IV. CONDENSED MATTER PHYSICS

Transcription

1 IV. CONDENSED MATTER PHYSICS UNIT I CRYSTAL PHYSICS Lecture - II Dr. T. J. Shinde Deprtment of Physics Smt. K. R. P. Kny Mhvidyly, Islmpur

2 Simple Crystl Structures Simple cubic (SC) Fce centered cubic (FCC) Body centered cubic (BCC) Hexgonl close pcked (HCP) Dimond structure Zinc blende structure NCl structure CsCl structure

3 NUMBER OF ATOMS PER UNIT CELL (n) The totl number of toms present in n unit cell is known s number of toms per unit cell. COORDINATION NUMBER (CN) It is the number of nerest neighboring toms to prticulr tom. ATOMIC RADIUS (r) It is the rdius of n tom. It is lso defined s hlf the distnce between two nerest neighboring toms in crystl. ATOMIC PACKING FACTOR (APF) It is the rtio of volume occupied by the toms or molecules in n unit cell (v) to the totl volume of the unit cell (V). APF APF Volume occupied by the toms in n unit cell Volume of the unit cell No. of toms present in the unit cell x Volume of the tom Volume of the unit cell

4 Simple Cubic Structure (SC) Ech nd every corner tom is shred by eight djcent unit cells. The contribution of ech nd every corner tom to one unit cell is 1/8. The totl number of toms present in unit cell 1/8 x 8 1. CORDINATION NUMBER (CN) For SCS tom, there re four nerest neighbours in its own plne. There is nother nerest neighbour in plne which lies just bove this tom nd yet nother nerest neighbour in nother plne which lies just below this tom. Therefore the co-ordintion number is 6. 4

5 ATOMIC RADIUS (R) Since the toms touch long cube edges, the tomic rdius for simple cubic unit cell is, r / ( 0.5) (where r, is the lttice constnt) ATOMIC PACKING FACTOR (APF) toms unit cell APF 1 volume 4 π (0.5) tom volume unit cell APF 0.56 Thus 5 percent of the volume of the simple cubic unit cell is occupied by toms nd the remining 48 percent volume of the unit cell is vcnt or void spce.

6 Fce Centered Cubic structure (FCC) Atoms re rrnged t the corners nd center of ech cube fce of the cell. The toms in FCC unit cell touches long the fce digonl. Ech nd every corner tom is shred by eight djcent unit cells. Therefore ech nd every corner tom contributes 1/8 of its prt to one unit cell. So the totl number of toms contributed by the corner toms is 1/8 8 1.

7 Two unit cells shre ech nd every fce centered tom. Therefore, the contribution of fce centered tom to unit cell is 1/. So, the totl number of toms contributed by the fce centered toms 1/ 6. The totl number of toms present in FCC unit cell COORDINATION NUMBER In its own plne it touches four fce centered toms. The fce centered toms re its nerest neighbors. In plne, which lies just bove this corner tom, it hs four more fce centered toms s nerest neighbors. Similrly, in plne, which lies just below this corner tom it hs yet four more fce centered toms s its nerest neighbors.

8 Therefore the number of nerest neighbours i.e., co-ordintion number for FCC unit cell ATOMIC RADIUS (R) Consider the figure. AB AC nd AC 4r. From the tringle ABC, AC AB + BC AC + ; AC AC i.e 4r Therefore tomic rdius 4

9 ATOMIC PACKING FACTOR (APF) toms unit cell APF 4 4 π ( /4) volume tom volume unit cell APF 0.74 Thus 74 percent of the volume of the FCC unit cell is occupied by toms nd the remining 6 percent volume of the unit cell is vcnt or void spce. Exmples :- Aluminum ( 0.405) Gold ( 0.408) ex: Al, Cu, Au, Pb, Ni, Pt, Ag

10 BODY CENTERED CUBIC STRUCTURE A body centred cubic structure hs eight comer toms nd one body centred tom. The tom t the centre touches ll the eight corner toms. In BCC unit cell, ech nd every corner tom is shred by eight djcent unit cells. So, the totl number of toms contributed by the corner toms is1/ A BCC unit cell hs one full tom t the centre of the unit cell. The totl number of toms present in BCC unit cell 1+1.

11 CO-ORDINATION NUMBER (CN) Let us consider body centred tom. The nerest neighbour for body centred tom is corner tom. A body centred tom is surrounded by eight corner toms. Therefore, the co-ordintion number of BCC unit cell 8. ATOMIC RADIUS (R) For BCC unit cell, the tomic rdius cn be clculted From figure AB BC AD nd CD 4r. From the tringle, ACD, CD AC + AD ; CD + (4r) ; 16r tomic rdius r 4 i.e. r 16

12 toms unit cell APF APF ATOMIC PACKING FACTOR (APF) 4 π π ( /4) volume unit cell volume tom Thus 68 percent of the volume of the BCC unit cell is occupied by toms nd the remining percent volume of the unit cell is vcnt or void spce. Exmples :- Chromium (0.89 nm) Iron (0.87 nm) Sodium (0.49 nm)

13 HEXAGONAL CLOSED PACKED STRUCTURE It consists of three lyers of toms. The bottom lyer hs six corner toms nd one fce centred tom. The middle lyer hs three full toms. The upper lyer hs six corner toms nd one fce centred tom. Ech nd every corner tom contributes 1/6 of its prt to one unit cell. The number of totl toms contributed by the corner toms of both top nd bottom lyers is 1/6 1. The fce centred tom contributes 1/ of its prt to one unit cell. ex: Cd, Mg, Ti, Zn

14 Since there re fce centred toms, one in the top nd the other in the bottom lyers, the number of toms contributed by fce centred toms is 1/ 1. Besides these toms, there re full toms in the middle lyer. Totl number of toms present in n HCP unit cell is CO-ORDINATION NUMBER (CN) The fce centered tom touches 6 corner toms in its plne. The fce centered tom touches 6 corner toms in its plne. The middle lyer hs toms.there re three more toms, which re in the middle lyer of the unit cell. Therefore the totl number of nerest neighbours is ATOMIC RADIUS (R) Consider ny two corner toms. Ech nd every corner tom touches ech other. Therefore r. i.e., The tomic rdius, r /

15 ATOMIC PACKING FACTOR (APF) v V APF Where, v 6 4/ πr Substitute r, v 6 4/ π π 8 AB AC BO. CX c/;where c height of the hcp unit cell. Are of the bse 6 re of the tringle ABO 6 1/ AB OO Are of the bse 6 1/ OO In tringle OBO cos0º OO ' BO O 'OB 0 OO ' Now, substituting the vlue of OO, OO cos 0º 1 Are of the bse 6 V Are of the bse height c APF v V π c π ( ) c

16 CALCULATION OF c/ RATIO In the tringle ABA, A ' AB 0 AB cos0 AA' 0 1 AA' AX But In the tringle AXC, AC AX + CX ; Substituting the vlues of AC, AX nd CX, + c 4 c + 4 c 8 c 8 c 0.74 π ) 8 ( π APF Atomic Pcking Frction 74%

17 Atom Positions in Cubic Unit Cells Crtesin coordinte system is use to locte toms. In cubic unit cell y xis is the direction to the right. x xis is the direction coming out of the pper. z xis is the direction towrds top. Negtive directions re to the opposite of positive directions. Atom positions re locted using unit distnces long the xes. -15

18 Directions in Cubic Unit Cells In cubic crystls, Direction Indicesre vectorcomponents of directions resolved long ech xes, resolved to smllest integers. Direction indices re position coordintesof unit cell where the direction vector emerges from cell surfce, converted to integers. -16

19 Procedure to Find Direction Indices Produce the direction vector till it emerges from surfce of cubic cell Determine the coordintes of point of emergence nd origin Subtrct coordintes of point of Emergence by tht of origin All re integers? YES Are ny of the direction vectors negtive? YES Represent the indices in squre brcket without coms with over negtive index (Eg: [11]) NO NO x z (1,1/,1) (0,0,0) (1,1/,1) - (0,0,0) (1,1/,1) x (1,1/,1) (,1,) The direction indices re [1] Convert them to y smllest possible integer by multiplying by n integer. Represent the indices in squre brcket without coms (Eg: [1] ) -17

20 Direction Indices - Exmple Determine direction indices of the given vector. Origin coordintes re (/4, 0, 1/4). Emergence coordintes re (1/4, 1/, 1/). Subtrcting origin coordintes from emergence coordintes, (1/4, 1/, 1/)-(/4, 0, 1/4) (-1/, 1/, 1/4) Multiply by 4 to convert ll frctions to integers 4 x (-1/, 1/, 1/4) (-,, 1) Therefore, the direction indices re [ 1 ] -18

21 Miller Indices Miller Indicesre used to refer to specific lttice plnes of toms. They re reciprocls of the frctionl intercepts (with frctions clered) tht the plne mkes with the crystllogrphic x,y nd z xes of three nonprllel edges of the cubic unit cell. z Miller Indices (111) -19 x y

22 Miller Indices - Procedure Choose plne tht does not pss through origin Determine the x,y nd z intercepts of the plne Find the reciprocls of the intercepts Frctions? Plce br over the Negtive indices Cler frctions by multiplying by n integer to determine smllest set of whole numbers Enclose in prenthesis (hkl)where h,k,l re miller indicesof cubic crystl plne forx,y nd z xes. Eg: (111) -0

23 Miller Indices - Exmples (100) x x z y Intercepts of the plne t x,y & z xes re 1, nd Tking reciprocls we get (1,0,0). Miller indices re (100). ******************* Intercepts re 1/, / & 1. tking reciprocls we get (, /, 1). Multiplying by to cler frctions, we get (6,,). Miller indices re (6). -1

24 Miller Indices - Exmples Plot the plne (101) Tking reciprocls of the indices we get (1 1). The intercepts of the plne re x1, y (prllel to y) nd z1. ****************************** Plot the plne ( 1) Tking reciprocls of the indices we get (1/ 1/ 1). The intercepts of the plne re x1/, y 1/ nd z1. -

25 Miller Indices - Exmple Plot the plne (110) The reciprocls re (1,-1, ) The intercepts re x1, y-1 nd z (prllel to z xis) To show this plne single unit cell, the (110) z origin is moved long the positive direction of y xis by 1 unit. y x -

26 Miller Indices Importnt Reltionship Direction indices of direction perpendiculr z to crystl plne re sme s miller indices of the plne. Exmple:- z [110] (110) y -4 Interplnr spcing between prllel closest plnes with sme miller indices is given by d hkl x h + k + l

27 Plnes nd Directions in Hexgonl Unit Cells Four indices re used (hkil) clled s Miller-Brvisindices. Four xes re used ( 1,, nd c). Reciprocl of the intercepts tht crystl plne mkes with the 1,, nd c xes give the h,k,i nd l indices respectively. -5

28 Bsl Plnes:- Intercepts 1 c 1 Hexgonl Unit Cell - Exmples (hkli) (0001) Prism Plnes :- For plne ABCD, Intercepts c (hkli) (1010) -6

29 Directions in HCP Unit Cells Indicted by 4 indices [u v t w]. u,v,t nd w re lttice vectors in 1,, nd c directions respectively. Exmple:- For 1,, directions, the direction indices re [ 1 1 0], [1 1 0] nd [ 1 1 0] respectively. -7

30 Summry of Crystl Structures

31 Volume density of metl Volume Density ρ v Mss/Unit cell Volume/Unit cell 1) Exmple:-Copper (FCC) hs tomic mss of 6.54 g/mol nd tomic rdius of nm. Find volume density nm na VN Where, n number of toms/unit cell; Atomic weight ;V Volume of unit cell for cubic ; N Avogdro s number 6.0 x 10 toms/mol 4 R 0.61 nm Volume of unit cell V (0.61nm) 4.7 x 10-9 m FCC unit cell hs 4 toms. Mss of unit cell m ρ v m V (4toms)(6.54g / mol) tmos / mol 8 9 Mg Mg 8.98 m m Mg g 6 g 8.98 cm 4. x 10-8 Mg -0

32 ) Chromium hs BCC structure nd density of 7.19 g/ml. Wht is the rdius of its tom? (Atomic mss of Cr mu) ) Copper hs FCC structure nd density of 8.96 g/ml. Wht is the volume of the unit cell? (Atomic mss of Cu mu) 4) Strontium hs BCC structure nd r 15 pm. Wht is the density? (Atomic mss of Sr 86.6 mu) 5) Nickel hs density of 8.9 g/ml. Its tomic rdius is 15 pm. Wht is the crystl structure? (Atomic mss of Ni mu). 6)Polonium metl crystllizes in simple cubic structure. Clculte the density of the polonium metl if the tom rdius is 176 pm.(atomic mss of Po09 mu). 7) Bsed on literture density of g cm -, wht is the rdius of Po? 8) The rdius of the copper tom is 17.8 pm, nd its density is 8.95 g/cm. Which unit cell is consistent with these dt: sc, bcc, or fcc? 9) Zinc hs HCP structure. The height of the unit cell is 4.95Å. Find (i). How mny toms re there in unit cell? nd (ii). Wht is the volume of the unit cell?

33 1) Height of the unit cell, c 4.95Å m In HCP structure, the number of toms present in the unit cell is 6. c 8 We know tht, the rtio c, Å 8 8 We lso know tht, volume of the unit cell for HCP structure is, V c or V ( ) V m 10. Sodium crystllises in cubic lttice. The edge of the unit cell is 4.16 Å. The density of sodium is 975kg/m nd its tomic weight is. Wht type of unit cell does sodium form? (Tke Avgdro number s toms (Kg mole)- 1

34 i) Edge of the unit cell, 4.16Å m ii) Density of the sodium, ρ 975 kg/m iii) Atomic weight of sodium, M iv) Avogdro's number, N toms/kg mole Density of the crystl mteril, ρ Number of toms in the unit cell, nm N n ρn M n ( ) 6 10 toms Since the body centred cubic cell contins toms in it, sodium crystllises in BCC cell.

35 11. A metllic element exists in cubic lttice. Ech side of the unit cell is.88 Å. The density of the metl is 7.0 gm/cm. How mny unit cells will be there in 100gm of the metl? - Edge of the unit cell,.88å m Density of the metl, ρ 7.0 gm/cm kg/m Volume of the metl 100gm 0.1kg Volume of the unit cell ( ) m Mss 0.1 Density Volume of 100gm of the metl Number of unit cells in the volume m