Condensed Matter Physics April, 8, 0 LUMS School of Science and Engineering PH-33 Solution of assignment 5 April, 8, 0 Interplanar separation Answer: To prove that the reciprocal lattice vector G = h b + k b + l b 3 is perpendicular to this plane, it suffices to show that G is perpendicular to two nonparallel vectors in this plane. For the plane (hkl), it intercepts axis a, a, and a 3 at a ratio h : k : l, thus the two vectors in the plane can be chosen as ( h a k a ) and ( h a l a 3). Obviously, they are not parallel to each other. From direct calculation: ( G h a ) ( k a = (h b + k b + l b 3 ) h a ) k a = 0 ( G h a ) ( l a 3 = (h b + k b + l b 3 ) h a ) l a 3 = 0 we know G is perpendicular to these two vectors, and hence perpendicular to the plane. (b). Among the indices hkl, at least one of them is non-zero, without lose generality, h 0. Since, by definition, this family of planes divide vector a into h parts with equal length, then the vector R = 0 ends on one plane, and R = a h ends on an adjacent plane. Therefore, the projection of R R = a h on to the direction perpendicular to this family of planes will be the distance between two neighboring planes. From part (a), G is perpendicular to these planes, hence: d(hkl) = h a G G = h a (h b + k b + l b 3 ) G = π G (c). For simple cubic, the reciprocal lattice has: b = π a ˆx b = π a ŷ b3 = π a ẑ where ˆx, ŷ, and ẑ are unit vectors along x, y, and z directions. Thus, and hence: G = (h b + k b + l b 3 ) = (π) a (h + k + l ) d(hkl) = (π) G = a h + k + l April, 8, 0
Condensed Matter Physics April, 8, 0 Reciprocal of Reciprocal Lattice Answer: (a). Let us take the LHS b ( b b 3 ) and substitute the given value of b, we get, Now use the vector calculus identity, last expression will become, Now using orthogonality condition, LHS = b ( b b 3 ) b ( b b 3 ) = π ( a a 3 ) a ( a a 3 ) ( b b 3 ) = π ( a a 3 ) ( b b 3 ) a ( a a 3 ) (a b) (c d) = (a c)(b d) (a d)(b c) b ( b b 3 ) = π ( a b )( a 3 b 3 ) ( a b 3 )( a 3 b ) a ( a a 3 ) bi a j = πδ ij where δ ij = 0 if i j and δ ij = if i = j We get, Hence proved. b ( b ((π)(π) (0)(0)) b 3 ) = π a ( a a 3 ) (π) 3 = a ( a a 3 ) (b). Let us suppose that c i are constructed from the b i in the same manner as the b i are constructed from a i. then, c c c 3 b = π b 3 b ( b b 3 ) b3 = π b b ( b b 3 ) b = π b b ( b b 3 ) April, 8, 0
Condensed Matter Physics April, 8, 0 Now we want to prove that c i are just a i. For this purpose consider c, b c = π b 3 b ( b b 3 ) Substitute of given value of b 3 in the last expression we get, c b π a a a = π ( a a 3 ) b ( b b 3 ) By using the following vector calculus identity, we have the last expression as, = (π) b ( a a ) [ b ( b b 3 )][ a ( a a 3 )] a (b c) = b(a c) c(a b) c = (π) a ( b a ) a ( b a ) [ b ( b b 3 )][ a ( a a 3 )] Again application of Orthogonality condition yields, From part(a), we know that, thus last expression will become, Hence Proved. (c). c = (π) π a [ b ( b b 3 )][ a ( a a 3 )] = (π) 3 a [ b ( b b 3 )][ a ( a a 3 )] b ( b b 3 ) = (π) 3 a ( a a 3 ) c = a ( a ( a a 3 )) a ( a a 3 ) = a Z S c Y b a X April, 8, 0 3
Condensed Matter Physics April, 8, 0 First, we note that if we construct the primitive cell using Σ 3 i=x i a i, where 0 x i we will always have a parallelepiped with a constant cross-sectional area. For any prism of this type, the volume is just the area of its base times the height of the prism. Lets say that a and a 3 span the base of the prism (you could choose any two). That means, by definition, the area of this base is A = a a 3 Now, the height of the prism is given by the absolute value of a, which is a = a cos θ So, the volume of the parallelepiped is A = a cos θ a a 3 = a ( a a 3 ) and the volume is proved. Reciprocal of Simple Hexagonal and Trigonal Bravais Lattice Answer: 3 (a). Given primitive vectors are, a a a = aˆx = a 3a ˆx + ŷ = cẑ While the reciprocal lattice vectors are, b b b3 a a 3 = π a ( a a 3 ) a 3 a = π a ( a a 3 ) a a = π a ( a a 3 ) Apply the given values of a, a and a 3, we get, b ( a = π ˆx + 3aŷ) (cẑ) [( ) ] a aˆx ˆx + 3a ŷ (cẑ) = π a ˆx a 3ŷ April, 8, 0 4
Condensed Matter Physics April, 8, 0 Similarly, b = π aˆx = 4π a 3ŷ (cẑ) (aˆx) [( ) a ˆx + 3a ŷ ] (cẑ) and (aˆx) ( a b3 = π ˆx + 3a [( ) ŷ) ] a aˆx ˆx + 3a ŷ (cẑ) = π c ẑ Hence Reciprocal of the simple hexagonal Bravais lattice is also simple hexagonal with lattice constant π and 4π c a. 3 Let θ be the angle between a real lattice vector a and a reciprocal lattice vector b, then, a b = a b cos θ ( π (aˆx) a ˆx ) a = (4π ) a 3ŷ a + 4π cos θ 3a π = 4π cos θ 3 ( ) 3 θ = arccos = 30 o Hence reciprocal lattice is rotated through 30 o about the c-axis with respect to the direct lattice. (b). In reciprocal lattice, c = π c and a = 4π a 3, c a = π c a 3 4π 3 a = c ( 3 c = a ) April, 8, 0 5
Condensed Matter Physics April, 8, 0 If c = c, then a a c a = ( ) c = a c If the direct lattice is ideal i.e. = a 8 3 c a = c a = ( ) 3 c a 3 3 c, then a 3 = 3 4 will be, ( ) 8 3 (c). Trigonal basis are, a = a(x), a = a(x) and a 3 = a(x). Primitive vectors of a trigonal will be, a a a 3 = a(xî + ĵ + ˆk) = a(î + xĵ + ˆk) = a(î + ĵ + xˆk) Where î, ĵ and ˆk are unit vectors along x, y and z directions. Length of all these three vectors is same i.e. a = a = a 3 = a ( + x ) Let θ be the angle between two vectors a and a, then, cos θ = a a a a ( + x) = ( + x ) Now Reciprocal lattice vectors b, b and b 3 will be, By symmetry, b b b3 = π a = π a = π a [(x )î + ( x)ĵ + ( x)ˆk] ( x) (x + ) [( x)î + (x )ĵ + ( x)ˆk] ( x) (x + ) [( x)î + ( x)ĵ + (x )ˆk] ( x) (x + ) April, 8, 0 6
Condensed Matter Physics April, 8, 0 Length of each b will be, + (x + ) b = b = b 3 = π a ( x)(x + ) Let θ be the angle between any two reciprocal lattice vectors say b and b, then, Now divide cos θ by cos θ, we get, Now, cos θ = b b b b = x + + ( + x) cos θ cos θ = x+ +(+x) (+x) (+x ) = ( + x ) + ( + x) = + (+x) (+x ) = + cos θ cos θ = cos θ + cos θ cos θ cos θ ( + x) = ( + x ) x + + ( + x) + cos θ cos θ = (x ) (x + ) ( + x )(x + x + 3) Which gives the required relation between a and b. Density of Lattice Points in a Lattice Plane Answer: 4 (a). Within each plane, the lattice is a d lattice. Let the area of the primitive cell for this d lattice be A, then since v is volume of primitive cell for 3 d lattice, it follows that A d = v. Since every primitive cell contains exactly one lattice points, the number density of lattice points on this plane ρ is: ρ = A = d v (b). Once the type of a lattice is given, the volume of primitive cell v is determined. Thus from part (a), the larger the inter-planer distance d is, the greater the density of points April, 8, 0 7
Condensed Matter Physics April, 8, 0 ρ is. Also the largest d corresponds to the family of planes for which the length of G, a reciprocal lattice vector perpendicular to them, is minimized. Thus, if we could find out a G o in reciprocal space with least length, then the real space lattice planes perpendicular to this G o will have largest density of lattice points. For fcc, choose: a = a (ŷ + ẑ), a = a (ẑ + ˆx), a 3 = a (ˆx + ŷ) then the reciprocal lattice has: b = π a (ŷ + ẑ ˆx), b = π a (ẑ + ˆx ŷ), b3 = π a (ˆx + ŷ ẑ), Above equations indicate that the reciprocal lattice of fcc is bcc. Since the shortest vector in bcc is the one from the corner of the cube to the body center, without lose generality, take: G o = b = π a (ŷ + ẑ ˆx) then in the real space lattice, the lattice planes perpendicular to this G o belong to family of {}, i.e. the lattice planes with the greatest density of points are the {} planes in a face-centered cubic Bravais lattice. For bcc, choose: a = π a (ŷ + ẑ ˆx), a = π a (ẑ + ˆx ŷ), a 3 = π a (ˆx + ŷ ẑ), then the reciprocal lattice has: b = a (ŷ + ẑ), b = a (ẑ + ˆx), b3 = a (ˆx + ŷ) Above equations indicate that the reciprocal lattice of bcc is fcc. Since the shortest vector in fcc is the one from the corner of the cube to the face center, without lose generality, take: G o b = a (ŷ + ẑ) then in the real space lattice, the lattice planes perpendicular to this G o belong to family of {0}, i.e. the lattice planes with the greatest density of points are the {0} planes in a body-centered cubic Bravais lattice. Miller Indices April, 8, 0 8
Condensed Matter Physics April, 8, 0 Answer: 5 For conventional unit cell two planes with indices {00} and {00} are given. For {00}, x =, y = z = 0 and for {00}, x = y == 0, z =. Now consider the fcc lattice as shown in the figure. z a3 a a y x In this fcc lattice we define primitive lattice vectors a, a and a 3. Let a be the lattice constant or space between two atoms. Location of face centered atom corresponding to primitive vector a is (a/, a/, 0). Similarly the location of face centered atom having primitive axis a is (0, a/, a/), and the location of face centered atom having primitive axis a 3 is (a/, 0, a/). Therefore, a = a (ˆx + ŷ) a = a (ŷ + ẑ) a 3 = a (ẑ + ˆx) For (00) plane: a = a, a = 0, a 3 = a Thus indices of (00) plane refereing to a, a and a 3 are ( a, 0, a) = a (0), or (0). For (00) plane: a = 0, a = a, a 3 = a Thus indices of (00) plane refereing to a, a and a 3 are (0, a, a) = a (0), or (0). Answer: 6 April, 8, 0 9
Condensed Matter Physics April, 8, 0 z z () [] y y x x z z _ () _ x y _ [] _ x y Symmetry Operations Answer: 7 (i). Glasses, bottles etc. (ii). Every rectangular shape object has a -fold symmetry. e.g. Book, door etc. (iii). Objects with higher rotational symmetry are fan, tyres of vehicles etc. April, 8, 0 0