the target plane from the origin 0 in Fig. 1. Therefore, the following a1 relationship is proved with respect to the vector . h a1 Hhkl 1 a1 a1 nD D .hb1 C kb2 C lb3 / h h jHhkl j jHhkl j h 1 h 1 .a1 b1 C 0 C 0/ D D jHhkl j h jHhkl j

dhkl D

(3)

Reference: Let us confirm some fundamental points for vector, area, volume, and the relevant matrices and determinants. Using any orthogonal set of unit vectors, e1 , e2 , and e3 , whose direction is the same as a coordinate axis, three vectors A, B, and C situated on a plane are expressed by the following equation. 9 A D a1 e1 C a2 e2 C a3 e3 = B D b 1 e1 C b 2 e2 C b 3 e3 ; C D c1 e1 C c2 e2 C c3 e3

(4)

In this case, the vector product of A and B is given as follows. A B D .a2 b3 a3 b2 /e1 C .a3 b1 a1 b3 /e2 C .a1 b2 a2 b1 /e3

(5)

5.6 Solved Problems

183

The area (S ) spanned by two vectors A and B situated on a plane corresponds to the case where the coefficients a3 and b3 for e3 of a unit vector perpendicular to the plane of the drawing are zero. Therefore, the following relationship may be obtained from (5). (6) A B D .a1 b2 a2 b1 /e3 D S e3 Equation (6) shows that the area of the parallelogram formed by two vectors A and B on a plane is expressed by .a1 b2 a2 b1 /, and if the direction of A!B corresponds to that of e1 ! e2 , it becomes positive sign, and if reverse, it becomes negative sign. It is also helpful to use a determinant for expressing these relationships. For example, in the case of (6), one obtains as follows: ˇ ˇ ˇ a1 a2 ˇ ˇ ˇ ˇ b1 b2 ˇ D a1 b2 a2 b1

(7)

On the contrary, the volume (V ) may be expressed in the following relation if three vectors A, B, and C are in the space. B C D .b2 c3 b3 c2 /e1 C .b3 c1 b1 c3 /e2 C .b1 c2 b2 c1 /e3

(8)

A .B C/ D a1 .b2 c3 b3 c2 / C a2 .b3 c1 b1 c3 / C a3 .b1 c2 b2 c1 / D V (9) Here, the relationships of e1 e1 D e2 e2 D e3 e3 D 1 and e1 e2 D e2 e3 D e3 e1 D 0 for unit vectors are utilized. ˇ ˇ ˇ a1 a2 a3 ˇ ˇ ˇ ˇ b1 b2 b3 ˇ D a1 .b2 c3 b3 c2 / a2 .b1 c3 b3 c1 / C a3 .b1 c2 b2 c1 / ˇ ˇ ˇc c c ˇ 1 2 3 D a1 b2 c3 C a2 b3 c1 C a3 b1 c2 a1 b3 c2 a2 b1 c3 a3 b2 c1

(10) (11)

Equation (9) shows that the direction of vector A, on the basis of a plane made by A and B, has positive sign if the vector A is situated at the same side of a plane made by vectors .B C/. If it is on the opposite side, the sign becomes negative. As mentioned previously, A .B C/ corresponds to a volume, and it is expressed by a determinant given by (10). Question 5.3 The direction of the nodal line of two planes, .hkl/ and .h0 k 0 l 0 /, which are not parallel to each other, is called a zone axis and is usually expressed as Œuvw. (1) Obtain a relationship with uvw, hkl, and h0 k 0 l 0 . (2) Check the Weiss zone law, hu C kv C lw D 0, using the fact that a reciprocal-lattice vector is perpendicular to the corresponding crystal plane.

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5 Reciprocal Lattice and Integrated Intensities of Crystals

Answer 5.3 (1) If the reciprocal-lattice vectors of the .hkl/ and .h0 k 0 l 0 / planes are Hhkl and Hh0 k 0 l 0 , respectively, a zone axis Œuvw may be expressed by the vector Zuvw whose direction is parallel to the vector product of two reciprocal-lattice vectors Hhkl and Hh0 k 0 l 0 . Such relationships are easily seen in Fig. 1. Hhkl Hh0 k 0 l 0 D .hb1 C kb2 C lb3 / .h0 b1 C k 0 b2 C l 0 b3 / D .hk 0 h0 k/b1 b2 C .kl 0 k 0 l/b2 b3 C .lh0 l 0 h/b3 b1

(1) (2)

Here, we use the relationship that the vector product of the same vector (exam3 ple: b1 b1 ) is zero. In addition, when using the relationships a1 D b2Vb , b3 b1 b1 b2 1 a2 D V , a3 D V , and V D V , (2) can be rewritten as follows. a3 a1 a2 Hhkl Hh0 k 0 l 0 D .hk 0 h0 k/ C .kl 0 k 0 l/ C .lh0 l 0 h/ V V V ˇ ˇ ˇ a1 a2 a3 ˇ ˇ 1 ˇ D ˇˇ h k l ˇˇ V ˇ 0 0 0ˇ h k l

(3) (4)

The vector Zuvw may be expressed in the following form, when the unit cell is defined by a1 , a2 , and a3 , called the primitive crystal-lattice vectors. Zuvw D ua1 C va2 C wa3 Therefore, the following relationship is obtained from (3) and (5). ˇ ˇ 0 0 1 ˇuˇ kl k 0 l ˇ ˇ ˇ v ˇ D @ lh0 l 0 h A ˇ ˇ ˇwˇ hk 0 h0 k

Zuvw

h'k'l'

hkl

Hhkl

Hh'k'l'

Fig. 1 The correlation between two reciprocal-lattice vectors and a zone axis

(5)

5.6 Solved Problems

185

(2) On the contrary, since a normal vector of a plane is perpendicular to any vector included in that plane and if the .hkl/ plane belongs to the zone axis Œuvw, the scalar product of the reciprocal-lattice vector Hhkl , corresponding to the plane, and the vector Zuvw showing the direction of the zone axis Œuvw should be zero i.e., Hhkl Zuvw D 0. Hhkl Zuvw D .hb1 C kb2 C lb3 / .ua1 C va2 C wa3 /

(6)

D hub1 a1 C hvb1 a2 C hwb1 a3 C kub2 a1 C kvb2 a2 C kwb2 a3 C lub3 a1 C lvb3 a2 C lwb3 a3

(7)

From the definition of the reciprocal-lattice vectors b1 , b2 , and b3 , the primitive crystal-lattice vectors a1 , a2 , and a3 , the following useful relationships are obtained, since b1 , for example, is known to be perpendicular to both the vectors a2 and a3 . bj ak D 1 .j D k/ (8) bj ak D 0 .j ¤ k/ By applying these relationships to Hhkl Zuvw , the Weiss zone law may be readily obtained. hu C kv C lw D 0 (9) Question 5.4 When two crystallographic directions are expressed by two vectors, Au D u1 a1 C u2 a2 C u3 a3 and Av D v1 a1 C v2 a2 C v3 a3 , the angle Av formed by two vectors Au and Av may be given by cos D jAAuujjA . vj Obtain an equation for providing cos in two cases, (1) orthorhombic system and (2) cubic system.

Answer 5.4 (1) Orthorhombic system ja1 j ¤ ja2 j ¤ ja3 j

a1 ? a2 ? a3

p p Au Au D ju1 a1 j2 C ju2 a2 j2 C ju3 a3 j2 p p jAv j D Av Av D jv1 a1 j2 C jv2 a2 j2 C jv3 a3 j2 jAu j D

Au Av D u1 v1 ja1 j2 C u2 v2 ja2 j2 C u3 v3 ja3 j2 u1 v1 ja1 j2 C u2 v2 ja2 j2 C u3 v3 ja3 j2 cos D p p ju1 a1 j2 C ju2 a2 j2 C ju3 a3 j2 jv1 a1 j2 C jv2 a2 j2 C jv3 a3 j2 (2) Cubic systems ja1 j D ja2 j D ja3 j D a jAu j D

q

u21 C u22 C u23 a

a1 ? a2 ? a3 jAv j D

q

v21 C v22 C v23 a

186

5 Reciprocal Lattice and Integrated Intensities of Crystals

Au Av D .u1 v1 C u2 v2 C u3 v3 /a2 u 1 v1 C u 2 v2 C u 3 v3 cos D q q u21 C u22 C u23 v21 C v22 C v23

Question 5.5 If the orthogonal vectors of unit length are set to ex , ey and ez , the primitive translation vectors of the hexagonal close-packed (hcp) structure may be given as follows. p 3 a a1 D aex C ey ; 2 2

p 3 a aex C ey ; a2 D 2 2

a3 D cez

(1) Obtain the unit vectors of reciprocal lattices b1 , b2 , and b3 . (2) Show the first Brillouin zone of the hcp lattice.

Answer 5.5 (1) For example, the definition of reciprocal-lattice vector b1 is as follows. b1 D

a2 a3 a1 a3 D V a1 .a2 a3 /

Therefore, when calculating a1 .a2 a3 / called the triple-scalar product using matrices, we obtain the following results. ˇ ˇ ˇ ex ey ez ˇ ˇ ˇ p ˇ ˇ .a2 a3 / D ˇˇ 3 a a 0 ˇˇ 2 ˇ ˇ 2 ˇ 0 0 cˇ ˇ p ˇa ˇ ˇ ˇ 0ˇ 3 ˇ ˇ ˇ D ex ˇ 2 ˇ ey ˇˇ 2 a ˇ 0 cˇ ˇ 0 p ac 3 D ex C acey 2 2

ˇ ˇ ˇ p ˇ ˇ 3 a ˇˇ ˇ ˇ 0ˇ C e ˇ a ˇ zˇ 2 2ˇ ˇ ˇ ˇ c 0 0ˇ (1)

! ! p p a ac 3 3 a1 .a2 a3 / D aex C ey ex C acey 2 2 2 2 p p 3 2 3 2 D a cex ex C a cey ey .* ex ey is zero/ 4 4 p 3 2 a c .* ex ex D ey ey D 1/ (2) D 2

5.6 Solved Problems

187

The vector products of .a3 a1 / and .a1 a2 / are also calculated as follows. p 3 ac acey .a3 a1 / D ex C 2 2 p 2 3a ez .a1 a2 / D 2

(3) (4)

From (1) to (4), the unit vectors of the reciprocal lattice b1 , b2 , and b3 can be obtained as follows. p 9 3 ac > > e ace C > x y > 1 a2 a3 1 > 2 2 D p p ex C ey > D b1 D > > > a1 .a2 a3 / a 3 2 3 > > a c > > > 2 = (5) > 1 1 a3 a1 > > b2 D p ex C ey D > > a1 .a2 a3 / a > 3 > > > > > > > 1 a1 a2 > ; D ez b3 D a1 .a2 a3 / c (2) In solid-state physics, the polyhedron called Wigner–Seitz cell is widely used. We take the central (Winger–Seitz) cell of the reciprocal lattice as the first Brillouin zone. Each such cell involves one reciprocal lattice point at the center of the cell. It may be suggested that the construction of the Wigner–Seitz type unit cell in crystal lattice is the same as that of the first Brillouin zone in reciprocal lattice. We may recall some fundamental points about the first Brillouin zone. In this unit cell, one reciprocal lattice point is set to the origin and connects with all points that are adjacent to the origin by a straight line. Next, normal planes perpendicular to the lines are constructed at their middle points, and we choose a polyhedron that has a minimum volume from the set of polyhedra enclosed by these planes. It is known that if the first Brillouin zone of the body-centered cubic lattice is bounded by the planes normal to the twelve vectors at their midpoints, then the resultant zone becomes a regular rhombic dodecahedron. With respect to the face-centered cubic lattice, the boundaries of the central cell are determined for the most part by eight planes normal to these vectors at their midpoints to form an octahedron. However, the corners of this octahedron are cut by planes that are the perpendicular bisectors of six other reciprocal-lattice vectors. The first Brillouin zone of the hcp lattice can be obtained by the following procedure. From (5), the arbitrary reciprocal-lattice vector Hpqr may be provided in the following form.

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5 Reciprocal Lattice and Integrated Intensities of Crystals

Hpqr D .pb1 C qb2 C rb3 / 1 1 r D p .p q/ex C .p C q/ey C ez a c 3a

(6) (7)

The shortest nonzero vectors may be given when the values of prq are any of the combinations of 100 or 010. All these results are summarized as follows. 9 1 1 1 1 > H100 D H100 p ex C ey ; D p C e e N x y > > > a a 3 3 > > > > > > = 1 1 1 1 H010 p ex C ey ; p ex ey H010 D (8) N D > a a 3 3 > > > > > > > > 2 1 1 2 > ; ; H D p D p H110 e e N x N x 110 a a 3 3 1 1 ez ; ez H001N D (9) c c The first Brillouin zone of the hcp lattice can be obtained from the perpendicular bisector planes of these eight reciprocal-lattice vectors given by (8) and (9). The results are shown in Fig. 1a. Namely, the first six reciprocal-lattice vectors make six side planes of the regular hexagonal prism and the two remaining reciprocal-lattice vectors provide both top and bottom planes. As shown in Fig. 1b, the first Brillouin zone of the hcp lattice is described, in another way, by the hexagonal prism with its height given by jb3 j D 1=c. Here, the hexagonal base is formed by lines that are obtained by drawing six vectors (solid lines) in the reciprocal-lattice space and further drawing the perpendicular bisector (broken lines) of each vector. H001 D

Fig. 1 The first Brillouin zone of the hcp lattice

Note: In the real hcp crystals, since the position of 23 31 21 is occupied by an p atom with the c/a value deviated from the ideal case .c=a D 8=3 D 1:633/, the Brillouin zone shows a more complicated form. Such information is available in other monographs on Solid-State Physics, see for example “Introduction to the Electron Theory of Metals” by Uichiro Mizutani, Cambridge University Press, (2001) Chap. 5.

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Question 5.6 Explain that the constructive interference is observed when the scattering vector corresponds to a reciprocal-lattice vector.

Answer 5.6 Let us consider the case where the incident X-ray beam with a wave vector s0 encounters the scatterer located at point B, which is defined by vector r in the real space, and produces the coherent scattering (see Fig. 1) by coupling the scatterer located at point A. It is also assumed here that the scattered waves in the direction of s are measured at the position P with a sufficiently long distance R from points A and B.

Fig. 1 The waves scattered from two points A and B and the scattering vector

The scattering vector q is given by q D s s0 , and the optical path difference produced from the scatterers at points A and B is D r q. Therefore, the scattered wave detected at the position of P is represented as superposition of the waves produced at points A and B, and it may be expressed in the following form.

D A C B D e2i.qRvt / C e2if.qRC/vt g De

2i.qRvt /

Œ1 C e

2iqr

(1) (2)

The first term to the right-hand side of (2) is equivalent to a common wave phase factor for all waves, and the second term corresponds to the amplitude of the scattered wave. Therefore, the interference effect proves only if the exponent portion of the second term is integer multiples of 2i. Namely, when the optical path difference is integer multiples of the wavelength, the scattered wave becomes in phase. In order to extend the result for two points A and B to a general case containing many scatterers, the scattering power of n-th scatterer is set to fn . Then, a generalized form of the amplitude of the scattered waves may be expressed as follows. X

n D e2i.qRvt / G.q/ (3)

D n

G.q/ D

X n

fn e2iqrn

(4)

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5 Reciprocal Lattice and Integrated Intensities of Crystals

With respect to the amplitude of the scattered waves, (4) shows that the maximum value is obtained when all the exponent portions are integer multiples of 2i. This is because the scattered waves by all the scatterers located at rn are in phase and reinforce one another (so-called constructive interference) to form a diffracted beam in the direction of q. Note that in all other directions of space, the scattered waves are out of phase and cancel one another. That is, rn is characterized by a certain periodicity (with a regular interval) so that q rn is expressed by integers. It may be helpful to recall the relationships between the crystal-lattice vector Rpqr showing arbitrary lattice points and the reciprocal-lattice vector Hhkl . Rpqr D pa1 C qa2 C ra3 Hhkl D hb1 C kb2 C lb3

(5) (6)

Here, a1 , a2 , and a3 are the primitive vectors of the crystal lattice, and b1 , b2 , and b3 are those of the reciprocal lattice, pqr and hkl are integers. If the scalar product between these two vectors is calculated, the following relationship may be confirmed. Rpqr Hhkl D .pa1 C qa2 C ra3 /.hb1 C kb2 C lb3 / D ph C qk C rl D integer

(7) (8)

Here, we use the vector property that bj ak is unity when j D k and that it is zero in the j ¤ k case. Equation (8) is one of the important relationships that always holds in a crystal. Accordingly, since the crystal-lattice vector Rpqr can be set to rn , the condition, for which all exponents of the exponential functions in (4) are integer multiples of 2i, may be given by the following equation. q D Hhkl

(9)

This means that the so-called constructive interference can be obtained only when the scattering vector corresponds to a reciprocal-lattice vector. Schematic diagram for this relationship is given in Fig. 2.

Fig. 2 Correlation between scattering vector and reciprocal-lattice vector

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Note: This is equivalent to the Bragg law. Further information is available in other monographs, see for example “Structure Determination by X-ray Crystallography” by Mark Ladd and Rex Palmer, Fourth Edition, Kluwer Academic/Plenum Publishers, London, (2003). Question 5.7 With respect to X-ray diffraction from a crystal, explain the relationships between the incident beam, the scattered beam, and the direction of a crystal for detecting the scattering intensity using the sphere of reflection (Ewald sphere) or the limiting sphere.

Answer 5.7 The present requirement is to consider the following point under the q D Hhkl condition. When the starting point of the scattering vector is set to the origin of the reciprocal-lattice vector, the wave vector s0 of the incident beam corresponds to the vector of .1=/, which points in the direction of q(D equivalent to the origin of the reciprocal-lattice vector). Here, is the wavelength of the incident beam. The scattered beam wave vector s turns into the vector orientated from the starting point of s0 to the terminal point of q, and the length of s0 and s is given by .1=/, and it may be simplified by q D Hhkl D .s s0 /=. These relationships may be expressed as shown in Fig. 1, explaining the conditions for diffraction graphically. The sphere of radius .1=/ in this figure is the Ewald sphere (it is also called Ewald reflection sphere). Constructive interference will be obtained if the condition of q D Hhkl is satisfied. This corresponds to the case where a reciprocal-lattice point hkl touches the surface of the Ewald sphere drawn around the origin. Accordingly, we can detect the diffracted intensity, when a detector is set to the terminal point of q. When any reciprocal lattice point of hkl does not touch the surface of the Ewald sphere, it is difficult to measure the diffraction intensity related to the hkl plane. Figure 1 shows the case where a crystal (real lattice) sample slightly rotates and also shows its relevance to the reciprocal-lattice points. However, it should also be kept in mind that even when a crystal sample is rotated, not all reciprocal-lattice points necessarily touch the surface of the Ewald sphere. This is also supported by the fact that only the reciprocal-lattice points, which are located at inside the diameter .2=/ of the Ewald sphere drawn around a point of (000), satisfy the condition of q D Hhkl as easily seen in Fig. 2 so that the detection of diffraction intensity is not always possible. In other words, the diffraction intensities for the reciprocal-lattice points can be observed when satisfying the following condition (see also Fig. 3) using the incident beam with the wavelength of . This sphere is called limiting sphere. 2 1 or dhkl (1) dhkl 2 When there are many small crystal grains and the direction of their crystal planes is distributed at random, the corresponding reciprocal-lattice points will be smeared out onto spheres around the point (000). As a result, with respect to the reciprocallattice points touching the surface of the Ewald sphere drawn by the radius of

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5 Reciprocal Lattice and Integrated Intensities of Crystals

.1=dhkl / around the origin, the higher the probability we obtain, the more frequently we will be able to detect the diffraction intensity (see Fig. 3). This is a fundamental principle for measuring the diffraction intensity using a goniometer for crystalline powder samples.

Fig. 1 Rotation of a crystal sample and its relevance to the reciprocal-lattice points

Fig. 2 Graphical representation for the relationship between the rotation of a crystal sample and the limiting sphere

Note: Considering the terminal point of the incident beam vector being the initial point of the reciprocal-lattice vector, Fig. 4 shows the relationships between the wavelength of the incident X-ray beam and the limiting sphere using the case of 1 and 2 . In this case, at the reciprocal-lattice points overlapped by the Ewald sphere with radius of 1=1 or 1=2 , the diffraction intensity will be detectable in the direction oriented from the center of the Ewald sphere to the respective reciprocallattice points. If the wavelength of the incident beam continuously changes from

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Fig. 3 Relationships between the wavelength of the incident beam and the limiting sphere for a polycrystalline powder sample

Fig. 4 Relationships between the terminal point of the incident beam and the initial point of the reciprocal-lattice vector when two beams with wavelengths of 1 and 2 are simultaneously incident on a sample

min to max , we will be able to detect the diffraction intensity with sufficiently high probability, since there are many Ewald spheres with various radii (Laue method). Question 5.8 Explain Huygens principle for a typical optical phenomenon where light passes through a small hole (or slit) and the relevant Kirchhoff theory of diffraction that handles it mathematically using wave equation.

Answer 5.8 As shown in Fig. 1a, when a plane wave perpendicularly encounters a plate with a small hole (or slit), a spherical wave is produced from the hole as a central point and propagates. Two or more spherical waves are produced from each hole if the plate has several small holes so that the interference of waves is observed

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5 Reciprocal Lattice and Integrated Intensities of Crystals

at a position apart from the plate (see Fig. 2b). Such interference behavior is known to depend on the periodicity of small holes. It may be suggested that Huygens principle is the generalized method for covering various wave phenomena including a relatively big hole. In such a case, a big hole is assumed to be the adjoined and connected small holes. Let us consider the optical phenomena with respect to a small hole located at a distance l0 from a light source P0 . This small hole is assumed to be a small lattice with dx and dy, and produces the spherical wave and its behavior is observed at a position P located at a distance l from the hole. For convenience of discussion by the Kirchhoff theory of diffraction, we set up the condition as shown in Fig. 2. Here, it is supposed that the plate is in the x y plane and that a small hole is located at the position characterized by the vector r from the origin. The incident wave hits a plate from the negative side of z-axis of Fig. 2, and only the wave encounters a small area characterized dr D dx dy will receive modulation. Here, dr is defined by the position with g.r/ D g.x; y; z D 0/ as a function of r. For example, considering that if a hole is open, it may be described as g.r/ D 1 and conversely, g.r/ D 0 for a closed hole. Then, we may express the case where a spherical wave is produced from a small hole dr by satisfying the condition g.r/ D 1. Next, let us consider that such a spherical wave is observed at the point P which is denoted by the vector l from dr, as well as the distance jl´j from the origin. Point P is characterized by the position using axes x, y, and z and angles ˛, ˇ, and , as shown in Fig. 2. According to the Kirchhoff theory of diffraction, if the wave 0 .l0 / from the light source is incident on a small hole dr and passes through it, and further reaches the point P, the contribution d (P) corresponding to that wave may be given by the following equations. d .P/ D is0 ˚ ˚D

e2is0 jlj g.r/ 0 .l0 /dr jlj

cos.n; l0 / cos.n; l/ 2

(1) (2)

Fig. 1 (a) The incident (plane) wave encounters one small hole and produces a spherical wave, (b) the interference effect found in two small holes and the related intensity distribution and (c) image of the interference effect found in a big hole

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As shown in Fig. 3, n in the small area defined by (dxdy) is a normal vector whose direction is oriented to the observation point P, and .n; l0 / and .n; l/ are the angles formed by the corresponding two vectors. It may be added, as readily seen in Fig. 3, that ˚ D 1 is well accepted when the light source P0 , the hole dr, and the observation point are located near a straight line.

Fig. 2 Spatial relationship between the small area dr in the x y plane and an observation point P

Fig. 3 Positional relationships between the normal vector and a light source, a hole and an observation point

Since the phase of the incident wave 0 .l0 / emitted from the light source is quite likely to be equal in a small area (dx dy) of the scatterer set in the x y plane, it is referred to as 0 .l0 / D 1. In addition, considering the relation of l D r l´ (see Fig. 2), the integration of (1), i.e., the amplitude of the wave observed at point P is given by the following equation. Z

.P/ D i s0

e2is0 jlj ˚g.r/dr jr l´j

(3)

jlj D jr l´j corresponds to a change in the location of the scatterer (plate) in which a small area (dx dy) is set up, and it varies significantly in comparison with the wavelength . Therefore, the term given by an exponential function in (3) is expected to show considerably large variation, whereas it is thought that 1=jr l´j, a denominator part of (3), shows monotonic change only. In addition, ˚ is considered

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5 Reciprocal Lattice and Integrated Intensities of Crystals

to be constant under the given condition of the scatterers. If these points are taken into account, (3) may be simplified in the following form: is0 ˚

.P/ D jlj

Z

e2is0 jlj g.r/dr

(4)

The coordinates of the observation point P are set to .xp ; yp ; zp /. As previously mentioned, the coordinates of the small area dr, i.e., a scattered object, can be expressed by (x, y, 0) so that we obtain the following equations (see to Fig. 2). jlj2 D .xp x/2 C .yp y/2 C z2p jl´j2 D xp2 C yp2 C z2p

)

9 2 jlj2 D jl´j2 y2 =

2.xp x C yyp / C x C xp x C yyp x2 C y 2 ; D jl´j2 1 2 C jl´j2 jl´j2 s xp x C yyp x2 C y 2 C jlj D jl´j 1 2 jl´j2 jl´j2

(5)

(6)

(7)

jl´j is sufficiently large in comparison with the absolute values of x and y. That is, since the relationships of jl´j x and jl´j y hold, the following approximation is obtained. 9 yp xp = y jlj D jl´j x jl´j jl´j (8) D jl´j x cos ˛ y cos ˇ ; Considering that g.r/dr is expressed by g.x; y/dx dy, the relationships of (8) can be rearranged by coupling (4) in the following form.

.P/ D

is0 ˚ 2is0 jl´j e jl´j

Z

g.x; y/e2is0 .x cos ˛Cy cos ˇ / dx dy

(9)

The term, which can be excluded from the integration of (9), is set to a constant C, and the components of x-axis and y-axis are described in the following equations when using the unit vector s0 of the incident wave. cos ˛ 9 = cos ˇ ; sy D s0 cos ˇ D sx D s0 cos ˛ D

(10)

Therefore, (9) is given in the following form. Z

.sx ; sy / D C

g.x; y/e2i.xsx Cysy / dx dy

(11)

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Here, .sx ; sy / corresponds to the two-dimensional Fourier transform of g.x; y/. Equation (11) may also be expressed as follows. Z

.sx ; sy / D C

g.x; y/e2i.x

cos ˇ cos ˛ Cy /

dx dy

(12)

Question 5.9 Calculate the diffraction intensity produced from the case where m slits with aperture width L are aligned at an interval d on a line (onedimensional array).

Answer 5.9 A schematic diagram of the given condition is illustrated in Fig. 1. Considering that if a hole is open, it may be described as g.r/ D 1 and in reverse g.r/ D 0 for a closed hole. Similar to this idea, a mathematical representation of slits is given by the following equations. 8 > = Ey 00 D Ey 0 cos 2 D Eoy cos 2 cos 2M > hE 002 i D hEx 002 i C hEy 002 i > ; 2 2 2 2 D hEox i C hEoy i cos 2 cos 2M

(4)

In this case, we obtain the intensity formula as follows. I 00 D KI0

1 C cos2 2 cos2 2M 2

(5)

If I0 is expressed by I 0 based on (3) and substituted into (5), the following equation is obtained. 2 2 00 0 1 C cos 2M cos 2 I DI (6) 1 C cos2 2M

Fig. 1 Schematic diagram of geometry for a diffractometer when placing a crystal monochromator in the incident beam

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5 Reciprocal Lattice and Integrated Intensities of Crystals

Question 5.14 Answer the following questions concerning the effect of an anomalous dispersion term on the structure factor of zinc blende (ZnS). Let set X ray atomic scattering factors of Zn and S be fZn and fS , and the real part 0 00 , fZn and fS0 , fS00 , and imaginary part of anomalous dispersion terms are fZn respectively. The positions of Zn in zinc blende are (0 0 0), (0 1/2 1/2), (1/2 0 1/2), (1/2 1/2 0), and those of S are (1/4 1/4 1/4), (1/4 3/4 3/4), (3/4 1/4 3/4), (3/4 3/4 1/4). (1) Obtain the structure factor Fhkl and jFhkl j2 including the anomalous dispersion terms. N (2) Obtain the structure factors for two diffraction peaks 111 and 1N 1N 1. (3) The values of the anomalous dispersion terms for Cu-K˛ radiation are 0 00 D 1:6, fZn D 0:68, fS0 D 0:32, and fS00 D 0:56. known to be fZn 2 N Compute the jFhkl j values for two cases, hkl D 111 and 1N 1N 1. (4) Compute the possible % difference detected in the diffraction intensity N between 111 and 1N 1N 1. Answer 5.14 (1) The structure factor of zinc blende is as follows (see Question 3.13).

i hCkCl 0 00 2 C ifZn C .fS C fS0 C ifS00 /e Fhkl D fZn C fZn .1 C ei.hCk/ C ei.kCl/ C ei.lCh/ /

i hCkCl 0 00 2 jFhkl j2 D fZn C fZn C ifZn C .fS C fS0 ifS00 /e .1 C ei.hCk/ C ei.kCl/ C ei.lCh/ /

i hCkCl 0 00 2 fZn C fZn ifZn C .fS C fS0 ifS00 /e .1 C ei.hCk/ C ei.kCl/ C ei.lCh/ / (2) The structure factors of 111 and 1N 1N 1N are described in the following equations. In the case of hkl D 111, 0 00 C fS00 / i.fS C fS0 fZn /g F111 D 4f.fZn C fZn 0 00 2 C fS00 /2 C .fS C fS0 fZn / g jF111 j2 D 16f.fZn C fZn

N In the case of hkl D 1N 1N 1, 0 00 F1N 1N 1N D 4f.fZn C fZn fS00 / C i.fS C fS0 C fZn /g 0 00 2 fS00 /2 C .fS C fS0 C fZn / g jF1N 1N 1N j2 D 16f.fZn C fZn

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p

(3) We can find sin D 2a3 in both cases of 111 and 1N 1N 1N by using the relation 2 p 2 2 Cl 2 3 ship of sin D h Ck . Then, it turns out to be sin D 25:4109 D 4a2 1

˚ . We also obtain fZn D 24:16 and fS D 11:86 for the atomic scatter0:16A ing factors based on the numerical data compiled in Appendix A.3. Substitute these numerical values for the corresponding terms in the equations acquired in question (2). Including the given values for the anomalous dispersion terms, we obtain the following results. jF111 j2 D 16f.24:16 1:6 C 0:56/2 C .11:86 C 0:22 0:68/2 g D 10669 jF1N 1N 1N j2 D 16f.24:16 1:6 0:56/2 C .11:86 C 0:32 C 0:68/2 g D 10390 jF111 j2 jF1N 1N 1N j2 279 D 0:026 D 2 jF111 j 10699

(4)

Therefore, this result implies that 2.6% of difference can be detected. It is also noteworthy that such difference in the intensities of two planes is observed because of the noncentrosymmetric nature of the zinc blende structure. Question 5.15 Cu3 Au is known to have a cubic lattice, and below the critical temperature 663 K, the Cu and Au atoms in Cu3 Au are arranged to form a perfectly ordered phase in which each unit cell contains one Au atom and three Cu atoms. Their positions are characterized as follows. The position of Au is 000, and the positions of Cu are 12 21 0, 12 0 21 , and 0 21 21 . On the contrary, in disordered phase, Cu and Au atoms are arranged at random on the atomic sites so that the probability that a particular site is occupied by a specific element is simply proportional to the atomic fraction. That is, each site is occupied by a statistical average of 14 Au and 34 Cu: (1) Derive the expression of F for the ordered phase. (2) Derive the expression of F for the disordered phase. (3) For what reflections will the F value be the same in both the ordered and disordered phases? Obtain also the reflections for not-equal case. (4) Derive the expression of F 2 for the ordered phase by introducing the anomalous dispersion terms f 0 and f 00 of both Cu and Au atoms.

Answer 5.15 (1) In the ordered phase, each unit cell contains the following particular atomic arrangement (see Fig. 1a). Au : One site 000 Cu : Three sites 12 21 0, 12 0 12 , 0 12 21 F D fAu C fCu Œei.hCk/ C ei.kCl/ C ei.lCh/

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5 Reciprocal Lattice and Integrated Intensities of Crystals

Fig. 1 Unit cells of Cu3 Au. (a) ordered phase and (b) disordered phase

(i) h; k; l : even/odd unmixed cos ! C1;

sin ! 0 .when h C k; k C l; l C h W All are even/ F D fAu C 3fCu

(ii) h; k; l : even/odd mixed cos ! 1;

sin ! 0

.h C k; k C l; l C h W Any of two is odd/ F D fAu fCu

(2) In the disordered phase (see Fig. 1b), the atomic scattering factor of the average Cu-Au atom is as follows. fav D

1 3 fAu C fCu 4 4

2i hCk 2i kCl 2i lCh 2 2 2 ) F D fav 1 C e Ce Ce (i) h; k; l : even/odd unmixed cos ! 1;

sin ! 0 .h C k; k C l; l C h W All are even/ F D fAu C 3fCu D 4fav

(ii) h; k; l : even/odd mixed cos ! 1;

sin ! 0

.h C k; k C l; l C h W Any of two is odd/: F D0

Note that no reflections of mixed indices are observed. (3) If h; k; l are not even/odd unmixed, the F value of the ordered phase is found to be equal to that of the disordered phase. On the contrary, the difference is

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found when h; k; l are even/odd number mixed. For this reason, the Bravais lattice of the ordered phase is simple cubic and that of the disordered phase is face-centered cubic. (4) F 2 of the ordered phase (i) even/odd unmixed 0 0 2 00 2 0 0 0 0 00 00 C fAu / C .fAu / C 6Œ.fAu C fAu / .fCu C fCu / C fAu fCu F 2 D .fAu 0 0 2 00 2 C9Œ.fCu C fCu / C .fCu /

(ii) even/odd mixed 0 0 2 00 2 0 0 0 0 00 00 F 2 D .fAu C fAu / C .fAu / 2Œ.fAu C fAu / .fCu C fCu / C fAu fCu 0 0 2 00 2 C.fCu C fCu / C .fCu /

Question 5.16 Answer the following questions concerning a Laue function in 2 / , using the case of d D 0:3 nm as an example. the form of sinsin2.NQd .Qd / (1) Explain the conditions where the Laue function gives a peak, its maximum value, and the reason why some small peaks appear in the region around the main peak. (2) Compute the height of the second peak in close vicinity of the main peak using the case of N D 20 as an example.

Answer 5.16 (1) In the Laue function, N is the number of unit cells, d is a lattice spacing, and Q is a wave vector. If we set x D Qd , both the denominator and the numerator of the Laue function will be zero for x D n (n is integer). The value of the Laue function for this limit can be found as follows. ( )

sin2 .N x/ 2N sin.N x/ cos.N x/ (1) D lim lim x!n x!n 2 sin x cos x sin2 x 1 2N fsin.2N x/g 2 D lim 1 x!n 2 fsin.2x/g 2 N sin.N 2x/ D lim D N2 x!n sin 2x where the following relationships are used.

(2)

(3)

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5 Reciprocal Lattice and Integrated Intensities of Crystals

sin.at/ cos.bt/ D

1 fsin.a C b/t C sin.a b/tg 2

sin t t D lim D1 t !0 t sin t at b b b sin bt sin bt D lim D11 D lim t !0 sin at t !0 bt sin at a a a lim

t !0

sin2 Nx sin2 x

Therefore, the Laue function has the limit of N 2 , corresponding to its maximum value, if x is zero, , 2, and 3 etc. For this reason, the Laue function is frequently represented in the normalized form by N 2 . The value of N itself stipulates the full peak width, i.e., it is deeply correlated with the peak sharpness in the vicinity of a reciprocal-lattice point. The larger the value of N is, the sharper the full peak width becomes. In other words, the peak width is inversely proportional to the number of unit cell found in the corresponding direction. Suppose d D 0:3 nm, peaks will be observed at the interval of Q D 3:33 nm1 in the reciprocal lattice. Since Q and d correspond to variables in reciprocal space and real space, respectively, they exhibit mutually reciprocal relationships. For example, for peaks at a relatively small interval of Q, the d -value is large and, inversely, we find peaks at a relatively large interval of Q if the d -value becomes small. Concerning the reason why a small peak appears in the region around the main peak, let us consider an extreme case where a sample consists only of three unit cells .N D 3/. As shown in Fig. 1, a small peak observed in the middle of the two peaks may be called the partial interference result, which is characterized by the phase difference of between the wave generated from a central unit cell and those of two unit cells of both ends.

N2

1

1 2 3

d = 0.3nm

2 Q1

3 nm–1 Q2

Q1

1 2 3

Q2

Fig. 1 Peak structure and its relevant factor appeared in the Laue function (N D 3)

(2) If the Laue function is calculated for the case of N D 20, we get the results illustrated in Fig. 2. The global maximum is found to be N 2 D 400. As a result, the local maximum of the 2nd peak is 18.7, that is about 4:7% of the global maximum.

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Fig. 2 Variations of the Laue function (N D 20)

Question 5.17 The function y D 2

N 2 sin 2

sin2 N x sin2 x

can be approximated by y D

in the close vicinity of x D 0, where D N x.

(1) Derive an equation for providing the full peak width at half maximum ordinate given by this function. (2) Derive an equation for providing the area under the peak given by this function. (3) Discuss “what is the dependence of N ,” with respect to the maximum ordinate, of the full peak width at half maximum ordinate and of the area under the peak. Answer 5.17 (1) Referring to the schematic diagram of Fig. 1, we compute the full width at half maximum ordinate of the following function y D N2

sin2 2

The purpose is to find , which gives y D N 2 and y D N 2 =2 when x D 0. sin2 N2 D 2 2 2 D0 sin2 2 sin C p sin p D0 2 2 N2

Since is also close to zero in the vicinity of x D 0, sin and are the same signs and we find sin C 2 ¤ 0.

216

5 Reciprocal Lattice and Integrated Intensities of Crystals

Therefore, sin p D 0 so that we may use the approximation of sin D x

x3 3Š

2

3 D0 6 2 ! p 2 21 D0 p 6 2 p p p 2 D 3 2. 2 1/ D 6 3 2 q p D ˙ 63 2

Therefore, it will be x D ˙

p

can be obtained as follows. x

p 63 2 N p , and the p 2 D 2 63 . N

corresponding peak width x

(2) Referring to the schematic diagram of Fig. 2, the area under the peak is estimated as follows. In the present case, the purpose is to find x, which gives 2 y D N 2 sinsin2Nxx D 0. From the condition .sin2 N x/=x 2 D 0, we find sin N x D 0. xD

m N

.m is integer/

It will be x D ˙ N as m D ˙1 in the close vicinity of x D 0. It is also found in this case that D ˙. On the contrary, since dx D d=N from the definition D N x, we obtain the area under the peak in the following equation.

Fig. 1 Schematic diagram of a diffraction peak and its half width

Fig. 2 Schematic diagram of a diffraction peak

5.6 Solved Problems

Z

=N =N

Note:

R 0

sin2 d 2

217

N2

sin2 N x dx D N 2x2

Z

N2

sin2 d D 2N 2 N

Z

sin2 d 2

is a constant that is independent on N .

(3) According to the above-mentioned results, the area under the peak is proportional to N , whereas the full peak width at half maximum ordinate is found to be proportional to N 1 . In addition, the maximum of the peak can be computed at the condition x D 0 . D 0/, and it is given in the following equation. sin2 sin lim N 2 2 D N 2 lim !0 !0

!2 D N2

Therefore, the maximum of the peak is dependent on N 2 . Question 5.18 Answer the following questions concerning the measurement of the diffraction intensity in the symmetrical transmission mode for a thin slab of crystalline powder sample of mass per unit area M 0 . Suppose the incident beam and diffracted beams form equal angles with the surface of the slab sample, and the incident beam intensity is expressed by I0 A0 D P0 . (1) Derive an equation providing the integrated intensity P 0 per unit length of the diffraction circle at a distance R from the thin slab sample. (2) Discuss what value of M 0 will maximize P 0 . Note: The integrated intensity P from the effective irradiated volume V is given by the following equation. P D I0

e4 m2e c 4

V 3 pF 2 4v2a

1 C cos2 2 2 sin 2

where is the wavelength, p is the multiplicity factor, F is the structure factor, and va is the volume of unit cell. The term given in parenthesis is the Lorentz polarization factor of a crystalline powder sample for an unpolarized incident X-ray beam I0 .

Answer 5.18 (1) The circumference of the Debye ring so as to form a cone of half apex angle located at distance R from a sample is 2R sin . Therefore, we find the following equation for the integrated intensity (see Fig. 5.4). I0 P P0 D D 2R sin 16R

e4 m2e c 4

V 3 pF 2 v2a

1 C cos2 2 sin sin 2

! (1)

218

5 Reciprocal Lattice and Integrated Intensities of Crystals

Here, V is the effective irradiated volume of a thin slab. By considering the geometrical relationship shown in Fig. 1 for the symmetrical transmission mode and setting the cross-sectional area to A0 , it can be replaced by the following equation. Z V D

t

e.t sec / A0 sec dx D A0 t sec e.t sec /

t D1

(2)

From (1) and (2), we obtain. I0 A0 P D 16R 0

e4 m2e c 4

3 pF 2 v2a

1 C cos2 2 cos sin2

te.t sec /

(3)

Fig. 1 Geometry for the symmetrical transmission mode

Since t corresponds to the thickness of the slab the sample is replaced with t D M 0 =0 , where M 0 is mass and 0 is density per unit area of the sample, P0 D

I0 A0 16R 0

e4 m2e c 4

3 pF 2 v2a

1 C cos2 2 cos sin2

m m sec e 0 0

(4)

m sec

(2) Setting g D M e 0 , P 0 has its maximum if g is maximized, so we need 0 to find the condition where the derivative of g is zero. 0 0 1 M M 0 sec M sec e 0 sec e 0 0 0 0 0 M 1 M sec 1 D sec e 0 0 0

g0 D

(5)

Based on this relationship, the maximum value of P 0 will be obtained at M 0 D 0 =. sec /. However, keep in mind that it is actually not so easy to control M 0 using this relationship.

Chapter 6

Symmetry Analysis for Crystals and the Use of the International Tables

6.1 Symmetry Analysis A crystal may be defined as a solid composed of atoms arranged on a regular threedimensional lattice and such periodicity in the atomic distribution features their structure. The geometry of atomic distributions in crystals is known to be characterized by the repetition, such as lattice translation (see Chap. 2). In addition to lattice translations, we find reflection and rotation. In these cases, an object is brought into coincidence with itself by reflection in a certain plane; rotation upon around a certain axis; or reflection in a certain plane. The repetition of a pattern by specific rules characterizes all symmetry operations and their fundamental points are given below. If a symmetry operation leaves a locus, such as a point, a line, or a plane unchanged (i.e., same atomic position), this locus is referred to as the symmetry element. For any operation excluding lattice translation for space group, the symmetry operation belongs to one of four cases: inversion (i ) expressed by a change from (x; y; z) to (x; y; z); rotatory-inversion (¯n); reflection (m: a mirror plane) expressed by a change from (x; y; z) to (x; y; z); and rotation (n, a rotation axis) expressed by a change (360ı=n) about an axis. In crystals, we may conclude that only one-, two-, three-, four-, and sixfold rotation axes can be accepted. Of course, the symmetry operations may be linked with one another. The symmetry operation called inversion relates a pairs of objects which are equidistant from and on opposite sides of a central point (called inversion center). That is, only a single point remains unchanged. Whereas, rotatory-inversion is one of the compound symmetry operations and it is frequently called roto-inversion. This operation is produced by the combination of a rotation of (360ı =n) around a certain axis followed by inversion through a point on the axis as a symmetry center. As shown in Fig. 6.1, the rotatory-inversion operation provides five cases denoted ¯ 2, ¯ 3, ¯ 4, ¯ and 6. ¯ However, the following three points are noteworthy. by symbols 1, (1) Since the rotatory-inversion operation of 1¯ is a rotation of 360ı followed by inversion through a point on the onefold roto-inversion axis, it is identical to inversion (i ), simply called center of symmetry or inversion center. (2) The rotatoryinversion operation of 2¯ is represented by rotation through an angle of 180ı followed by inversion to take one point into an equivalent one. However, these two points are

219

220

6 Symmetry Analysis for Crystals and the Use of the International Tables

1

2=m

3=3+1

4

6=3+m

Fig. 6.1 Examples of rotatory-inversion operation

also related to one another by reflection in a plane normal to the axis, so that the rotatory-inversion of 2¯ is identical to a mirror reflection (m). (3) As easily seen in Fig. 6.1, successive applications of the rotation-inversion operation of 3¯ alter a point into altogether six equivalent positions. This variation can be reproduced by combining operations with a threefold rotation axis and inversion (i ). Similarly, the rotation-inversion operation of 6¯ is also represented by combining operations with a threefold rotation axis and a twofold roto-inversion axis, or mirror plane perpendicular to the axis. These three points suggest that the rotatory-inversion operations ¯ 2, ¯ 3, ¯ and 6¯ are not included in the except for 4¯ result in no new operation, so that 1, independent symmetry operations. For convenience, some additional details are given for understanding the ¯ using two different cases as an example. In rotatory-inversion operation of 4, Fig. 6.2a, we easily find the results obtained by symmetry operations of fourfold rotation and inversion, because successive operations about the fourfold axis move a point from 1 to 2, 3, 4, and back to 1. On the other hand, the inversion center alters it from each of those positions to 7, 8, 5, and 6, respectively. This combination of symmetry operations results in a mirror plane normal to the axis. In this case, two individual symmetry operations are found to be linked which are themselves symmetry operations. Whereas, in Fig. 6.2b illustrating the results obtained by symmetry operation of fourfold rotation about an axis followed by inversion through a point on its axis. Successive applications of these operations move a point at 1 to 2, 3, 4, and back to 1. In this case, we can find neither the inversion center nor the fourfold rotation axis (see auxiliary points denoted by open circles in Fig. 6.2b. Namely, two symmetry operations in this case are made in sequence as a single matter referred to as a new symmetry operation, which is called compound symmetry operation.

6.1 Symmetry Analysis

221

a

b

4 3

1

3

1

2

2 8

5 6

7

4

Fig. 6.2 Applications of the fourfold rotation and inversion. (a) Combination of independent symmetry operation, (b) Compound operation (roto-inversion)

In conclusion, the independent symmetry operations for discussing the symmetry of the three-dimensional atomic arrangement are eight: inversion (i ), reflection ¯ This means that the whole (m), rotation (1, 2, 3, 4, 6), and rotatory-inversion of 4. periodic array observed in crystals can be covered by repeating the parallel translation (translational operation) of the structure derived from these eight kinds of symmetry element. In other words, as already mentioned in Chap. 2, there are seven crystal systems for classification, which consist of 14 kinds of Bravais lattices and a crystal is known to be classified into 32 point groups on the basis of eight kinds of symmetry element. In addition, when it is extended to include space groups, by adding point groups, Bravais lattices, screw axis, and glide reflection plane, it will be classified into 230 in total. For this analysis, we have to include two compound symmetry operations; rotation and translation (screw axis) and reflection and translation (glide plane). A brief description for screw axes and glide reflection planes is given below. The symmetry operation of the so-called “screw rotation” consists of a rotation of 360ı =n where n is 2, 3, 4, and 6 and a translation by a vector parallel to the axis. The screw axis is expressed by nm and its operation is to translate by (m=n) times the length of a unit lattice vector along the direction of a rotation axis every one operation with respect to n-axis of rotation. All of the screw rotation axes are shown in Fig. 6.3 (see also Question 6.1). Although the direction of rotation itself is not so important in the screw rotation, the definition is illustrated in Fig. 6.4 using a right-handed axial system as an example. This case shows an operation which is a rotation around a c-axis from the a-axis toward the b-axis by an angle followed by a positive translation along a c-axis, called the motion of a right-handed screw. The compound symmetry operation of a glide reflection consists of a reflection and a translation by the vector qg parallel to the plane of reflection. For convenience, Fig. 6.5 shows a comparison of the operation of a glide plane with that of a mirror plane on a point lying off the planes.

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6 Symmetry Analysis for Crystals and the Use of the International Tables

21

2

4

6

3

41

61

31

42

62

63

32

43

64

65

Fig. 6.3 Examples of possible screw axes Fig. 6.4 A right-handed screw system

The description of this symmetry element may also be simplified by reference to the unit lattice vectors a, b, and c. For example, with respect to a-glide plane perpendicular to the b-direction, the reflection operation is made through the glide plane and then displaced by the vector, corresponding to one-half of a lattice translation 1 a , parallel to the glide plane. Similarly, we may obtain b- and c-glide planes per2 pendicular to one of the other directions. There is also a diagonal glide plane n by

6.1 Symmetry Analysis

223

Fig. 6.5 Comparison of the operation of (a) a glide plane and (b) a mirror plane

translation of the diagonal direction. That is, the n-glide plane, if it is perpendicular to c, gives a glide component of 12 a C 12 b . Furthermore, we have one additional case; the diamond glide plane denoted by d which can be featured by one quarter (1=4) of a lattice translation along the line parallel to the body-centered direction. Then, there are five kinds of glide reflection planes in all. As mentioned previously, all crystal structures can be classified into seven crystal systems using parameters of three vectors a, b, and c (or those lengths a, b, and c) and the interaxial angles between them, ˛, ˇ, and . The relationships between crystal systems and symmetry elements are summarized in Table 6.1. The atomic distribution in crystals is characterized by its periodicity in a regular three-dimensional lattice and it is known to be classified into 32 point groups using eight symmetry elements (see Fig. 2.1). In addition, the periodicity in regular threedimensional lattice can be analyzed by the concept of symmetry. Particularly, if we use eight kinds of symmetry element; “reflection (m),” “rotation (1, 2, 3, 4 and ¯ together with eleven “screw axes” 6),” “inversion (i ),” and “rotatory-inversion (4)” and five “glide planes,” all the possible geometric arrangement of atoms in threedimensional lattice space can be classified into 230 in all, called “space groups.” This implies that the number of geometric arrangements with periodicity is limited in three-dimensional lattice space. In other words, any crystal can be described only by one of the 230 space groups. In addition, the real crystal structures are not evenly distributed over these 230 space groups. It is rather unevenly distributed, so that there are many space groups that do not represent any real crystal structure. The relationships of crystal systems with point group and space groups are summarized in Table 6.2.

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6 Symmetry Analysis for Crystals and the Use of the International Tables

Table 6.1 Seven crystal systems, symmetry elements, and relevant data Crystal systems

Subunit

a¤b¤c ˛ ¤ ˇ ¤ ¤ 90ı Monoclinic a¤b¤c ˛ D D 90ı ˇ ¤ 90ı Orthorhombic a ¤ b ¤ c Triclinic

Minimum symmetry elements

Point groups

None

1, 1N

0000

One diad axis or mirror plane Three orthogonal diad or inverse diad axis One tetrad or inverse tetrad axis Form triad axes

2, m, 2/m

1000

1

222; mm2, mmm

3000

3

4; 4N ; 4=m; 422; 4mm; 4N 2m; 4=mmm 23; m3; 432; 4N 3m m3m 3; 3N ; 32; 3m; 3N m

4010

5

6430

9

3100

3

6001

7

Tetragonal

aDb¤c ˛ D ˇ D D 90ı

Cubic

aDbDc ˛ D ˇ D D 90ı aDb¤c One triad or inverse triad ˛ D ˇ D 90ı D 120ı or axis aDbDc ˛ D ˇ D ¤ 90ı aDb¤c One hexad or inverse hexad ˛ D ˇ D 90ı D 120ı axis

Trigonal

Hexagonal

6; 6N ; 6=m622; 6mm; 6N m2; 6=mmm

Number of Number rotation of mirror axes 2346 plane

6.2 International Tables It is not necessary to know all (230) the space groups individually, because we can easily get many of the most important information of the space groups from “International Tables for Crystallography Vol. A, Fifth Edition (2002)” published by the International Union of Crystallography (referred to as IUCr). Although two methods, Sch¨onflies symbols and Hermann–Mauguin symbols have been used for describing the 32 point groups (see Question 6.2), the IUCr suggests the use of Hermann–Mauguin symbols. Note that Sch¨onflies symbols are widely used in respect to spectroscopy, especially of organic molecules. This handbook is quite useful for determining or interpreting the structure of crystals of interest. Although the contents cover only 24 space groups frequently found in real crystals, a textbook is also available for beginners titled on Brief Teaching Edition of Volume A Spacegroup Symmetry, International Tables for Crystallography, Fifth Revised Edition, edited by T. Hahn, Kluwer Academic Publishers, Dordrecht, and Holland (2002). International Tables for Crystallography (often referred to as International Tables) provides the following information: (1) Short space group symbol, Sch¨onflies symbol, point group, crystal system, number of the space group, full space group symbol, and Patterson symmetry symbol.

6.2 International Tables

225

Table 6.2 Crystal systems and relations to point groups and space groups Crystal system Point Space groups Triclinic 1 P1 1N P 1N Monoclinic

2 m 2/m

P 2; P 21 ; C 2 Pm, Pc,Cm, Cc P2/m ,P 21 =m, C2/m, P2/c, P 21 =c ,C2/c

Orthorhombic

222 mm2

P222, P2221 ; P21 21 2; P21 21 21 ; C2221 ; C222; F222; I222; I21 21 21 Pmm2; Pmc21 ; Pcc2; Pma21 ; Pca21 ; Pnc21 ; Pmn21 ; Pba2; Pna21 ; Pnn2; Cmm2; Cmc21 ; Ccc2; Amm2; Abm2; Ama2; Aba2; Fmm2; Fdd2, Imm2, Iba2, Ima2, Aba2, Fmm2, Fdd2, Imm2, Iba2, Ima2 Pmmm, Pnnn, Pccm, Pban, Pmma, Pnna, Pmna, Pcca, Pbam, Pccn, Pbcm, Pnnm, Pmmn, Pbcn, Pbca, Pnma, Cmcm, Cmca, Cmmm, Cccm, Cmma, Ccca, Fmmm, Fddd, Immm, Ibam, Ibca, Imma

mmm

Tetragonal

P 4; P 41 ; P 42 ; P 43 ; I 4; I 41 P 4N ; I 4N P 4=m; P 42 =m; P 4=n; P 42 =n; I 4=m; I 41 =a P 422; P 421 2; P 41 22; P41 21 2; P 42 22; P42 21 2 P 43 22; P 43 21 2; I 422; I 41 22 4mm P 4mm; P 4bm; P 42 cm; P 42 nm; P 4cc; P 4nc, P 42 mc; P 42 bc; I 4mn; I 4cm; I 41 md; I 4cd 4N 2m P 4N 2m; P 4N 2c; P 4N 21 m; P 4N 21 c; P 4N m2; P 4N c2, P 4N b2; P 4N n2; I 4N m2; I 4N c2; I 4N 2m; I 4N 2d 4/mmm P 4=mmm; P 4=mcc; P 4=nbm; P 4=nnc; P 4=mbm, P 4=mnc; P 4=nmm; P 4=ncc; P 42 =mmc; P 42 =mcm, P 42 =nbc; P 42 =nnm; P 42 =mbc; P 42 =mnm; P 42 =nmc, P 42 =ncm; I 4=mmm; I 4=mcm; I 41 =amd; I 41 =acd 4 4N 4=m 422

Trigonal–hexagonal 3 3N 32 3m 3N m 6 6N 6=m 622 6mm 6N m 6/mmm

P 3; P 31 ; P 32 ; R3 P 3N ; R3N P 312; P 321; P 31 12; P 31 21; P 32 12; P 32 21; R32 P 3m1; P 31m; P 3c1; P 31c; R3m; R3c P 3N 1m; P 3N 1c; P 3N m1; P 3N c1; R3N m; R3N c P 6; P 61 ; P 65 ; P 63 ; P 62 ; P 64 P 6N P 6=m; P 63 =m P 622; P 61 22; P 65 22; P 62 22; P 64 22; P 63 22 P 6mm; P 6cc; P 63 cm; P 63 mc P 6N 2m; P 6N c2; P 6N 2m; P 6N 2c, P 6=mmm; P 6=mcc; P 63 =mcm; P 63 =mmc

Cubic

P 23; F 23; I 23; P 21 3; I 21 3 Pm3N ; Pn3N ; Fm3N ; Fd3N ; Im3N ; Pa3N ; Ia3N P 432; P 42 32; F 432; F 41 32; I 432; P 43 32; P 41 32; I 41 32 P 4N 3m; F 4N 3m; I 4N 3m; P 4N 3n; F 4N 3c; I 4N 3d Pm3N m; Pn3N n; Pm3N n; Pn3N m; Fm3N m; Fm3N c, Fd3N m; Fd3N c; Im3N m; Ia3N d

23 m 3N 432 4N 3m m3N m

226

6 Symmetry Analysis for Crystals and the Use of the International Tables

(2) Projection of the symmetry elements of the space group along special axes (high symmetry). The origin is in the upper left corner. (3) Projection of a general position. (4) Information for the selection of the origin. (5) Asymmetric unit. (6) Symmetry operations of the space group. (7) General and special positions, multiplicity Wyckoff letter, site symmetry, and coordinates of equivalent positions. Patterson symmetry is the symmetry of the Patterson function for Fourier transformation and Wyckoff letter provides the equivalent positions in a unit cell. More details are obtained from the International Tables for Crystallography. The first alphabetical capital letter is to show the lattice symbol of the Bravais lattices (P, F, I, A, B, C, and R) as summarized in Table 6.3 and next three characters indicate symmetry elements related to the particular orientation in crystal systems as summarized in Table 6.4. For example, we find Cmm2 (orthorhombic) in the International Tables, Vol.A, p.238. This Cmm2 shows that space lattice is base-centered (C) and next three characters of mm2 inform us the symmetry elements with respect to directions of [100], [010], and [001], respectively. That is, this orthorhombic space lattice shows mirror plane m, perpendicular to both a- and b-axes and a twofold rotation axis along the c-axis. For another example, P121 /c1 (monoclinic) found in the International Tables, Vol.A, p.184 shows that space lattice is primitive (simple) and it has the 21 screw axis parallel to b-axis and the c-glide plane which is perpendicular to the 21 screw axis. Since the description of space groups is generally used in a simplified form as much as possible, so-called short space group symbol, for example, P121 =c1 ! P21 =c and F 4=m3N 2=m ! Fm3N m (see Vol. A, p. 688). Then, the users are requested to get familiar with the relationships between symbols and crystal systems including image of atomic positions. Practice makes perfect (see Questions 6.1–6.6).

Table 6.3 Number and coordinates of the lattice points in the unit cells of Bravais lattices Lattice symbols No. of lattice points Coordinates of lattice in a unit cell points in a unit cell P 1 0,0,0 1 1 A 2 0,0,0 ; 0, , 2 2 1 1 B 2 0,0,0 ; ,0, 2 2 1 1 C 2 0,0,0 ; , ,0 2 2 1 1 1 I 2 0,0,0 ; , , 2 2 2 2 1 1 1 2 2 R 3 0,0,0 ; , , ; , , 3 3 3 3 3 3 1 1 1 1 1 1 F 4 0,0,0 ; , ,0 ; ,0, ; 0, , 2 2 2 2 2 2

6.2 Solved Problems

227

Table 6.4 The order of Hermann–Mauguin symbols and their relation to directions in a crystal Crystal systems 1st index 2nd index 3rd index Triclinic None Monoclinic

[010](b-axis)* [001](c-axis)*

Orthorhombic

[100]

Tetrogonal

[001]

Trigonal Referred to hexagonal axes

[001]

Trigonal Referred to rhombohedral axes

[111]

Hexagonal

[001]

[010]

[001]

Cubic

Orthogonal axis D Unique axis.

The International Tables do not cover only Vol.A focusing on symmetry of space groups, but also Vol.B published in 2001 covering information about the reciprocal lattice, structure factor, Fourier transform, and others including structural analysis by diffuse scattering, dynamical theory, and its applications. One can also find Vol.C, Third Edition in 2004 providing mathematical, physical, and chemical tables including absorption coefficients and X-ray atomic scattering factors. This volume includes sample preparation techniques, methods for the determination of lattice parameters, refining techniques for the structure determination. Further, Vol. D published in 2003 Physical Properties of Crystals, Vol.E in 2002, Sub-periodic Groups, Vol. F in 2001 Crystallography of Biological Macromolecules, and Vol.G in 2005, Definition and Exchange of Crystallographic Data. There is also A1, “Symmetry relations between space groups” published in 2004. It should be suitably selected, depending on the purpose.

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6 Symmetry Analysis for Crystals and the Use of the International Tables

6.3 Solved Problems (8 Examples)

Question 6.1 Explain screw axes and glide planes which are important for analyzing the three-dimensional regular array in crystal lattice.

Answer 6.1 A three-dimensional periodic array in crystals is known to be reproduced by the infinite repetition of identical structural units, but we find the following problem. When supposing an operation around a point brings it to self-coincidence, it is difficult to distinguish between the result obtained by the one cycle operation and that where it returns to the original position from another lattice point separated with several cycles. For this purpose, the operations of rotation and translation may be linked with one another. This is particularly true for the space groups of centered lattices. That is, it is necessary to introduce screw rotation and glide reflection. The screw axis may be denoted by nm and its operation is to translate by (m=n) times the length of a unit lattice vector along the direction of a rotation axis every one operation about the n-axis of rotation. As a specific example, a comparison is made in Fig. 1 using the relationships between twofold rotation axis and twofold screw rotation axis. In the operation of twofold screw rotation axis (21 ), the point alters from A to B due to a translation operation applied by half of a unit lattice translation along the direction parallel to the long-axis after the 180ı rotation as similar to the twofold rotation axis case and if the same operation is repeated, the point A does not return to its starting point. This operation does not come into coincidence with itself. The point A moves the point corresponding to the position applied by a unit lattice translation along the direction parallel to the long axis.

a

b

B π

B One cycle

A

λ/2 A

π

Fig. 1 Twofold rotation axis (a) and twofold screw rotation axis (b)

6.3 Solved Problems

229

Fig. 2 Threefold screw axes

Fig. 3 Example of a glide reflection operation denoted by “a-glide plane”

In the operation of threefold screw rotation axis, a translation operation is made by 1/3 of a unit lattice translation along the direction of a rotation axis after every 120ı rotation. However, as shown in Fig. 2, keep in mind that there are two ways like the treads of a spiral staircase, clockwise and counterclockwise. In order to distinguish these two cases, the threefold screw rotation axis describes clockwise (31 ) and counterclockwise (32 ). It may be added that there are three fourfold screw rotation axes and five sixfold screw rotation axes (see Fig. 6.3) and screw rotation axes are only allowed in crystals parallel to those directions possibly accepted for rotation axes in the corresponding point group. Glide plane is divided into three categories; axial glide plane and diagonal and diamond glide plane. Reflection across the so-called mirror plane is followed by translation parallel to the plane by one-half of the length of a unit lattice translation vector. For example, “a-glide plane” is the case where it projects on a mirror surface in pairs and a translation operation by ( 12 a) is made along the direction parallel to the mirror surface (see Fig. 3). Similarly, there are axial b-glide plane and c-glide plane.

230

6 Symmetry Analysis for Crystals and the Use of the International Tables

In other words, glide planes are designated by symbols suggesting the relationships of their glide components to the unit lattice vectors. The n-glide plane is characterized by a translation operation along the diagonal direction. In addition, the diamond glide plane denoted by d -glide plane features one-quarter (1/4) of a lattice translation along the line parallel to the body-centered direction. Symbols and relevant information of these glide planes including a mirror plane are summarized in Table 1. Table 1 Symbols and their relevance for glide planes including a mirror plane

Question 6.2 A crystal structure is known to be characterized by points in an infinite three-dimensional regular array. Explain the geometry of crystals using symmetry elements. Keep in mind the viewpoint of symmetry given in a crystal structure, not the mathematical issue.

6.3 Solved Problems

231

Answer 6.2 In crystals, a three-dimensional periodic arrangement of atoms is always present and such nature is represented by points. The concept of a lattice corresponding to a three-dimensional periodic arrangement of points is a purely mathematical subject, but it is well accepted as the space lattice. In this case, we use lattice translation for the repetition operation. However, we find other repetition methods called symmetry operations. For example, it is relatively easy to understand the rotation operation as illustrated in Figs. 1 and 2. An operation is required to bring a point into coincidence with itself, such as rotation about an axis and reflection in a plane. When a symmetry operation gives a locus, such as a point, a line, or a plane which is left unchanged by this operation, the locus is referred to as the symmetry element. Note that symmetry may be defined as that spatial property of a pattern or body by which the pattern or body can be brought from a starting state to another indistinguishable state by a certain operation. A symmetry element is also considered to a geometrical entity (point, line, or plane) in a pattern or body, A point is a dimensionless entity in space with coordinates that specify its position. An axis is a line joining two points and a plane is formed by two intersecting lines.

Fig. 1 Point groups having only rotation as symmetry element

Seven crystal systems, each related to the type of unit cell, are combined with 32 point groups which are associated with elements of symmetry in the unit cell itself. Note that the symmetry elements enable us to represent all the possible point arrangements, although a real crystal is a single, unrepeated object. Anyway, with respect to the components of symmetry element, we may suggest a center of sym¯ (or a roto-inversion center), a mirror plane, glide planes, rotation axes, metry, 1, screw axes, and inversion axes. For example, the point which does not move in the inversion operation is called “inversion center.” The operation-related mirror plane is given as follows. Any point on one side of a mirror plane is matched with an equivalent point on the other side at the same distance from the plane along a line

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6 Symmetry Analysis for Crystals and the Use of the International Tables

Fig. 2 Fourfold rotation axis (a) and its notation (b)

normal to it. In other words, the mirror reflection is to match any point or any object with the case where it is reversed after rotating 180ı about a twofold axis perpendicular to the target plane. Similarly, the inversion operation can be combined with other axes of rotation called “rotatory-inversion” and this case is also called “onefold rotatory-inversion.” Considering all these factors, we may conclude for symmetry in the following. For the symmetry elements allowed in a three-dimensional periodic arrangement of points, we find ten different ways of “rotation and rotatory inversion” without translation; namely, the rotation axes of one, two, three, four, and sixfold rotation ¯ 3, ¯ 4, ¯ 6¯ axes, as expressed by n and the rotatory inversion (roto-inversion) axes of 2, denoted by n. N However, it is rather stressed here that there are only EIGHT inde¯ and pendent symmetry elements by excluding threefold rotatory-inversion axis (3) ¯ sixfold rotatory-inversion axis (6). In other words, it is found out that there are 32 point groups for covering a three-dimensional periodic arrangement of points in space lattice using these eight symmetry elements and their combinations. Some key points for point groups are summarized as follows. (1) Point groups where only rotation is recognized as a symmetry element (five cases: 1, 2, 3, 4, and 6). (2) Point groups where only axis of rotatory-inversion is recognized as symmetry ¯ 2¯ D m, 3, ¯ 4, ¯ and 6D ¯ 3=m ). element (five cases: 1, (3) Point groups where n-fold rotation axis is perpendicular to twofold rotation axis (four cases: 222, 32, 422, and 622). (4) Point groups which have a mirror plane perpendicular to the n-fold rotation axis (three cases: 2=m, 4=m, and 6=m). (5) Point groups which have a mirror plane parallel to the n-fold rotation axis (four cases: 2mm, 3m, 4mm, and 6mm). (6) Point groups of 222, 32, 422 and 622 (corresponding to the No.3 cases), when further considering a mirror plane perpendicular to the n-fold rotation axis (four ¯ cases: mmm, 6m2, 4=mmm and 6=mmm). (7) Point groups of 222 and 32 (corresponding to the No.3 case), when further considering a mirror planes so as to bisect an angle formed by twofold rotation axes ¯ ¯ parallel to the plane of drawing (two cases: 42m and 3m).

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(8) Point group where four threefold rotation axes mutually intersect to make the tetrahedral angle of 109:471ı (one case only: 23). Similarly, point group where four threefold rotation axes mutually intersect to make the octahedral angle of 70:529ı (one case only: 432). (9) Point groups of 23 and 432 (corresponding to the No.8 case), when further ¯ considering several mirror planes (three cases: m3, 43m, and m3m). In order to represent the 32 point groups, we find two methods, Sch¨oenflies and Hermann–Mauguin symbols. Although Sch¨onflies symbols are widely used in the field of spectroscopy, especially of organic molecules, the use of Hermann–Mauguin symbols become popular in crystallography, because of the IUCr suggestion. If need, we can use the results summarized in Table 1. It may also be noteworthy that the abbreviation form is generally used in cases possibly described by com¯ for a cube bination of higher symmetry. For example, in the point group of 432 described by a combination of symmetry operations of fourfold, threefold, and twofold axes, the best description for symmetry is given by two mirror planes; one is 4=m perpendicular to fourfold axis and another is 2=m perpendicular both to threefold rotatory-inversion axis 3¯ and twofold axis. In this case, the original ¯ is widely employed. Some full-notation is m4 3¯ m2 , but the abbreviation form of m3m helpful information can be obtained from the results of Table 2. The following information may be convenient in the symmetry operations for crystallography. When including translation, it is necessary to consider “screw axes” with translation to the direction of rotation and its axis and a mirror plane and a “glide plane” with translation parallel to it. Keeping these factors in mind, it is necessary to stipulate the direction of an axis as well as the direction of the translational operation, so that the combination of eleven screw axes and five glide planes as listed in Table 3 may be linked with one another.

Table 1 Hermann–Mauguin and Sch¨oenflies symbols for the 32 crystallographic point groups Harmann– Sch¨oenflies Harmann– Sch¨oenflies Harmann– Sch¨oenflies Mauguin symbols Mauguin symbols Mauguin symbols symbols symbols symbols 1 C1 422 D4 6=m C6h 1 Ci 4mm C4v 622 D6 4N 2 m D2˛ 2 C2 6mm C6v 6N m 2 D3h m Cs 4=mmm D4h 2=m C2h 3 C3 6=mmm D6h 222 D2 3 C3i 23 T mm 2 C2v 32 D3 m N 3 Th mmm D2k 3m C3v 432 O 4 C4 3m D3˛ 43m T˛ 4 S4 6 C6 m3m Oh 4=m C4h 6 D3h

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6 Symmetry Analysis for Crystals and the Use of the International Tables

Table 2 The 32 point groups and their relation to the crystal systems Crystal system Point groups Triclinic 1N 1 Monoclinic 2=m m,2 Orthorhombic 2=m 2=m 2=m mm2; 222 (mmm) Tetragonal 4=m 2=m 2=m 4N 2m, 4mm, 422 (4=mmm) 4=m, 4N , 4 Trigonal 3N 2=m 3m, 32, 3N , 3 (3N m) Hexagonal 6=m 2=m 2=m 6N m2, 6mm, 622 (6=mmm) 6=m, 6N , 6 Cubic 4=m, 3N 2=m 4N 3m, 432, 2=m 3N , 23 (m3N m) (m3N ) ( ): abbreviated symbols. Table 3 Symbols and their relevance of symmetry elements including translation Symbol Symmetry elements Graphical symbol Translation 21 31 32

2-fold screw 3-fold screw

c=2; a=2 or b=2 c=3 2c=3

41

4-fold screw

c=4 2c=4

42 43

3c=4

61

6-fold screw

c=6

62

2c=6

63 64

3c=6 4c=6

65 a; b

n

Diagonal glide plane

5c=6 Translation parallel to the plane of paper (a=2; b=2 etc.) Translation perpendicular to the plane of paper (c=2 etc.) .a C b/=2 etc.

d

Diamond glide plane

.a C b/=4 etc.

Glide plane

c

Question 6.3 The mathematical concept of space groups is used for structural analysis of crystals. Explain the space groups using a monoclinic system as an example.

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Answer 6.3 Usually more than one symmetry element is present in crystals. For crystals, only the rotational values n D 1; 2; 3; 4; and 6 are permitted and 32 crystallographic point groups are generated. Point groups are the symmetry groups of a finite body, but space groups should be described as an infinite set of symmetry elements. That is, space groups provide the symmetry not only of crystal lattices but also of crystal structures. When 32 point groups are analyzed and arranged on the basis of degrees of rotation axis, all crystals are known to be classified into one of the seven crystal systems. Furthermore, the periodic sequence found in crystals has been systematically analyzed using all symmetry operations possible; four symmetry operations of mirror reflection, rotation, inversion and rotatory-inversion and eleven screw axes and five translational operations of glide planes and their combinations and then we obtain the 230 space groups. Combinations of these 230 space groups are proved from purely mathematical point of view. In other words, any periodic arrangement found in crystals can be expressed by one of the 230 space groups. In the monoclinic system, the lattices are characterized by two unit cell descriptions; simple lattice (P ) which has only one lattice point in a unit cell and one centered lattice (C ) which has a lattice point at the center of the ab-plane. Point groups are 2, m, and 2/m. For example, let us consider the case of P and C with a point group of 2, we obtain four combinations such as P 2, C 2, P 21 , and C 21 . However, C 2 is equivalent to C 21 , as readily seen from the results of Fig. 1, so that this combination provides three space groups denoted by P 2, C 2, and P 21 . a C2

c

1 2

+

+

+ + 12

+

1 2

+

+

+

+

+ + 12

+

Fig. 1 The operation of the space group C2

Since one screw axis and one glide-plane (c-glide plane in this case) will be taken into consideration, we have to check the following five cases of 21 , c, 21 =m, 2=c, and 21 =c. Taking all these factors, the 13 combinations are found to be possible as the space lattices P and C of the monoclinic system. The results are summarized in Table 1 together with subgroups of P2/m and C2/m. Note that we also find the relationships of C 21 D C 2=m, C 21 =c D C 2=c, and C 21 D C 2. All possible space groups of the monoclinic system can be derived in a slightly different way. Starting from the two monoclinic space groups of highest symmetry of P2=m and C2=m. In C2=m, there are a-glide planes at (x, 1=4, z) and (x, 3=4, z) and 21 -axes at (1=4, y, 0), (1=4, y, 1=2), (3=4, y ,0), and (3=4, y, 1=2).

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6 Symmetry Analysis for Crystals and the Use of the International Tables

The monoclinic subgroups for the point group 2=m are m and 2. Then, the replacement of point symmetry elements of 2 and m can be done by using 21 and a-glide plane, respectively. Since m is parallel to the plane denoted by (010), only a-, c-, and n-glide planes are possible. However, a different selection of the a- and c-axes will convert either a-glide or n-glide into c-glide plane. For this reason we need to take only the c-glide plane into account. Accordingly, the replacement of 2 and m by 21 and c results in the 13 monoclinic space groups as summarized in Table 1. Table 1 Space and point groups for the monoclinic crystal system Point groups Space groups 2=m

P 2=m P 21 =m P 2=c P 21 =c Pm Pc P2 P 21

m 2

C 2=m C 21 =m C 2=m C 2=c C 21 =c C 2=c Cm Cc C2 C 21 C 2

When the point symmetry elements 2 and m is replaced by 21 , a screw axis always appears between the twofold rotation axes, so that the symmetry element of 21 has been excluded in the centered-lattice (C ) case. In other words, a- and nglide planes occur in the C -centered space group case, so that the pairs of symbols C 21 =m D C 2=m, C 21 =c D C 2=c, and C 21 D C 2 give only a single space group each. Similarly, the same combination may be considered about other crystal systems and it is quite complicated as actual work. Since this point is purely handled as mathematical issue and the answer (restricted to 230 cases) has already come out, we should use the result. Detailed information of these 230 space groups are available in the International Tables, Volume A with the chart showing an equivalent positions in a unit cell. For convenience, Table 2 shows the space group symbols for the 14 Bravais lattices. Table 2 The space group symbols for the 14 Bravais lattices P Triclinic

P 1N

Monoclinic

P 2=m

C

I

F

C 2=m

Orthorhombic P 2=m 2=m 2=m C 2=m 2=m 2=m I 2=m 2=m 2=m F 2=m 2=m 2=m Tetragonal

P 4=m 2=m 2=m

I 4=m 2=m 2=m

Trigonal

P 6=m 2=m 2=m

R3N 2=m

Hexagonal Cubic

P 4=m3N 2=m

I 4=m3N 2=m

F 4=m3N 2=m

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Question 6.4 Explain the Laue groups which are important as an indicator, when determining the space group of the structure of a desired substance.

Answer 6.4 The periodicity in regular three-dimensional lattice can be analyzed by the concept of symmetry, but one of our main interests in X-ray crystallography is to reveal the structure of the desired substances. This suggests “how to obtain information that the atomic distribution in a crystal of interest is described by one of the 230 space groups and by one of the 32 point groups with sufficient reliability.” Since the measured X-ray diffraction data enable us to provide information of the reciprocal lattices, the symmetry of reciprocal lattices is examined first and then the symmetry of crystals are extracted. The use of centrosymmetric point groups called “Laue groups” is known to be quite useful for this purpose. Figure 1 shows the experimental set-up for taking a Laue photograph on a flat-plate film using a white X-ray source. The crystal is stationary with respect to the X-ray beam, so that the crystal acts as a kind of filter for selecting the correct wavelength for each reflection under the Bragg law. The resultant spots lie on ellipses and all of which have one end of their major axis at the center of the photographic film. All spots on one ellipse arise through reflections from planes that lie in one and the same zone. A diffraction pattern formed by spots is centrosymmetric.

Zone axis f

Crystal Film

Fig. 1 Schematic diagram for the experimental condition for taking a Laue photograph on a flatplane film (Transmission type)

Laue photographs using a white X-ray source providing information about symmetry of the weighted reciprocal lattice will reveal the presence of all the symmetry elements associated with the various point groups, but will add a center of symmetry (for noncentrosymmetric point groups). Namely, the arrangement of spots obtained on the Laue photograph exhibits only the symmetry that would be found from a crystal having the corresponding centrosymmetric point groups. There are only eleven possible symmetries and they are called “Laue groups.” It may be added that the Laue group assigned to a crystal of interest gives the symmetry of the complete diffraction pattern from that crystal. Thus, the classification of the 32 point

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6 Symmetry Analysis for Crystals and the Use of the International Tables

groups is possible by means of the Laue diffraction symmetry as shown in Table 1, where the symmetry of the Laue photographs on a flat-plate film can be described for directions of the incident X-ray beam normal to the crystallographic forms as listed. Note that point group projection symmetry corresponds to the symmetry of the projection of the general form of a point group on to a plane. Table 1 Crystal systems, Laue groups, and Laue projection symmetry

a

Referred to hexagonal axis.

A rotation axis of crystal lattice becomes that of the corresponding reciprocal lattice and the distribution of the weighted reciprocal lattice points usually has a symmetry center. Therefore, a point group appeared in the reciprocal lattice points can be attributed to one of the Laue groups. In this respect, symmetry elements and equivalent positions of the Laue groups are very important fundamental information for structural analysis of crystals by X-ray diffraction. Such information is summarized in Fig. 2a using the method of projection. These results are obtained in the following procedure. Let us consider the spherical surface passing through a symmetry center and mark the position where a symmetry element and this spherical surface intersect. Then, as shown in Fig. 2b, project the marked positions from right above on the equatorial plane of the sphere.

6.3 Solved Problems

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a

1

2/m

mmm

4/mmm

3

3m

6/mmm

m3

4/m

6/m

m3m

Projection

2-folol rotation axis

b

e

an

l rp

o

irr

M

Fig. 2 Symmetry elements and equivalent positions of the Laue group (a). Example of projection using the 2=m case (b)

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6 Symmetry Analysis for Crystals and the Use of the International Tables

The equivalent positions are displayed by a huge comma, and its top surface is colored in white and its bottom in black. If the mirror plane is situated on the equatorial plane, two marks overlap each other upside down on the plane as they are projected. Such overlap is illustrated by black and white overlaid in the figure. Note that the circumference of a circle drawn by thick line indicates that a mirror plane is on the paper surface, the lines other than the circumference show the crossing section between the mirror plane and the sphere by projection. Question 6.5 Show the variation of Hermann–Mauguin symbol, when the axes of the space group of Pnma (orthorhombic) are altered within a unit cell. In the same way, show the results of the space group given by Pna21 (orthorhombic).

Answer 6.5 Let us consider the variation expressed by abc ! bNac ! cab ! cN ba ! bca ! aNcb, when the starting axis is set to abc. In this case, keep in mind that the origin, O, should be positioned at the top-left corner and further set b-axis to abscissa (horizontal direction), a-axis to ordinate (vertical direction), and c-axis to perpendicular direction to the plane of the drawing in the righthanded system. As shown in Fig. 1, this corresponds to the case of watching a crystal from the right-hand side. While keeping this condition and if every rotatabc

b

abc

b -a c

O

a a

a’b’c’ b -a c

O

b

Watch a crystal from the right-hand side. b’

Rotate the diagram above-mentioned in a clockwise direction by 90° and determine the axes.

a’ Fig. 1 Fundamentals for changing axes with respect to the orthorhombic system

6.3 Solved Problems

241

ing the drawing clockwise by a quarter turn (90ı) as shown in Fig. 1, decide axes of a0 , b0 , c0 . Such processes are summarized in Fig. 2. For convenience, Fig. 3 also shows the results using the conversion relationships between Pnma and Pbnm when changing the axes from abc in Pnma to cab in Pbnm, as an example.

Fig. 2 Variations due to abc ! b ac N ! cab ! cba N ! bca ! acb N

Fig. 3 The relationships between Pnma and Pbnm due to the change in axes of abc ! cab

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6 Symmetry Analysis for Crystals and the Use of the International Tables

The results are summarized as follows. abc ! bNac ! cab ! cN ba ! bca ! aNcb Pnma; Pbnm; Pmcn; Pnam; Pmnb; Pcmn Pna21 ; P21 nb; Pc21 n; Pn21 a; Pbn21 ; P21 cn When the crystal axes are altered, we find changes for not only atomic coordinates .x; y; z/ but also the basis vectors of reciprocal space .a ; b ; c / as well as the reflection indices .hkl/ and lattice points .u; v; w/. Since information about the axial transformation in each crystal system is provided using matrices in the International Tables, Volume A (see pp.77–89), some additional details are given below. The general transformation of the coordinate system consists of two parts; a linear part and a shift of origin. The linear part suggests a variation of orientation or length or both of the basis vectors a; b; c. Note that the shift vector is zero for a purely linear transformation. The 3 rows 3 columns matrices P of the transformation from a; b; c to a0 ; b0 ; c0 is given using row matrices .a; b; c/ as follows. .a0 ; b0 ; c0 / D .a; b; c/P 0

1

P P P B 11 12 13 C B D .a; b; c/ @ P21 P22 P23 C A P31 P32 P33

(1)

(2)

Miller indices of a plane .h k l/ are also given in the same way. .h0 ; k 0 ; l 0 / D .h; k; l/ P

(3)

Note that the Miller indices are usually made relative prime before and after the transformation. If using the inverse matrices of P to Q.D P1 /, the transformation of reciprocal lattice axes, atomic coordinates, and lattice points will be given by the following equation using row matrices, .a =b =c /, .x=y=z/, and .u=v=w/. 0

a0

1

a

1

B C B C B b0 C D Q B b C @ A @ A c0 c 10 1 0 Q11 Q12 Q13 a CB C B CB C DB @ Q21 Q22 Q23 A @ b A Q31 Q32 Q33 9 .x0 =y0 =z0 / D Q.x=y=z/ = .u0 =v0 =w0 / D Q.u=v=w/ ;

c

(4)

(5)

(6)

6.3 Solved Problems

243

For convenience, a general comment is supplemented with respect to the case where the axes of a space group are changed within a unit cell. A space group symbol is usually described by four symbols such as Pnma and Cmca, and they provide the relationship of each symmetry element with respect to the direction of a crystal axis. In other words, a space group symbol such as monoclinic and orthorhombic crystal systems is found to depend on the choice of a crystal axis. However, for crystal systems of relatively high symmetry the selection of a crystal axis does only a small variation in the space group symbol, because the selection of an axis is rather limited. In addition, since the triclinic crystal system has no axial N symmetry element, the space group symbol is limited to only P1 and P1. If b-axis is fixed in a monoclinic crystal system, the selection of other two axes is limited to a two-dimensional plane. Thus, it is relatively easy to understand that three kinds of notations given by P21 =c; P21 =a, and P21 =n are the same space groups. Whereas, it is not optional how to select an axis as for the orthorhombic crystal system characterized by the three axes which are at right angles to one another (mutually perpendicular). Nevertheless, there are 3 ways of selecting a-axis and 2 ways for b-axis after fixing a-axis, which results in six combinations even if the selection is limited to the right-handed system. Note that the space group symbols related to the axis are just three and the first being the notation to represent centered lattices, the second being the order of appearance of the axis symbol or plane one, and the third being the forward direction of a glide plane. This is applicable to the unit cell transformations. For example, when the directions of three axes in space are numbered as 1, 2, and 3 and each direction is named by a, b, and c, respectively, one can obtain six combinations as summarized in Table 1. Table 1 Transformations of the crystal axes in the orthorhombic system 1

2

3

3m31

(1)

a

b

c

Cmca

(2)

a

c

b

Bmab

(3)

b

c

a

Abma

(4)

b

a

c

Ccmb

(5)

c

a

b

Bbcm

(6)

c

b

a

Acam

Some additional details for this combinations are given. Let us temporarily assume that the case (1) of Table 1 is given by Cmca which represents the symbol of space lattice, the first symmetry element, the second symmetry element, and third symmetry element. The symbol of Cmca is, of course, a temporary one, so that it depends on the method “how to give name an axis.” For this reason, 3m31 related to the directions of three fixed axes 1, 2, and 3 is provided in the top right end of Table 1. As setting is made in this way, six variations will be decided in turn and the resultant symbols will be obtained as shown in the right-end column of Table 1.

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6 Symmetry Analysis for Crystals and the Use of the International Tables

For example, since the axis of 3 is given as a for the case (3), its symbol will be Abma with A for the space lattice, b for the first symmetry element which is the direction of the glide plane for a-axis, m for the second symmetry element as b-axis matches the axis of 1 and then no change, and a for the third symmetry element which corresponds to the direction of the glide plane for c-axis. More information can be obtained from the International Tables for Crystallography. Question 6.6 Obtain the structure factor of Fhkl and jFhkl j2 and show the condition where the diffraction intensity can be observed in the following three cases: (1) The fourfold rotation axis along the a3 -axis in a unit cell. (2) The 41 screw axis along the a3 -axis in a unit cell. (3) The 42 screw axis along the a3 -axis in a unit cell.

Answer 6.6 (1) The atomic positions in a unit cell with the fourfold rotation axis are given by x; y; zI y; N x; zI x; N y; N z; and y; x; N z. Then, the structure factors Fhkl and jFhkl j2 in this case are computed in the following equations. n o Fhkl D f e2ilz e2i.hxCky/ C e2i.kxhy/ C e2i.hxCky/ C e2i.kxhy/ D 2f e2ilz fcos 2.hx C ky/ C cos 2.kx hy/g D 4f e2ilz cos f.h C k/x .h k/yg cos f.h k/x C .h C k/yg jFhkl j2 D 16f 2 cos2 f.h C k/x .h k/yg cos2 f.h k/x C .h C k/yg Therefore, there is no condition where the diffraction intensity cannot be detected with respect to the atoms in a unit cell with the fourfold rotation axis. (2) The atomic positions in a unit cell with the 41 -screw axis are given by x; y; zI N y; N z C 12 ; and y; x; N z C 34 . The structure factors for this case are y; N x; z C 14 I x; computed as follows. Fhkl D f e2ilz fe2i.hxCky/ C eil=2 e2i.kxhy/ C eil e2i.hxCky/ C ei3l=2 e2i.kxhy/ g Here, let us consider the case of l D 4n, n o Fhkl D f e2ilz e2i.hxCky/ C e2i.kxhy/ C e2i.hxCky/ C e2i.kxhy/ jFhkl j2 D 16f 2 cos2 f.h C k/x .h k/yg cos2 f.h k/x C .h C k/yg

6.3 Solved Problems

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There is no condition where the diffraction intensity cannot be detected, along the way similar to the fourfold rotation axis case. On the other hand, let us consider the case of l D 4n ˙ 1, one obtains the following results. Fhkl D 2f e2ilz fi sin 2.hx C ky/ sin 2.kx hy/g ˚ jFhkl j2 D 4f 2 sin2 2.hx C ky/ C sin2 2.kx hy/ In addition, the following results are obtained for the case of l D 4n C 2. Fhkl D 4f e2ilz sin f.h C k/x .h k/yg sin f.h k/x C .h C k/yg jFhkl j2 D 16f 2 sin2 f.h C k/x .h k/yg sin2 f.h k/x C .h C k/yg It is noteworthy from these two results that the diffraction intensity for a peak whose Miller indices are given by (0 0 l) can be observed only under the condition l D 4n. (3) The structure factor Fhkl is computed from the atomic positions in a unit cell with the 42 -screw axis; x; y; zI y; N x; z C 12 I x; N y; N zI y; x; N z C 12 in the following form. n o Fhkl D 2f e2ilz cos 2.hx C ky/ C eil cos 2.kx hy/ When considering l D 2n, Fhkl D 2f e2ilz fcos 2.hx C ky/ C cos 2.kx hy/g jFhkl j2 D 4f 2 fcos 2.hx C ky/ C cos 2.kx hy/g2 When considering l D 2n C 1, Fhkl D 2f e2ilz fcos 2.hx C ky/ cos 2.kx hy/g jFhkl j2 D 4f 2 fcos 2.hx C ky/ cos 2.kx hy/g2 In conclusion, the diffraction intensity for a peak whose Miller indices are given by (0 0 l) can be observed only under the condition l D 2n. Question 6.7 Let us consider the monoclinic space group denoted by P21 /c, showing the primitive lattice with the 21 -fold screw axis parallel to b-axis, the c-glide plane being perpendicular to b-axis with glide of 2c . Answer the following questions.

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6 Symmetry Analysis for Crystals and the Use of the International Tables

(1) Illustrate the symmetry of this space lattice such as a symmetry center and show both general position and special positions. (2) Find the extinction condition. Answer 6.7 (1) Assuming the 21 -fold screw axis and the glide plane being perpendicular to b-axis as shown in Fig. 1. In addition, the center of point symmetry (a symmetry center) is indicated by a solid circle at 14 or 34 along the c-axis. If the position of .0; 14 ; 14 / is set to the symmetry center, we obtain the results of Fig. 2. Therefore, the general positions can be given as follows. x; y; zI x; N y; N zNI x; N

1 1 1 1 C y; zI x; y; C z 2 2 2 2

Fig. 1 Monoclinic space lattice. Solid circles indicate symmetry centers

The special positions corresponding to the symmetry center are also given as follows (see solid circles in Fig. 2) .0 0 0 0

0/ W 0 1 2

W 0

1 1 ; 2 2

1 2 1 2

1 0 ; 2

1 0 0 W 2 0

1 2

1 W 2

1 2

1 2 1 2

(2) In order to find the extinction condition, the following formulas for trigonometric functions were used.

6.3 Solved Problems

247

Fig. 2 Monoclinic space lattice when setting a symmetry center to 0; 14 ; 14 . Solid circles indicating a symmetry center at 0 or 12 along the c-axis

8 ACB AB ˆ ˆ sin A C sin B D 2 sin cos ˆ ˆ ˆ 2 2 ˆ < AB ACB cos cos A C cos B D 2 cos ˆ 2 2 ˆ ˆ ˆ ˆ A C B A B ˆ : cos A cos B D 2 sin sin 2 2 The structure factor is computed as follows. F D e2i.hxCkyClz/ C e2i.hxkylz/ f 1 1 1 1 C e2ifhxCk . 2 Cy /Cl . 2 z/g C e2ifhxC. 2 y /kC. 2 Cz/l g D cos 2.hx C ky C lz/ i sin 2.hx C ky C lz/ C cos 2.hx ky lz/ i sin 2.hx ky lz/ 1 1 Cy kC z l C cos 2 hx C 2 2 1 1 Cy kC z l i sin 2 hx C 2 2 1 1 C cos 2 hx C y kC Cz l 2 2 1 1 y kC Cz l i sin 2 hx C 2 2 D 2 cos .2; 0/ cos 2 C 2 cos 2

2.hx C ky C lz/ 2

kCl cos 2.hx ky C lz/ 2

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6 Symmetry Analysis for Crystals and the Use of the International Tables

i sin 2

kCl cos .hx ky C l/ 2

D 2 cos 2.hx C ky C lz/ C 2 cos 2

kCl cos 2.hx ky C lz/ 2

(i) In the case of k C l D 2n, F D 4 cos 2.hx C lz/ cos 2ky f The extinction does not take place. (ii) In the case of k C l D 2n C 1, F D 4 sin 2.hx C lz/ sin 2ky f The extinction takes place under the condition of h D l D 0 or k D 0. Question 6.8 Information about the space group Pnma, No.62 (orthorhombic) can be obtained from the International Tables for Crystallography, Volume A, page 298-299. Explain the key points.

Answer 6.8 The symbol Pnma described by a shortened form (notation) of Hermann–Mauguin method describes that Bravais lattice is a simple lattice(P) with three symmetry elements mnm with respect to the direction of Œ100, Œ010, and Œ001 (see Table 6.4). Namely, this orthorhombic has mirror planes perpendic16 provides the Sch¨oenflies’ description ular to these directions. The symbol of D2h and No.62 corresponds to the number allocated in the 230 space groups. Next, P 21 =n=21=m=21=a is full expression of three symmetry elements and they are as follows. 21 =n shows the 21 screw axis and a diagonal plane of glide reflection perpendicular to it for the direction of a-axis, the 21 =m indicates the 21 screw axis and a mirror plane vertical to it for the direction of b-axis, and the 21 =a shows the 21 screw axis and a mirror plane perpendicular to it for the direction of c-axis, respectively. The space group of the Patterson function is given by Pmmm. The Patterson function corresponds to the Fourier transform of the square of the structure factor and it is widely used to obtain a map of interatomic distances in the unit cell. In other words, the Patterson function is a superposition of peaks derived from all atomic pairs in the unit cell directly related to the measured diffraction data. (Refer to other textbooks for details of the Patterson function, for example, M.M. Woolfson, An Introduction to X-ray Crystallography, 2nd Edition, Cambridge University Press (1997).) Four figures are the so-called standard setting of the space group or the spacegroup tables and first three figures are characterized by the Hermann–Mauguin symbols in the headline. The space-group tables for each orthorhombic space group is known to consist of three projections of the symmetry elements along the c-axis

6.3 Solved Problems

249

(upper left), the a-axis (lower left), and the b-axis (upper right), in addition to the general position diagram. For example, with respect to the upper left corner as its origin, the projection is made along the c-axis, setting a-axis and b-axis to the horizontal (abscissa) and the vertical (ordinate) axes, respectively. For convenience, diagrams for the standard setting are shown in Fig. 1 using the orthorhombic and tetragonal space groups as an example, where G = general position diagram. Note that all these figures are described in the right-handed coordinate system.

Fig. 1 Diagrams for the standard setting as described in the space-group tables; orthorhombic and tetragonal space groups

For each orthorhombic space group, there are six different ways of assigning the labels a, b, c to the three orthorhombic symmetry directions. These settings correspond to the six conversions of the labels of the axes including identity conversion. Three space group symbols written on the horizontal axes in three figures of projection are Pmnb, Pbnm, and Pmcn which are corresponding to the axial conversions of abc (fundamental axis), acb, N and cba. N Similarly, Pmnb, Pbnm, and Pmcn are related to the axial conversions of bac, N cNab, N and bca in the vertical axes case. In addition, the numerical value 1/4 placed beside the symmetry element symbols shows the inner height of a unit cell in the projection direction. The figure at the lower right is corresponding to the general position diagram, which is given only in the projection along c-axis. It may also be suggested that this figure shows the equivalent positions in the same projection as the upper left and provides information about what arrangement of atoms in the general positions will , show the equivalent positions and be possible in a unit cell. Both marks and , and are related by a mirror plane. They are called enantiomorphs, so that , is given in a right-handed system. represents the left-handed system, if The signs and numerical values, C; and 12 C, 12 placed beside are the coordinates to the projection direction of the equivalent positions, and the present case suggests Cz, z, 12 C z, 12 z. Some other information are summarized as follows. Origin: The determination of crystal structures is facilitated by the selection of a suitable origin. The line of “Origin” provides the origin selected in the space group

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6 Symmetry Analysis for Crystals and the Use of the International Tables

table. According to the International Tables, the position of symmetry center is set to the origin for space groups with a symmetry center and the position of the highest site symmetry becomes the origin for space groups without a symmetry center. In the space group Pnma, the origin is set to a symmetry center denoted by 1N on the twofold screw axis (as given by “on 1 21 1”) which is equal to the b-axis. Asymmetric unit: An asymmetric unit of a space group is considered to be the smallest closed part of space from which by using all symmetry operations, the whole space can be filled. In other words, the line of “Asymmetric unit” shows that the independent area in a unit cell is given in the region; 0 x 12 , 0 y 14 , and 0 z 1 in Pnma. Symmetry operations: The geometric description of the symmetry operations is given in the space-group tables under the heading “Symmetry operations.” These information give a link between the space group diagrams and the general positions. The line of “Symmetry operations” shows the symmetry elements as well as positions related to symmetry operations. In the space group Pnma, symmetry operations of (1)–(8) correspond to eight equivalent positions. Generators: The line of “Generators” provides all symmetry operations required to generate all equivalent positions of the general positions from coordinates x; y; z. For example, t.1; 0; 0/, t.0; 1; 0/, and t.0; 0; 1/ indicate the translational operation that moves the coordinates x; y; z described by (1) of general positions to the directions of a,- b-, and c-axes by one unit cycle. In the space group Pnma, although symmetry operations of (1)–(8) are cited, all equivalent positions can be generated by operations of (1), (2), (3), and (5). Accordingly, the operations of (4), (6), (7), and (8) are excluded. Positions: The column of “Position” more explicitly called Wyckoff positions provide information of the equivalent positions when considering site symmetry and are defined as a group of crystalline positions. The following information (a) to (e) classified into general positions and special positions are provided in the space group table. (a) Multiplicity: This is the number of equivalent points per unit cell and keep in mind it differs from the number of the equivalent lattice planes in one plane of a form called multiplicity factor. The multiplicity of the general position is equal to the order of the point group to which the space group under consideration belongs. On the other hand, the multiplicity of the special position is given by the divisor of multiplicity of the general position. For example, the number of equivalent point in the special position 4c is one half of the general equivalent position. This is attributed to the condition that one special position is formed by overlapping two general equivalent positions. (b) Wyckoff letter: This is simply a coding scheme for the Wyckoff positions. Usually, the notation is made due to the higher degree of site symmetry, starting with a at the bottom column for the special position and upward in the alphabetical order traced back to the general position. For example, the general position of

6.3 Solved Problems

251

atomic position in the space group Pnma is 8d and its special positions are 4c, 4b, 4a in order. (c) Site symmetry: The site symmetry groups of the different points of the same special position are symmetrically equivalent subgroups of the space group and then all points of one special position can be described by the same site symmetry symbol. The column of “Site symmetry” provides the symmetry which the atomic position has. There are two ways in the space group Pnma where 4 atoms occupy 8 symmetry center sites and the special positions 4a and 4b correspond to them. There is the mirror symmetry (m) in the special position 4c. In addition, in order to clarify symmetry directions, the irrelevant axial directions are shown by dots such as .m. Here this .m. indicates a mirror plane perpendicular to the b-axis. (d) Coordinates: The sequence of the coordinate triplets is based on the Generators. centered space groups, the centering translations such as .0; 0; 0/C 1 1 1For C are given above the coordinate triplets. The symbol C indicates that 222 the components of the centering translations have to be added to the listed coordinate triplets for obtaining a complete Wyckoff position. Coordinates corresponding to a-, b-, and c-axes with a parallelepiped as a unit cell are referred to as x; y; z. The length of this unit cell is normalized as 1. When one atom exists in the space group Pnma, we find eight atoms without exception at the following positions in a unit cell. .x; y; z/, x C 12 ; y; z C 12 , x; y C 12 ; z , x C 12 ; y C 12 ; z C 12 , .x; y; z/, x C 12 ; y; z C 12 , x; y C 12 ; z , x C 12 ; y C 12 ; z C 12 , where x is x. N (e) Reflection conditions: Information of the extinction rule is given for the case where atoms are located at the general positions. For example, 0kl W k Cl D 2n (n is integer) shows, with respect to the 0kl reflection, that when k C l is an odd number, its crystal structure factor is zero and if k Cl is an even number, it is not zero. On the other hand, when atoms are located only at the special positions, new information of the extinction rule appears in the column listed as Special for every Wyckoff sign, in addition to the conditions given for the general positions. For example, if an atom is only at the 4a position, the structure factor becomes zero, when h C l or k is an odd number. Symmetry of special projections provides information corresponding to twodimensional space groups and for example, this is used to project the crystal structure with respect to the direction perpendicular to the reciprocal lattices using the two-dimensional intensity data of a zero layer. In each space group table, three different projections are given with respect to the direction indicated by “Along” which is the projection to the plane perpendicular to this direction. Projections depend on crystal system and such information is as follows. Projections are made to the directions of c-, a-, and b-axes for triclinic, monoclinic, and orthorhombic systems. Similarly, the directions of c- and a-axes and Œ110 for tetragonal system, c- and a-axes and Œ210 for hexagonal sysN and Œ21N 1 N for trigonal system and Œ001, Œ111, and Œ110 for tem, Œ111, Œ110, cubic system, respectively, are used for the projections. Following the projection

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6 Symmetry Analysis for Crystals and the Use of the International Tables

direction, information related to the plane groups generated by the projection of the space groups are provided as the Hermann–Mauguin symbol, p2gm, at the present case. In the following line, the relationship with the basic axes a0 , b0 and in the line after that, the origin of plane groups, for example, .0; 0; z/ is given using the unit cell coordinates of the space group.

Chapter 7

Supplementary Problems (100 Exercises)

Exercise 1.1 When accelerating an electron by 1 kV, compute the values of energy, momentum and wavelength using the de Broglie relation. Exercise 1.2 Calculate the values of mass absorption coefficient of gallium arsenide (GaAs) and barium titanate (BaTiO3 ) for Cu-K˛ radiation. Exercise 1.3 Air composition can be considered to be mostly 80mass%N2 and 20mass%O2 and its density is 1:29 103 Mg=m3 D g=cm3 at normal temperature (273.15 K) and pressure (101,325 Pa). If Cu-K˛ radiation passes through 360 mm of normal air, by what percentage is the intensity reduced. Perform a similar calculation for the Fe-K radiation. Exercise 1.4 3:138 1021 atoms are found to be included in 0.5 g of a metallic substance consisting of a single element. Calculate the atomic weight of this substance. Exercise 1.5 What is the relationship between the wavelengths K of absorption edge found in each atom at the characteristic energies and the critical excitation voltage VK for the case where the K-shell electron is removed. Estimate the excitation voltage of Mo-K˛ radiation assuming that the wavelength of K-absorption edge of molybdenum is 0.06198 nm. Exercise 1.6 In the photoelectric absorption process where a photoelectron is released from an atom, priority is given to inner-shell electrons; K-shell electrons are released before L-shell electrons. The recoil of atom is necessarily produced in the photoelectric absorption process, but its energy variation is known to be negligibly small (see Question 1.6). In other words, photoelectric absorption by free electrons does not occur. Explain why this is so using the law of conservation of momentum. When a lead plate was irradiated with X-rays with an energy of 150 keV, K-shell photoelectric absorption occurred at the surface of the plate. In this case, the speed of the photoelectrons was found to be 1:357108 m/s. Calculate the K-absorption edge of lead, assuming that no energy loss occurs.

253

254

7 Supplementary Problems

Exercise 1.7 When a tungsten plate was irradiated by X-rays with the energy of 150 keV, the photoelectric absorption of K-shell was made at the surface of plate and the photoelectron was ejected. Assuming that there is no energy loss in this phoW D 69:52 toelectric absorption process and the K absorption edge of tungsten is EK keV, find the velocity of photoelectron, at ejection. Exercise 1.8 Compute the energy of K˛ radiation emitted from tungsten (atomic number 74), under the condition that Moseley’s law is obeyed and the shielding constant for generating the K˛ radiation is set to zero and the Rydberg constant is 1:097 107 (m1 ). Exercise 1.9 In the operation of X-ray tube, if setting applied voltage to V, excitation voltage to VK and current to i, the number of photons of the characteristic radiation emitted, IK , is approximated by the following equation, called “Storm formula”. IK D BS i.V VK /n Where BS is a proportionality constant and n D 1.5 is widely used. The value of a proportionality constant is set to 4:25 108 (As1 sr1 ) and the excitation voltage of tungsten is given by 69.5 keV. When a tungsten X-ray tube is operated at 100 kV and 1 mA, answer the following questions. (1) Compute the number of photons of the characteristic X-rays released per unit solid angle using the Storm formula. (2) In this case, the average energy of the characteristic X-rays was also found to be 60.7 keV. Compute the intensity. Exercise 1.10 The transmission rate was found to be 0.65, when an iron plate was irradiated by X-rays with the energy of 100 keV. The mass absorption coefficient and density of iron for X-rays with the energy of 100 keV are 0.215 cm2 =g and 7.87 g=cm3 , respectively. (1) Estimate the thickness of this iron plate. (2) Check the amount of intensity loss in this iron plate as a function of the incident X-rays. Exercise 1.11 The thickness of a substance required to make intensity of the incident X-rays down to 50% is called “the half value layer”. Calculate the thickness of a layer required to decrease the incident X-ray intensity to 10% when the half value layer of iron to X-rays with a certain energy is known to be 0.041 mm. Exercise 1.12 The minimum energy required to release an electron from an atom in a solid substance is called the work function. During experiments involving light exposure of the tungsten used for the filament of an X-ray tube, the threshold wavelength for photoelectron emission is found to be 274.3 nm. Calculate the work function of the metal. On the other hand, applying a reverse voltage to the filament can prevent electron emission. Calculate the reverse voltage required to prevent electron emission when tungsten is irradiated by light with a wavelength of 100 nm.

7 Supplementary Problems

255

Exercise 1.13 The wavelengths of K˛ and Kˇ radiation for magnesium are 0.9890 and 0.9521 nm, respectively. With reference to the K-Shell, describe the atomic core level of the L- and M-shells in magnesium. N N .102/; .020/; .220/; N .112/ N and Exercise 2.1 Illustrate the planes of .110/; .111/; N the directions of Œ001; Œ100; Œ110; Œ210; Œ122; Œ120 for a cubic lattice. Exercise 2.2 Identify the closest packing plane for bcc and fcc structures and estimate the atomic density on these planes. Exercise 2.3 Graphite, sometimes called “black lead”, is known to have a layered structure in which carbon atoms occupy the vertex positions of a regular (equilateral) hexagon. Draw the unit cell in the plane of this layered structure. Exercise 2.4 Potassium has a bcc structure with a lattice parameter of a D 0:520 nm. (1) Compute the interatomic distances of the 1st and 2nd nearest neighbors. (2) Estimate the coordination numbers of the 1st and 2nd nearest neighbors. (3) Estimate the density based on the crystal structure. Exercise 2.5 It is known that the hexagonal close-packed lattice will be materialp ized in the case of c=a D 8=3 D 1:633, and magnesium is close to this ideal case, because of c=a D 1:624. The values of density and molar mass of magnesium are .1:74 106 g/m3 / and 24.305 g, respectively. (1) Compute the volume of a unit cell. (2) Estimate the lattice parameter a as well as the 1st nearest neighbor interatomic distance. Exercise 2.6 Typical crystal structures of metallic elements are known to be fcc, hcp and bcc. Compute the packing fraction of these three structures. For comparison, compute the packing fraction of simple cubic structure. Exercise 2.7 CaS and MgS have a NaCl-type structure. The radii of Ca2C and Mg2C ions are 0.099 and 0.065 nm, respectively. On the other hand, the radius of S2 ion is known to be 0.182 nm. Describe the stability of CaS and MgS crystals using the difference in size of positive and negative ions. Exercise 2.8 Aluminum has a fcc structure with a lattice parameter of a D 0:40497nm. Compute the interatomic distances in the planes (100) and (111). Exercise 2.9 The coordination polyhedra can be defined as a group of the nearest neighbor atoms surrounding a central atom. The number of such nearest neighbors is called the coordination number. If the coordination number decreases, the volume of the coordination polyhedra will decrease. Calculate the reduction percentage in atomic radius when the coordination number changes from 12 to 8 or from 12 to 4.

256

7 Supplementary Problems

Exercise 2.10 Supposing that a body-centered cubic (bcc) structure is filled with hard spheres, obtain the maximum radius of the sphere which can fit into octahedral and tetrahedral voids. Exercise 2.11 Iron exhibits an ˛-phase at temperatures below 1,183 K and a ı-phase at temperatures above 1673 K. Both of these phases are characterized by a bcc structure, whereas the -phase found at temperatures between 1,183 and 1,673 K has a fcc structure. It is also known that only the -phase exhibits relatively high solubility for carbon. (1) Explain the difference in carbon solubility of these phases from a structural point of view. (2) Titanium is known to have a hcp structure at room temperature. Explain the difference in formation of titanium hydride and titanium carbide. Exercise 2.12 Sodium chloride is known to contain four molecules in a unit cell. The molecular weight of sodium chloride per mole and the value of density at 298 K are 58.44 g and 2:164 106 g/m3 , respectively. (1) Compute the specific volume and molecular volume of sodium chloride crystal. (2) Estimate the distance between sodium ion and chlorine ion in sodium chloride crystal. Exercise 2.13 With respect to caesium chloride crystal, (1) estimate the specific volume and molecular volume. (2) Also find the distance between cesium ion and chlorine ion. The molecular weight of causium chloride per mole and the density value at 298 K are 168.5 g and 3:970 106 g/m3 , respectively. Exercise 2.14 Potassium bromide (KBr) has a NaCl-type structure. The ionic radii of KC and Br are 0.133 and 0.195 nm, respectively. The molecular weight of KBr is 119.00 g. (1) Determine the lattice parameter a and the density of a KBr crystal by assuming the additivity of ionic radii. (2) Find the ratio of r C =r required to prevent the direct contact of anions. (3) Potassium halides also have a NaCl-type structure. For the same halogen, the lattice parameter of rubidium halide is 0.028 nm larger than that of potassium halide. Estimate the ionic radius of RbC from this difference. Exercise 2.15 In the zinc blende (ZnS) structure, one element occupies the corner and the face-centered positions of a unit cell, and the other element occupies the tetrahedral positions of the diamond structure. In addition, the ionic radii of Zn2C and S2 are 0.074 and 0.184 nm, respectively. (1) Find the nearest-neighbor coordination number of Zn2C and S2 . (2) Estimate the angle formed by a S2 and consecutive nearest-neighbor Zn2C ions. (3) Estimate the radius ratio .r C =r / required to avoid direct contact between anions .S2 / by assuming that positive and negative ions touch each other. (4) Give a possible reason why ZnS does not have a NaCl-type structure.

7 Supplementary Problems

257

Exercise 2.16 A unit cell of a caesium chloride (CsCl) crystal with a lattice parameter of a D 0:4123 nm has a Cl ion at each vertex and CsC ion at the center of cube. The molecular weight of CsCl is 168.36 g. Compute the distance between CsC and Cl ion and the density of this ionic crystal. Exercise 2.17 The simplest crystal structure for zinc sulfide is the diamond-like cubic structure. The diamond structure may be visualized as follows; four atoms are added every other one to the tetrahedral site which exists at the center of eight “octants” contained in a unit cell of fcc structure. Explain the relationships among the face-centered cubic, diamond and zinc sulfide structures by illustrating their unit cells and the atomic arrangements along the vertical-axis for zinc sulfide structure. Note: An octant is the name of a small cubic unit cell. Exercise 2.18 In annealing twins of a metallic element with fcc structure, it is likely to find the deformation in the (111) plane so as to form twin structure in the N direction. Illustrate the movement of each atom during such deformation with Œ112 twin structure. Exercise 2.19 Draw the atomic arrangement in unit cells of fcc, hcp and bcc lattices which are the typical crystal structures of metallic elements. Show atomic arrangement at different distances along z-axis (vertical axis). Exercise 2.20 Draw the layer-by-layer atomic arrangement in a unit cell along the vertical (z-) axis for of diamond and NaCl structures.

Exercise 2.21 Draw the layer-by-layer atomic arrangement in a unit cell along the vertical (z-) axis for zinc blende (ZnS), fluorite (CaF2 ) and rutile (TiO2 ) structures.

258

7 Supplementary Problems

Exercise 2.22 A four-index system (H K i L) is often used to identify planes in a hexagonal lattice, in addition to the usual three-index (h k l) system. A similar method is frequently used for describing directions. (1) Using the three-index system, planes, (100), (010) and (001) may appear equivalent to one another. However, show that (001) is not equivalent to other two N planes by the use of four indices. Further, show that (110) is equivalent to .120/. (2) Show that the directions [100], [010], [001] are not equivalent by the use of N direction is perpendicular to the four indices. Also demonstrate that the Œ1010 N plane. .1010/ Exercise 2.23 The lattice points having the symmetry elements characteristic of the rhombohedral system may also be referred to a hexagonal cell. Explain the transformation between rhombohedral axes and hexagonal axes. Exercise 2.24 Find the geocentric angle between Sendai, Japan (38ı of north latitude, 141ı of east longitude) and Seattle, U.S.A. (47ı of north latitude, 123ı of west longitude). Exercise 2.25 Draw the standard stereographic projection of a cubic crystal when projected on the plane of (001) or (011). Exercise 2.26 Show the direction equivalent to perpendicular axis (1) and hexagonal axis (2) in the stereographic projection. Exercise 2.27 Show the stereographic projection of two cases; (1) tetragonal pyramid and (2) tetragonal dipyramid. Also show the angular coordinates with respect to a certain plane in the tetragonal pyramid case.

a

b

Exercise 2.28 Explain how to determine the position of a rotating axis of a cube using the Wulff net. Exercise 3.1 Find the number of molecules and electrons contained in one cubic meter (m3 ) water .H2 O/ at 273 K.

7 Supplementary Problems

259

Exercise 3.2 Differential cross section de =d˝ of one free electron for coherent X-ray scattering, which is generated from the area per unit solid angle in the direction of scattering angle , is given by the following equation. re de D .1 C cos2 / d˝ 2

.m2 =sr/

where re is the classical electron radius .2:8179 1015 / m and sr is steradian. (1) Estimate the value of de =d˝ in the direction of scattering angle 45ı with respect to unit solid angle and unit scattering angle. (2) Estimate the values of de =d˝ in the area from scattering angle zero to radian (180ı) and show the results in the graphical form. Exercise 3.3 Compute the number of photons produced per unit area by the incoherent scattering, when the X-ray beam with 100 keV energy and the number of photons per unit area is given by Nx D 2 1012 .m2 / transmits through 20 mm thick water layer. Note that the value per electron of the incoherent scattering crosssection for the incident X-rays with energy of 100 keV is e D 70:01 1030 m2 . Exercise 3.4 Compute how much the scattering intensity from one electron is at the distance of 0.01 m away from an electron, assuming that the polarization factor is negligible. Next estimate the number of electrons per unit mass for magnesium. The atomic number of magnesium is 12 and its molar mass is 24.305 g. In addition, confirm the scattering intensity from 1 g of Mg, at the distance of 0.01 m. Exercise 3.5 Let us consider that X-rays with the wavelength of 0.01 nm collide with a free electron and produce incoherent scattering at scattering angle of 60ı . (1) Find the wavelength of the scattered photon. (2) Estimate the energy of the scattered photon in keV. Exercise 3.6 The incoherent scattering was obtained at 180ı the scattering angle; the energy of the recoil electron was found to be 30 keV. Compute the scattered photon energy, by considering conservation law of both energy and momentum. Note that if the momentum of an electron is set to p, the rest mass to me and the speed of light to c, the relationship of .E=c/2 p 2 D .me c/2 shall be approved with respect to the total energy of an electron E. Exercise 3.7 Calculate the so-called Compton wavelength accompanying the Compton scattering of an electron. Exercise 3.8 The electron density of a hydrogen atom is given as follows. D

.e2r=a / a2

Z ˚ a D 0:053 nm D 0:53 A

dV D 1

260

7 Supplementary Problems

(1) Derive the atomic scattering factor f and the incoherent scattering intensity per atom i(M) as a function of .sin =/. (2) Calculate the values of f and i(M) at values of .sin =/ = 0.0, 0.2 and 0.4. Exercise 3.9 Let us consider that X-rays with an energy of 100 keV collide with a free electron and are incoherently scattered. Find the energy of the scattered photon under the condition that the recoil angle ( ) is zero. Hint: A value can not be assigned to 1= tan because of the zero in the denominator, so use that 2 ! 2 as ! 0, where 2 is the scattering angle. Exercise 3.10 Cuprous chloride (CuCl) has ZnS type structure and the molecular weight per mole is 99.00 g and the density is 4:135 106 g/m3 at 298 K. When the Mo-K˛ radiation with the wavelength of D 0:07107 nm is used, find the angle at which a strong peak corresponding to the reflection from the (111) plane will be observed. Exercise 3.11 The molar mass of magnesium oxide having NaCl type structure is 40.30 g and the density is 3:58 106 g/m3 at 298 K. (1) Compute the values of .sin =/ at which peaks corresponding to the reflection from the planes of (100), (110) and (111) may appear. (2) Explain the condition for detecting these three peaks by the crystallographic structure factor. Exercise 3.12 AgCl is known to have NaCl type structure. Which of the following indices are allowed in the X-ray diffraction pattern? 100, 010, 001, 110, 101, 011, 111, 200, 020, 002, 120, 102, 012, 210, 201, 021, 220, 202, 022, 211, 121, 112, 221, 212, 122 and 222. Exercise 3.13 Uranium is known to have an orthorhombic crystal lattice with four atoms in a unit cell. When the length of the unit cell is a standard unit, the positions of the four atoms u v w are expressed as 0y 14 I 0 y 34 I 12 21 C y 14 and 12 21 y 34 , where y expresses an arbitrary location. (1) Find the Bravais lattice of uranium in addition to the structure factor Fhkl . (2) Determine the plane indices for which the diffracted intensity cannot be observed. Exercise 3.14 In the calcium fluorite .CaF2 / structure, atoms occupy the position of 34 43 43 , which is not filled in the so-called zinc blende (ZnS) structure. That is, Ca occupies the positions of 000I 12 21 0I 0 21 21 I 12 0 12 , and F occupies the positions of 14 41 41 I 34 43 41 I 14 43 43 I 34 41 43 I 34 43 43 I 14 41 43 I 34 41 41 I 14 43 41 . (1) Obtain the structure factor Fhkl of CaF2 and also express it using the structure factor F .f cc/ of a face-centered cubic lattice.

7 Supplementary Problems

261

(2) Determine the plane indices for which the diffracted intensity can and cannot be observed.

CaF2 ZnS

Ca

Fa

Exercise 3.15 Graphite has hexagonal crystal lattice which contains four atoms per unit cell at positions of 000I 13 32 0I 00 21 and 23 31 21 . Show that the structure factor is given by the following equations. In addition, find the condition of h k l in which the diffracted intensity cannot be observed. l D even W

h C 2k F D 4f cos2 3

l D odd

F D i2f sin 2

h C 2k 3

Exercise 3.16 Explain characteristic features of the structure of caesium chloride (CsCl) crystal. In addition, estimate the diffracted intensity values for peaks from the (100), (110) and (111) planes by assuming that the atomic scattering factor can be approximated by the atomic number of each element, that is 55 for Cs and 17 for Cl, respectively. Exercise 4.1 Derive an equation of absorption factor for measuring the diffraction intensity from a slab sample with thickness t by the symmetry-transmission method. Exercise 4.2 Thermal expansion coefficient of copper is 16:6 106 /K. Compute how much the temperature control is required for a sample in order to obtain the lattice parameter of copper within the uncertainty of ˙0:00001 nm at 293 K. Exercise 4.3 The lattice parameter of aluminum at room temperature is a D 0:4049 nm and its Debye–Waller factor is BT D 8:825 103 nm2 . Let us consider the intensity measurements of aluminum for peaks corresponding to the 111, 311 and 420 planes. What is the intensity at room temperature as a percentage of that at absolute zero Kelvin.

262

7 Supplementary Problems

Exercise 4.4 A diffraction pattern of a powder crystalline sample with a very small volume was recorded using a Debye–Scherrer camera with a flat-plate film positioned at a distance D from the specimen and perpendicular to the incident beam. Find the intensity P 0 per unit length of the diffraction circle on the flat-plate film as a function of D. Exercise 4.5 For two metallic samples, the following numerical data (Tables A and B) were obtained from X-ray diffraction patterns using Cu-K˛ radiation ( D 0:1542 nm). By considering the fact that fcc or hcp-type feature is observed in the diffraction pattern, determine the crystal structure and also estimate the lattice parameter. Table A 1 2 3 4 5

2q 43.16 50.28 73.97 89.86 95.05

d(A) 2.0761 1.8148 1.2816 1.0917 1.0453

Table B I / I0 100 48 26 24 7

1 2 3 4 5 6 7 8 9 10 11 12

2q 36.30 38.97 43.22 54.32 70.07 70.61 77.04 82.11 83.70 86.53 89.90 94.92

d(A) 2.4751 2.3114 2.0935 1.6890 1.3430 1.3341 1.2380 1.1739 1.1556 1.1249 1.0913 1.0464

I / I0 53 40 100 37 48 35 6 26 4 16 11 6

Exercise 4.6 For two unknown samples, the following numerical data (Tables A and B) were obtained from X-ray diffraction patterns using Cu-K˛ radiation ( D 0:1542 nm). By assuming that the unknown sample is a single phase, identify each sample using the Hanawalt method. Table A 1 2 3 4 5 6 7 8

2q 36.90 42.77 62.14 74.50 78.51 93.95 105.74 109.92

d(A) 2.436 2.114 1.494 1.274 1.218 1.055 0.967 0.942

Table B I / I0 20 100 55 5 15 5 2 15

1 2 3 4 5 6 7 8

2q 35.83 41.74 60.45 72.33 76.05 90.77 101.68 105.47

d(A) 2.506 2.164 1.532 1.307 1.252 1.083 0.994 0.969

I / I0 85 100 60 35 20 10 15 25

7 Supplementary Problems

263

Exercise 4.7 X ray diffraction pattern (Fig. A) and the relevant numerical data (Table A) for the unknown sample were obtained by Cu-K˛ radiation. Identify the sample by applying the Hanawalt method.

Table A

Fig.1

Exercise 4.8 X ray diffraction pattern (Fig. A) and the relevant numerical data (Table A) for the unknown sample were obtained by Cu-K˛ radiation. Identify the sample by applying the Hanawalt method.

Fig. A

264

7 Supplementary Problems

1 2 3 4 5 6 7 8 9 10 11 12

2q 20.80 25.59 26.66 35.16 36.56 37.77 39.46 40.27 42.42 43.36 45.78 46.15

d (A) 4.271 3.482 3.344 2.553 2.458 2.382 2.284 2.240 2.131 2.087 1.982 1.967

Table A

I / I1 17 62 100 95 7 36 5 1 6 90 2 2

13 14 15 16 17 18 19 20 21 22 23 24

d (A) 1.816 1.742 1.673 1.661 1.605 1.543 1.512 1.453 1.406 1.384 1.375 1.371

2q 50.25 52.54 54.88 55.31 57.42 59.96 61.32 64.10 66.51 67.71 68.21 68.44

I / I1 10 41 3 2 80 8 7 2 28 4 40 3

Exercise 4.9 A diffraction experiment was carried out on a powder sample of silicon using Cu-K˛ radiation. The structure of silicon is the same as that of diamond crystal lattice. In the high angle region, the peak splitting attributed to the difference between K˛1 and K˛2 is clearly observed in Fig. A and the relevant angular data are summarized in Table A for the (440), (531), (620), and (533) planes. Compute the lattice parameter using the extrapolation method.

Table A

Fig. A

Exercise 4.10 Ultra fine particles of iron were produced by the evaporation method and their average sizes are estimated 25, 50, 90 and 120 nm using the laser particle analyzer. Assuming that the resultant particles are strain free and the peak broadening detected in the measurement with Cu-K˛ radiation ( D 0:15406 nm) arises

7 Supplementary Problems

265

only from variation of size of the crystallites, estimate the peak width (the value of FWHM) of the (110) plane possibly detected by using the Scherrer equation. Exercise 4.11 The peak broadening as well as the decrease in peak height was observed, during a diffraction experiment by Cu-K˛ ( D 0:1542 nm) radiation on the cold worked aluminum sample. The results are summarized in Table A where the FWHM values for peaks corresponding to the (111), (200), (220), and (311) planes. The aluminum sample was fully annealed in order to remove the stress and strain induced by cold-work and similar measurements were done for comparison. The results are summarized in Table B with the corresponding FWHM values. Compute the average size of crystallites in the cold-worked aluminum sample. Table A 1 2 3 4

Table B

hkl

2q (degree)

FWHM (degree)

111 200 220 311

38.47 44.72 65.13 78.27

0.188 0.206 0.269 0.303

1 2 3 4

hkl

2q (degree)

111 200 220 311

38.47 44.70 65.10 78.26

FWHM (degree) 0.102 0.065 0.089 0.091

Exercise 4.12 The diffraction experiment using Cu-K˛ ( D 0:1542 nm) radiation was carried out for the mixed powder sample of MgO and CaO and two or more clearly separated diffraction peaks were obtained. According to the preliminary analysis, the diffraction peak of the (111) plane of MgO is found to overlap with that of the (200) plane of CaO near 2 D 37ı (see Fig. A) and it is difficult to separate. Other peaks can be identified with each component. The measured integrated intensities of the corresponding peaks are summarized in Table A. Calculate the contents of MgO and CaO. Both of MgO and CaO have NaCl type structure

Fig. A

266

7 Supplementary Problems

and the lattice parameters are 0.42112 nm for MgO and 0.48105 nm for CaO (see Appendix A.9). Table A

Exercise 4.13 The diffraction experiment using Cu-K˛ line ( D 0:1542 nm) radiation was made for a mixed powder sample of Si containing Cu and four clearly separated diffraction peaks were obtained. These four peaks can be attributed to peaks of the (111) and (220) planes of Si and the (111) and (200) planes of Cu. The integrated intensities of these four peaks are summarized in Table A. Calculate the contents of Cu and Si using the direct method. Table A

1 2 3 4

2 (degree) 28.40 43.28 47.31 50.43

Integrated intensity I 162.3 359.7 87.2 120.4

hkl Si(111) Cu(111) Si(220) Cu(200)

Exercise 4.14 Slags used in steel-making process dissolve various elements in the glassy phase and the melilite component is known to be frequently involved. After griding this slag sample in a ball mill, the diffraction peak corresponding to the (211) plane of the melilite component was observed at about 32ı scattering angle, as shown in Fig. A. With increasing milling time, the peak broadening as well as the decrease in peak height was observed as summarized in Table A. Compute the variation in average size of crystallites by assuming that measured FWHM values arise only from change of the crystallite size.

7 Supplementary Problems

267 6 CaCO3 Fe

4

Melilite

Table A Melilite (211) 2q FWH degree M degree

31.24

0.059

1

31.25

0.105

2

31.27

0.272

4

31.30

0.319

6

31.31

0.483

2 Intensity [a.u.]

Milling time (hour)

0H

1H

2H

4H

0 25

6H 30

35 40 2q [degree]

45

50

Fig. A

Exercise 5.1 For a rectangular (or orthogonal) coordinate system XY, let us consider that the unit vectors, taken from a point O along the positive directions of the individual axes, are ex and ey . If the coordinates of two vectors from O to the points P1 and P2 are given by P1 .x1 ; y1 / and P2 .x2 ; y2 /, obtain the relationship for the coordinates P.x; y/ which divides the distance between P1 and P2 with the ratio of m : n. Exercise 5.2 In a rectangular coordinate system XY, when the angle between a straight line and the positive X -axis is , the angle to the positive Y -axis is given by f.=2/ g. Accordingly, the two directional cosines are D cos and

D cosf.=2/ g D sin , respectively. In this case, m D . =/ D .sin = cos / D tan is called the directional coefficient of the straight line. Obtain the relationships of the directional coefficients and directional cosines when a straight line q1 is perpendicular to a straight line q2 , where their directional coefficients are given by m1 and m2 , respectively. Exercise 5.3 Let us consider that a two-dimensional lattice is expressed by the basic vectors a D 2ex and b D ex C 2ey . Obtain the primitive vectors of the reciprocal-lattices, A* and B*. Exercise 5.4 Show that the magnitude of the reciprocal-lattice vector b1 is equal to the reciprocal of a spacing of the (100) plane.

268

7 Supplementary Problems

Exercise 5.5 Real crystal lattice vectors a1 and a2 are parallel to the drawing plane, whereas a3 is perpendicular to the drawing plane. When the lengths of the real crystal-lattice vectors are ja1 j D 3:0; ja2 j D 2:0; ja3 j D 1:0 and the angle formed by a1 and a2 is ˛12 D 60ı , solve the following problems. (1) Draw the real crystal-lattice vectors a1 and a2 , and also the reciprocal-lattice vectors b1 and b2 for the above conditions. (2) Draw the (110), (210) and (310) planes in the real crystal-lattice based on the definition of Miller indices hkl, and also show H110 ; H210 and H310 in the reciprocal-lattice when Hhkl D hb1 C kb2 C lb3 . Exercise 5.6 Let us consider that a crystallographic direction may be expressed by Auvw D ua1 C va2 C wa3 and the direction perpendicular to the .hkl/ plane by Hhkl D hb1 C kb2 C lb3 . If the angle between Auvw and Hhkl is , express cos for the orthorhombic system in terms of u; v; w; h; k; l; a1 ; a2 and a3 . Exercise 5.7 It is known that the trigonal (rhombohedral) lattice unit cell is the primitive unit cell for the fcc structure. Similarly, the orthorhombic lattice unit cell is the primitive lattice unit cell for the hexagonal structure. Find the method of conversion for the equivalent plane between these two primitive unit cells using the fact that only one reciprocal-lattice vector can have a real physical meaning. Exercise 5.8 For a face-centered cubic lattice, answer the following questions. (1) Obtain the primitive (translation) vectors of the reciprocal-lattices. (2) Show the first Brillouin zone. Exercise 5.9 Obtain the so-called Bragg equation or Laue equation by using the condition that the strongest scattering amplitude from crystals is obtained when the scattering angle is in agreement with the reciprocal-lattice vector. Exercise 5.10 Show that a given formula for the summation of the real crystal lattice vectors is not zero only if the vector q is equal to the reciprocal-lattice vector. G.q/ D

X

e2iqrn

n

Exercise 5.11 If the sin = value is very small, show that both the atomic scattering factor f and the incoherent scattering intensity i.M/ are given by a parabolic function of sin =. Exercise 5.12 In the classical theory, the amplitude of vibration of atomic electrons is considered to be continuously and infinitely variable. On the other hand, from a quantum-mechanical point of view, the angular frequency is that of an X-ray photon with just enough energy to eject the electron from the atom. This is the case when the wavelength of the incident X-rays is close to the absorption edge and it is

7 Supplementary Problems

269

known as anomalous scattering (or resonance scattering) phenomenon. Considering this phenomenon as the interaction between X-rays and a harmonic oscillator with an attenuation term, explain the variation of the atomic scattering factor f. Exercise 5.13 With respect to a diffraction phenomenon from one-dimensional slit called the Fraunhofer diffraction which is attributed only to the aperture width L, obtain the diffraction intensity at the point P which is at distance R sufficiently far from the slit. Exercise 5.14 Obtain the diffraction intensity produced from a two-dimensional lattice described by an interval a repeated m times in the x-direction and an interval b repeated n times in the y-direction, respectively. Here, m and n take sufficiently large values. Exercise 5.15 Let us consider a one-dimensional lattice in which identical scattering centers are at positions rn D na (n: integer). The Xscattering amplitude G in this one-dimensional lattice is proportional to G D e naq , and the sum over m lattice points is given by. GD

m1 X nD0

e i naq D

1 e i m.aq/ 1 e i.aq/

(1) Obtain the diffraction intensity I. (2) In this case, the maximum value of the diffraction intensity can be obtained when a q is given by integer multiples of 2. Let us consider the case of a small change in q such that a q D 2h C ı.ı > 0/. Calculate the minimum value of ı for which the intensity becomes zero. Exercise 5.16 In the model system of carbon tetrachloride, four chlorine atoms are located at the positions of regular tetrahedron and the carbon atom occupies the center of the tetrahedron. By applying Debye’s equation, obtain the approximate expression for the diffraction intensity from carbon tetrachloride molecules constructed with a total of five atoms with two distinct elements as the scatterer. Exercise 5.17 Consider diffraction from a single crystal which N1 a, N2 a, and N3 a primitive cubic lattices arranged along the three coordinate axis, < 100 > direction, with the lattice parameter of a D 0:25 nm. Explain features of the reciprocal-lattices for the following three cases with different numbers of unit cells along the three directions. (1) N1 D N2 D N3 D 104 (2) N1 D N2 D 104 ; N3 D 10 (Reciprocal-lattice of a thin film layer) (3) N1 D 104 ; N2 D N3 D 10 (Reciprocal-lattice of a peace of string sample) Exercise 5.18 Calcium fluorite (CaF2 ) has a face-centered cubic lattice containing four (CaF2 ) per unit cell and the lattice parameter is a D 0:5463 nm. The atomic

270

7 Supplementary Problems

positions, 000 for Ca, 14 , 41 , 14 , 34 , 43 , 34 for F and other positions can be obtained by the face-centering translations. (1) Obtain the structure factor F. (2) Compute the values of the squares of the scattering factor F 2 for peaks corresponding to the planes 111 and 222 using the numerical values of f from Appendix A.3. (3) Compute the F 2 values including the anomalous dispersion factors f and f 00 for Ca.

Exercise 6.1 When the axes of the orthorhombic space group Pbcn are altered within the same unit cell, show the changes that occur to Hermann–Mauguin symbols. Repeat for the orthorhombic space group Cmca. Exercise 6.2 Show the coordinates of the equivalent positions when the symmetry operation is made for the general positions x, y, and z expressed by C2=c (International Tables: No.15, c-axis is taken on a unique axis) and P21 21 21 (International Tables: No.19). Exercise 6.3 If the (100) plane is set to a glide plane, explain the effect of the glide-reflection operation of c=2, with respect to the (100) plane, on the structure factor Fhkl and jFhkl j2 . In addition show the conditions which enable us to detect the diffraction intensity. Exercise 6.4 Explain the method for determining the screw axis including translational operation or a plane of gliding-reflection by applying the extinction law to diffraction peaks. Exercise 6.5 Rutile-type titanium dioxide (TiO2 ) belongs to the tetragonal system containing two TiO2 formula units per unit cell and the atomic positions are given as; Ti W

000

W

111 222

OW

uu0

W

u u0

W

11 1 uu C 2 22

W

uC

11 1 u 22 2

(1) Obtain the structure factor F by considering the equivalent lattice points. (2) Obtain the conditions for which F becomes zero without reference to the value of u. (3) Explain which of a simple tetragonal or body-centered tetragonal lattice is a Bravais lattice. Exercise 6.6 Supposing an incident beam is reflected from a plane h1 k1 l1 , it is possible to coincide this beam with the direction of the reflected beam from another plane h2 k2 l2 .

7 Supplementary Problems

271

(1) Show that the direction of the second reflection is equivalent to that due to direct reflection from a plane h3 k3 l3 such that: h3 D h1 C h2 ; k3 D k1 C k2 ; l3 D l1 C l2 N and .h2 k2 l2 / planes of a diamond crys(2) When double reflections from the .1N 11/ tal with a lattice parameter of a D 0:3567 nm are produced using Cu-K˛ . D 0:1542 nm) radiation, it is possible to obtain a so-called pseudo-reflection from the (222) plane. Identify such .h2 k2 l2 / planes. In addition, show the direction of the incident beam by the directional cosine S0 along three orthogonal axes a1 ; a2 and a3 . Exercise 6.7 Following data give the lattice parameters and the basis of the rutile structure of titanium dioxide (TiO2 ). Draw both the perspective drawing and the projection on the (x; y; 0) plane. In addition, show a stereographic projection. Lattice

Basic atom position

a

Tetragonal P Ti W 0; 0; 0 a D 0:459 nm

1 1 1 ; ; 2 2 2 O W 0:3; 0:3; 0

c D 0:296 nm

Atom positions

Space groups

Ti W 0; 0; 0 1 1 1 ; ; 2 2 2

P 42 =mnm

f

O W x; x; 0

1 2 1 0:2; 0:8; 2

1 1 1 C x; x; 2 2 2 x D 0:3 1 1 1 x; C x; 2 2 2

0:7; 0:7; 0

x; N x; N 0

0:8; 0:2;

Exercise 6.8 Pyrite is known to have NaCl type structure with four FeS2 formula units per unit cell. The lattice parameter is a D 0:5408 nm. In addition, the space group symmetry suggests the following atomic positions; 11 1 1 11 I 0 I 0 22 2 2 22 1 1 S W u; u; u I C u; u; uN I 2 2 1 1 uN ; uN ; uN I u; C u; uI 2 2 1 1 1 1 uN ; C u; u I C u; uN ; C u 2 2 2 2

Fe W 000 I 0

272

7 Supplementary Problems

1 1 1 1 uN ; u I C u I C u; u; u; 2 2 2 2 (1) Obtain the structure factor F for the unmixed h k l by excluding the anomalous dispersion factors. (2) For a powder sample of pyrite, the integrated intensities of the diffraction peaks for the (111) and (200) planes were found to be 69.2 and 277.5 in arbitrary units, respectively. Compute the value of u from the ratio of the integrated intensities of two peaks, when considering only the u-range between 0.30 and 0.40. N on a general pole Exercise 6.9 Show the rotatory inversion operation of 5N and 10 of the stereographic projection. Exercise 6.10 With respect to the space group P21 =c, which is frequently found in real crystals, information can be obtained from the International Tables for Crystallography, Volume A, page 184–185. Explain the key points. Exercise 6.11 Information on the space group Cmm2, which is frequently found in real crystals, can be obtained from the International Tables for Crystallography, Volume A, page 238–239. Explain the key points.

Chapter 8

Solutions to Supplementary Problems

Exercise 1.1 E D 1:602 1016 J; p D 1:708 1023 kg m/s; D 3:879 1011 m Exercise 1.2 D 68:62 .cm2 =g/; D 234:8 .cm2 =g/ GaAs BaTiO3 Exercise 1.3 32%.Cu K˛/; 54%.Fe K˛/ Exercise 1.4 5:95 g Exercise 1.5 VK .kV/ D

1:240 ; For Mo-K˛; VK D 20:01 kV K .nm/

Exercise 1.6 EKPb D 88 keV Exercise 1.7

1:510 108 m/s

Exercise 1.8 55:9 keV Exercise 1.9 Photon numberIK W 1:28 1011ŒmAs1 sr1 ; IntensityI W 1:24 103 .J=mAs sr/

273

274

8 Solutions to Supplementary Problems

Exercise 1.10 2:55 mm Exercise 1.11 0:14 mm Exercise 1.12 Work function 4:52 eV; Reverse voltage Vi W 7:88 Volt Exercise 1.13 1290 eV .L-shell/ and 48 eV .M-shell/ higher than the K-shell energy: Exercise 2.2 bcc W 1:06; fcc W 1:15 Exercise 2.3 Rhombuses obtained by drawing lines between the centers of regular hexagons. Exercise 2.4 1. r1 D 0:450 nm, r2 D 0:520 nm 2. n1 D 8, n2 D 6 3. Density : 0:923 106 g/m3 Exercise 2.5 1. 46:39 1030 m3 2. a D 0:3207 nm r1 D a Exercise 2.6 fcc : 0.7405, hcp : 0.7405, bcc : 0.6802, simple cubic : 0.5236 Exercise 2.7 See the coordination number of 6 in Question 2.15. The sphere radius of octahedron formed by six S2 ions is about 0.075 nm which is smaller than the ionic radius of Mg2C ion. Exercise 2.8 0.2864 and 0.4050 nm for (100) plane, 0.2864 nm for (111) plane. Exercise 2.9 Reduction percentage from 12 to 8: 26.8% and from 12 to 4: 77.5%

8 Solutions to Supplementary Problems

275

Exercise 2.10 Octahedral void: 0.155r, Tetrahedral void: 0.291r Exercise 2.11 1. Tetrahedral void in fcc: 0.107 nm (0.414r), Octahedral void in fcc: 0.058 nm (0.225r). Tetrahedral void in bcc: 0.072 nm (0.291r). Carbon atom (size: 0.140.15 nm) is found to be relatively easy to occupy the tetrahedral void in fcc structure. 2. Packing condition of hcp in the ideal state is equal to the fcc case. Exercise 2.12 1. 0:4621 106 (m3 /g), 27:01 106 (m3 /mol) 2. 0.2820 nm Exercise 2.13 1. 0:2519 106 (m3 /g), 42:41 106 (m3 /mol) 2. 0.3577 nm Exercise 2.14 1. a D 0:656 nm, Density D 2:80 106 g/m3 2. r C =r D 0:414 3. Ionic radius of RbC : 0.147 nm Exercise 2.15 1. 2. 3. 4.

4 D 109:48ı r C =r D 0:225 r C =r D 0:402 for ZnS. Since the radius of Zn2C cation is smaller than the maximum radius fitting into the vacant space formed by NaCl structure, the direct contact between anions of S2 is possible, so that the structure is unstable.

Exercise 2.16 Distance : 0.3571 nm, Density : 3:989 106 g/m3 Exercise 2.18 When applying the annealing process to metals with fcc structure after coldworking, we can frequently observe the twin crystal growth, called “annealing twins.” Note that “deformation twins” are also known in metals with bcc or hcp structure. In order to explain the formation of the annealing twin structure, the positions of about 67% (D2=3) of total atoms are unchanged and we take a uniform ¯ direction into account. This enables us shear motion of the (111) layer to the [112] to move each layer only the quantity proportional to the distance of twins.

276

8 Solutions to Supplementary Problems

Exercise 2.22 Three indices $ Four indices .hkl/ .HKiL/ N .100/ .1010/ N .010/ .0110/ .001/ .0001/ N N .020/ .0210/ N .110/ .1120/

Three indices $ Four indices Œuvw ŒU V tW N Œ100 Œ21N 10 N N Œ010 Œ1210 Œ001 Œ0001

Exercise 2.23 The relationship between (hkl) for rhombohedral structure and (HKiL) for the hexagonal case is H C K C L D 3k (k : integer). If .H C K C L/ is not given by integer multiple of 3, the system is considered to be hexagonal. Exercise 2.24 68ı Exercise 2.28 In cubic, the zone pole and the corresponding zone circle are at right angles to one another, so that the angle between the planes of two zone circles is equal to that formed by two poles of the corresponding zones. Exercise 3.1 Nm D 3:34 1028 molecules/m3, Ne D 3:34 1029 electrons/m3 Exercise 3.2 D 5:96 1030 m2 /sr. Estimate the differential cross section using the 30 2 e relation of d˝ D 2 sin d. d m /rad at (sin 45ı D 0:7071). d˝ D 2:65 10 2. The values of differential cross section per unit solid angle are obtained as a function of angle . Note that the value at 90ı is found half of the 180ı case.

1.

de d˝

.degree/ 0 30 90

cos2 1:0 0:750 0

de .degree/ d˝ 7:94 1030 100 6:95 1030 130 3:97 1030 180

cos2 0:030 0:413 1:0

de d˝ 4:09 1030 5:61 1030 7:94 1030

8 Solutions to Supplementary Problems

277

Exercise 3.3 If the number of electrons in water layer is set to Ne , the number of photons per unit area Nip due to the incoherent scattering is given as Nip D e Ne Nx D 6:95 109 m2 . Exercise 3.4 Intensity scattered from one electron : Ie D 7:95 1026 I0 , Intensity scattered from electrons of 1 g Mg : Ie0 D 0:024I0 , so that it is measurable. Exercise 3.5 (1) D 0:0112 nm,

(2) E D 110:7 keV

Exercise 3.6 E D 73:815 keV Exercise 3.7 D 0:002426.1 cos 2/ nm Exercise 3.8 ˚ and k D 4 sin = a D 0:53 A

1.

f D fe D

16 1 D 2 2 C 4/ f1 C .1:06 sin =/2 g

.a2 k 2

i.M / D 1 fe2 D 1

1 f1 C .1:06 sin =/2 g

2. sin =

Exercise 3.9 E D h D 71:7 keV Exercise 3.10 D 6:52ı

f

i.M /

0:0

1:0

0:2

0:48

0:77

0:4

0:13

0:98

4

278

8 Solutions to Supplementary Problems

Exercise 3.11 For (100) : 0:119 1010 m1 , For (110) : 0:168 1010 m1 and For (111) : 0:206 1010 m1 . Intensity is not detected if (hkl) is mixed. Exercise 3.12 Four indices : 111, 200, 220 and 222 are allowed. Exercise 3.13 .h C k/ F 2 D 4fU 2 1 C cos f1 C cos .4yk C l/g 2 F D 0; if .h C k/ is odd number. Exercise 3.14 n oi h F D fCa C 2fF cos .h C k C l/ 2 cos .h C k C l/ 1 F .f cc/ 2 2 hCkCl 4n 4n ˙ 1 4n ˙ 2

h; k; l Mixed Unmixed Unmixed Unmixed

F 0 4.fCa C 2fF / 4fCa 4.fCa 2fF /

Exercise 3.15 F D 0, when l D 2n C 1, h D 2k D 3n Exercise 3.16 jF100 j2 D .55 17/2 D 382 ;

jF111 j2 D .55 17/2 D 382

jF110 j2 D .55 C 17/2 D 722

Exercise 4.1 AD

sec es ts .1sec /

Exercise 4.2 T D ˙1:67K

8 Solutions to Supplementary Problems

279

Exercise 4.3 111 : 92.3%, 311 : 74.4%, 420 : 58.4% Exercise 4.4 I0 I P D D jF j2 p 2D tan 2 2D 0

1 C cos2 2 sin2 cos tan 2

Exercise 4.8 Note: Not only three d values, but also eight d values are used. When using the Hanawalt method, preference of d values is suggested in comparison to those of the relative intensity ratio. Exercise 4.9 a D 0:54305 nm (in average), a D 0:54302 nm (least-squares method). Exercise 4.10 0:34ı for t D 25 nm, 0:17ı for t D 50 nm, 0:10ı for t D 90 nm and 0:07ı for t D 120 nm. Exercise 4.11 93 nm (Hall method), 110 nm (least-squares method). Exercise 4.12 RMgO cCaO ICaO 106:0 0:97 107 D D D 1:03 cMgO IMgO RCaO 74:5 1:34 107 cMgO D 0:49 and cCaO D 0:51 Exercise 4.13 1:74 107 cCu ICu RSi 359:7 D D D 0:228 cSi ISi RCu 162:3 16:87 107 cSi D 0:81 and cCu D 0:19 Exercise 4.14 Br D 2 .FWHM/ D

0:9 0:9 !"D " cos Br cos

280

8 Solutions to Supplementary Problems

Milling time hour 1 2 4 6

2 degree 31.25 31.27 31.30 31.31

cos 0.9630 0.9630 0.9629 0.9629

( m) 0.095 0.031 0.026 0.017

Exercise 5.1 xD

nx1 C mx2 mCn

yD

ny1 C my2 mCn

Exercise 5.2 The angle formed by the straight lines q1 and q2 is given by cos ˛ D 1 2 C 1 2 . If these two lines are mutually perpendicular, we find cos ˛ D 0. From the relationships of 21 C 21 D 1 and sin2 ˛ C cos2 ˛ D 1, one obtains sin ˛ D .2 1 1 2 / 1 and tan ˛. The direction coefficients of two straight lines are expressed by m1 D 1 2 and m2 D 2 respectively, then one obtains; tan ˛ D

1 1

1C

2 2 1 2 1 2

D

m1 m2 1 C m1 m2

The required condition is given by 1 C m1 m2 D 0. Exercise 5.3 2ex ey 1 bc 1 D D ex ey a .b c/ 4 2 4 2ey 1 ca D D ey B D a .b c/ 4 2 A D

Exercise 5.4 ˇ ˇ jb1 j D ˇˇ

ˇ 1 1 a2 a3 a2 a3 ˇˇ D ! D a1 .a2 a3 / ˇ ja1 jja2 a3 j cos a1 cos d100

Exercise 5.5 b1 D

a2 a3 1 p D p 3; 1; 0 ; a1 a2 a3 3 3

b2 D

a3 a1 1 D p .0; 3; 0/ a1 a2 a3 3 3

8 Solutions to Supplementary Problems

281

Exercise 5.6 cos D

h2 a12

C

k2 a22

hu C kv C lw 12 1 l2 C a2 Œu2 a12 C v2 a22 C w2 a32 2 3

Exercise 5.7 For vector A, we obtain the following relations: A D m11 a C m12 b C m13 c .m11 a C m12 b C m13 c/ .ha C kb C lc / D m11 h C m12 k C m13 l D H Similarly, we obtain for vectors B and C and they are summarized as follows: 0 1 0 1 0 10 1 H h m11 m12 m13 h @ K A D @ k A D @ m21 m22 m23 A @ k A L l m31 m32 m33 l Exercise 5.8 The unit vectors of reciprocal lattices b1 , b2 and b3 are as follows: 9 1 a2 a3 > D .ex C ey ez / > > > > V a > = 1 a3 a1 D .ex C ey C ez / b2 D > V a > > > > 1 a1 a2 ; D .ex ey C ez / > b3 D V a

b1 D

The arbitrary reciprocal lattice vector Hpqr is given by the following: Hpqr D pb1 C qb2 C rb3 D Œ.p q C r/ex C .p C q r/ey C .p C q C r/ez Then we obtain the shortest non-zero vectors which consist of eight f111g planes of the equilateral hexagon and six f002g planes of the square. They give the first Brillouin zone of a face-centered cubic lattice and this result is known to correspond to the Wigner Seitz unit cell of a body-centered cubic lattice. Exercise 5.9 The scalar products of crystal lattice vector Rpqr and its reciprocal lattice vector Hhkl is known to be always an integer. If atoms in a crystal are located only at the lattice point Rpqr , a scatterer may be set with rn D Rpqr . Therefore, the condition which enables us to detect a diffraction peak with sufficient intensity is given by q D Hhkl . The absolute value of both sides of this equation is taken.

282

8 Solutions to Supplementary Problems

jqj D jHhkl j !

1 2 sin D dhkl

) 2dhkl sin D

.Bragg condition/

When considering the reciprocal lattice vectors and the scalar products of crystal lattice vectors, we readily obtain a formula of Laue condition, such as a1 .ss0 / D h. Exercise 5.10 In a sufficiently large real space lattice, the summation of the given formula is unchanged even if substituting r0n D rn C n for rn . Such particular relation is used. G.q/ D

X

e2iqrn ) .1 e2iqn /G.q/ D 0

n

Exercise 5.11 1 Discussion may be possible if using Taylor’s expansion of f .x/ D .1Cx 2 /2 and in the very small value of sin sin 1 , the atomic scattering factor fn can be 2 . approximated by fn 1 2 2an sin

Exercise 5.12 The anomalous dispersion terms are given by following equations (for details, refer to a monograph for the related subjects, such as R.W. James: Optical Principles of the Diffraction of X-rays, G. Bell & Sons, London (1954)). 1 f .!/ D 2 0

Z

dgoj d!jo

f 00 .!/ D

( !jo

1 2

Z

) !jo ! !jo C ! C d!jo 2 2 .!jo !/2 C oj =4 .!jo C !/2 C oj =4 dgoj d!jo

!jo

oj =2 d!jo 2 .!jo !/2 C oj =4

Where ! is the photon energy and its subscript denotes the state of photon such as the initial (o) and the j -th scattering process. oj shows the convoluted width of states of o and j and goj is the so-called characteristic oscillatory strength. There are some methods of computing a function of .dgoj =d!jo /, but the procedure of Cromer and Liberman’s scheme using the relativistic wave function is probably the best at the present time. Information of the anomalous dispersion terms including mass absorption coefficient of various elements in the wide energy region is available in the SCM-Database, URL: http://res.tagen.tohoku.ac.jp/˜waseda/scm/AXS/index.html

8 Solutions to Supplementary Problems

283

Exercise 5.13 Considering that is the angle between the vector showing the direction of propagation of the wave and z-axis and s0 is the unit vector of the incident wave, respectively, the diffracted intensity IP is given as follows. 2

IP D L

sin.L s0 sin / L s0 sin

2

Exercise 5.14 Considering the two-dimensional slit system arrayed repeatedly in a-period to x-direction and b-period to y-direction and the diffraction intensity is estimated if a small slit may be expressed by a wave vector sx sy of each direction. ˇ ˇ ˇ ˇ ˇ sin.msx a/ ˇ2 ˇ sin.nsy b/ ˇ2 ˇ ˇ ˇ ˇ I Dˇ sin.sx a/ ˇ ˇ sin.sy b/ ˇ Exercise 5.15 1. The diffraction intensity I is obtained by multiplying the amplitude of a scattered wave G and its complex conjugate.

I DG GD

sin2

˚1

m.a q/ ˚2

2 1 sin 2 .a q/

2. Setting to a q D 2h C ı.ı > 0/, the sine function of the numerator is taken into account. Under the condition of ı > 0, the minimum value is obtained in the m case of ı D . 2 Exercise 5.16 I.q/ D fC2 C 4fCl2 C 12fCl2

sin.2qrClCl / sin.2qrCCl / C 8fC fCl 2qrClCl 2qrCCl

Exercise 5.17 The form of a diffraction peak may be discussed with Laue function, because the scattering intensity is proportional to the square of its amplitude (see Question 5.9). Let us to set the number of a unit cell to N and the wave vector to Q. For example, a peak is found if Qa is given by integer multiple of 2 or if the relation of Qa D 2 n C 2=N (N is an integer) is satisfied, Qa becomes zero for the first time, so that the value of Q, which the Laue function becomes zero, is given by Q D .2=aN /. Therefore, with respect to the case of a D 0:25 nm, we obtain Q D 2:513 103 nm1 for N D 104 and Q D 2:513 nm1 for N D 10. For

284

8 Solutions to Supplementary Problems

discussion, we also include that the length of the reciprocal lattice vector is equal to the reciprocal of the spacing dhkl ; 4 nm1 and a D 0:25 nm in the present case. 1. Each reciprocal lattice point is located at the vertices of a cube with the side of 4 nm1 and each lattice point is very sharp of the order of 2:5 103 nm1 . 2. In this case, we may find the streaked reciprocal lattice extended to the direction of the N3 in a thin film by about 1,000 times in comparison to those of N1 and N2 . 3. This case corresponds to a narrow string-like sample. We may find a very sharp reciprocal lattice along the direction of the narrow string-like sample, whereas the reciprocal lattice is rather spread with the order of 2.513 nm1 in the plane perpendicular to the string-like sample. Exercise 5.18 1.

2.

n oi h F D fCa C 2fF cos .h C k C l/ 2 cos2 .h C k C l/ 1 2 2 f1 C cos .h C k/ C cos .h C l/ C cos .l C h/g

2 2 2 D 16fCa ; F222 D 16.fCa 2fF /2 F111

2 F111 D 16 .15:43/2 D 3809:4 .fCa D 15:43/

2 F222 D 16 .11:24 2 4:76/2 D 47:3 .fCa D 11:24; fF D 4:76/

3. If mixed, F D 0 and if unmixed, the structure factor is as follows. n o2 0 002 C2fF cos .hCk Cl/ 2 cos2 .hCk Cl/ 1 CfCa F 2 D 16 f0Ca C fCa 2 2 Exercise 6.1 Set the standard to abc, obtain the variation of Hermann–Morguin symbols when changing cab ! bca ! a¯cb ! ba¯c ! c¯ ba. Exercise 6.3 Fhkl D f e2i.kyClz/ .e2ihx C eil e2ihx / jFhkl j2 D 4f 2 cos2 2hx.for l D 2n/ jFhkl j2 D 4f 2 sin2 2hx.for l D 2n C 1/ With respect to the 0kl peaks, the intensity is detected only when l D 2n. Exercise 6.4 For example, let us consider the case where there is the 21 -screw axis parallel to b-axis through the origin. In this case, if an atom is found in (x, y, z), there is

8 Solutions to Supplementary Problems

285

1 necessarily an atom in x; y C ; z . In this case the structure factor is given 2 in the following. F .hkl/ D

N=2 X

˚

fi exp 2i.hxj C kyj C lzj /

j D1

k Cexp 2i hxj C kyj C lzj C 2

When considering that both h and l are zero, one obtains; F .0k0/ D

N=2 X

fi exp.2ikyj / C f1 C exp.ik/g

j D1

The extinction conditions are given as follows. N=2 X

F .0k0/ D

j D1

F .0k0/ D 0

fi exp.2ikyj /

k D 2n

9 > > =

> > k D 2n C 1 ;

Thus, with respect to the 21 -screw axis, we can not detect the diffraction intensity from the 0k0 peak where k is odd number and it corresponds to the direction of a screw axis. Discussion is possible for other screw axes, along the way similar to the 21 -screw axis case. Similarly, the extinction condition appears for a plane of glide reflection. For example, when there is a c-glide plane perpendicular to b-axis, if there is an atom at .x; y; z/, we always find an atom at x; y; z C 12 . F .h0l/ D 2 F .h0l/ D 0

N=2 X j D1

˚

fi exp 2i.kyj C lzj /

9 > > l D 2n = > > ; l D 2n C 1

Thus, the diffraction intensity is not detected when the index for giving the direction perpendicular to a glide plane is zero and the index for the direction of a translation axis is odd number. Exercise 6.5 1.

h 2i hCkCl 2 C fO e2i.uhCuk/ C e2i.uhuk/ F D fTi 1 C e 2i hCkCl uhCuk 2i hCkCl Cuhuk 2 2 Ce Ce

286

2.

8 Solutions to Supplementary Problems

hCkCl hCkCl i sin 2 F D fTi 1 C cos 2 2 2 hCkCl C fO 2 cos 2u.h C k/ C cos 2 2 hCkCl C i sin 2 2 cos 2u.h C k/ 2 (i) h C k C l D 2n F D fTi C fO Œ2 cos 2u.h C k/ C 2 cos 2u.h k/ 6D 0 (ii) h C k C l D 2n C 1 F D fO Œ2 cos 2u.h C k/ 2 cos 2u.h k/ 6D 0

However, it should be kept in mind that F D 0 in the case of hCk Cl D 2nC1 with h or k is zero. 3. The structure factors can be estimated from the given atomic positions. (i) h C k C l D 2n n oi h F D 2 fTi C fO e2i.uhCuk/ C e2i.uhuk/ (ii) h C k C l D 2n C 1

F D0

In conclusion, Bravais lattice is considered to be body-centered tetragonal. Exercise 6.6 1. For example, Bragg equation of the plane h1 k1 l1 is as follows. S1 S0 D h1 b1 C k1 b2 C l1 b3 2. According to the results obtained in (1), 2 D 1 C h2 ; 2 D 1 C k2 ; 2 D 1 C l2 ! .h2 k2 l2 / D .331/ N (331) and (222) In the orthogonal-axes, a1 a2 and a3 , the formulas of .1N 11/, N N planes are given as follows. For .111/ plane; a2 a3 a1 C C D1 a a a

) a1 C a2 a3 C a D 0

Similarly, for (331) plane; 3a1 C 3a2 C a3 a D 0 and for (222) plane; 2a1 C 2a2 C a3 a D 0.

8 Solutions to Supplementary Problems

287

The direction cosine of S0 is set as x; y; z, the formula of S0 is given as follows. a2 a3 a1 C C ; ) x 2 C y 2 C z2 D 1 x y z N plane is set as 1 , the following relation When the angle formed by S0 and .1N 11/ is readily obtained. xCyz sin 1 D p 2 1 C 12 C .1/2

and 2d.1N 11/ N sin 1 D

xCyzD

3 2a

6 a 3 9 Then one obtains; z D , x C y D 4a 4a Using the values of a D 0:3567 nm, D 0:1542 nm and x 2 C y 2 C z2 D 1, .x; y; z/ can be estimated. For S0 and (222);

2x C 2y C 2z D

.x; y; z/ D .0:946; 0:027; 0:324/ .0:027; 0:946; 0:324/ Exercise 6.8 1.

2i hCk 2i kCl 2i hCl 2 2 2 F D fFe 1 C e Ce Ce 2i hCk 2 C 2fS e2iuŒhCkCl C e e2iuŒhkl Ce

2i kCl 2

e2iuŒhCkl C e

2i hCl 2

e2iuŒhkCl

For the unmixed case, F D 4Fe C 2fS f2 cos 2ul cos 2u.h C k/ C 2 cos 2ul cos 2u.h k/g 2. The area of a diffraction peak A is proportional to the intensity per unit length P 0 which is given by the multiplicity factor m and the structure factor F . Then the following relation is obtained. A200 D A111

0 P200 0 P111

D

m200 m111

F200 4fFe C8fS F111 4fFe C8fS

2 2

288

8 Solutions to Supplementary Problems

The structure factors are given as follows, F200 D 4fFe C 8fS cos 4u F111 D 4fFe C 4fS cos 2uŒcos 4u C 1 For cubic systems m111 D 8 and m200 D 6. Estimate the atomic scattering factors of f .S/ and f .Fe/ for two peaks of 200 and 111 from Appendix A.3 and the lattice parameter. f .Fe/111 D 20:6

f .S/111 D 12:0

f .Fe/200 D 19:8

f .S/200 D 11:6

If we use the given values of A111 D 69:2 and A200 D 277:5 as well as m111 =m200 D 1:33, we obtain u D 0:387.

Appendix A

A.1 Fundamental Units and Some Physical Constants SI: LeSyst`emac Internation d’Unit`es Seven SI base units

Derived SI units with a specific name

Time: minute and hour, Plane angle: degree, minute, second, Volume: liter and Mass: metric ton. These units are non-SI units, but they are accepted for use with the SI units.

289

290 Physical constants

a b

One twelfth of mass of 12 C: Temperature 273.15 K, Pressure 101325 Pa(1 atm).

Units frequently used with SI units

a

˚ is also used in comparison to the electric current A. A

A Appendix

A.2 Atomic Weight, Density, Debye Temperature and Mass Absorption Coefficients

A.2 Atomic Weight, Density, Debye Temperature and Mass Absorption Coefficients (cm2 =g) for Elements

Å

Θ (K)

Å

Θ (K)

Å

Θ (K)

: Debye temperature, Unit of density: Mg/m3 :

291

292

A Appendix Å

Θ (K)

Å

Θ (K)

Å

Θ (K)

: Debye temperature, Unit of density: Mg/m3 :

A.2 Atomic Weight, Density, Debye Temperature and Mass Absorption Coefficients Å

Θ (K)

Å

Θ (K)

Å

Θ (K)

: Debye temperature, Unit of density: Mg/m3 :

293

294

A Appendix Å

Θ (K)

(Å)

Θ (K)

Å

Θ (K)

: Debye temperature, Unit of density: Mg/m3 :

A.3 Atomic Scattering Factors as a Function of sin =

295

A.3 Atomic Scattering Factors as a Function of sin =

(Continued)

296

A Appendix

(Continued)

A.3 Atomic Scattering Factors as a Function of sin =

297

298

A.4 Quadratic Forms of Miller Indices for Cubic and Hexagonal Systems

A Appendix

A.5 Volume and Interplanar Angles of a Unit Cell

299

A.5 Volume and Interplanar Angles of a Unit Cell Cellvolumes Cubic : V D a3 Tetragonal :V D a2 c Hexagonal : V D

p

3a2 c 2

p 3

D 0:866a2 c

Trigonal : V D a 1 3 cos2 ˛ C 2 cos3 ˛ Orthorhombic : V D abc Monoclinic : V D abc sin ˇ p Triclinic : V D abc 1 cos2 ˛ cos2 ˇ cos2 C 2 cos ˛ cos ˇ cos Interplanar angles The angle between the plane .h1 k1 l1 / of spacing d1 and the plane .h2 k2 l2 / of d2 is estimated from the following equation, where V is the volume of a unit cell: Cubic : cos D q

h1 h2 C k1 k2 C l1 l2 .h1 2 C k1 2 C l1 2 /.h2 2 C k2 2 C l2 2 /

Tetragonal : cos D r

Hexagonal : cos D q

Trigonal : cos D

h1 h2 Ck1 k2 a2 h1 2 Ck1 2 a2

l1 l2 c2

h2 2 Ck2 2 a2

l1 c2

C

l2 2 c2

h1 h2 C k1 k2 C 12 .h1 k2 C h2 k1 / C

3a2 l l 4c 2 1 2

2

.h1 2 Ck1 2 Ch1 k1 C 3a l 2 /.h2 2 Ck2 2 C h2 k2 C 4c 2 1

3a2 2 l / 4c 2 2

a 4 d1 d2 Œsin2 ˛.h1 h2 C k1 k2 C l1 l2 / V2 C .cos2 ˛ cos ˛/.k1 l2 C k2 l1 C l1 h2 C l2 h1 C h1 k2 C h2 k1 /

Orthorhombic : cos D r

d1 d2 Monoclinic : cos D sin2 ˇ Triclinic : cos D

C

C 2

h1 h2 a2 h1 2 a2

"

C

k1 2 b2

C

C

k1 k2 b2

l1 2 c2

C

l1 l2 c2

h2 2 a2

C

k2 2 b2

C

l2 2 c2

h1 h2 k1 k2 sin2 ˇ l1 l2 .l1 h2 C l2 h1 / cos ˇ C C 2 a2 b2 c ac

#

d1 d2 ŒS11 h1 h2 C S22 k1 k2 C S33 l1 l2 V2 C S23 .k1 l2 C k2 l1 / C S13 .l1 h2 C l2 h1 / C S12 .h1 k2 C h2 k1 /

S11 D b 2 c 2 sin2 ˛ S22 D a2 c 2 sin2 ˇ S33 D a2 b 2 sin2

S12 D abc 2 .cos ˛ cos ˇ cos / S23 D a2 bc.cos ˇ cos cos ˛/ S13 D ab 2 c.cos cos ˛ cos ˇ/

300

A Appendix

A.6 Numerical Values for Calculation of the Temperature Factor Values of .x/ D

1 x

Z

x 0

e

d 1

xD

, : Debye temperature T

For x lager than 7, .x/ values are approximated by .1:642=x/. Debye temperatures are compiled in Appendix A.2 using the following reference: (C.Kittel: Introduction to Solid State Physics, 6th Edition, John Wiley & Sons, New York (1986), p.110.)

A.7 Fundamentals of Least-Squares Analysis

301

A.7 Fundamentals of Least-Squares Analysis Let us consider that the number of n-points have coordinates(x1; y1 ), (x2 ; y2 ) (xn ; yn ), and the x and y are related by a straight line with the form of y D a C bx. Our problem is to find the best value of a and b which makes the sum of the squared errors a minimum by using the least-squares method. In this case, we use the following two normal equations: X

X aCb x; X X X xy D a xCb x2 : yD

X

(1) (2)

For given n-points, the following four steps are suggested: (i) Substitute the experimental data into y D a C bx for obtaining n-equations. 9 y1 D a C bx1 > > > y2 D a C bx2 = : :: > : > > ; yn D a C bxn

(3)

(ii) Multiply each equation by the coefficient of a (1 in the present case) and add for obtaining the first normal equation. y1 D a C bx1 ; y2 D a C bx2 ; :: : yn D a C bxn ; n X

yD

X

aCb

X

x:

(4)

(iii) Multiply each equation by the coefficient b and add for obtaining the second normal equation. x1 y1 D x1 a C bx1 2 ; x2 y2 D x2 a C bx2 2 ; :: : xn yn D xn a C bxn 2 n X

xy D a

X

xCb

X

x2

(5)

(iv) Simultaneous solution of the two equations of (4) and (5) yields the value of a and b.

302

A.8 Prefixes to Unit and Greek Alphabet

Greek alphabet

A Appendix

A.9 Crystal Structures of Some Elements and Compounds

A.9 Crystal Structures of Some Elements and Compounds

These data are taken from the following references. B.D. Cullity: Elements of X-ray Diffraction (2nd Edition), Addison-Wesley (1978). F.S. Galasso: Structure and Properties of Inorganic Solids, PergamonPress (1970).

303

Index

Abscissa, 76 Absorption edge, 173, 254 Absorption factor, 261 Accelerating voltage, 3 a-glide plane, 229 Air, 253 Aluminum, 155 Amplitude, 78 Angular coordinates, 62 Angular dispersion, 108 Angular width, 124 Anomalous dispersion, 173 Anomalous scattering, 269 Aperture, 204, 269 Aperture width, 197 Applied voltage, 254 Asymmetric unit, 226, 250 Atomic scattering factor, 71, 78, 95, 173, 295 Attenuation term, 269 Auxiliary points, 220 Average mass absorption coefficients, 122 Axial ratios, 116

Back-reflection, 151 Barium titanate, 59 Barn, 81 Black lead, 255 Body-centered cubic, 187 Bohr radius, 96 Bonding, 24 Bragg angle, 176 Bragg condition, 74, 116 Bragg law, 74, 78 Bravais lattice, 23, 79, 171, 226 Brillouin zone, 186, 268

Caesium chloride, 55, 256, 257

Calcium fluorite, 260, 269 Calcium oxide, 161 Calibration curve, 122, 158, 162 Carbon dioxide, 118 Carbon solubility, 256 Carbon tetrachloride, 269 Cellvolumes, 299 Centered lattices, 243 Center of symmetry, 219 Centrosymmetric, 237 Characteristic radiation, 172 Circular arcs, 33 Classical electron radius, 81, 173, 259 Clockwise, 229 Coaxial cone, 80 Coherent scattering, 70, 172 Cold-work, 265 Columns matrices, 242 Common quotient, 135 Complex conjugate, 78, 89, 199 Complex exponential function, 76, 89 Complex number, 76, 88 Complex plane, 76 Components of symmetry element, 231 Compound symmetry operation, 220 Compton equation, 85 Compton scattering, 68, 172 Compton shift, 85, 87 Compton wavelength, 69, 259 Conservation of momentum, 84 Constructive interference, 176, 189, 190 Conversion relationships, 241 Coordinates, 28, 231, 242 Coordinate triplets, 251 Coordination number, 255 Coordination polyhedra, 255 Coplanar, 74 Copper, 261 Counterclockwise, 88, 229

305

306 Cristobalite, 51 Cross product, 180 Cross-sectional area, 153 Crystal imperfection, 175 Crystal lattice, 169 Crystallinity, 158 Crystallites, 121, 123, 166, 266 Crystal monochromator, 109, 208 Crystal orientation, 110 Crystallographic forms, 238 Cubic, 115, 172, 185 Cuprous chloride, 260 Cuprous oxide, 143

De Broglie relation, 1, 253 Debye approximation, 128 Debye characteristic temperature, 129 Debye formula, 207 Debye ring, 178, 217 Debye-Scherrer camera, 151, 262 Debye’s equation, 269 Debye temperature, 291 Debye-Waller factor, 113, 128 Delta function, 95, 175, 205 Denominator, 199, 213 Density, 291 Destructive interference, 91, 124, 175 Determinants, 182 d -glide plane, 230 Diagonal direction, 223 Diamond, 101, 104, 114 Diamond glide plane, 223 Dielectric constant of vacuum, 81 Diffraction angle, 74 Diffractometer, 107, 208 Dihedral angle, 33 Dimensionless entity, 231 Dipole approximation, 174 Direct comparison method, 153 Direct contact, 256 Directional coefficient, 267 Directional cosine, 267, 271 Directions of a form, 28 Disordered phase, 211 Distortion, 126 Divergent slit, 108 Double angle of the cosine formula, 100

Eccentric, 119 Edge-line, 103 Effective element number, 18 Eight symmetry elements, 223, 232

Index Einstein relation, 82, 83, 85 Electric capacity, 81 Electric field, 90 Electron charge, 81 Electron radius, 68 Electron rest mass, 81 Electron unit, 70, 72 Electron wave functions, 71 Eleven screw axes, 235 Ellipses, 237 Elliptic polarization, 202 Elliptic polarized wave, 202 Ellipticity, 204 Equatorial circle, 32 Equatorial plane, 32, 238 Equivalent positions, 236, 238, 270 Ewald sphere, 178, 191 Excitation voltage, 4, 253 Exponent portion, 189 Exponential functions, 199 Extinction law, 270 Extrapolation, 120

Face center, 105 Face-centered cubic, 187 Face-centering translations, 104 Film shrinkage, 151 Filter, 109 First Brillouin zone, 187 First-order reflection, 74 Five translational operations, 235 Fluorescent, 118 Focusing geometry, 107 Forbidden, 109 Form factor, 71 Forward direction, 243 Four-index system, 258 Fourier transform, 97 Fractional coordinates, 76, 78 Fractional error, 145, 151 Fraunhofer diffraction, 269 Free electron, 259 Full space group symbol, 224 FWHM, 165

gamma-phase, 256 General position, 226, 246, 270 Generators, 250 Geocentric angle, 64, 258 Geometrical entity, 231 Geometric progression, 198 Glide plane, 221, 228

Index Glide-reflection, 270 Global maximum, 214 Gold, 41 Goniometer center, 119 Grain size, 123 Great circles, 33 Grid net, 62

Half apex angle, 178 Half maximum ordinate, 215 Half value layer, 254 Hall method, 126, 166 Hanawalt method, 118, 139 Harmonic oscillator, 269 Hematite, 17 Hermann-Mauguin symbols, 224, 233 Hexagonal, 116, 134, 172, 255, 258, 261 Hexagonal close-packed, 99, 186 Hexagonal prism, 188 Highest symmetry, 235 Horizontal dispersion, 108 Huge comma, 240 Huygens principle, 193 Hydrogen, 97, 259

Ice, 57 Ideally mosaic, 110 Ideally perfect, 110 Ideal mosaic single crystal, 178 Imaginary number, 88 In phase, 73, 93, 108 Incoherent scattering, 69, 172, 259 Incoherent scattering intensity, 72 Indexing, 120 Individual symmetry operations, 220 Infinitesimally thin layer, 111 Infinite thickness, 111, 179 Inhomogeneous strain, 126, 166 Inner shell electron, 172 Instrumental broadening factor, 126, 165 Integer multiple, 73, 175, 189, 200 Integral width, 124, 126 Integrand, 97, 98 Integrated intensity, 110, 153, 176, 272 Inter-axial angles, 21, 223 Interference effect, 94 Intermediate state, 177 Intermetallic compounds, 58 Internal standard method, 120, 158 International Tables, 227, 272 Interplanar angles, 299 Interplanar spacing, 75

307 Interstitial, 46 Intra-molecular correlations, 207 Inverse Fourier transform, 97 Inverse matrices, 242 Inversion, 23, 219 Inversion center, 219 Ionic crystals, 25 Isotropic distribution, 205 IUCr, 224, 233

JCPDS cards, 118, 140

K˛ doublet, 148 Kirchhoff theory, 193, 197

Latitude circles, 33 Latitude lines, 33 Lattice parameters, 120 Lattice plane, 26 Lattice strain, 164 Lattice symbol, 226 Lattice translation, 219 Laue equation, 268 Laue function, 175, 199, 213 Laue groups, 237 Laue photographs, 170, 237 Law of conservation of energy, 84 Law of conservation of momentum, 253 Law of cosines, 84 Least-squares method, 162, 301 Limiting sphere, 178, 191 Limit value, 200 Linear absorption coefficient, 5, 111 Linearity, 162 Lithium niobate, 16 Locus, 231 Longitude lines, 33 Lorentz factor, 110 Lorentz-polarization factor, 113, 127 Lorentz transformation, 2

Magnesium, 255, 259 Magnesium oxide, 137, 150, 161, 164, 260 Mass absorption coefficient, 5, 153, 291 Mean square displacement, 113 Mechanical grinding, 124 Melilite, 266 Meridian circles, 33 Method of Nelson-Riley, 120 Middle-point of full width, 149

308 Miller-Bravais indices, 28, 48 Miller indices, 26, 170, 242 Mirror plane, 220, 221 Missing reflections, 79 Modified scattering, 172 Modulation, 194 Momentum of a photon, 83 Monoclinic, 234, 243, 245 Monoclinic sub-groups, 236 Mosaic structure, 175 Moseley’s law, 3, 12, 254 Multiplicity factor, 27, 110, 250

Nearest-neighbor, 40, 255 n-glide plane, 223, 230 Nodal line, 183 Non-centrosymmetric, 211 Non-parallel planes, 65 Nonprimitive, 171 Non-uniform strain, 126 Normal equations, 301 Normal planes, 187 Normal vector, 182 Numerator, 199, 213

Octahedral void, 43 Octants, 257 Odd multiple, 106 Off-centering, 151 One-dimensional lattice, 269 One unit cycle, 250 Optical path difference, 70, 189 Optical path distance, 94 Ordered phase, 211 Ordinate, 76 Organic molecules, 224 Origin, 249 Orthogonal, 201, 267, 271 Orthorhombic, 185, 240, 268, 270 Outer electron shell, 172 Out of phase, 91 Oval, 202

Packing fraction, 35, 255 Para-focussing, 108 Parallelepiped, 180 Parallelogram, 180, 183 Parallelogram law, 76 Parallel translation, 221 Partial integration, 98 Partial interference, 214

Index Particle size, 119, 123, 126 Particular orientation, 226 Path difference, 74, 76, 92 Path length, 73 Patterson symmetry symbol, 224 Peak broadening, 123 Peak splitting, 264 Peak width, 124, 215 Perfect crystal, 177 Periodic sequence, 235 Permittivity of free space, 81 Perovskite, 58 ı-phase, 256 Phase, 78 Phase difference, 73, 76, 91, 92, 175, 214 Photoelectric absorption, 5, 253 Photoelectron, 11, 254 Plane angle, 80 Plane groups, 252 Plane-polarized, 201 Planes of a zone, 30 Plane spacing, 114, 126 Plane spacing equation, 131 Plane wave approximation, 174 Point group, 221, 223 Polar coordinates, 205 Polar net, 33 Polarization factor, 68, 111 Poles, 31 Porosity, 35 Position vector, 174 Potassium chloride, 138 Potassium halide, 256 Power-series expansions, 78, 88 Precision measurements, 121 Preferred orientation, 118 Primitive, 171, 226 Projection direction, 249 Projection sphere, 31 Pseudo-reflection, 271 Pyrite, 271

Quantitative analysis, 153 Quarter turn, 241 Quotient, 129

Rachinger method, 149 Real number, 88 Real space lattice, 169 Receiving slit, 108 Reciprocal lattice, 169, 238 Reciprocal of a spacing, 267

Index Reciprocal of the volume, 181 Reciprocal space, 97, 242 Recoil angle, 86, 260 Recoil electron, 85, 86 Recoil of atom, 11 Recoil phenomenon, 68 Rectangular, 267 Reference sphere, 31 Reference substance, 123 Reflection, 23, 219 Regular tetrahedron, 50 Relative intensity ratio, 114 Relativistic, 82 Repeated-reproducibility, 148 Rhombic dodecahedron, 187 Rhombohedral, 258, 268 Right-handed coordinate, 249 Right-handed screw, 221 Rotation, 23, 219 Rotatory inversion, 22, 219 Roto-inversion, 22, 219 Rowland circle, 108 Rubidium halide, 256 Rydberg constant, 4, 13

Scalar product, 180 Scattering amplitude, 70 Scattering coefficient, 14 Scattering phase shift, 94 Scattering slit, 108 Scattering vector, 70, 94, 107, 207 Scherrer’s equation, 125 Sch¨onflies symbols, 224, 233 Screw axis, 221, 228 Screw rotation, 221 Search manual, 118 Self-coincidence, 228 Semi-infinite, 112 Shortest non-zero vectors, 188 Shortest wavelength limit, 8 Short space group symbol, 224 Silicon, 264 Site symmetry, 250 Small circles, 33 Solid angle, 80 Space groups, 223, 234 Space group symbols, 236 Space-group tables, 248 Space lattice, 232 Spatial resolution, 108 Special positions, 226, 246 Speed of light, 82 Spherical polar coordinates, 95

309 Spherical symmetry, 70, 173 Spherical wave, 193 Square of the amplitude, 89 Standard projection, 34 Standard substance, 162 Starting point, 191 Stereographic projection, 31, 258 Storm formula, 254 Strain, 126 Structure factor, 78 Sub-grains, 176 Substitutional, 46 Substructure, 175 Super-lattice, 47 Superposition of waves, 91 Symmetry element, 219, 223, 231, 238 Symmetry operation, 219 Symmetry-transmission method, 218, 261

Temperature factor, 129 Terminal point, 191 Terrestrial globe, 31 Tetragonal, 116 Tetragonal dipyramid, 258 Tetragonal pyramid, 258 Tetrahedral void, 43 Theorem of conformal mapping, 34 Theorem of corresponding circle to circle, 34 Thermal expansion coefficient, 261 Thermal vibration, 113, 128 Thomson equation, 69, 173 Titanium carbide, 256 Titanium dioxide, 270 Titanium hydride, 256 Total diffraction intensity, 108 Translational operation, 221, 270 Translational positions, 105 Transmission rate, 254 Transmittivity, 6 Trigonal, 268 Trigonometric function, 91, 100, 199, 246 Triple-scalar product, 169, 180 Tungsten, 147, 254 Twin, 257 Two-dimensional lattice, 267, 269 Two-dimensional space groups, 251

Ultra fine particles, 264 Uncertainty, 120 Unit lattice translation, 228 Unmodified scattering, 172 Unpolarized incident X-ray beam, 72

310 Uranium, 260 Vector product, 180, 187 Volume fraction, 153 Volume of unit cell, 156 Wave function, 98 Wave number vector, 94 Wave-particle duality, 171 Wave vector, 70, 96, 191 Weight fraction, 122, 160 Weight ratio, 122 Weiss rule, 65

Index Weiss zone law, 183 Wigner-Seitz cell, 187 Work function, 254 Wulff net, 33 Wyckoff letter, 226 Wyckoff position, 251

X-ray analysis, 118

Zinc blende, 104, 210, 256, 260 Zinc sulfide, 257 Zone axis, 30, 66, 183