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 let c ˆk + b + c Now we hve primitive trnsltionl vectors î, b ĵ, c + b + c Where c + b + c is the body centered position of the cubic unit cell defined by, b nd c. The Brvis lttice is therefore bcc. Volume of conventionl unit cell will be, vol of conventionl cell Å 7 Å Since the volume of primitive unit cell is hlf of the volume of the bcc (since the conventionl unit cell contins two lttice points while primitive unit cell contins only one), thus volume of primitive unit cell will be, Lttice Types in -D Answer: vol.ofprimitivecell vol. of conventionl cell 7 Å.5 Å (i) Since b c nd γ β γ 90 o, i.e. three primitive vectors re unequl nd there is lck of symmetry. This is chrcteristic of triclinic structure. Therefore the unit cell belongs to triclinic system. (ii) Since b c, γ β 90 o nd γ 0 o, This is chrcteristic of
Hexgonl structure. Therefore the unit cell belongs to Hexgonl system. Tetrhedrl Angles Answer: The body digonls of the cubes shown in the figure re vectors tht connect the lttice points t the origin to the lttice points t the body centers. The primitive cell obtined on completing the rhombohedron. In terms of the cube edge, the primitive trnsltionl vectors re, (ˆx + ŷ ẑ) (ŷ + ẑ ˆx) (ẑ + ˆx ŷ) Let the ngle between nd be θ, then ( ) ( (ˆx + ŷ ẑ). (ŷ + ẑ ˆx) Wigner Seitz Cell Answer: 4 cos θ ) (ˆx + ŷ ẑ) (ŷ + ẑ ˆx) cos θ ( ) ( ) ( + + ( + ) 4 4 cos θ cos θ ( θ cos ) 09.5 o ) ( ) + (). By construction every vector in the Brvis lttice hs n inverse vector in the sme Brvis lttice. Similrly following the Wigner Seitz cell s construction in which we use perpendiculr bisectors,we find tht for every side of the Wigner Seitz cell there is n inverse of tht side i.e.wigner Seitz cell consists of pirs ( ) + ( ) cos θ
of opposite sides. Thus the number of sides of the Wigner Seitz cell re even. Tht is why tringulr Wigner Seitz cell is meningless i.e Wigner Seitz cell cnnot hve three sides. Since Wigner Seitz cell is n re enclosed by the perpendiculr bisectors it cnnot hve two sides becuse two sides cnnot form closedre.thus Wigner Seitz cell hs minimum 4 sides. Number of sides of Wigner Seitz cell re lwys even. Therefore Wigner Seitz cell be closed re with 4 sides(rectngle), 6 sides (Hexgon) or more. Now let us prove it by constructing the Wigner Seitz cell. Let us consider Brvis lttice shown in figure (). B -A O A -A A -B O -C -B B C fig() ( b ) Tke Brvis lttice point s origin O. The nerest Brvis lttice point to the right side of the origin point is A. There is negtive or inverse point A on the left side of origin, such tht mesure of OA is equl to the mesure of O( A). similrly we hve points B nd B tht re opposite to ech other. Drw perpendiculr lines joining A to A nd B to B through O. The perpendiculr bisectors to these lines will give us the Wigner Seitz cell tht is rectngulr. Now consider figure (b). Tke Brvis lttice point s origin O. The nerest Brvis points re A nd A. Tke the closest point to O not on the line OA, cll this point B. Drw perpendiculr to OA through O. Let point B is on left side of this perpendiculr line. Now there will be Brvis lttice point C on the line through B nd prllel to OA, such tht BC OA. Any other Brvis lttice point on BC will not contribute to this side of Wigner Seitz cell, since it hs to be further out thn A, B, C or A. Similrly on the other side of OA we will find the points B nd C. The Wigner Seitz cell is Hexgon, unless B is on the norml line, in which cse the cell is rectngle. (b). The digonl of prllelogrm is the line joining the opposite corners of the prllelogrm. The opposite corners of prllelogrm fce of Wigner Seitz cell for the fcc lttice hve points (/4, /4, 0) nd (/, 0, /). While the other digonl connects the points (/4, /4, /4) nd (/4, /4, /4). If is the
length of one side of the given fcc, then, Similrly, AC (/, 0, /) (/4, /4, 0) BC (/4, /4, /4) (/4, /4, /4) Thus rtio will be, : : HCP Structure Answer: 5 C + A + B Let us consider n idel hcp structure in which nd c with α β 90 o while γ 0 o. Let use spheres of rdius R on ech lttice point of hcp structure such tht R. Since hcp consists of two interpenetrting hexgonl Brvis lttices, displced from one nother by ā / + ā / + ā /, thus Thus ā + ā + ā R Also c ā (ā + ā ) + ā R (ā + ā ) + c R Also from the figure it is cler tht, (ā + ā ) + c (R) (ā + ā ) + 4 c 4R 4
Since ngle between ā nd ā is 60 o, from the figure it is cler tht, Where AB BC nd Consider tringle ABD, AC AB + BC AB cn be clculted s follows. tn 0 o / x x tn 0 o AB BC AC + ā + ā Using the vlue of (ā + ā ) we hve, Answer: 6 ( ) + 4 c 4R + c 4 c 4 c 8 c.6 In the simple cubic ech tom is surrounded by 8 other toms. There re 8 corners in simple cube. Ech tom contributes its /8 prt in simple cube. Therefore, totl number of toms 8 8 tom Let lttice constnt is, then rdius of ech tom will be /. volume of one tom in simple cube 4 π(/) 4 π 8 π 6 totl volume of cube 5
Thus the coverge of simple cubic lttice will be, coverge of SC volume occupied by n tom totl volume π 6 0.5 %coverge 0.5 00 5 % 6