1 1. Crystallography 1.1 Introduction 1.2 Crystalline and Non-crystalline materials crystalline materials single crystals polycrystalline material

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1 P g e. Crystllogrphy. Introduction Crystllogrphy is the brnch of science tht dels bout the crystl structures of elements. The crystl structures of elements re studied by mens of X-ry diffrction or electron bem diffrction or neutron bem diffrction. Among these three methods, X-ry diffrction is mostly used becuse it is powerful experimentl tool. The X-ry diffrction is used to determine the structure of new mteril, or it is used to identify the chemicl composition of common mteril hving known structure. The study of the crystl structure of mteril gives some dvntges to us. By studying the crystl structure of mteril, (i) one cn know bout whether the mteril is crystlline or non-crystlline nd (ii) the structure dependent properties of the mterils.. Crystlline nd Non-crystlline mterils In nturl crystl, the toms or ions re rrnged in regulr nd periodic mnner in three dimensions. Such crystls re sid to be crystlline mterils. The exmples for the crystlline mterils re the metls like silver, copper, gold, luminium etc. The crystlline mterils re further clssified into single crystls nd polycrystlline mterils. In solid, if the periodic rrngement of toms extends up to the entire specimen without ny interruption, then it is sid to be single crystls. Single crystls my exist in nture or it cn be synthesized rtificilly. It is generlly, very difficult to grow the single crystls becuse the environment should be crefully controlled. When crystlline solid is composed of number of mny smll crystls or grins, then it is sid to be polycrystlline mteril. In some mterils there is no regulr nd systemtic periodic rrngement of toms reltively over lrge tomic distnce. The toms re rrnged in rndom mnner. Such mterils re clled s non-crystlline mterils. They re glss, rubber, polymers etc. The non-crystlline mterils re lso clled s morphous mterils.. Isotropy nd nisotropy The properties of certin mterils depend on its crystllogrphic direction of mesurement. The properties such s modulus of elsticity, refrctive index, electricl conductivity of some mterils vries for [00] nd [] crystllogrphic directions. The substnce in which the mesured properties depend on its

2 P g e crystllogrphic direction of mesurement is clled s nisotropic nd the directionlity of properties is clled s nisotropy. The properties of some substnces do not depend upon the direction of mesurement. They re sid to be isotropic. The triclinic structure is normlly nisotropic. The mgnetic mteril, such s iron gets esily mgnetized long [00] directions thn [0] nd [] directions. Similrly, Ni gets esily mgnetized long [] thn [00] nd [0] directions. This shows the directionl properties of mgnetic mterils.. Lttice, bsis nd crystl structure.. Lttice The toms or ions re rrnged in nturl crystl in systemtic periodic rrngement. In order to study the rrngement of toms in nturl crystl, the points (dots) re rrnged. The rrngement of points is clled s lttice. If the points re rrnged in two dimension, then it is sid to be two dimensionl lttice. If the points re rrnged in three dimension, then it is sid to be three dimensionl lttice. Consider two dimensionl lttice s shown in Fig... Ech nd every point is clled s lttice point. The distnces between ny two lttice points long X nd Y directions re represented s nd b respectively. They re clled s lttice trnsltionl vectors. In three dimensionl lttice, one cn need three lttice trnsltionl vectors, b nd c. Fig.. Lttice, bsis nd crystl structure The term lttice cn be defined in nother wy. In n rrngement of points, if the rrngement looks like sme when it is viewed from different positions, then tht rrngement is clled s lttice. Consider the set of lttice points A,B, C, D, E, & F nd A',B', C', D', E', & F'. The distnces, AB=A'B'=, AC=A'C'=,

3 P g e AD=A'D'= 5, AE=A'E'= nd AF=A'F'=. This shows tht this rrngement of points looks like sme, when it is viewed either from the set of lttice points A, B,C, D, E, F or A', B', C', D', E', F'. Therefore, this rrngement is sid to be lttice... Bsis A smllest rrngement or group of toms to be fixed in ech nd every lttice point is sid to be bsis. Consider two dimensionl lttice s shown shown in Fig... The group of toms tht is ssocited with every lttice point is clled s bsis. The bsis is rrnged in ech nd every lttice points so s to get structure. In the cse of nturl crystl lso, group of toms re repetedly rrnged so s to form crystl structure. Consider sodium chloride unit cell. The N nd Cl ions re repetedly rrnged in NCl unit cell. The N nd Cl ions combined together re clled s bsis for NCl unit cell. Fig.. Lttice, bsis nd crystl structure.. Crystl structure A crystl structure is formed by rrnging the bsis in ech nd every lttice points. It cn be expressed s Lttice + bsis crystl structures (.) The eqution represents crystl structure is obtined by rrnging the bsis in ech nd every lttice points. Consider the bsis shown in Fig.. (b) is rrnged in ech nd every lttice points in Fig.. (), structure s shown in Fig..(c) is obtined. Therefore, the crystl structure is the result of two quntities; nmely, lttice nd bsis..5 Unit cells The regulr nd periodic rrngement of toms in nturl crystl shows tht smll group of toms form repetitive pttern. The repetitive pttern formed by rrnging smll group of toms one bove the other is clled s unit cell. It is similr to the rrngement of bricks one bove the other while constructing wll.

4 P g e The unit cell is the bsic structurl unit or the building block of crystl structure. The unit cell is chosen to represent the symmetry of crystl structure nd it is shown in Fig... There re fourteen different types of unit cells in three dimensions. Fig.. Unit cell The xes of the unit cell re constructed by drwing prllel lines to three mutully perpendiculr edges. The prllel lines drwn to three mutully perpendiculr edges re clled s crystllogrphic xes. Fig.. Crystllogrphic xes They re represented s X, Y nd Z xes. The intercepts mde by the unit cell long the crystllogrphic xes re clled s primitives. In Fig.., the intercepts re OA, OB nd OC. These intercepts re represented s, b nd c. These re lso clled s lttice constnts. The ngle between the X nd Y xes is represented s, the ngle between Y nd Z xes is represented s nd the ngle between Z nd X xes is represented s. These three ngles,, nd re clled s interfcil ngles or interxil ngles. The interfcil ngles re shown in Fig..5. To construct unit cell, three interfcil ngles (,, nd nd three lttice prmeters (, b, nd c) re needed. These six prmeters re clled s lttice prmeters. A primitive

5 5 P g e cell is minimum volume occupying cell nd it contins only one tom per cell nd it is shown in Fig... Fig..5 Interfcil ngles Fig.. Primitive cell. Crystl systems The crystls re clssified into seven crystl systems bsed on the unit length of the xis nd reltion between the interfcil ngles. They re cubic, tetrgonl, orthorhombic, monoclinic, triclinic, trigonl nd hexgonl. The trigonl crystl system is lso clled s rhombohedron... Cubic system: In cubic crystl, ll the lengths of the unit cell re equl, i.e.,=b=c nd the interfcil ngles re equl to 90 O, i.e., ===90 O. The cubic systems consist of three unit cells. They re simple cubic, body centered cubic nd fce centered cubic unit cells... Tetrgonl system: In tetrgonl crystl, ll the lengths of the unit cell long the X nd Y xes re equl nd they re not equl to the length of unit cell long Z xis. i.e., =b c. nd the interfcil ngles re equl to 90 O, i.e., ===90 O. The tetrgonl system consists of two unit cells. They re simple tetrgonl nd body centered tetrgonl unit cells... Orthorhombic systems: In n orthorhombic system, ll the lengths of the unit cell re not equl, (i.e., b c) nd the interfcil ngles re equl to 90 O, (i.e., ===90 O ). The orthorhombic system consists of four unit cells. They re simple orthorhombic, body centered orthorhombic, fce centered orthorhombic nd bse centered orthorhombic unit cells... Monoclinic systems: In monoclinic system, ll the lengths of the unit cell re not equl, i.e., b c. nd the interfcil ngles nd re equl to 90 O, wheres is not equl to 90 O. i.e., ==90 O, 90 O. The monoclinic system consists of two unit cells. They re simple monoclinic nd bse centered monoclinic unit cells.

6 6 P g e..5 Triclinic systems: In triclinic system, ll the lengths of the unit cell re not equl, i.e., b c. nd ll the interfcil ngles re not equl i.e.,. The triclinic system consists of only one unit cell... Trigonl systems: In trigonl system, ll the lengths of the unit cell re equl, i.e.,=b=c. nd the interfcil ngles re equl, but they re other thn 90 O, i.e., == 90 O. The trigonl system consists of only one unit cell...7 Hexgonl systems: In hexgonl system, the lengths of the unit cell =b, but they re not equl to c, i.e.,=b c. nd the interfcil ngles ==90 O but =0 O.. The hexgonl system consists of only one unit cell..7 Brvis lttices In 88, Brvis rrnged the points in different wys in three dimensions so tht the environment looks the sme from ech point. This rrngement provides different types of unit cells in three dimensions. These fourteen different types of unit cells in three dimensions re known s Brvis lttices. The fourteen different types of Brvis lttices (unit cells) re shown in Fig..7. The crystl systems, lttice prmeters, lttice symbols nd number of unit cells re displyed in Tble.. Tble.: Crystl systems nd number of unit cells S.No. Crystl systems Lttice prmeters Lttice symbols Number of unit cells. Cubic =b=c, ===90 O P, I, F. Tetrgonl =bc, ===90 O P,I. Orthorhombic bc, ===90 O P,I,F,C. Monoclinic bc, ==90 O 5. Triclinic bc, P,C P. Trigonl =b=c, ==90 O P 7. Hexgonl =bc, =90 O =0 O P

7 7 P g e Fig.7. Brvis lttices.8 Cubic unit cells There re three unit cells in cubic crystl structures. They re, simple cubic unit cell, body centered cubic unit cell nd fce centered cubic unit cell. In this section let us discuss bout the number of toms in the unit cell, tomic rdius, coordintion number nd tomic pcking fctor for these three unit cells.

8 8 P g e.8. Atomic rdius: The tomic rdius is defined s the hlf of the distnce between ny two successive toms in the unit cell. For simple cubic unit cell the tomic rdius, r=..8. Coordintion number: The number of nerest neighbouring toms to prticulr tom is known s the coordintion number. For simple cubic unit cell the coordintion number is six..8. Pcking density: The pcking density is used to find how much volume of the unit cell is occupied by toms. It is defined s the rtio between the totl volumes of the toms nd the volume of the unit cell. Pcking density = Number of toms per unit cell Volume of one tom Volume of the unit cell (.) The pcking density of simple cubic unit cell is Simple cubic (SC) unit cell A simple cubic unit cell consists of eight corner toms. In simple cubic unit cell, corner tom touches with nother corner tom. The simple cubic unit cell is shown in Fig..8. () Fig..8 Simple cubic unit cell: () reduced sphere unit cell nd (b) the closely pcked Simple cubic unit cell with hrd spheres. Number of toms in simple cubic unit cell A simple cubic unit cell consists of 8 corner toms. Ech nd every corner tom is shred by eight djcent unit cells. Therefore, one corner tom contributes th of its prts to one unit cell. Since, there re eight corner toms in unit cell, 8 (b) the totl number of toms is 8 x 8=. Therefore, the number of toms in simple cubic unit cell is one. Atomic rdius

9 9 P g e In simple cubic lttice, corner tom touches with nother corner tom. Therefore, r=. So, the tomic rdius of n tom in simple cubic unit cell is Coordintion number Consider corner tom in simple cubic unit cell. It hs four nerest neighbours in its own plne. In lower plne, it hs one more nerest neighbour nd in n upper plne, it hs one more nerest neighbour. Therefore, the totl number of nerest neighbour is six. Pcking density The pcking density of simple cubic unit cell is clculted s follows: Pcking density = = Substituting, r= Number of toms per unit cell Volume of one tom Volume of the unit cell r, we get, Pcking density = = The pcking density of simple cubic unit cell is 0.5. It mens, 5% of the volume of the unit cell is occupied by toms nd the remining 8% volume is vcnt..8.5 Body centered cubic (BCC) unit cell A body centered cubic unit cell hs eight corner toms nd one body centered tom. In body centered cubic unit cell, the toms touches long the body digonl. The body centered cubic unit cell is shown in Fig..9.. () (b)

10 0 P g e Fig..9 Body centered cubic unit cell: () reduced sphere unit cell nd (b) the closely pcked BCC unit cell with hrd spheres. Number of toms in simple cubic unit cell A corner tom in body centered cubic unit cell is shred by eight djcent unit cells. Therefore, one corner tom contributes 8 th of its prts to one unit cell. Since, there re eight corner toms in unit cell, the totl number of tom contributed by the corner tom is 8 x 8=. In ddition, body centered cubic unit cell hs body centered tom t the centre of the unit cell. Therefore, the totl number of toms present in body centered cubic unit cell is two. Atomic rdius In body centered cubic unit cell, the corner toms touches long the body digonl. From Fig..0, the length of the body digonl is r. From tringle ABD, AD =AB +BD (.) For cubic unit cell, the cube edge is. Therefore, AB=BD=. Substituting, the vlues of AB, nd BD we get, AD = AD= = Consider the tringle ADH, AH =AD +DH (.) Substituting the vlues of AH=r, AD=, nd DH= r Fig..0 Atomic rdius clcultion in BCC unit cell

11 P g e 6r r (.5) The tomic rdius of n tom in BCC unit cell is Coordintion number r In BCC unit cell, body centered tom is surrounded by eight corner toms. For body centered tom, corner tom is the nerest neighbour. Therefore, the number of nerest neighbour is eight. Pcking density The pcking density of body centered cubic unit cell is clculted s follows: Pcking density = Pcking density = = Totl volume of the toms Volume of the unit cell Number of toms per unit cell Volume of one tom Volume of the unit cell r Substituting, r=, we get, = Pcking density = The pcking density of body centered cubic unit cell is 0.8. It mens, 8% of the volume of the unit cell is occupied by toms nd the remining % volume is vcnt..8. Fce centered cubic (FCC) unit cell A fce centered cubic unit cell consists of eight corner toms nd six fce centered toms. In fce centered cubic unit cell, n tom touches with nother tom long the fce digonl. The fce centered cubic unit cell is shown in Fig...

12 P g e () (b) Fig.. Fce Centered cubic unit cell: () reduced sphere unit cell nd (b) the closely pcked FCC unit cell with hrd spheres. Number of toms in fce centered cubic unit cell In FCC unit cell, there re eight corner toms nd six fce centered toms. A corner tom is shred by eight djcent unit cells. Therefore, one corner tom contributes 8 th of its prts to one unit cell. Since, there re eight corner toms in unit cell, the totl number of toms contributed by the corner toms is 8 x 8=. There re six fce centered toms. A fce centered tom is shred by two unit cells. Therefore, one fce centered tom contributes hlf of its prts to one unit cell. So, the totl number of toms contributed by the fce centered toms is totl number of toms present in fce centered cubic unit cell is four. x= nd the Atomic rdius In fce centered cubic unit cell, the toms touches long the fce digonl. Therefore the fce digonl is equl to r. To find the tomic rdius of the tom in fce centered cubic unit cell, consider the tringle ABC. From tringle ABC, AC =AB +BC For cubic unit cell, the cube edge is. Therefore, AB=BC=. Substituting the vlues of AB, BC nd AC we get, (r) = r = r 8.

13 P g e Fig.. Atomic rdius clcultion for FCC unit cell The tomic rdius of n tom in FCC unit cell is Coordintion number r. (.) Consider fce centered tom in the upper plne of FCC unit cell. It is surrounded by four corner toms. These corner toms re nerest neighbours for this fce centered tom. There re four more fce centered toms tht re nerest neighbours for this reference fce centered tom in lower plne. Similrly this reference fce centered tom hs four more fce centered toms s its nerest neighbours in n upper plne. Therefore the totl number of the nerest neighbour is. Pcking density The pcking density of fce centered cubic unit cell is clculted s follows: Pcking density = Pcking density = Substituting, r= = Totl volume of the toms Volume of the unit cell Number of toms per unit cell Volume of one tom Volume of the unit cell r, we get, = Pcking density = The pcking density of fce centered cubic unit cell is 0.7. It mens, 7% of the volume of the unit cell is occupied by toms nd the remining % volume is

14 P g e vcnt. The pcking density 0.7 is the mximum vlue nd hence this unit cell is sid to be cubic closed pcked (CCP) structure..9 Hexgonlly closed pcked (HCP) structure A hexgonlly closed pcked structure consists of three lyers of toms, nmely bottom lyer, middle lyer nd upper lyer. The middle lyer lies just bove the bottom lyer t distnce of c from the bottom lyer. The upper lyer lies t distnce of c from the bottom lyer. The bottom lyer consists of six corner toms nd one fce centered tom. The middle lyer hs three toms. The upper lyer hs six corner toms nd one fce centered tom. The lttice prmeters for HCP structure is given below: =b c nd ==90 O nd =0 O. A hexgonlly closed pcked unit cell is shown in Fig... Fig.. HCP unit cell The rrngement of toms in hexgonlly closed pcked structure is explined using simple figure shown in Fig... The bottom lyer is rrnged by rrnging seven toms hving equl rdius in such wy tht one tom is surrounded by six other toms. The centre of these toms re mrked s A nd the centre of the surrounding toms re joined, it will constitutes hexgon. These six toms re the corner toms for the bottom lyer nd the middle tom is the fce centered tom. The second lyer is rrnged over the first lyer of toms. There re six possible plces to rrnge the second lyer. Among these six plces, three lternte plces re mrked s B nd the remining three plces re mrked s C. Consider the second lyer is plced over the plces mrked s B. There re two possible wys to rrnge the third lyer of tom. In the first wy, the third lyer is rrnged directly over the first lyer. This will constitute stcking sequence of ABABAB nd it will form HCP unit cell. In nother wy the third lyer is plced over the plces mrked s C. The stcking sequence goes on s ABCABCABC nd it will constitute FCC structure.

15 5 P g e Fig.. Arrngement of toms in HCP unit cell Number of toms in HCP unit cell The bottom lyer of HCP unit cell consists of six corner toms nd one fce centered tom. Ech nd every corner tom in the bottom lyer is shred by six unit cells. Therefore, corner tom contributes 6 th of its prts to one unit cell. Since there re six corner toms in the bottom lyer, the totl number of toms 6 contributed by the corner toms is 6. The fce centered tom in the bottom lyer is shred by two unit cells. Therefore, it contributes cell. The totl number of toms present in the bottom lyer is + = of its prts to one unit. The upper lyer lso hs toms. The middle lyer hs three toms. Therefore, the totl number of toms present in the unit cell is, ++ = Fig..5 HCP structure- tomic rdius

16 6 P g e Atomic rdius Consider the bottom lyer of toms. A corner tom touches with nother corner tom s shown in Fig..5. Therefore, r= nd hence Coordintion number r. Consider the fce centered tom in the bottom lyer. It is surrounded by six corner toms. So, there re six nerest neighbours for fce centered tom in its own plne. The middle lyer of toms lso touches the bottom lyer. So, the three middle lyers of toms re the nerest neighbours lying in the upper plne. The unit cell tht lies below this unit cell lso hs three more toms in the middle lyer nd these three toms re the nerest neighbour to the fce centered tom. Therefore, the totl number of nerest neighbour is ++=. Pcking density The pcking density of fce centered cubic unit cell is clculted s follows: Pcking density = Number of toms per unit cell Volume of one tom Volume of the unit cell The volume of the unit cell is clculted using the reltion, Volume of hexgon = re of the bottom surfce X height = x re of n equilterl tringle x height = 6 c Pcking density of HCP unit cell= Substituting, r= Pcking density =, we get, 6 6 c = c 6 r 6 c Substituting c 8 we get,

17 7 P g e Pcking density = 8 = =0.7. The pcking density of hexgonl unit cell is 0.7 nd it is the mximum vlue. Therefore, hexgonl unit cell is lso sid to be hexgonlly closed pcked (HCP) structure..0 Reltion between c nd Consider the bottom surfce of hexgonl unit cell. It hs six corner toms nd one fce centered tom. Let A, B, C, D, E nd F be the corner toms nd O be the fce centered tom. The middle lyer of toms is plced over the first lyer. Let I, G, nd H be the second lyer of toms. Consider the tringle AFO. It is n equilterl tringle. In the tringle, AFO, A nd F re corner toms nd O is the fce centered tom. Let us bisect the fces AF nd AO. Let OE, nd FJ be the perpendiculr bisectors drwn to the lines AF nd AO. X is the centroid. The second lyer of toms lies exctly t distnce of c from the bottom lyer. Fig.. Bottom lyer of HCP unit cell In Fig..7, OG= nd XG= c tringle, XJO re 0 O, 0 O nd 90 O. In the tringle, XJO, XJ sin 0 = OX OX sin0= Fig..7 Reltion between c nd of HCP unit cell. Consider the tringle, XJO. The ngles of the OX

18 8 P g e OX=. Consider the tringle, XOG. Applying Pythgors theorem, we get OG =XG +OJ (.7) c c c 8 c 8.6 (.8) The rtio between c nd in hexgonl unit cell is,. c 8. Comprison of tomic rdius, coordintion number nd pcking density of SC, BCC, FCC nd HCP unit cells The tomic rdius, coordintion number, pcking density, number of toms in unit cell for simple cubic (SC), body centered cubic (BCC), fce centered cubic (FCC) nd hexgonlly closed pcked unit cells re listed in Tble.. Crystl Number of toms systems per unit cell Simple cubic Body centered cubic Fce Centered cubic Atomic Coordintion Pcking rdius number density Hexgonlly 0.7 closed pcked structure Tble. Pcking density, tomic rdius, coordintion number of SC, BCC, FCC nd HCP unit cells From Tble., one cn infer tht the pcking density increses with the increse of coordintion number.. Reltion between the tomic weight A nd intertomic distnce

19 9 P g e The mss of the unit cell is given by Mss = volume of the unit cell x density m = V x (.9) Let A be the tomic weight. The tomic weight of substnce represents the mss of one tom of tht substnce. Let n be the number of toms in unit cell. The mss of the unit cell is na. Usully, the tomic weight is given in tomic mss unit (.m.u.). To convert it into kilogrm, it should be divided by the Avogdro s constnt, NA. Therefore, the mss of the unit cell is, na m (.0) N A From Eq.(.9) nd Eq.(.0), we get, na V N A For cubic unit cell, V=. Therefore, the bove eqution becomes, na N A Eq.(.) gives the reltion between the intertomic distnce nd tomic weight.. Crystl plnes nd Miller indices rrngement. (.) In nturl crystls, the toms or ions re rrnged in regulr nd periodic This periodic rrngement of toms or ions in nturl crystl produces prllel equidistnt plnes. These prllel nd equidistnt plnes formed by the periodic rrngement of toms or ions in nturl crystl is sid to be crystl plnes. The crystl plnes with lttice spcings d, d nd d re shown in Fig..8. Fig..8 Crystl plnes Willim Hllowes Miller (80-880) devised method to represent crystl plnes. According to Miller the crystllogrphic plnes re specified in terms of indexing schemes. The plnes re indexed using the reciprocls of the xil

20 0 P g e intercepts. The Miller indices re the set of numbers, used to represent crystl plne, obtined from the reciprocls of the intercepts mde by the crystl plnes... Procedure used to find the Miller indices of plne In order to find the Miller indices of plne, the following steps to be followed: i. The intercepts mde by the plne re noted nd they should be written in terms of the lttice constnts,, b nd c. ii. The coefficients of the intercepts re noted iii. Find the inverse for these coefficients, iv. Find LCM nd then multiply the frctions by LCM. v. Write the integers within the prenthesis vi. These integers written within the prenthesis represents the Miller indices of the given plne. Fig..9 Miller indices of plne Consider plne ABC s shown in Fig..9. It mkes intercepts long the X, Y nd Z xes. The intercepts re, b, nd c. The coefficients of the intercepts re,,,. The inverses re, nd.. The Lest common multiplier (LCM) is. Multiplying these frctions by LCM, we get,,. These three vlues re written within the prenthesis s (). It represents the Miller indices of the plne ABC. A set of plnes, which re structurlly equivlent re clled s fmily of plnes. They re represented by brces { }. The {0} plnes re (0), (0), (0), ( 0 ), ( 0), ( 0), ( 0), ( 0), ( 0 ), ( 0 ), ( 0 ) nd ( 0 ).

21 P g e.. Slient fetures of Miller indices The importnt fetures of the Miller indices re, i. Miller indices represent set of prllel plnes. It does not represent single ii. iii. plne. If plne lies long n xis, it is not possible to find the intercepts mde by the plne correctly. To find the Miller indices of plne tht lies long n xis, consider prllel plne nd find the Miller indices of tht plne. The Miller indices of these two prllel plnes re sme. If plne does not mke ny intercept long prticulr xis, then it is ssumed tht it will meet tht xis t infinity. The intercepts for tht xis is tken s. iv. The negtive intercept is lso tken into ccount. For exmple, the Miller index nottion ( 00 ) represents, the plne hs negtive intercept t Y xis. v. The X, Y nd Z xes re represented s (00), (00) nd (00) respectively vi. vii. The Miller indices of plne (0) is red s one zero two nd it should not be red s one hundred nd two. There is no comm, or ny other specil chrcters should be introduced in between the integers while writing the Miller indices of plne... Advntges of finding Miller indices The dvntges of finding the Miller indices of the plne re given below: i. If Miller indices of plne is known, then the intercepts mde by the plne is ii. iii. b c, nd., where (hkl) re the Miller indices of the plne nd, b, nd c h k l re the lttice constnts. The Miller indices of plne (hkl) nd the Miller indices of the direction of tht plne [hkl] re sme. If Miller indices of two plnes re known, then the ngle between these two plnes cn be determined using the reltion, uvw cos (.) uvwuvw iv. where (uvw) nd (uvw) re the Miller indices of the plnes. Miller indices of plne is used to find the reltion between interplnr distnce, d nd intertomic distnce,.

22 P g e For cubic unit cell, d (.) hk l.. Miller indices determintion for the plnes in cubic unit cells Consider cubic unit cell. Let us find the Miller indices of the plnes, (i) BCFG, (ii) ADFG, nd (iii) AFH shown in Fig..0. To find the Miller indices of the bove plnes, the crystllogrphic xes X, Y nd Z re drwn by drwing prllel lines to the edges EF, EH nd EA. They re represented s X, Y, nd Z xes respectively. The Miller indices of the plnes re determined by finding the intercepts mde by the plnes s follows: (i) The plne BCFG The intercepts mde by the plne BCFG re, b, c. The coefficients of the intercepts re,,. The inverse for these coefficients re,,,. The Miller indices re (00). (ii) The plne ADFG The intercepts mde by the plne ADFG re, b, c. The coefficients of the intercepts re,,. The inverse for these coefficients re,, 0,. The Miller indices re (0). (i) The plne AFH The intercepts mde by the plne AFH re, b, c. The coefficients of the intercepts re,,. The inverse for these coefficients re,,,. The Miller indices re (). Fig..0 Miller indices determintion for cubic unit cell. Reltion between the interplnr distnce nd intertomic distnce The distnce between ny two djcent sme kinds of toms is known s intertomic distnce. It is represented by the letter,. The distnce between ny

23 P g e two successive prllel plnes is known s interplnr distnce. It is represented by the letter d. Consider cubic unit cell. Consider plne ABC in the cubic unit cell. Let (hkl) be the Miller indices of the plne ABC. The intercepts mde by the plne long b c X, Y nd Z xes re, nd h k l. Consider nother prllel plne OPQ is pssing through the origin, O. Let ON is the perpendiculr line drwn between O nd the plne ABC. It represents the interplnr distnce, d. Let the ngle between ON nd X xis s, the ngle between ON nd Y xis s nd the ngle between ON nd Z xis s. From Fig... Fig.. Reltion between interplnr nd intertomic distnce ON d dh cos ' (.) OA h ON d dk cos ' (.5) OB b b k nd ON d dl cos ' (.) OC c c l From the properties of the direction cosines cos +cos +cos = (.7) dh dk b dl c d h k b l c For cubic unit cell, =b=c. Eq.(.8) cn be written s (.8)

24 P g e h k d d h l (.9) k l Eq.(.9) gives the reltion between the interplnr distnce nd intertomic distnce..5 Direction of plne nd Miller indices A perpendiculr line drwn to plne is clled s its direction. The direction of plne is represented by set of three numbers written within squre brckets, which re obtined by identifying the smllest integer position intercepted by the line from the origin of the crystllogrphic xes. The direction [ ] represents the line originting from the origin psses through the points,,-,. The direction of plne is pssing through negtive xis, it is represented s br over the index like [ ]. A set of directions, which re structurlly equivlent re clled s fmily of directions. They re represented by ngulr brckets < >. The <0> plnes re [0], [0], [0], [ 0 ], [ 0], [ 0], [ 0], [ 0], [ 0 ], [ 0 ], [ 0 ] nd [ 0 ]..5. Procedure to drw the direction of plne in cubic unit cell To drw the direction of plne in cubic unit cell, sy for exmple [], mrk the point,,, in the cubic unit cell. Drw line between the origin 0,0,0 nd the point,,. This line represent the [] direction of the cubic unit cell. For exmple to drw the [00], [0] nd [] directions in cubic unit cell, drw cubic unit cell. Mrk the X, Y nd Z xes for the cubic unit cell. Mrk the points,,0,0. Drw line between the origin nd the point,0,0. This line represents the direction [00]. Then to drw the direction [0] mrk the points,,,0. Drw line between the origin nd the point,,0. This line represents the direction [0]. The directions, [00], [00], [00], [0], [0], [0] nd [] re shown in Fig... Fig.. Direction of plne

25 5 P g e. Liner density nd plnr density The number of toms per unit length long given direction in crystl structure is known s liner density [LD]. It is represented in m -. Liner density = Number of toms centered on direction vector length of the direction vector (.0) Let us determine the liner density long the [0] direction of FCC unit cell. The [0] direction is shown in Fig... The toms X, Y, nd Z lies long the [0] plne. The tom X contins only hlf of the tom, wheres the remining hlf is shred by nother unit cell, Y hs full tom nd Z hs only hlf of the tom. Totlly, [0] direction hs toms. The length of [0] direction is Liner density=. Therefore, Number of toms lying on direction vector = length of the direction vector Fig... [0] direction in the FCC unit cell for liner density clcultion () (b) Fig.. Plnr density of [0] direction in FCC unit cell () (0) plne nd (b) clcultion of number of toms in (0) plnes of FCC

26 6 P g e The number of toms per unit re in given plne of crystl structure is clled s plnr density. It is represented in m -. Plnr density= Number of toms lying on plne re of the plne (.) Consider (0) plne of FCC unit cell s shown in Fig... The (0) plne hs toms. Its re is Plnr density=. Therefore, the plnr density is Number of toms lying on plne = re of the plne.7 Some specil crystl structures.7. Dimond unit cell Dimond crystllizes in cubic crystl structure. In dimond unit cell ll the toms re crbon. The dimond unit cell consists of two interpenetrting FCC unit cells. The dimond lttice my be considered s two identicl toms t 000, positions ssocited with ech point of the FCC lttice. The unit cell hs eight corner toms, six fce centered toms nd four toms on the body digonls. The body digonl tom lies t distnce of one fourth of the body digonl. The body digonl toms re tetrhedrlly bonded with corner toms nd hence touch with the corner toms. The dimond unit cell is shown in Fig..5. Fig..5 Dimond unit cell Number of toms in unit cell There re eight corner toms. Ech nd every corner toms re shred by eight djcent unit cells. Therefore, one corner tom contributes 8 th of its prts to

27 7 P g e one unit cell. Since there re eight corner toms, the number of toms contributed by the corner tom is 8 8 one. There re six fce centered toms. Ech nd every fce centered tom is shred by two djcent unit cells. So, fce centered tom contributes of its prts to one unit cell. Therefore, the totl number of toms contributed by the fce centered toms is 6 three. The body digonl toms lies within the unit cell. There re four body digonl toms. Therefore, the totl number of toms in dimond unit cell is (++=8) eight. Atomic rdius The toms tht lie long the body digonl of the unit cell re tetrhedrlly bonded with corner toms nd hence touch with the corner toms. The body digonl tom lies t distnce of one fourth of the body digonl. Therefore, r x body digonl of the unit cell r = r (.) 8 Coordintion number The body digonl tom is tetrhedrlly bonded with other toms. Therefore, the number of nerest neighbour is four. Therefore, the coordintion number is four. Pcking density The pcking density of dimond unit cell is clculted s follows: Pcking density= = Number of toms per unit cell Volume of one tom Volume of the unit cell 8 r = 8 8

28 8 P g e = 6 =0. The pcking density of dimond unit cell is 0.. This shows tht nerly % of the dimond unit cell is vcnt. Dimond is one of the hrdest mterils. The high vlue of hrdness of dimond is due to its crystl structure nd the strong intertomic covlent bonds. Some of the elements tht crystllize in dimond unit cell re C, Si, Ge nd tin..7. Cesium chloride crystl structure Cesium chloride crystllizes in cubic unit cell. The Cs + ion lies in the body centered position, wheres the Cl - ions occupy the corner tom position. Ech nd every Cl - ion t the corner is shred by eight unit cells. contributes 8 corner tom position is Therefore, one Cl - ion th of its prts to one unit cell. The totl number of Cl - ions t the 8 8 one. The number of Cs + ion present in the unit cell is one. Therefore, totlly the unit cell hs only one CsCl tom. For Cs + ion, the Cl - ions re the nerest neighbours. Therefore the coordintion number is eight. This unit cell is not true body centered cubic unit cell, becuse toms of two different kinds re involved. Some compounds exhibiting this crystl structure is CsCl, CsBr, CsI, TiI, BeCu, AlNi. The CsCl unit cell is shown in Fig... Fig.. CsCl unit cell.7. Zinc blende structure The ZnS srystl structure is presented in Fig..7. This crystl structure is clled s zinc blende or sphlerite structure, fter the minerlogicl term for zinc sulphide. The S tom occupies the corner position nd the fce centered positions. The Zn tom occupies the interior tetrhedrl position. If the Zn tom nd S toms

29 9 P g e re reversed vice vers, the sme type of crystl structure is obtined. This structure is lso the interpenetrtion of two FCC lttices. If only Zn toms re considered, it constitutes the FCC lttice. Similrly, if only the S toms re considered it constitutes FCC lttice. There is tetrhedrl bonding between the Zn nd S toms. The coordintion number is four, becuse of the tetrhedrl bonding Zn tom is surrounded by four S toms nd vice vers. In unit cell, there re ZnS toms. Some of the common mterils, which crystllizes in this crystl structure re, ZnS, CdS, ZnSe, GAs, CuCl, SiC, nd ZnTe. Fig..7 ZnS unit cell.7. Sodium chloride unit cell Sodium chloride crystllizes in cubic unit cell. The sodium nd chlorine ions re lterntively rrnged in the sodium chloride unit cell. The unit cell looks like the interpenetrtion of two FCC unit cells. If only the N + ions considered, it constitutes FCC unit cell. Similrly, if Cl - ions re considered, it constitutes nother FCC unit cell. The unit cell hs N + ions nd Cl - ions nd hence totlly NCl ions re present in the unit cell. The crystls such s NCl, MnS, LiF, MgO, CO, FeO nd,nio crystllizes in this unit cell. The sodium chloride unit cell is shown in Fig..8.

30 0 P g e Fig..8 NCl unit cell.7. Grphite structure The grphite structure is shown in Fig..9. In grphite, the crbon toms re connected together in hexgonl pttern by covlent bonding in two dimensionl pttern. Tht is, the crbon toms re rrnged in the form of lyers or sheet structures. Ech crbon tom is connected by three other crbon toms in hexgonl pttern by shring of electrons. Crbon hs four vlence electrons. Three electrons re used for forming covlent bond in hexgonl pttern nd the remining one electron is free. The fourth electron is deloclized nd resontes between the three covlent bonds. Therefore, the grphite hs very good electricl conductivity. The bond length between ech crbon tom is. Å. In grphite the sheets re held together by vn der Wls bond. The distnce between ech sheet is. Å. This wek inter-sheet bonding is the reson for soft chrcteristics of grphite. Fig..9 Grphite structure

31 P g e.8 Polymorphism nd llotropy The bility of solid mterils to exist in more thn one form or crystl structure by the chnge of pressure or temperture is known s polymorphism. The internl rrngement of the solid chnges due to the pplictions of pressures or tempertures nd hence the solid mteril exists in different crystl structure. If the chnge in the crystl structure is reversible these polymorphic chnge is sid to be llotropy. (i) Allotropy in Iron The crystl structure of iron chnges with temperture nd it gets reverted bck on cooling. Therefore, iron is sid to hve llotropic form. The chnge of crystl structure of iron with respect to temperture is given below: Temperture Form Structure Lttice constnt Property Up to 78 o C -iron BCC - mgnetic 78 o C to 90 o C -iron BCC 0.9 nm Non-mgnetic 90 o C to 0 o C -iron FCC 0. nm Prmgnetic 0 o C to 55 o C -iron BCC 0.9 nm Prmgnetic Above 55 o C Molten form (ii) Allotropy of crbon Crbon hs four llotropic forms, nmely, dimond, grphite, fullerence, nd crbon nnotubes nd they re shown in Fig..0. Dimond crystllizes in cubic unit cell nd it hs two interpenetrting FCC structures in which the toms re connected by covlent bonds. It hs no free electrons nd hence it is best insultor. () Dimond (b) Grphite

32 P g e (c) C0 (d) Crbon nnotube Fig..0 Allotropy of crbon Grphite crystllizes in lyer or sheet structure. The crbon toms in the lyer re connected together in hexgonl structure by covlent bonding. It hs free electrons nd its electricl conductivity is very high. The lyer or sheet structures re connected by vn der Wls bond. The third llotropy of crbon is fullerence. Fullerence re crbon-0 clusters with sphericl shpe nd formed like bll with fces. Of these fces, were pentgons nd 0 were hexgons exctly like soccer bll. This soccer bll shped C 0 molecule ws nmed s buckerminsterfullerence or buckybll in short. C, C 70, C 7 nd C 8 re the other relted compounds found to be composed only of crbon. These molecules nd crbon-0 re clled s llotropy of crbon. These new crbon molecules re clled s fullerene. Fullerene hs only pentgons nd hexgons. The fourth llotropy of crbon is crbon nnotubes. The crbon nnotubes re tube like structures hving dimeter of nerly to 0 nm nd length in the order of mm. In crbon nnotubes, ll the toms re mde up of crbons. There re two types of crbon nnotubes, nmely, single wll crbon nnotube nd multiwll crbon nnotubes. (iii) Allotropy in other mterils Coblt crystllizes in HCP structure in ordinry temperture but on heting bove 77 o C, it chnges to FCC structures. Mngnese exists in four forms,,, nd. Tin exists in metllic form in ordinry temperture (white or ) but nonmetllic form (grey or ) with the structure of dimond, below bout o C. The metls like clcium, lnthnum nd scndium re in HCP structure t low temperture nd trnsform to FCC structure t high tempertures. The metls like zirconium, titnium, thllium nd beryllium re in HCP structure t low temperture nd trnsform to BCC structure t high tempertures.

33 P g e SOLVED PROBLEM. From the knowledge of crystl structure, clculte the density of dimond (r=0.07 nm). Solution mss density volume mss of one crbon tom density 8 volume 8 mss of one crbon tom density 8 r 8 density density kg m -. The density of dimond = 50.8 kg m -.. Determine the percentge volume chnge tht occurs when Ti chnges from BCC structure to HCP structure. With the BCC structure the lttice prmeter =0. nm nd with the HCP structure =0.9 nm nd c=0.8 nm. Solution Volume of the BCC unit cell = =(0. x 0-9 ) =.59 x 0-9 m - Volume of HCP unit cell = c 6 = =.059 x 0-8 m Chnge in volume =.05 x x 0-9 =.999 x Volume chnge in percentge = 00% =9.% The percentge volume chnge =9.%. 9 9

34 P g e. Copper hs FCC structure nd its tomic rdius is.78 Å. Clculte its density. The tomic weight of copper is.5. Given dt tomic rdius of copper =.78 Å. Atomic weight of copper =.5 Solution The density nd the tomic weight re relted using the eqution, na V N A Since, copper crystllizes in cubic unit cell, substituting V=, we get, na N The density is given by, For FCC, r A na N A i.e., r = x x.78 x 0-0 = For FCC, n=. Substituting the vlues of n, A, NA, nd, we get, na N = A The density of copper is 8980 kg m = 8980 kg m -. NCl crystllizes in FCC structure. The density of NCl is 80 kg m -. Clculte the distnce between djcent toms. Given dt Solution Density of NCl = 80 kg m - Moleculr weight of NCl = tomic weight of N + tomic weight of Cl =+5.5 =58.5 The density nd the tomic weight re relted using the eqution, na V N A Since, NCl is cubic unit cell, substituting V=, we get,

35 5 P g e The intertomic distnce is given by, na N A na N A Substituting the vlues of n, A, NA,, we get, = 5.8 x 0-0 m. The intertomic distnce in NCl crystl is 5.8 x 0-0 m. 5. Clculte the interplnr spcing for (0) nd () plnes in simple cubic lttice whose lttice constnt is 0. nm. Given dt: Lttice constnt =0. nm Solution The reltion between interplnr nd intertomic distnce is given by For (0) plne For () plne d h k l 9.00 d = d = =0.99 nm. =0. nm. The interplnr spcing for (0) plne is 0.99 nm nd for () is 0. nm 6. Identify the xil intercepts mde by the following plnes: (i) (0), (ii) () (iii) ( ). Solution

36 6 P g e If the Miller indices of the plne is (hkl), then the intercepts mde by the b plne is, h k nd c. nd l c b For the plne () the intercepts re, For the plne ( ) the intercepts re. For the plne (0), the intercepts re nd b, b c c. nd c. b 0, 7. Find the ngle between two plnes () nd () in cubic lttice. Solution The ngle between ny two plnes hving Miller indices (uvw) nd (uvw) is uvw cos uvwuvw cos 5 cos.096 cos.0( 96 ) =5.8 O =5 O 50.9 The ngle between the plnes () nd () is 5.8 O. 8. Sketch the following crystllogrphic plnes for the cubic systems (00), (0), (), (00), (0) nd (). Solution (i) For plne with Miller indices (hkl), the intercepts re h b c, k l,. The intercepts of the plne (00) re,,. The plne (00) is shown in Fig.. (i). (ii) The intercepts of the cubic plne (0) re,,. The plne (0) is shown in Fig.. (ii). (iii) The intercepts of the plne () re,,. The plne () is shown in Fig.. (iii).

37 7 P g e (iv) (v) (vi) The intercepts of the plne (00) re /,,. The plne (00) is shown in Fig.. (iv). The intercepts of the plne (0) re, /,. The plne (0) is shown in Fig.. (v). The intercepts of the plne () re /,,. The plne () is shown in Fig.. (vi). (i) (00) plne (ii) (0) plne (iii) () plne (iv) (00) plne (v) (0) plne (vi) () plne Fig.. (i) (00) plne, (ii) () plne, (iii) () plne, (iv) (00) plne, (v) (0) plne nd (vi) () plne.

38 8 P g e 9. The interplnr distnce between the plnes ( ) in luminium (FCC structure) is 0.8 nm. Wht is the lttice constnt? Given dt Solution interplnr distnce, d=0.8 nm Miller indices of the crystl plne = ( ) d hk l The interplnr distnce is given by, d hk l The lttice constnt is 0.09 nm m 0. Show tht for simple cubic system, d00:d0:d= 6 : : Solution d hk l The interplnr distnce is given by, d 0 0 The interplnr spcing for d00 plne= = The interplnr spcing for d0 plne= d = 0 The interplnr spcing for d plne= d = The rtio, d00 : d0 : d= : d00 : d0 : d= : : : Multiplying the RHS by 6, we get d00 : d0 : d= 6 : :. Find the rtio of the intercepts mde by () plne in simple cubic crystl.

39 9 P g e Solution b c The intercepts mde by the plne (hkl) is,, h k l b c rtio of the intercepts is,, the intercepts re,,. For the plne () the. For cubic unit cell, =b=c. Therefore,. The rtio of the intercepts, l:l:l = : : = : : The LCM is. Multiplying by the LCM, we get l:l:l = : : The rtio of the intercepts mde by () plne in simple cubic crystl is l:l:l = : :. In crystl, whose primitives re 0.8Å,.Å,.5Å plne () cuts n intercepts 0.8Å long X-xis. Find the lengths of the intercepts long Y nd Z xes. Solution b c The intercepts mde by the plne (hkl) is,, h k l b c rtio of the intercepts is,, The rtio of the intercepts, l:l:l = b c : : It is given tht, l= 0.8Å. Therefore, 0.8Å:l:l = b c : : Substituting the vlues of, b nd c, we get,. 0.8Å:l:l = 0.8Å: Å:.5 Å Solving, we get, l=0.å nd l=0.5å... For the plne () the. Find the nerest neighbour distnce in simple cubic, body centered cubic nd fce centered cubic unit cells. (i) Simple cubic unit cell

40 0 P g e In simple cubic unit cells, corner tom is the nerest neighbour to nother corner tom. The distnce between ny two corner toms is, i.e., the nerest neighbour distnce is. (ii) Body centered cubic unit cell In BCC unit cell, the body centered tom is the nerest neighbour to corner tom. The distnce between body centered tom nd corner tom is r= (iii) Fce centered cubic In FCC unit cell, the fce centered tom is the nerest neighbour to corner tom. The distnce between fce centered tom nd corner tom is, r= 8. For simple cubic lttice of lttice prmeters.0 Å, clculte the spcing of the lttice plne (). Solution d hk l The interplnr distnce is given by, 0.00 = =0.8 Å 5. The rdius of copper is.78 Å. It crystllizes in FCC unit cell. The tomic weight of copper is.5. Clculte the number of toms per unit cell. Its density is 8980 kg m -. Given dt Rdius of copper =.78 Å Atomic weight of copper =.5 Solution The tomic rdius of FCC crystl, r= 8 The intertomic distnce, = = 8 r =. x0-0 m The density nd tomic weight re relted by

41 P g e nm N A n N M A = =.07 The number of toms per Cu unit cell is = 6. The rtio of the intercepts of n orthorhombic crystl re :b:c=0.9::0.79. Wht re the Miller indices of the fces with the following intercepts: (i) 0.::0.88, (ii) 0.858::0.75, (iii) 0.9: :0. Solution (i) Given tht :b:c=0.9:: The intercepts 0.::0.88= 0. :: The coefficients re = c : b : ::. The inverses re,,,. Miller indices for the given plne is () ii) Given tht :b:c=0.858:: The intercepts 0.858::0.75= :: =:b:c The coefficients re,,. The inverses re, LCM is. Multiplying by the LCM, we get,,, Miller indices for the given plne is () iii) Given tht :b:c=0.9: : :: 0.79 The intercepts 0.9::0.= 0. 9 : : 0. 6 = c : b:

42 P g e The coefficients re : :. The inverses re,,0,. Miller indices for the given plne is (0) 7. How mny toms per unit re re there in (i) (00), (ii) (0) nd (iii) () plne in mteril tht crystllizes in fce centered cubic (FCC) unit cell. (i) (00) plne The number of toms presents per unit cell in (00) plne is +=. Number of toms per m = = 8r = r. Fig.. FCC unit cell (00) plne (ii) (0) plne The number of toms presents per unit cell in (0) plne is Number of toms per m = = 8 r r =. Fig.. FCC unit cell (0) plne

43 P g e (iii) () plne The number of toms presents per unit cell in () plne is Number of toms per m = = 8 r r =. 6 Fig.. FCC unit cell () plne 8 Clculte () the tomic pcking frction for FCC metls, (b) the ionic pcking frction of FCC NCl. Solution () Atomic pcking fctor (APF) = The pcking density of fce centered cubic unit cell is clculted s follows: Pcking density = Pcking density = = Totl volume of the toms Volume of the unit cell Number of toms per unit cell Volume of one tom Volume of the unit cell r Substituting, r= 8, we get, = Pcking density = (b) Ionic pcking fctor of NCl=

44 P g e Volume of N toms Volume of Volume of the unit cell Cl toms = r rr R where r nd R re the rdii of N + nd Cl - ions respectively. = r R rr Substituting the vlues of r nd R, we get, = ( 0) ( 0 ) =0.9=0.7. The ionic pcking fctor of NCl crystl is 0.7.

45 5 P g e SHORT QUESTIONS. Wht is ment by crystllogrphy?. Wht re crystlline mterils?. Wht re polycrystlline mterils?. Define the term non-crystlline mterils. 5. Wht re isotropic substnces? 6. Define the term nisotropy. 7. Define lttice. 8. Define bsis. 9. Define the term crystl structure. 0. Define the term unit cell.. Wht do you men by crystllogrphic xes?. Wht re primitives?. Define interfcil ngles.. Wht is primitive cell? 5. Wht re lttice prmeters? 6. Write the nmes of seven crystl systems. 7. Wht re Brvis lttices? 8. Define the term tomic rdius. 9. Wht do you men by coordintion number? 0. Wht is pcking density?. Determine the pcking density of simple cubic unit cell.. Obtin the tomic rdius of the tom in BCC unit cell.. Clculte the tomic rdius of the tom in FCC lttice.. Deduce the pcking density of simple cubic lttice. 5. Derive the pcking density of body centered cubic unit cell. 6. Obtin the pcking density of fce centered cubic unit cell. 7. Explin the rrngement of toms in HCP structure. 8. Clculte the number of toms in HCP unit cell. 9. Deduce the pcking density of HCP unit cell. 0. Deduce the reltion between intertomic distnce nd the tomic weight of substnce.. Wht re crystl plnes?. Wht re Miller indices?. List out the procedure used to find the Miller indices of crystl plne.

46 6 P g e. Mention ny four slient fetures of Miller indices. 5. Mention ny four dvntges of finding Miller indices. 6. Wht is fmily of plnes? Give exmples. 7. Wht is ment by direction of plne? 8. Wht is fmily of directions? Give exmples. 9. Wht re the steps to be followed to drw the direction of plne? 0. Define the term liner density.. Define the term plnr density>. Determine the pcking density of the dimond unit cell.. Determine the tomic rdius of the dimond unit cell.. Wht is llotropy? Give exmples. 5. Wht re polymorphisms? Give exmples. DESCRIPTIVE QUESTIONS. Explin simple cubic unit cell nd hence find the tomic rdius, number of toms present in simple cubic unit cell, coordintion number, nd pcking density of simple cubic unit cell.. Explin body centered cubic unit cell. Determine the tomic rdius, number of toms in unit cell, pcking density d coordintion number for body centered cubic unit cell.. Explin fce centered cubic unit cell, with net sketch. Obtin the number of toms in unit cell, tomic rdius, coordintion number, pcking density of fce centered cubic unit cell.. Explin the rrngement of toms in hexgonl unit cell. Determine the coordintion number, pcking density, tomic rdius nd number of toms in HCP structure. 5. Determine the pcking density of BCC, FCC nd HCP unit cells. 6. Determine the tomic rdius, nd coordintion number of SC, BCC, FCC nd HCP unit cells. 7. Show tht hexgonl unit cell demnds n xil rtio of Determine the pcking density of HCP unit cell. 8. Describe dimond unit cell nd hence determine the number of toms in unit cell, tomic rdius, coordintion number nd pcking density of dimond unit cells. c 8.