THE SPACE LATTICE AND UNIT CELL
|
|
- Holly Hopkins
- 5 years ago
- Views:
Transcription
1 6 Introution of Crystl Struture All the hlies MCl hve negtive enthlpies of formtion, whih inites tht thermoynmilly (tht is in terms of energy) it is fesile to form the ompouns MCl from the elements.. The most negtive enthlpies of formtion our with the fluories. For ny given metl, the vlues erese in the sequene fluorie > hlorie > romie > ioie. Thus the fluories re the most stle, n the ioies the lest stle.. The enthlpies of formtion for the hlories, romies n ioies eome more negtive on esening the group. This tren is oserve with most slts, ut the opposite tren is foun in the fluories. Ioni ompouns my lso e forme in solution, when similr yle of energy hnges must e onsiere, ut the hyrtion energies of the positive n negtive ions must e sustitute for the lttie energy. THE SPACE LATTICE AND UNIT CELL The regulr rrngement of n infinite set of points (toms, ions or moleule ) in spe is lle lttie or spe lttie. The spe lttie my e one imensionl, two imensionl n three imensionl epening upon the numer of prmeters require to efine it Tle Ltties n the Prmeters Require for Defining Spe Lttie Type of lttie Repet istnes Interfil ngles No. of prmeters require for efining the lttie -Dimensionl Only one repet istne -Dimensionl, 3-Dimensionl,,,, Two repet istnes n n interfil ngle Three repet istnes n three interfil ngles A spe lttie : The regulr rrngement of onstituent prtiles (i.e., toms, ions or moleules) of rystl in three imensionl spe is lle rystl lttie or spe lttis. Unit ell: The smllest three imensionl portion of omplete spe lttie when repete over n gin in ifferent iretions proues the omplete spe lttie is lle the unit ell. For esriing unit ell, we must know: () the istnes, n i.e; the lengths of the eges of unit ell. () the ngles, n etween the three imginry xes, Clssifition of Unit Cell : On the sis of the lotion of the lttie points within unit ell, we n lssify unit ells into the following types : Primitive unit ell Non-primitive unit ell () Simple or primitive: In this type, points (i.e, toms, ions or moleules) re present only t the orners of the unit ell. E.g. SCC (Simple Cui Cell). () Non-primitive : In this type of unit ell, the lttie points re present not only t orners ut lso t some other speifi positions. (i) Fe-entre: In this type, points re present t the orners s well s the entre of eh of the six fes.
2 Introution of Crystl Struture (ii) Boy-entre: In this type, points re present t the orners n n itionl point is present t the entre of the unit ell. 7 (iii) En fe entre : In this type, points re present t the orners n t the entre of the two enfes. (iv) Ege entre unit ell : Lttie points t ll orners + t eh ege entre. SOLVED EXAMPLES. The rystl lttie struture of soium, vnium n molyenum is BCC (Boy entere ui). Whih of the following metlli mixtures re most likely to form soli solution? [TIFR 00] () V n Mo () V n N () Mo n N () N, V, Mo Atomi rii : N = 0, V = 35, Mo = 45. Metlli mixtures of element hving omprle size form soli solution. Corret option is (). Whih of the prllelogrm in the figure elow re vile unit ells? [TIFR 00] i () i () ii ii iii () iii () ll of the ove Corret option is () Brvis lttie: Brvis showe tht the unit ells n e rrnge in regulr three imensionl orer in, the seven types of rystl systems in 4 ifferent wys. These rrngements re lle spe ltties orbrvis ltties. The 4 Brvis ltties my elong to either of the ove mentione four types of unit ells. Seon Clssifition of Unit Cell : On the sis of xil length n inter-fil ngle, the unit ells n e lssifie into the following 7 types whih re known s 7-rystl systems. Crystl Systems: At first sight there seems to e infinite numer of shpes of rystls. However, reful exmintion of severl thousn rystls of vrious sustnes hs revele tht only seven possile rystl symmetries re possile. The seven ifferent omintions of symmetry elements re lle rystls systems. These rystl systems iffer in the length of the unit ell eges (, n ) n the ngle etween the unit ell eges. Simple or Primitive (P) Boy-entere (I) Figure-() : Cui Spe Ltties Fe-entere (F)
3 Introution of Crystl Struture Simple or Primitive (P) Boy-entere (I) En-entere (C) Figure-() : Orthorhomi Spe Ltties Fe-entere (F) Simple (P) Boy-entere (I) Simple (P) En-entere (C) Figure-() : Tetrgonl n Monolini Spe Ltties Trilini Hexgonl Rhomoherl or Trigonl (R) Figure-() : Trilini, Hexgonl n Rhomoherl Spe Ltties Spe Groups : The olletion of totl numer of symmetry opertions of ll the symmetry elements in rystl is lle its spe group. Chrteristis of Seven types of Crystl Systems System No. of Brvis Ltties (4) Mximum symmetry elements. Cui 3 9 Plnes 3 xes. Tetrgonl 5 Plnes 5 xes 3. Hexgonl 7 Plnes 7 xes 4. Trigonl or 7 Plnes Rhomoherl 7 xes Axes n Angles = = = = = = = = = = Exmples NCl, KCl, CF, ZnS, Cu O, Dimon Alums, P, Ag, Au, Hg SnO, ZnO, TiO, NiSO 4, ZrSiO 4, White Sn ZnO, Pl, CS, HgS, Grphite, Ie Mg Zn C NNO 3, CSO 4, Clite, ICl, Qurtz, As, S, Bi Spe Group Orthorhomi (Rhomi) 4 3 Plnes 3 xes 6. Monolini Plnes xes 7. Trilini No Plnes No xes = = = = KNO 3, K SO 4, PCO 3, BSO 4, Rhomi Sulphur, MgSO 4 7H O N SO 4 0H O, N B 4 O 7 0H O, CSO 4 H O, Monolini Sulphur CuSO 4 5H O, K Cr O 7, H 3 BO
4 Introution of Crystl Struture SOLVED EXAMPLES 9. A rystl system hrterise y n 90º, 90º is [BHU-05] () trilini () monolini () rhomi () trigonl Corret option is (). Whih of the following is true for n orthorhomi lttie? [BHU 07] (), 90º (), 90º (), 90º, 90º (), Corret option is () 3. Wht is the unit ell hving imension n? [CUCET 07] () orthorhomi () monolini () trilini () rhomoherl Corret option is () 4. In tetrgonl rystl (), 90º (), 90º (), 90º (), 0º, 90º Corret option is () [DU 06] 5. Assign the Brvis lttie type for the following unit-ell struture [DU 07] () Tetrgonl I () Cui I () Orthorhomi I () Monolini 90 Therefore, tetrgonl rystl system. Corret option is () 6. One of ngle in monolini rystl system is [DU 07] () less thn 90 egree () greter thn 90 egree () less thn 30 egree () less thn 0 egree Monolini rystl system, Therefore, either greter or less then 90º. Corret option is (, ) Interplnr Distne: The perpeniulr seprtion etween jent plnes of rystl lttie is lle inter plnr sping, or the length of the perpeniulr from the origin on the unit plne represente y set of miller inies is lso known s interplnr sping. For n orthorhomi system, h k l So, hkl h k l ; hkl
5 0 Introution of Crystl Struture where h, k, l e Miller inies of the plne n,, re the imensions of the ell. For ui system, So, hkl h k l For tetrgonl system, So, hkl h k l SOLVED EXAMPLES. How o the spings of the three plnes 00, 0 n of ui lttie vry? hkl ( h k l ) ; 00 ( 0 0 ) 0 ( 0 ) ; ( ) 3 Thus, 00 : 0 : : : : : Potssium hlorie rystllises with oy entre ui lttie. Clulte the istne etween the 00, 0 n plnes. The length of the sie of the unit ell is 5. 34Å Sine we know tht for ui system, hkl ( h k l ) For 00 plne; 00.67Å ( 0 0 ) For 0 plne; Å ( 0 ) For plne,.54 Å ( ) 3. Consier rystl with simple ui lttie. If on heting, the unit ell volume inreses y 5% isotropilly (eqully in ll iretions), the perentge inrese in the (0) interplnr istne is [HCU 0] ().64 ().67 ().50 () % inrese 00
6 Introution of Crystl Struture Corret option is () %. 4. For simple ui system the sping of (00), (0) n () plnes re in the rtio of () : : 3 () : : 3 () :: () 3 : : [BHU-0] (00) (0) () Rtio : : : :. 3 3 Corret option is () LATTICE POINT The points representing the toms, moleules or ions in unit ell re known s lttie points. In the vrious unit ells, there re three kins of lttie points. () Points lote t the orners of unit ell. Sine suh point lies t the orner of unit ell, it is shre etween eight suh unit ells. Thus only (/) th prt of eh suh point ontriutes to ny one unit ell. () Points lote t the entre of fe of unit ell. Sine suh point is shre etween two suh unit ells, only / of eh suh point ontriutes to ny unit ell. Points lote t the entre of the unit ell. This point, eing present entirely within unit ell, wholly elongs to this unit ell. Contriution of lttie point t prtiulr lotion (per unit ell) : Lotion. Boy entre Contriution. Fe entre / 3. Ege entre /4 4. Corner / The numer of toms elonging to unit ell is lso terme s lttie sites n is enote y Z. Z = T.L.P. (Totl Lttie Points) Contriution Thus the totl numer of lttie points (or numer of toms) per unit ell in the four types of unit ells my e lulte s elow. () Simple or primitive: () Fe-entre: Z Z () Boy-entre: Z () En-entre : In ition to points present on the orners, suh unit ell lso hs points only on the en
7 Introution of Crystl Struture (two) fes. Sine eh point of the ltter type is shre etween two, unit ells, it ontriutes to to eh unit ell. Suh points re present only in en fe-entre unit ells. Thus the numer of toms per unit ell =. (e) Ege-entre : Z 4 4 The numer of prtiles immeitely jent to eh prtile in the rystl lttie is known s the oorintion numer for tht lttie. In other wors, oorintion numer is the nerest neighours surrounings prtiulr lttie point. It is hrteristi of given lttie. In simple ui lttie, eh prtile is joine y 6 other prtiles n so the oorintion numer of simple ui lttie is 6. Similrly, the oorintion numer for oy entre n fe entre ui ltties re n respetively. Lttie Z C.N.. SCC 6. BCC 3. FCC 4 4. En CC 4 5. HCP 6 SOLVED EXAMPLES. A ompoun hving BCC lttie struture is forme y elements X n Y Atoms X re present t the orners n tom Y t the entre of the ue. Derive the fornul of the ompoun? Sine orner toms re shre etween eight orners, therefore, the numer of toms, X t eight orners = tom The tom Y present t the entre of ue is not eing shre i.e. one tom; unit ue Hene the formul of the ompoun is XY.. A ompoun, forme y elements X n Y, rystllises in the ui struture, where Y toms re t the orner of ue n X toms re t lternte fes. Wht is the formul of the ompoun? Numer of toms Y t eight orners of ui unit ell = = Numer of toms X present on lternte fe of unit ell = = Formul of ompoun = XY 3. In rystlline soli nions A, re rrnge in p mnner. 50% of the otherl vois re oupie y B n 50% of the tetrherl vois re oupie y toms C. Wht is the formul of the soli? We know tht the numer of otherl vois re equl in numer to the toms forming p, while the numer of tetrherl vois re oule in numer. Rtio of the numer of otherl vois n tetrherl vois is : Now 50% of the otherl vois re oupie, therefore the rtio of A n B is A : B The 50% of the tetrherl vois re oupie y C so, the rtio of A n C is : i.e. A : C or A : C, Hene the rtio of A, B n C is A : B : C So, the formul of the ompoun is A BC.
8 Introution of Crystl Struture 4. Clulte the numer of toms present in the unit ell of monotomi sustne (element) of () simple ui lttie-() oy-entre ui () fe-entre ui. () Numer of toms in unit ell of Simple Cui Lttie. A simple ui lttie hs only eight toms on the orners. As ontriution y eh therefore numer of toms present in the unit ell = () Numer of toms in unit ell of Boy Centre Cui (BCC). This littie hs toms on the orners n one tom within the oy. Therefore, Contriution y toms present on the orners = Contriution y the tom present within the oy = Therefore, Numer of toms present in the unit ell = + =. () Numer of toms in unit ell of the Fe Centre Cui (FCC). This lttie hs toms on the orners n 6 toms on the fes (one on eh fe). Contriution y toms on the orners = Contriution y toms on the fes 6 = 3 Therefore, Numer of toms present in the unit ell = + 3 = Clulte the numer of toms in ui se unit ell hving one tom on eh orner n two toms on eh oy : igonl. There re four oy igonls. Thus there re toms within the oy of the unit ell whih re not shre y ny other unit ell. Contriution y toms unit ell t the orners =. Hene totl tom/unit ell = + = If three elements P, Q n R rystllize in ui soli lttie with P toms t the orners, Q toms t the ue entre n R toms t the entre of the fes of the ue, then write the formul of the ompoun. Atoms P per unit ell =. Atoms Q per unit ell = Atoms R per unit ell = 6 3 Hene the formul is PQR In fe entre rrngement of A n B toms. Where A toms re t the orners of the unit ell n B toms re t the fe entres. For eh unit ell, one A tom is missing from orner position n one B toms is missing from one fe position. The simplest formul of the resulting ompoun will e () A 4 B 40 () A 7 B 0 () A -x B 3-x () AB [TIFR 06] 7 One A tom is missing from orner position = 7. 5 One B tom is missing from one fe position = 5. A7/ B5/ Multiply y : A7 B 0. Corret option is (). A ompoun lloy of metls P n Q hs unit ell ontining P toms t the orners, while tom Q re present t the fe entres. The formul of the ompoun shoul e [BHU-0] () PQ () PQ () PQ 3 () P 3 Q P Corners Formul PQ3. Corret option is () Q Fe entres 6 3 3
9 4 Introution of Crystl Struture 9. An element rystllizes in n FCC ltttie. How mny toms re there per unit ell? [BHU-0] () () () 3 () 4 For FCC lttie, Z = 4. Corret option is () 0. An element rystllizes in BCC lttie. How mny toms re there per unit ell? [BHU-03] () () () 4 () 9 For BCC lttie, Z =. Corret option is () NUMBER AND LOCATION OF VOIDS IN A CRYSTAL In lose-pke struture (p or hp), if there re N spheres (toms or ions) in the pking, then Numer of otherl vois = N, Numer of tetrherl vois = N For exmple, in the ui lose pking (p) i.e. fe-entre ui (f) unit ell there re 4 toms or ions, therefore, there re 4 otherl vois. () Otherl vois: One otherl voi is present t the oy entre of the ue n otherl vois re present on the entres of the eges of the ue. But eh voi on the ege entre is shre y 4 unit ells. Hene its ontriution in the unit ell = 4 Hene the effetive numer of otherl vois in the p struture () Tetrherl vois: The tetrherl vois present in the p rise from the ft tht there re spheres t the orners of the unit ell n eh sphere touhes three spheres present on the fe entres of the three joining fes, eh giving rise to one tetrherl voi. Note: (i) Numer of otherl vois = Numer of toms present in the lose pke rrngement n (ii) Numer of tetrherl vois = Numer of otherl vois = Numer of toms.
10 Introution of Crystl Struture 5 SOLVED EXAMPLES. Rtio of the numer of tetrherl vois to otherl vois in ui lose pking (p) rrngement is () : () : () : () 3 : [HCU-03] p rrngement (Z = 4) Tetrherl vois = Otherl vois = 4 Rtio = :. 4 Corret option is (). In rystlline soli, nions B re rrnge in ui lose pking. Ctions A re eqully istriute etween otherl n tetrherl vois. If ll the otherl vois re oupie, wht is the formul of the soli? Suppose the numer of nions B = 00 Then Numer of otherl vois = 00 Numer of tetrherl vois = 00 As otherl n tetrherl vois re eqully oupie y tions A n ll the otherl vois re oupie (given), therefore 00 tions A re present in otherl vois n 00 tion A re present in tetrherl vois. In other wors, orresponing to 00 nions B, there re 00 tions A i.e. tion A n nions B re in the rtio :. Hene the formul of the soli will e A B. 3. In orrunum, oxie ions re rrnge in hexgonl lose pking n luminium ions oupy two-thir of the otherl vois. Wht is the formul of orrunum. Suppose oxie ions = 90. Then otherl vois = 90. Hene A 3+ ions 3 90 = 60. Therefore, rtio A 3+ : O = 60 : 90 = : 3 i.e. formul is A O In soli, oxie ions re rrnge in p.one-sixth of the tetrherl vois re oupie y the tions A while one-thir of the otherl vois re oupie y the tion B. Wht is the formul of the ompoun? Suppose O ions = 90. Then otherl vois = 90 n tetrherl vois = 0. Ctions A 0 = 30 6 n tion B 90 = Therefore, rtio A : B : O = 30 : 30 : 90 = : : 3 i.e. formul is ABO A soli is me up of two elements P n Q. Atoms Q re in p rrngement while toms P oupy ll the tetrherl sites. Wht is the formul of the ompoun? Sol. Suppose numer of toms Q = n Then numer of tetrherl sites = n Numer of toms P = n Rtio P : Q = n : n = : i.e. formul is P Q. STRUCTURES OF IONIC COMPOUNDS It hs lrey een isusse tht ioni ompouns onsist of positive n negtive ions rrnge in mnner so s to quire minimum potentil energy (mximum stility). To hieve the mximum stility, ions in rystl shoul e rrnge in suh wy tht fores of ttrtion re mximum n fores of repulsion re minimum. Hene, for mximum stility the oppositely hrge ions shoul e s lose s possile to one nother n similrly hrge ions s fr wy s possile from one nother.
11 6 Introution of Crystl Struture Among the two ions onstituting the inry ompouns, the lrger ions (usully nions) form lose-pke rrngement (hp or p) n the smller ions (usully tions) oupy the interstitil vois. Thus in every ioni ompoun, positive ions re surroune y negtive ions n vie vers. Normlly eh ion is surroune y the lrgest possile numer of oppositely hrge ions. This numer of oppositely hrge ions surrouning eh ion is terme its oorintion numer. The oorintion numer of positive n negtive ions of ompoun re sme when the two types of ions re equl in numer (e.g., NCl, ZnS, et.). On the other hn, when n ioni ompoun ontins ifferent numer of positive n negtive ions (s in CC, N S et.), the oorintion numers of positive n negtive ions re ifferent. For exmple, in CC sine Cl ions re twie the numer of C + ions, the oorintion numer of lium ion is twie the oorintion numer of hlorie ion. Simple ioni ompouns re of two types, i.e., AB n AB type. From the knowlege of losepke strutures n the vois evelope therein, we n hve n ie out the struture of simple ioni ompouns. () If the nions (B ) onstitute the rystl lttie n ll otherl vois re oupie y tions (A + ), then the formul of the ioni soli is A + B. () Similrly, if hlf of the tetrherl vois re oupie y tions, then the formul of the soli rystl eomes A + B. () When the nions (B ) re onstituting spe lttie n ll the tetrherl vois re oupie y the tions (A + ), then the formul of the soli rystl will e A B. () When the nions (B ) re present t the lttie points n ll the otherl vois re oupie y the tions (A + ), then the formul of rystlline soli will eome A B. AB type ioni solis NCl, CsCl, ZnS AB type ioni solis CF, SrF Ioni Compouns of AB Type:. Struture of soium hlorie (NCl) or Rok Slt rystl : It hs f rrngement lso lle ui lose pke (p). Slient fetures of NCl rystl struture re summrise elow: (i) The nions (Cl ) re present t the lttie points of fe entre ui lose struture. (ii) N + ions re oupying ll the otherl vois, so eh N + ion is surroune y six Cl ions. (iii) Sine there will e six otherl vois (holes) roun eh hlorie ion, so eh hlorie ion is surroune y six soium ions. (iv) The o-orintion numer of Cl s well s of N + ion is six. Therefore, it is terme s 6:6 o-orintion rystl. (v) On pplying pressure, NCl struture (6 : 6 oorintion hnges to CsCl struture ( : oorintion). (vi) The ioni rius of N + ion (r = 95 pm) n the rius of Cl ion (R = pm) gives rius rtio r/r = This vlue of rius rtio suggest n otherl rrngement of ions. (vii) The totl numer of N + tion n Cl nion present in one unit ell re lulte s elow. Numer of soium ions : In NCl, the N + ions hve een shown y rk spheres), = ( ions t the ege entre ) + ( ion t oy entre ) = 3 + = 4 4 Numer of hlorie ions : = ( ions t the orner ) + (6 ions t the fe entres ) = + 3 = 4 Therefore, the unit ell of NCl hs ontriution of 4 N + ions n 4 Cl ions i.e. 4 NCl formul unit per unit ell. The soium hlorie struture is lso lle s rok-slt struture. (viii) The ege length of the unit ell of NCl type of rystl (r + R). Ioni rius Ege length () of unit ell of the NCI type = of the tion Ioni rius of the nion
12 Introution of Crystl Struture 7 Or ( r r ) Thus the istne etween N + n Cl ions Exmples hving rok-slt struture i.e. NCl type of struture re Alkli hlies : NCI, LiCl, KBr, RI, AgCl, AgBr, AgI, NH 4 Cl, NH 4 Br, NH 4 I Oxies of lkline erth metls : MgO, CO, TiO, FeO, NiO N Cl. Potssium Chlorie (KCl): * KCl is metl hlie slt. * In its pure stte, it is oorless n hs white or olourless rystl pperne with rystl struture tht leves esily in three iretion. * It is fe entrere ui rystl. * Struture similr to tht of NCl. K + Figure 3. Cesium hlorie (CsCl) type struture : In this type of ioni rystl, the size of Cs + ion is quite ig s ompre to Cl ion. Therefore, it quires the oy entre ui lttie struture
13 Introution of Crystl Struture (i) The Cl ions re present t the orners of the ui unit ell n the Cs + ions re t the oy entre of the unit ell. (ii) Eh Cs + ion in this moe of pking is touhing hlorie ions n eh Cl ion is touhing eight Cs + ions. Therefore, this struture will hve : o-orintion. (iii) At high temperture, CsCl struture ( : o-orintion) hnges to NCl struture (6 : 6 oorintion). (iv)the ioni rii of Cs + n Cl ions re 69 pm n pm respetively. This gives rius rtio s r 69 pm 0.93 R pm (v) This vlue of rius rtio orrespons to BCC type of unit ell. (vi) Eh unit ell will hve the ontriution of one Cs + n one Cl ion s shown elow Numer of Cs + ion = (ion t the oy entre) = Numer of Cl ion = (ions t the orner) / = Therefore, the unit ell hs ontriution of one Cs + ion n one C ion per unit ell (vii) Reltion etween rius of tion, nion n ege length of the ue 3 r r Cs Cl The unermentione ioni rystls re the exmples of CsCl type of rystl lttie. CsCl, CsBr, CsI, TlCl, TlBr, CS. 4. Zin sulphie struture : Zin sulphie exists in two forms: (i) ZnS Blene (ii) Zns Wurtzite (i) zin lene or Sphlerite: Struture of zin lene is ui lose pking (p). (i) The sulphie ions (S ) oupy fe entre ui lttie points. n the zin ions (Zn + ) oupy hlf of the totl numer of tetrherl vois (ii) Eh sulphie ion is surroune y 4 Zn + ions n eh Zn + ion is surroune y 4S ions. Therefore, ZnS hs 4 : 4 oorintion. (iii) The ioni rii of zin ion (Zn + ) n sulphie ions (S ) re 74 pm n 4 pm respetively. Therefore, the rius rtio r/r = 0.40 It suggests the tetrherl rrngement of ions in the-rystl lttie. (iv) The ontriution of Zn + n S ions per unit ell is lulte s isusse elow. Numer of sulphie ions = ( ions t the orners ) + (6 ions present t the fes )= + 3 = 4 Numer of zin ions = 4 (ions within the oy of unit ell) = 4 Thus the numer of ZnS units per unit ell is 4. The following ioni solis re oserve to form ZnS type of rystl struture. ZnS, CuCl, CuBr, CuI, US, AgI, BeS (ii) Wurtzite Struture. The unit ell for wurtzite struture is shown in following figure in whih eh Zn + ion is represente y hollow irle n eh S ion is represente y soli irle. The wurtzite struture my e esrie s follows:. The sulphie ions re rrnge in hexgonl lose pke (hp) rrngement.. The zin ions oupy hlf of the tetrherl sites.
14 Introution of Crystl Struture 9 3. There re two tetrherl sites for eh S ion in the lttie. Sine only hlf of them re oupie y Zn + ions, the stoihiometry of the ompoun is :. 4. Eh S ion is surroune y four Zn + ions whih re ispose towrs the orners of regulr tetrheron. Similrly, eh Zn + ion is surroune y four S ions. The rrngement roun one Zn + ion n one S ion is shown in following Figure. Therefore, the oorintion numers of Zn + n S ions in wurtzite struture re 4 : The rius rtio r + / r in this struture is 0.40 whih lso suggests tht the oorintion numer is 4 n tht the rystl hs tetrherl struture. Thus, the rius rtio rule supports the ove struture. The ompouns suh s AgI, ZnO, NH 4 F n AlN hve wurtzite struture. It will e note tht the si ifferene etween zin lene n wurtzite struture is tht wheres in the former se the sulphie ions re rrnge in p type of pking, in the ltter se they re rrnge in hp type of pking. 5. Dimon Cui Strutures: The imon lttie n e onsiere to e forme y interpenetrting two f ltties long the oy igonl y (/4) th ue ege. One sulttie hs its origin t the point (0, 0, 0) n the other t point qurter of the wy long the oy igonl (t the point /4, /4, /4). The si imon lttie n the tomi positions in the ui ell of imon projete on ue fe re shown in figure elow. The frtions enote height out the se in units of ue ege. The point t 0 n / re on the f lttie, those t /4 n 3/4 re on similr lttie isple mong the oy igonl y /4 of the ue ege. The pking ftor of this struture is thus lulte s follows: XY 4 4 XZ XY YZ But, XZ r Therefore, r 3 6 (or) the nerest neighour istne, r Figure: Dimon Struture 3 4 Z
15 30 Introution of Crystl Struture Lttie onstnt, r 3 -Cristollite Struture : SiO Si 4+ = In DCC lttie. 4 3 r r Pking ftor v/v 0.34 or 34% r 6 O = Eh oxie ion in etween ny Si-Si on. 4 Z :Si 6 4. O 6 6. F. 4 Si 4+ O C.N.:Si 4 n O. Effet of Pressure on Crystl Struture: Inrese of pressure inreses the o-orintion numer uring rystlliztion e.g. y pplying high pressure, the NCl rystl struture hving 6 : 6 o-orintion numer hnges to CsCl rystl struture hving oorintion numer :. high pressure NCl type rystl CsCl type rystl (6:6 o-orintion) (: o-orintion) Effet of Temperture on Crystl Struture : Inrese of temperture, however, ereses the o-orintion numer e.g., upon heting to 760 K, the CsCl rystl struture hving o-orintion of : hnges to NCl rystl strutures hving o-orintion 6 : 6 high tempressure CsCl type rystl NCl type rystl (: o-orintion) (6:6 o-orintion) SOLVED EXAMPLES. Upon heting to out 500ºC CsCl rystl hnges its struture to rok slt struture. Wht hppens to the oorintion numer of Cs? [BHU-04] () hnges from 6 to () hnges from to () hnges from to 6 () oes not hnge high tempressure CsCl type rystl NCl type rystl (: o-orintion) (6:6 o-orintion) Corret option is () Ioni Compouns of AB Type: Fluorite struture (rystl struture of CF ) : The lium fluorie rystl is ompose of lium ions (C + ) n fluorie ions (F ).
16 Introution of Crystl Struture 3 It hs Cui Close pke (p) rrngement. (i) The C + ions re present s the fe entre ui lttie n the fluorie ions (F ) oupy ll the tetrherl vois. (ii) The rius rtio, lulte from the ioni rii of C + ion (99 pm) n F ion (36 pm), is This shows tht the oorintion numer of lium ion is eight, i.e. eh lium tion is surroune y eight fluorie nions in oy entre ui rrngement. Eh fluorie ion is in ontt with four lium ions. Thus CF hs : 4 o-orintion. (iii) Eh unit ell of CF rystl is oserve to hve 4C + ions n F ions. Numer of C + ions: = ( ions t the orners ) + (6 ions t the fe entre ) = + 3 = 4 Numer of F ions = ( ions within the oy ) = Therefore, eh unit ell hs 4 CF moleules. Solis forming fluorite type of struture re CF, SrF, BF, BCl, CF, HgF, PF, CuF, SrCl et. Ioni Compouns of A B Type Antifluorite struture (rystl struture of Li O): Antifluorite struture is hving rrngement of tions n nions opposite to fluorite struture. (i) In the rystl struture of Li O, the O ions onstitute ui lose pke lttie (FCC struture) n the Li + ions oupy ll the tetrherl vois (ii) Eh oxie ion, O ion, is in ontt with Li + ions n eh Li + ion is hving ontt with 4 oxie ions. Therefore, Li O hs 4 : o-orintion. Exmples forming ntifluorite struture inlue N O, K O, K S n N S. Struture of ferrite n relte ompouns: 3 Ferrite (Fe 3 O 4 ) n relte ompouns of generl formul M Fe O4 re ferrimgneti in nture. The M + n e Fe +, Mn +, Zn +, Mg +, Co +, Ni + or some omintion of these ions. Ferrite (Fe 3 O 4 ) is the oule oxie hving omposition FeO Fe O 3. This shows, tht the numer of Fe 3+ ions is twie the numer of Fe + ions. Their respetive rrngement in the rystl lttie my e illustrte s elow. (i) In, Fe 3 O 4, the oxie ions onstitute the ui lose pke (FCC) rrngement. (ii) All the Fe + ions n hlf of the Fe 3+ ions oupy otherl vois, while the remining hlf of the Fe 3+ ions oupy tetrherl vois. Struture of Titnium Dioxie, TiO (Rutile Struture) : The unit ell of Rutile struture is shown in following figure in whih titnium ions re represente y hollow irles n oxie ions re represente y soli irles. The struture my e esrie s follows : Ti 4+ = In istorte oy entere ui. O = oxie ions re present t the entres of the tringle generte y ny opposite eges t the oy entre Ti. 4 oxie ions re present in plne whih is perpeniulr to the plne initil oxie ions. These 4 oxie ions re present on opposite fes (two on eh fe). Eh of these 4 oxie ions is present t /4 th istne of the fe igonl from orner long the fe igonl. These 4 oxie ions generte squre plnr struture, n ll the 6 oxie ions generte n otherl struture t the oy entre Ti.
17 3 Introution of Crystl Struture Z 4 Ti F O C.N. Ti 6 n O 3. Struture of Clium Crie (CC ) : The unit ell of lium rie is shown in following figure. The struture of lium rie is similr to tht of soium hlorie. Here soium ions re reple y lium ions n hlorie ions y C groups. The ron toms re ssoite in pirs, s shown in following figure, forming the C groups. The C groups re ligne in prllels s shown. The ui symmetry oserve in soium hlorie is, however, istorte onsierly in the se of lium rie for ovious resons. The struture of FeS is similr to tht of CC. However, the S units re not ligne in prllels s re the C units in the se of CC. C + = At oy entre + t eh ege entre Oh voi. C At orners + t eh fe entres FCC C 4 Z 4 C 6 4 F 4 C.N. C 6 n C 6 Struture of Cmium Ioie (CI ) : The unit ell of mium ioie is shown in following figure in whih C + ions re represente y hollow irles n ioie ions re represente y soli irles. The slient fetures of this struture re esrie elow:. The ioie ions re rrnge in hexgonl lose pke (hp) rrngement while mium ions lie in otherl sites etween every two lyers of ioie ions.. Eh C + ion is surroune otherlly y six I ions, s shown in following figure. On the other hn, eh I ion hs three C + ions s nerest neighours. The three C + ions surroun the I ion in n rrngement in whih C + ions form the se of trigonl pyrmi whose pex is the I ion. 3. The rrngement of the three lyers orrespons to the omposition CI. This rrngement is ontinue s shown n the struture is referre to s lyer lttie. The fores operting in etween the I C I lyers to hol them together re wek. As result, this ompoun is flky n n e esily leve.
18 Introution of Crystl Struture 33 The Ilmenite Struture: Ilmenite is the minerl FeTiO 5. Its struture is losely relte to the orunum struture exept tht the tions re of two kins. In ilmenite the tions re Fe + n Ti 4+, ut mny sustnes with the ilmenite struture hve tions with hrges of (+, +5) or (+3, +3). The Perovskite Struture: Perovskite is the minerl CTiO 3. Its struture, shown in following figure, is se on p rry of oxie ions together with lrge tions, similr in size to the oxie ion. The smller tions lie in otherl holes forme entirely y oxie ions. Agin, the iniviul tion hrges re not importnt so long s their sum is +6. The struture is opte y mny fluories with tions of ifferent sizes, suh s KZnF 3. C + = At oy entre Ti 4+ = At ll orners C O 4 = At eh ege entre Z Ti O 3 F Chrteristi properties of vrious types of ioni solis C.N. From C O C.N. 4 Ti 6 O 4 O Ti
19 34 Introution of Crystl Struture S. No. Struture Ction Anion Numer of ions F C.N. Ction Anion Totl Ction Anion. NCl,KCl Oh voi FCC CC Oh voi FCC CF, BF, FCC All T voi SrF 4. N O All T FCC voi 5. CTiO 3, C + : Ege entre 3 5 CTiF 3 Boy entre Ti 4+ : Corner 6 6. ZnS(B) Hlf T FCC voi 7. ZnS(W) Hlf T voi HCP ol new SiO DCC Si O Si 9. TiO Distorte BCC CsCl Boy entre Corner SOLVED EXAMPLES. Wht is the empiril formul for the following unit ell: [TIFR 00] Ti O () CTi O () CTi O () CTiO 3 () CTi O 3 C Boy entre Ti Corners 4 O Ege entre 3 4 CTiO 3 Corret option is ()
20 Introution of Crystl Struture 35. NCl, KCl, NBr n KBr rystllize in FCC ltties. Their noin n tion touh long the ege of the unit ell. The imensions of their unit ells re 56. pm, 67.7 pm, 596. pm n 65.6 pm, respetively. From these t, wht n you sy out the size of the ioni rii (within n error of out 5%)? () Ioni rii of the tions epen on the nture of the nions [TIFR 05] () Ioni rii of the nions epen on the nture of the tions () Both () n () () Ioni rii re inepenent of the ions Corret option is () 3. KCl rystllizes in ui unit ell with Cl ions t eh vertex n fe entre. How mny K + ions n Cl ions re there is eh unit ell of KCl? [HCU-0] () K + ion n Cl ion () K + ions n Cl ions () 4 K + ions n 4 Cl ions () K + ions n Cl ions Cl t eh vertex n fe entre K + is present in otherl vois = 4. 4K + ions n 4Cl ions re there in eh unit ell of KCl. Corret option is () 4. The Perovskite struture hs the formul A II B IV O 3. The A tions form primitive ui lose pking with the oxie ions oupying the entres of the fes n B tions t the oy entre. The totl numer of ions of A, B n O in the unit ell re, respetively. [HCU-04] (), n 3 (), n 6 (), n 3 (), n 3 Perovskite struture A II B IV O 3. A II tion = only orners s O oupy fe entre position. O ion = entre of fes 6 3. B IV tion = oy entre =. Corret option is () 5. Perovskite is the minerl CTiO 3. The Perovskite rystl struture is opte y severl oxies s well s some fluries. Whih one, mong the given formule; most likely represents known fluorie hving the perovskite struture? [BHU-0,5] () CTiF 3 () KZnF 3 () CTiF 5 () CMgF 4 Corret option is () 6. CsF opts the NCl rystl struture. If the unit ell ege is of length 4.0 A, wht is the shortest istne etween the tion n nion in the rystl? [BHU-0] ().0 A ().4 A () 3.4 A () 4.0 A 4.0 = 4.0,.0. Corret option is () 7. Perovskite is the minerl CTiO 3. The perovskite rystl struture is opte y severl oxies s well s some fluories. Whih one, mong the given formule, most likely represents known fluorie hving the perovskite struture? [BHU-04] () CTiF 3 Corret option is () () KMnF 3 () NMnF 4 () CFeF 3
21 36 Introution of Crystl Struture. A ompoun lloy Cu 3 Au rystllizes in ui lttie with Cu t the fe enters n Au t the orners. How mny formul units of the ompoun re there in eh unit ell? [BHU 0] () 4 () 3 () () Cu = Fe entre 6 3. Au = Corners. Therefore, nswer is AuCu 3. i.e., formul unit =. Corret option is () 9. How mny toms re there in unit ell for the imon lttie? [BHU 0] () () 4 () 6 () Corret option is ()
ENERGY AND PACKING. Outline: MATERIALS AND PACKING. Crystal Structure
EERGY AD PACKIG Outline: Crstlline versus morphous strutures Crstl struture - Unit ell - Coordintion numer - Atomi pking ftor Crstl sstems on dense, rndom pking Dense, regulr pking tpil neighor ond energ
More informationNumbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point
GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply
More informationChapter 4rth LIQUIDS AND SOLIDS MCQs
Chpter 4rth LIQUIDS AND SOLIDS MCQs Q.1 Ioni solis re hrterize y () low melting points () goo onutivity in soli stte () high vpour pressure () soluility in polr solvents Q.2 Amorphous solis. () hve shrp
More information1 This diagram represents the energy change that occurs when a d electron in a transition metal ion is excited by visible light.
1 This igrm represents the energy hnge tht ours when eletron in trnsition metl ion is exite y visile light. Give the eqution tht reltes the energy hnge ΔE to the Plnk onstnt, h, n the frequeny, v, of the
More informationMid-Term Examination - Spring 2014 Mathematical Programming with Applications to Economics Total Score: 45; Time: 3 hours
Mi-Term Exmintion - Spring 0 Mthemtil Progrmming with Applitions to Eonomis Totl Sore: 5; Time: hours. Let G = (N, E) e irete grph. Define the inegree of vertex i N s the numer of eges tht re oming into
More informationSolutions for HW9. Bipartite: put the red vertices in V 1 and the black in V 2. Not bipartite!
Solutions for HW9 Exerise 28. () Drw C 6, W 6 K 6, n K 5,3. C 6 : W 6 : K 6 : K 5,3 : () Whih of the following re iprtite? Justify your nswer. Biprtite: put the re verties in V 1 n the lk in V 2. Biprtite:
More informationIdentifying and Classifying 2-D Shapes
Ientifying n Clssifying -D Shpes Wht is your sign? The shpe n olour of trffi signs let motorists know importnt informtion suh s: when to stop onstrution res. Some si shpes use in trffi signs re illustrte
More informationEigenvectors and Eigenvalues
MTB 050 1 ORIGIN 1 Eigenvets n Eigenvlues This wksheet esries the lger use to lulte "prinipl" "hrteristi" iretions lle Eigenvets n the "prinipl" "hrteristi" vlues lle Eigenvlues ssoite with these iretions.
More informationComparing the Pre-image and Image of a Dilation
hpter Summry Key Terms Postultes nd Theorems similr tringles (.1) inluded ngle (.2) inluded side (.2) geometri men (.) indiret mesurement (.6) ngle-ngle Similrity Theorem (.2) Side-Side-Side Similrity
More informationBravais lattices and crystal systems
3 Brvis ltties nd rystl systems 3. Introdution The definitions of the motif, the repeting unit of pttern, nd the lttie, n rry of points in spe in whih eh point hs n identil environment, hold in three dimensions
More informationCHEM1611 Answers to Problem Sheet 9
CEM1611 Answers to Prolem Sheet 9 1. Tutomers re struturl isomers whih re relte y migrtion of hyrogen tom n the exhnge of single on n jent oule on. Compoun Tutomer 2 2 2 2 2 2 2 2 2 2 2 2. () Whih pir
More information912 o C 1400 o C 1539 o C α iron γ iron δ iron. liquid iron BCC FCC BCC
Polymorphism or Allotropy Mny elements or ompounds exist in more thn one rystlline form under different onditions of temperture nd pressure. This phenomenon is termed polymorphism nd if the mteril is n
More informationCIT 596 Theory of Computation 1. Graphs and Digraphs
CIT 596 Theory of Computtion 1 A grph G = (V (G), E(G)) onsists of two finite sets: V (G), the vertex set of the grph, often enote y just V, whih is nonempty set of elements lle verties, n E(G), the ege
More informationCS 491G Combinatorial Optimization Lecture Notes
CS 491G Comintoril Optimiztion Leture Notes Dvi Owen July 30, August 1 1 Mthings Figure 1: two possile mthings in simple grph. Definition 1 Given grph G = V, E, mthing is olletion of eges M suh tht e i,
More informationSurds and Indices. Surds and Indices. Curriculum Ready ACMNA: 233,
Surs n Inies Surs n Inies Curriulum Rey ACMNA:, 6 www.mthletis.om Surs SURDS & & Inies INDICES Inies n surs re very losely relte. A numer uner (squre root sign) is lle sur if the squre root n t e simplifie.
More informationQUB XRD Course. The crystalline state. The Crystalline State
QUB XRD Course Introduction to Crystllogrphy 1 The crystlline stte Mtter Gseous Stte Solid stte Liquid Stte Amorphous (disordered) Crystlline (ordered) 2 The Crystlline Stte A crystl is constructed by
More informationProportions: A ratio is the quotient of two numbers. For example, 2 3
Proportions: rtio is the quotient of two numers. For exmple, 2 3 is rtio of 2 n 3. n equlity of two rtios is proportion. For exmple, 3 7 = 15 is proportion. 45 If two sets of numers (none of whih is 0)
More informationfor 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
Applitions of Integrtion Are of Region Between Two Curves Ojetive: Fin the re of region etween two urves using integrtion. Fin the re of region etween interseting urves using integrtion. Desrie integrtion
More informationTrigonometry Revision Sheet Q5 of Paper 2
Trigonometry Revision Sheet Q of Pper The Bsis - The Trigonometry setion is ll out tringles. We will normlly e given some of the sides or ngles of tringle nd we use formule nd rules to find the others.
More information1 This question is about mean bond enthalpies and their use in the calculation of enthalpy changes.
1 This question is out men ond enthlpies nd their use in the lultion of enthlpy hnges. Define men ond enthlpy s pplied to hlorine. Explin why the enthlpy of tomistion of hlorine is extly hlf the men ond
More informationLESSON 11: TRIANGLE FORMULAE
. THE SEMIPERIMETER OF TRINGLE LESSON : TRINGLE FORMULE In wht follows, will hve sides, nd, nd these will e opposite ngles, nd respetively. y the tringle inequlity, nd..() So ll of, & re positive rel numers.
More informationSTRUCTURAL ISSUES IN SEMICONDUCTORS
Chpter 1 STRUCTURAL ISSUES IN SEMICONDUCTORS Most semiconductor devices re mde from crystlline mterils. The following gures provide n overview of importnt crystlline properties of semiconductors, like
More informationChem 130 Second Exam
Nme Chem 130 Second Exm On the following pges you will find seven questions covering vries topics rnging from the structure of molecules, ions, nd solids to different models for explining bonding. Red
More informationQUADRATIC EQUATION. Contents
QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,
More informationGM1 Consolidation Worksheet
Cmridge Essentils Mthemtis Core 8 GM1 Consolidtion Worksheet GM1 Consolidtion Worksheet 1 Clulte the size of eh ngle mrked y letter. Give resons for your nswers. or exmple, ngles on stright line dd up
More informationProbability. b a b. a b 32.
Proility If n event n hppen in '' wys nd fil in '' wys, nd eh of these wys is eqully likely, then proility or the hne, or its hppening is, nd tht of its filing is eg, If in lottery there re prizes nd lnks,
More information18.06 Problem Set 4 Due Wednesday, Oct. 11, 2006 at 4:00 p.m. in 2-106
8. Problem Set Due Wenesy, Ot., t : p.m. in - Problem Mony / Consier the eight vetors 5, 5, 5,..., () List ll of the one-element, linerly epenent sets forme from these. (b) Wht re the two-element, linerly
More informationTranslation symmetry, Space groups, Bloch functions, Fermi energy
9/7/4 Trnsltion symmetry, Spe groups, Bloh funtions, Fermi energy Roerto Orlndo Diprtimento di Chimi Università di Torino Vi ietro Giuri 5, 6 Torino, Itly roerto.orlndo@unito.it Outline The rystllogrphi
More informationAnalytical Methods for Materials
Anlyticl Methods for Mterils Lesson 7 Crystl Geometry nd Crystllogrphy, Prt 1 Suggested Reding Chpters 2 nd 6 in Wsed et l. 169 Slt crystls N Cl http://helthfreedoms.org/2009/05/24/tble-slt-vs-unrefined-se-slt--primer/
More informationFactorising FACTORISING.
Ftorising FACTORISING www.mthletis.om.u Ftorising FACTORISING Ftorising is the opposite of expning. It is the proess of putting expressions into rkets rther thn expning them out. In this setion you will
More informationArea and Perimeter. Area and Perimeter. Solutions. Curriculum Ready.
Are n Perimeter Are n Perimeter Solutions Curriulum Rey www.mthletis.om How oes it work? Solutions Are n Perimeter Pge questions Are using unit squres Are = whole squres Are = 6 whole squres = units =
More informationIntermediate Math Circles Wednesday 17 October 2012 Geometry II: Side Lengths
Intermedite Mth Cirles Wednesdy 17 Otoer 01 Geometry II: Side Lengths Lst week we disussed vrious ngle properties. As we progressed through the evening, we proved mny results. This week, we will look t
More informationSIMPLE NONLINEAR GRAPHS
S i m p l e N o n l i n e r G r p h s SIMPLE NONLINEAR GRAPHS www.mthletis.om.u Simple SIMPLE Nonliner NONLINEAR Grphs GRAPHS Liner equtions hve the form = m+ where the power of (n ) is lws. The re lle
More informationSolids of Revolution
Solis of Revolution Solis of revolution re rete tking n re n revolving it roun n is of rottion. There re two methos to etermine the volume of the soli of revolution: the isk metho n the shell metho. Disk
More information5. Every rational number have either terminating or repeating (recurring) decimal representation.
CHAPTER NUMBER SYSTEMS Points to Rememer :. Numer used for ounting,,,,... re known s Nturl numers.. All nturl numers together with zero i.e. 0,,,,,... re known s whole numers.. All nturl numers, zero nd
More informationIV. CONDENSED MATTER PHYSICS
IV. CONDENSED MATTER PHYSICS UNIT I CRYSTAL PHYSICS Lecture - II Dr. T. J. Shinde Deprtment of Physics Smt. K. R. P. Kny Mhvidyly, Islmpur Simple Crystl Structures Simple cubic (SC) Fce centered cubic
More information1 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.
1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the
More information3 Angle Geometry. 3.1 Measuring Angles. 1. Using a protractor, measure the marked angles.
3 ngle Geometry MEP Prtie ook S3 3.1 Mesuring ngles 1. Using protrtor, mesure the mrked ngles. () () (d) (e) (f) 2. Drw ngles with the following sizes. () 22 () 75 120 (d) 90 (e) 153 (f) 45 (g) 180 (h)
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More informationMCH T 111 Handout Triangle Review Page 1 of 3
Hnout Tringle Review Pge of 3 In the stuy of sttis, it is importnt tht you e le to solve lgeri equtions n tringle prolems using trigonometry. The following is review of trigonometry sis. Right Tringle:
More informationCrystalline Structures The Basics
Crystlline Structures The sics Crystl structure of mteril is wy in which toms, ions, molecules re sptilly rrnged in 3-D spce. Crystl structure = lttice (unit cell geometry) + bsis (tom, ion, or molecule
More informationPAIR OF LINEAR EQUATIONS IN TWO VARIABLES
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES. Two liner equtions in the sme two vriles re lled pir of liner equtions in two vriles. The most generl form of pir of liner equtions is x + y + 0 x + y + 0 where,,,,,,
More informationLecture 6: Coding theory
Leture 6: Coing theory Biology 429 Crl Bergstrom Ferury 4, 2008 Soures: This leture loosely follows Cover n Thoms Chpter 5 n Yeung Chpter 3. As usul, some of the text n equtions re tken iretly from those
More informationSection 1.3 Triangles
Se 1.3 Tringles 21 Setion 1.3 Tringles LELING TRINGLE The line segments tht form tringle re lled the sides of the tringle. Eh pir of sides forms n ngle, lled n interior ngle, nd eh tringle hs three interior
More informationParticle Physics. Michaelmas Term 2011 Prof Mark Thomson. Handout 3 : Interaction by Particle Exchange and QED. Recap
Prtile Physis Mihelms Term 2011 Prof Mrk Thomson g X g X g g Hnout 3 : Intertion y Prtile Exhnge n QED Prof. M.A. Thomson Mihelms 2011 101 Rep Working towrs proper lultion of ey n sttering proesses lnitilly
More informationAlgebra 2 Semester 1 Practice Final
Alger 2 Semester Prtie Finl Multiple Choie Ientify the hoie tht est ompletes the sttement or nswers the question. To whih set of numers oes the numer elong?. 2 5 integers rtionl numers irrtionl numers
More informationCounting Paths Between Vertices. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs. Isomorphism of Graphs
Isomorphism of Grphs Definition The simple grphs G 1 = (V 1, E 1 ) n G = (V, E ) re isomorphi if there is ijetion (n oneto-one n onto funtion) f from V 1 to V with the property tht n re jent in G 1 if
More informationLUMS School of Science and Engineering
LUMS School of Science nd Engineering PH- Solution of ssignment Mrch, 0, 0 Brvis Lttice Answer: We hve given tht c.5(î + ĵ + ˆk) 5 (î + ĵ + ˆk) 0 (î + ĵ + ˆk) c (î + ĵ + ˆk) î + ĵ + ˆk + b + c î, b ĵ nd
More informationLecture 2: Cayley Graphs
Mth 137B Professor: Pri Brtlett Leture 2: Cyley Grphs Week 3 UCSB 2014 (Relevnt soure mteril: Setion VIII.1 of Bollos s Moern Grph Theory; 3.7 of Gosil n Royle s Algeri Grph Theory; vrious ppers I ve re
More informationPYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS
PYTHGORS THEOREM,TRIGONOMETRY,ERINGS ND THREE DIMENSIONL PROLEMS 1.1 PYTHGORS THEOREM: 1. The Pythgors Theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides
More information2.4 Theoretical Foundations
2 Progrmming Lnguge Syntx 2.4 Theoretil Fountions As note in the min text, snners n prsers re se on the finite utomt n pushown utomt tht form the ottom two levels of the Chomsky lnguge hierrhy. At eh level
More information22: Union Find. CS 473u - Algorithms - Spring April 14, We want to maintain a collection of sets, under the operations of:
22: Union Fin CS 473u - Algorithms - Spring 2005 April 14, 2005 1 Union-Fin We wnt to mintin olletion of sets, uner the opertions of: 1. MkeSet(x) - rete set tht ontins the single element x. 2. Fin(x)
More informationMaintaining Mathematical Proficiency
Nme Dte hpter 9 Mintining Mthemtil Profiieny Simplify the epression. 1. 500. 189 3. 5 4. 4 3 5. 11 5 6. 8 Solve the proportion. 9 3 14 7. = 8. = 9. 1 7 5 4 = 4 10. 0 6 = 11. 7 4 10 = 1. 5 9 15 3 = 5 +
More informationLogic, Set Theory and Computability [M. Coppenbarger]
14 Orer (Hnout) Definition 7-11: A reltion is qusi-orering (or preorer) if it is reflexive n trnsitive. A quisi-orering tht is symmetri is n equivlene reltion. A qusi-orering tht is nti-symmetri is n orer
More informationSOME COPLANAR POINTS IN TETRAHEDRON
Journl of Pure n Applie Mthemtis: Avnes n Applitions Volume 16, Numer 2, 2016, Pges 109-114 Aville t http://sientifivnes.o.in DOI: http://x.oi.org/10.18642/jpm_7100121752 SOME COPLANAR POINTS IN TETRAHEDRON
More informationActivities. 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
MEP: Demonstrtion Projet UNIT 4: Trigonometry UNIT 4 Trigonometry tivities tivities 4. Pythgors' Theorem 4.2 Spirls 4.3 linometers 4.4 Rdr 4.5 Posting Prels 4.6 Interloking Pipes 4.7 Sine Rule Notes nd
More informationSTRAND I: Geometry and Trigonometry. UNIT 32 Angles, Circles and Tangents: Student Text Contents. Section Compass Bearings
ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet ontents STR I: Geometry n Trigonometry Unit 32 ngles, irles n Tngents Stuent Tet ontents Setion 32.1 ompss erings 32.2 ngles n irles 1 32.3 ngles
More informationMomentum and Energy Review
Momentum n Energy Review Nme: Dte: 1. A 0.0600-kilogrm ll trveling t 60.0 meters per seon hits onrete wll. Wht spee must 0.0100-kilogrm ullet hve in orer to hit the wll with the sme mgnitue of momentum
More informationI 3 2 = I I 4 = 2A
ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo 2.13 We re ske to use KCL to fin urrents 1 4. The key point in pplying KCL in this prolem is to strt with noe where only one of the urrents
More informationApplied. Grade 9 Assessment of Mathematics. Multiple-Choice Items. Winter 2005
Applie Gre 9 Assessment of Mthemtis Multiple-Choie Items Winter 2005 Plese note: The formt of these ooklets is slightly ifferent from tht use for the ssessment. The items themselves remin the sme. . Multiple-Choie
More informationMatrix- System of rows and columns each position in a matrix has a purpose. 5 Ex: 5. Ex:
Mtries Prelulus Mtri- Sstem of rows n olumns eh position in mtri hs purpose. Element- Eh vlue in the mtri mens the element in the n row, r olumn Dimensions- How mn rows b number of olumns Ientif the element:
More informationProject 6: Minigoals Towards Simplifying and Rewriting Expressions
MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy
More informationChemistry Practice Exam
Chemistry Prtie Exm 1 2 3 Whih of the following is n element? A rop of merury. A splinter of woo. A rystl of sugr. A rop of wter. Whih of the following nswers ontins only elements? Soium, soium hlorie,
More informationCS 2204 DIGITAL LOGIC & STATE MACHINE DESIGN SPRING 2014
S 224 DIGITAL LOGI & STATE MAHINE DESIGN SPRING 214 DUE : Mrh 27, 214 HOMEWORK III READ : Relte portions of hpters VII n VIII ASSIGNMENT : There re three questions. Solve ll homework n exm prolems s shown
More informationApril 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then.
pril 8, 2017 Mth 9 Geometry Solving vetor prolems Prolem Prove tht if vetors nd stisfy, then Solution 1 onsider the vetor ddition prllelogrm shown in the Figure Sine its digonls hve equl length,, the prllelogrm
More informationA Primer on Continuous-time Economic Dynamics
Eonomis 205A Fll 2008 K Kletzer A Primer on Continuous-time Eonomi Dnmis A Liner Differentil Eqution Sstems (i) Simplest se We egin with the simple liner first-orer ifferentil eqution The generl solution
More informationF / x everywhere in some domain containing R. Then, + ). (10.4.1)
0.4 Green's theorem in the plne Double integrls over plne region my be trnsforme into line integrls over the bounry of the region n onversely. This is of prtil interest beuse it my simplify the evlution
More informationMathematical Proofs Table of Contents
Mthemtil Proofs Tle of Contents Proof Stnr Pge(s) Are of Trpezoi 7MG. Geometry 8.0 Are of Cirle 6MG., 9 6MG. 7MG. Geometry 8.0 Volume of Right Cirulr Cyliner 6MG. 4 7MG. Geometry 8.0 Volume of Sphere Geometry
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationSEMI-EXCIRCLE OF QUADRILATERAL
JP Journl of Mthemtil Sienes Volume 5, Issue &, 05, Pges - 05 Ishn Pulishing House This pper is ville online t http://wwwiphsiom SEMI-EXCIRCLE OF QUADRILATERAL MASHADI, SRI GEMAWATI, HASRIATI AND HESY
More informationEXTENSION OF THE GCD STAR OF DAVID THEOREM TO MORE THAN TWO GCDS CALVIN LONG AND EDWARD KORNTVED
EXTENSION OF THE GCD STAR OF DAVID THEOREM TO MORE THAN TWO GCDS CALVIN LONG AND EDWARD KORNTVED Astrt. The GCD Str of Dvi Theorem n the numerous ppers relte to it hve lrgel een evote to shoing the equlit
More informationSolid State Electronics EC210 Arab Academy for Science and Technology AAST Cairo Spring 2016 Lecture 1 Crystal Structure
Solid Stte Electronics EC210 AAST Ciro Spring 2016 Lecture 1 Crystl Structure Dr. Amr Byoumi, Dr. Ndi Rft 1 These PowerPoint color digrms cn only be used by instructors if the 3 rd Edition hs been dopted
More informationInspiration and formalism
Inspirtion n formlism Answers Skills hek P(, ) Q(, ) PQ + ( ) PQ A(, ) (, ) grient ( ) + Eerise A opposite sies of regulr hegon re equl n prllel A ED i FC n ED ii AD, DA, E, E n FC No, sies of pentgon
More informationLecture 8: Abstract Algebra
Mth 94 Professor: Pri Brtlett Leture 8: Astrt Alger Week 8 UCSB 2015 This is the eighth week of the Mthemtis Sujet Test GRE prep ourse; here, we run very rough-n-tumle review of strt lger! As lwys, this
More informationExercise sheet 6: Solutions
Eerise sheet 6: Solutions Cvet emptor: These re merel etended hints, rther thn omplete solutions. 1. If grph G hs hromti numer k > 1, prove tht its verte set n e prtitioned into two nonempt sets V 1 nd
More informationA Study on the Properties of Rational Triangles
Interntionl Journl of Mthemtis Reserh. ISSN 0976-5840 Volume 6, Numer (04), pp. 8-9 Interntionl Reserh Pulition House http://www.irphouse.om Study on the Properties of Rtionl Tringles M. Q. lm, M.R. Hssn
More information1 1. Crystallography 1.1 Introduction 1.2 Crystalline and Non-crystalline materials crystalline materials single crystals polycrystalline material
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
More information2. There are an infinite number of possible triangles, all similar, with three given angles whose sum is 180.
SECTION 8-1 11 CHAPTER 8 Setion 8 1. There re n infinite numer of possile tringles, ll similr, with three given ngles whose sum is 180. 4. If two ngles α nd β of tringle re known, the third ngle n e found
More informationPart I: Study the theorem statement.
Nme 1 Nme 2 Nme 3 A STUDY OF PYTHAGORAS THEOREM Instrutions: Together in groups of 2 or 3, fill out the following worksheet. You my lift nswers from the reding, or nswer on your own. Turn in one pket for
More informationSection 2.1 Special Right Triangles
Se..1 Speil Rigt Tringles 49 Te --90 Tringle Setion.1 Speil Rigt Tringles Te --90 tringle (or just 0-60-90) is so nme euse of its ngle mesures. Te lengts of te sies, toug, ve very speifi pttern to tem
More informationAPPROXIMATION AND ESTIMATION MATHEMATICAL LANGUAGE THE FUNDAMENTAL THEOREM OF ARITHMETIC LAWS OF ALGEBRA ORDER OF OPERATIONS
TOPIC 2: MATHEMATICAL LANGUAGE NUMBER AND ALGEBRA You shoul unerstn these mthemtil terms, n e le to use them ppropritely: ² ition, sutrtion, multiplition, ivision ² sum, ifferene, prout, quotient ² inex
More informationMatrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix
tries Definition of tri mtri is regulr rry of numers enlosed inside rkets SCHOOL OF ENGINEERING & UIL ENVIRONEN Emple he following re ll mtries: ), ) 9, themtis ), d) tries Definition of tri Size of tri
More informationNecessary and sucient conditions for some two. Abstract. Further we show that the necessary conditions for the existence of an OD(44 s 1 s 2 )
Neessry n suient onitions for some two vrile orthogonl esigns in orer 44 C. Koukouvinos, M. Mitrouli y, n Jennifer Seerry z Deite to Professor Anne Penfol Street Astrt We give new lgorithm whih llows us
More informationAnswers and Solutions to (Some Even Numbered) Suggested Exercises in Chapter 11 of Grimaldi s Discrete and Combinatorial Mathematics
Answers n Solutions to (Some Even Numere) Suggeste Exercises in Chpter 11 o Grimli s Discrete n Comintoril Mthemtics Section 11.1 11.1.4. κ(g) = 2. Let V e = {v : v hs even numer o 1 s} n V o = {v : v
More informationGrade 6. Mathematics. Student Booklet SPRING 2008 RELEASED ASSESSMENT QUESTIONS. Assessment of Reading,Writing and Mathematics, Junior Division
Gre 6 Assessment of Reing,Writing n Mthemtis, Junior Division Stuent Booklet Mthemtis SPRING 2008 RELEASED ASSESSMENT QUESTIONS Plese note: The formt of these ooklets is slightly ifferent from tht use
More informationPythagoras Theorem PYTHAGORAS THEOREM.
Pthgors Theorem PYTHAGORAS THEOREM www.mthletis.om.u How oes it work? Solutions Pthgors Theorem Pge 3 questions Right-ngle tringles D E x z Hotenuse is sie: F Hotenuse is sie: DF Q k j l Hotenuse is sie:
More informationLinear Algebra Introduction
Introdution Wht is Liner Alger out? Liner Alger is rnh of mthemtis whih emerged yers k nd ws one of the pioneer rnhes of mthemtis Though, initilly it strted with solving of the simple liner eqution x +
More information9.1 Day 1 Warm Up. Solve the equation = x x 2 = March 1, 2017 Geometry 9.1 The Pythagorean Theorem 1
9.1 Dy 1 Wrm Up Solve the eqution. 1. 4 2 + 3 2 = x 2 2. 13 2 + x 2 = 25 2 3. 3 2 2 + x 2 = 5 2 2 4. 5 2 + x 2 = 12 2 Mrh 1, 2017 Geometry 9.1 The Pythgoren Theorem 1 9.1 Dy 2 Wrm Up Use the Pythgoren
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors
More informationCARLETON UNIVERSITY. 1.0 Problems and Most Solutions, Sect B, 2005
RLETON UNIVERSIT eprtment of Eletronis ELE 2607 Swithing iruits erury 28, 05; 0 pm.0 Prolems n Most Solutions, Set, 2005 Jn. 2, #8 n #0; Simplify, Prove Prolem. #8 Simplify + + + Reue to four letters (literls).
More informationGeometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272.
Geometry of the irle - hords nd ngles Geometry of the irle hord nd ngles urriulum Redy MMG: 272 www.mthletis.om hords nd ngles HRS N NGLES The irle is si shpe nd so it n e found lmost nywhere. This setion
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationFarey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University
U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions
More informationAP CALCULUS Test #6: Unit #6 Basic Integration and Applications
AP CALCULUS Test #6: Unit #6 Bsi Integrtion nd Applitions A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS IN THIS PART OF THE EXAMINATION. () The ext numeril vlue of the orret
More informationCHENG Chun Chor Litwin The Hong Kong Institute of Education
PE-hing Mi terntionl onferene IV: novtion of Mthemtis Tehing nd Lerning through Lesson Study- onnetion etween ssessment nd Sujet Mtter HENG hun hor Litwin The Hong Kong stitute of Edution Report on using
More informationChapter One Crystal Structure
Chpter One Crystl Structure Drusy Qurtz in Geode Tbulr Orthoclse Feldspr Encrusting Smithsonite Peruvin Pyrite http://www.rockhounds.com/rockshop/xtl 1 Snow crystls the Beltsville Agriculturl Reserch Center
More informationLesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem.
27 Lesson 2: The Pythgoren Theorem nd Similr Tringles A Brief Review of the Pythgoren Theorem. Rell tht n ngle whih mesures 90º is lled right ngle. If one of the ngles of tringle is right ngle, then we
More informationAP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals
AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into
More informationTrigonometry and Constructive Geometry
Trigonometry nd Construtive Geometry Trining prolems for M2 2018 term 1 Ted Szylowie tedszy@gmil.om 1 Leling geometril figures 1. Prtie writing Greek letters. αβγδɛθλµπψ 2. Lel the sides, ngles nd verties
More informationy = c 2 MULTIPLE CHOICE QUESTIONS (MCQ's) (Each question carries one mark) is...
. Liner Equtions in Two Vriles C h p t e r t G l n e. Generl form of liner eqution in two vriles is x + y + 0, where 0. When we onsier system of two liner equtions in two vriles, then suh equtions re lle
More information