Lecture 5: Crystal planes and Miller Indices

Similar documents
1.3 SCALARS AND VECTORS

2. There are an infinite number of possible triangles, all similar, with three given angles whose sum is 180.

GM1 Consolidation Worksheet

April 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then.

THREE DIMENSIONAL GEOMETRY

H (2a, a) (u 2a) 2 (E) Show that u v 4a. Explain why this implies that u v 4a, with equality if and only u a if u v 2a.

m m m m m m m m P m P m ( ) m m P( ) ( ). The o-ordinte of the point P( ) dividing the line segment joining the two points ( ) nd ( ) eternll in the r

( ) { } [ ] { } [ ) { } ( ] { }

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

QUB XRD Course. The crystalline state. The Crystalline State

m A 1 1 A ! and AC 6

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point

Comparing the Pre-image and Image of a Dilation

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 H 3 CH3 C. NMR spectroscopy. Different types of NMR

A Study on the Properties of Rational Triangles

VECTOR ALGEBRA. Syllabus :

ENERGY AND PACKING. Outline: MATERIALS AND PACKING. Crystal Structure

6.3.2 Spectroscopy. N Goalby chemrevise.org 1 NO 2 CH 3. CH 3 C a. NMR spectroscopy. Different types of NMR

Bravais lattices and crystal systems

QUADRATIC EQUATION. Contents

Intermediate Math Circles Wednesday 17 October 2012 Geometry II: Side Lengths

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

Section 1.3 Triangles

Introduction to Olympiad Inequalities

Proving the Pythagorean Theorem

Vectors. Chapter14. Syllabus reference: 4.1, 4.2, 4.5 Contents:

PYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS

912 o C 1400 o C 1539 o C α iron γ iron δ iron. liquid iron BCC FCC BCC

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

Chapter 8 Roots and Radicals

Polynomials. Polynomials. Curriculum Ready ACMNA:

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

Miller indices and Family of the Planes

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1. 1 [(y ) 2 + yy + y 2 ] dx,

LESSON 11: TRIANGLE FORMULAE

5. Every rational number have either terminating or repeating (recurring) decimal representation.

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

SECTION A STUDENT MATERIAL. Part 1. What and Why.?

Translation symmetry, Space groups, Bloch functions, Fermi energy

Probability. b a b. a b 32.

CHENG Chun Chor Litwin The Hong Kong Institute of Education

SECOND HARMONIC GENERATION OF Bi 4 Ti 3 O 12 FILMS

Particle Physics. Michaelmas Term 2011 Prof Mark Thomson. Handout 3 : Interaction by Particle Exchange and QED. Recap

INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable

THE PYTHAGOREAN THEOREM

Lecture 1 - Introduction and Basic Facts about PDEs

UNIVERSITY OF OSLO. Faculty of Mathematics and Natural Sciences

Trigonometry and Constructive Geometry

Maintaining Mathematical Proficiency

Exercise sheet 6: Solutions

Computing data with spreadsheets. Enter the following into the corresponding cells: A1: n B1: triangle C1: sqrt

Polyphase Systems. Objectives 23.1 INTRODUCTION

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES

3.15 NMR spectroscopy Different types of NMR There are two main types of NMR 1. C 13 NMR 2. H (proton) NMR

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e

Thermodynamics. Question 1. Question 2. Question 3 3/10/2010. Practice Questions PV TR PV T R

Reflection Property of a Hyperbola

H 4 H 8 N 2. Example 1 A compound is found to have an accurate relative formula mass of It is thought to be either CH 3.

Linear Algebra Introduction

Light and Optics Propagation of light Electromagnetic waves (light) in vacuum and matter Reflection and refraction of light Huygens principle

f (x)dx = f(b) f(a). a b f (x)dx is the limit of sums

Polyphase Systems 22.1 INTRODUCTION

Spacetime and the Quantum World Questions Fall 2010

2.1 ANGLES AND THEIR MEASURE. y I

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx

Appendix C Partial discharges. 1. Relationship Between Measured and Actual Discharge Quantities

NON-DETERMINISTIC FSA

Symmetrical Components 1

y z A left-handed system can be rotated to look like the following. z

Similar Right Triangles

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL:

where the box contains a finite number of gates from the given collection. Examples of gates that are commonly used are the following: a b

] dx (3) = [15x] 2 0

CALCULATED POWDER X-RAY DIFFRACTION LINE PROFILES VIA ABSORPTION

8 THREE PHASE A.C. CIRCUITS

Activities. 4.1 Pythagoras' Theorem 4.2 Spirals 4.3 Clinometers 4.4 Radar 4.5 Posting Parcels 4.6 Interlocking Pipes 4.7 Sine Rule Notes and Solutions

Section 6: Area, Volume, and Average Value

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)

Something found at a salad bar

Algorithm Design and Analysis

Electromagnetism Notes, NYU Spring 2018

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

PROPERTIES OF TRIANGLES

Review of Gaussian Quadrature method

Factorising FACTORISING.

Two Triads of Congruent Circles from Reflections

DEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

Math Lesson 4-5 The Law of Cosines

Can one hear the shape of a drum?

Trigonometry Revision Sheet Q5 of Paper 2

Mathematics. Area under Curve.

The Ellipse. is larger than the other.

System Validation (IN4387) November 2, 2012, 14:00-17:00

Physics 505 Homework No. 11 Solutions S11-1

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

Transcription:

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Leture 5: Crystl plnes nd Miller Indies Index system for rystl diretions nd plnes Crystl diretions: Any lttie vetor n e written s tht given y Eq.(.). The diretion is then speified y the three integers [nnn3]. If the numers nnn3 hve ommon ftor, this ftor is removed. For exmple, [] is used rther thn [], or [00], rther thn [400]. When we spek out diretions, we men whole set of prllel lines, whih re equivlent due to trnsntionl symmetry. Opposite orienttion is denoted y the negtive sign over numer. For exmple: Crystl plnes: The orienttion of plne in lttie is speified y Miller indies. They re defined s follows. We find interept of the plne with the xes long the primitive trnsltion vetors, nd 3. Let s these interepts e x, y, nd z, so tht x is frtionl multiple of, y is frtionl multiple of nd z is frtionl multiple of 3. Therefore we n mesure x, y, nd z in units, nd 3 respetively. We hve then triplet of integers (x y z). Then we invert it (/x /y /z) nd redue this set to similr one hving the smllest integers y multiplying y ommon ftor. This set is lled Miller indies of the plne (hkl). For exmple, if the plne interepts x, y, nd z in points, 3, nd, the index of this plne will e (33). The orienttion of rystl plne is determined y three points in the plne, provided they re not olliner. If eh point ly on different rystl xis, the plne ould e speified y giving the oordintes of the points in terms of the lttie onstnts,,. A nottion onventionlly used to desrie lttie points (sites), diretions nd plnes is known s Miller Indies. A rystl lttie my e onsidered s n ssemly of equidistnt prllel plnes pssing through the lttie points nd re lled lttie plnes. In order to speify the orienttion one employs the so lled Miller indies. For simpliity, let us strt with two dimensionl lttie nd then generlized to three dimensionl se. 7

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 The eqution of plne in -D nd 3D hving the interepts, nd,, respetively re x y x y z nd Crystl diretion is the diretion (line) of xes or line from the origin nd denoted s [], [00], [00] et. How to find Miller Indies: To determine the indies for the plne p in Figure, -first we hve to find the interepts with the xes long the sis vetor x y z e x, y, z. We form the frtionl triplet,,. -Tke reiprol to this set.,,. Let these interepts -Then redue this set to similr one hving the smllest integers multiplying y ommon ftor. To determine the Miller indies: (i) Find the interepts on the xes long the sis vetor,, in terms of the lttie onstnts, nd. The xes my e those of primitive or nonprimitive ell. x y z Let these interepts e x, y, z. We form the frtionl triplet,,. (ii) Tke the reiprols of these numers. (iii) Redue the numers to three smllest integers y multiplying the numers with the sme integrl multipliers. This lst set is enlosed in prentheses (h k l), is lled the index of the plne or Miller Indies. The Miller indies speify not just one plne ut n infinite set of equivlent plnes. Note tht for ui rystls the diretion [hkl] is perpendiulr to plne (hkl) hving the sme indies, ut this is not generlly true for other rystl systems. Exmples of the plnes in ui system: 8

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Exmple: Let the interepts re x =, y = 3/, z. x y z We first form the set,, 3, Then invert it y, 3,,, nd finlly multiply y ommon (ftor) denomentor. Whih is 6, to otin the miller indies (3 4 6). Exerise: x =, y = 3, z = 6 (3). The indies of some importnt plnes in ui rystl 9

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Reltion etween interplnr sping nd Miller indies: Let us onsider three mutully perpendiulr oordinte xis, OX, OY, nd Oz nd ssume tht plne (hkl) prllel to the plne pssing through the origin mkes interepts /h, /k nd /l on the three xes t A. B nd C respetively s shown in figure. Let OP =, the interplner sping e norml to the plne drwn from the origin nd mkes ngle,, nd with the three xes respetively. Therefore, OA, OB, OC h k l From OPA we get, OP os OA Similrly, from OPB we get nd from OPC we get h OP os OB OP os OC l But, for retngulr oordinte system, using diretionl osine we hve k os + os + os = () Sustituting the vlues of os, os nd os in Eq. we get, h k h l k () This is the generl formul nd is pplile to the primitive lttie of orthorhomi, tetrgonl nd ui systems. i) Orthorhomi system: h k l ii) Tetrgonl system: = # = =.4Å nd =.74Å then d 0 =.4Å h k l l O Z C A P X B 30 Y

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 ii) Cui system: = = h k l # = Å, d = 3Å # = 4Å, d 3 =.0 Å # d 00 =, d 0 = Å d = 3 Å d 00 : d 0 : d = : : 3 For d 00 = (d 00,s ) = nd d 0 = (d 0,s ) = d = (d,s ) = 3 d 00 : d 0 : d = :: 3 For d 00 = (d 00,s ) = d 0 = (d 0,s ) = d = (d,s ) = 3 nd d 00 : d 0 : d = : : 3 Ex: Determine the Miller Indies of plne whih is prllel to x-xis nd uts interepts of nd, respetively long y nd z xes. Solution: i) Interepts ii) Division y unit trnsltion iii) Reiprols iv) After lering frtion 0 4 3 Therefore the required Miller indies of the plne (04) 3

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Ex: Determine the M. I. of plne thet mkes interepts of Å, 3 Å, 4 Å on the o-ordinte xes of n orthorhomi rystl with :: = 4:3: Solution: Here the unit trnsltions re = 4, = 3 nd = following the sme proedure i) Interepts 3 4 ii) Division y unit trnsltion 4 3 3 4 iii) Reiprols iv) After lering frtion 4 Therefore the Miller indies of the pln is (4) 3

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Leture 6: X-ry diffrtion nd Brgg s lw: The inter-tomi sping in rystls is of the order of Å. Beuse of the short wvelength (omprle to the inter-plner distne), X-rys re sttered y djent toms in rystls whih n interfere nd give rise to diffrtion effets. When X-rys enter into rystl, eh tom ts s diffrtion entre nd rystl s whole ts like three dimensionl diffrtion grting. The diffrtion pttern so produed n tell us muh out the internl rrngement of toms in rystl. Let us onsider rystl mde up of equidistnt prllel plnes of toms with the inter-plner sping. Further, onsider monohromti x-ry em of wvelength hving ommon wve front, flls t n ngle on the plnes s shown in Figure. Eh tom stter the x-rys more or less uniformly in ll diretions, ut euse of the periodi rrngement of toms, the sttered rdition from ll toms in set of plnes is in phse where they interfere onstrutively. In ll other diretions, there is destrutive interferene. Consider two of the inoming x-ry OA nd O E inlined t n ngle with the topmost plne of the rystl nd re sttered in the diretions AP nd EP, lso t n ngle with tht plne. Sine the pth length of the rys OEP nd O AP re the sme, they rrive t P nd P respetively in phse with eh other nd gin form ommon wvefront. This is the ondition for sttering in phse y single plne of the rystl. Now, let us onsider X-ry sttering from two djent plnes (hkl) nd (hkl) s shown in Figure. If EB nd ED re prllel to the inident nd sttered wvefront respetively, the totl pth O CP is longer thn the pth OEP y n mount = BCD = BC + BD () 33

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Now, from the right ngle tringle EBC nd EDC, we hve BC = dsin = BD So, = d sin () If two onseutive plnes sttered in phse with eh other then we know tht the pth differene must e equl to n integrl multiple of wvelength, i.e. = n, where n = 0,,... gives the order of refletion. Thus the ondition for onstrutive interferene (in-phse sttering) y set of equidistnt prllel plnes in rystl is given y d sin = n (3) This is the well known Brgg s lw, whih ws first derived y the English physiists Sir W.H. Brgg nd his son Sir W.L. Brgg in 93. Thus diffrtion (onstrutive) ours for ertin disrete vlues of for whih the Brgg s ondition is fulfilled. n As (sin) mx =, we get, d. Tht is, must not e greter thn twie the interplner sping, otherwise no diffrtion will our. This oservtion is n exmple of X-ry wve interferene, ommonly known s X-ry diffrtion (XRD), nd ws diret evidene for the periodi tomi struture of rystls postulted for severl enturies. The Brggs were wrded the Noel Prize in physis in 95 for their work in determining rystl strutures eginning with NCl, ZnS nd dimond. Although Brgg's lw ws used to explin the interferene pttern of X-rys sttered y rystls, diffrtion hs een developed to study the struture of ll sttes of mtter with ny em, e.g., ions, eletrons, neutrons, nd protons, with wvelength similr to the distne etween the tomi or moleulr strutures of interest. Willim Henry Brgg nd Willim Lwrene Brgg were the first nd (so fr) the only fther-son tem to hve jointly won the prize. Other fther/son luretes inlude Niels nd Age Bohr, Mnne nd Ki Sieghn, J. J. Thomson nd George Thomson, Hns von Euler-Chelpin nd Ulf von Euler, nd Arthur nd Roger Kornerg, who were ll wrded the prize for seprte ontriutions. W. L. Brgg ws 5 yers old t the time, mking him the youngest Noel lurete to dte. For ertin speifi wvelengths nd inident ngles, intense peks of refleted rdition (known s Brgg peks) were produed. The onept of Brgg diffrtion pplies eqully to neutron diffrtion nd eletron diffrtion proesses. When x-rys re inident on n tom, they mke the 34

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 eletroni loud move s does ny eletromgneti wve. The movement of these hrges rerdites wves with the sme frequeny (lurred slightly due to vriety of effets); this phenomenon is known s Ryleigh sttering (or elsti sttering). The sttered wves n themselves e sttered ut this seondry sttering is ssumed to e negligile. A similr proess ours upon sttering neutron wves from the nulei or y oherent spin intertion with n unpired eletron. These re-emitted wve fields interfere with eh other either onstrutively or destrutively (overlpping wves either dd together to produe stronger peks or sutrt from eh other to some degree), produing diffrtion pttern on detetor or film. The resulting wve interferene pttern is the sis of diffrtion nlysis. When the energeti eletrons strike the trget, whih is pure metl suh s opper or molydenum, nd remove inner (K) shell eletrons. When this hppens, other eletrons from higher level shells drop into the vnt K-shell nd in so doing emit photon (X-ry) whose wvelength (energy) is hrteristi of the metl trget mteril. In order to remove the inner shell eletron, the inoming eletron must hve n energy greter thn the differene in energy etween the inner (K) shell eletron nd free eletron in the ondution nd of the trget metl. This energy differene is referred to s the sorption edge energy. Both KCl nd KBr hve sodium hloride struture shown in the figure. In this struture the two types of toms re rrnged lterntively t the lttie sites of simple ui lttie. The spe lttie is f with sis of two non-equivlent toms t 000 nd ½½½. In KCl the numer of eletrons of K + nd Cl - ions re equl nd the hrge distriution is similr. Therefore, the form ftors for K + nd Cl - re lmost extly equl, so tht the rystl looks to x-rys s if it were montomi simple ui lttie of lttie onstnt /. Only even integers our in the refletion indies when these re sed on ui lttie of lttie onstnt. In KBr the form ftor of Br - is quite different thn tht of K +, nd ll refletions of the f lttie re present. Figure: Comprison etween X-ry refletions from KCl nd KBr. 35

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Exerise: Determine the ngle through whih n X-ry of wvelength 0.440Å e refleted from the ue fe of rokslt rystl (d =.84Å). n Solution: Given = 0.440Å. d =.84Å sin d st 0.440 order refletion, n =, sin sin(0.078).84 nd o / order refletion, n =, sin 0.078 8 59 3 rd o / order refletion, n = 3, sin 3 0.078 3 3 34, et. 4 o 9 / Exerise: Determine the wvelength of the diffrtion em, when em of X-ry hving wvelengths in the rnge 0.Å to Å inident t n ngle of 9 with the ue fe of rokslt rystl (d =.84Å) Solution: n = = (.84) Sin 9 = 0.8804 Å n = = 0.8804Å = 0.440 Å n = 3 3 3 = 0.8804Å 3 = 0.935 Å n = 4 4 4 = 0.8804Å 4 = 0.0 Å n = 5 5 5 = 0.8804Å 5 = 0.760 Å < 0. Å whih shows the wvelength of the X-rys re 0.8804, 0.440, 0.935 nd 0.0 Å. Experimentl x-ry diffrtion Methods: To stisfy Brgg s lw, it is neessry to vry either the ngle of inlintion of the speimen to the em or the wvelength of the rdition. The three stndrd methods of X-ry rystllogrphy re ) Lue Method: A sttionry single rystl is irrdited y rnge of X-ry wvelengths. ) Rotting rystl Method: A single rystl speimen is rotted in em of monohromti X- rys. ) Powder Method: A polyrystlline powder speimen is kept sttionry in em of monohromti rdition. Of these tehniques, Lue method is used only for known rystl orienttion mesurement. 36

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 The powder method ssumes tht ll orienttions re present in the smple, so tht regrdless of the ngle of inidene, there will e grin in the proper orienttion for eh refletion (diffrtion). The ptterns re very useful for identifition of unknowns. There re ompiled indexes of powder diffrtion dt for minerls, s well s inorgni ompounds nd orgni ompounds. If the Miller indies of the diffrtion peks re known, it is possile to determine the unit ell prmeters of the mteril from the pek positions. Cell prmeters n then e used to determine omposition if the ell vrition with omposition is known. If more thn one minerl is present in the smple it is possile, lthough not esy, to determine the reltive proportions of the minerls. To do this one must hve stndrd pttern for eh pure minerl to otin the reltive internities of the peks form eh minerl. It is then possile to use the reltive intensities of non-overlpping peks to give n estimte of the minerl proportions, lled mode". Exerise: Compute the lttie sping for the () refletion of olivine with = 4.830 Å = 0.896 Å nd = 6.88 Å: /d = h / + k / + l / /d = (/4.830) + (/0.896) + (/6.88) d =.077 Å Wht is the ngle for this refletion using Cu k-lph rdition ( lmd =.5405 Å) nlmd = d sin thet thet = sin - (lmd /d) thet = 0.43º thet = 40.85º 37

Leture Notes on Struture of Mtter y Mohmmd Jellur Rhmn, Deprtment of Physis, BUET, Dhk-000 Compute the sping for (3) grnet with =.46Å nd Cu k-lph rdition ( l =.5405 Å) d = / ( h + k + l ) / d =.46/() / = 3.455 Å n lmd = d sin thet thet = sin - (n lmd /d) = sin - (.54/(3.455)) thet =.876º thet =5.754º Wht is the energy, in joules, of n X-ry photon of Cu k-lph rdition ( lmd =.540 Å)? E = h nu n = /l h = Plnk's Constnt = 6.6 x 0-34 j se E = h / l = speed of light = 3 x 0 8 m/se ev =.609 x 0-9 E = (6.6 x 0-34 ) j s (3 x 0 8 m/se) /.5405 x 0-0 m E =.86x0-5 j /.609x0-9 j/ev in eletron volts E = 806 ev Wht frequeny is Mo k-lph rdition ( lmd= 0.7096 Å)? n = / lmd n = (3.0 x 0 8 )m/se / (.7096 x 0-0 ))m n = 4.97 x 0 8 se - = 4.97x0 8 htz (htz = hertz = se - ) The sorption edge of Cu k-series rdition is.380 Å. Wht is the minimum KV setting on the X- ry genertor required to produe Cu k-series rdition? E = h / l = (6.6x0-34 jse)(3.0x0 8 m/se) (.380 x 0 - ))m / (.609x0-9 ) j/ev E = 8956 ev E = 8.956 Kev = 8.9 KV 5.4. X-ry Powder Diffrtion Instruments 38