reciprocal lattice fcc bcc sc
DESCRIPTION
reciprocal latticeTRANSCRIPT
Direct space to reciprocal space
ijji aa πδ2* =•
Reciprocal spaceReal (direct) space
Note: The real space and reciprocal space vectors are not necessarily in the same direction
Laue Equations� Another way of expression the diffraction condition ∆k =
G is by the Laue equations, which can be derived by taking the scalar product of ∆k and G with a1, a2, and a3
� These equations have a simple interpretation: for a Bragg reflection, ∆k must lie on a certain cone about the direction of a1 (for example). Likewise, it must be on a cone for a2 and a3. Thus, a Bragg reflection which satisfies all three conditions must lie at the intersection of these three cones, which is at a point in reciprocal space.
a1 � ∆k = 2πν1 a2 � ∆k = 2πν2 a3 � ∆k = 2πν3
(Laue Equations)
Single crystal diffractionEwald sphere
kk�
Each time ∆k = G, a reciprocal
lattice vector, you get a Bragg reflection. This is a point of intensity at some 2θ angle, and some Φ angle in space
Bragg reflections
Powder vs. Single crystals� So, Bragg peaks are usually at a single spot in real space for single
crystals. For powder samples, however, which are composed of many tiny single crystals randomly oriented, the ∆G vectors exist on a cone of scattering (think of taking the Ewald circle, and rotating about k).
k
k�∆G
Another picture of powder x-ray diffraction
∆Gk�
2θ
k
Whenever Bragg�s law is satisfied (2dsinθ = nλ), we get a diffraction peak. Since there
is usually some crystallite which has it�s planes orienting in the proper direction, we get a cone of scattering about the
angle 2θ
Brillouin ZonesPoint D in reciprocal
space
½ GD
� A Brillouin Zone is defined as a Wigner-Seitz primitive cell in the reciprocal lattice.
� To find this, draw the reciprocal lattice. Then, use the same algorithm as for finding the Wigner-Seitz primitive cell in real space (draw vectors to all the nearest reciprocal lattice points, then bisect them. The resulting figure is your cell).
� The nice result of this is that it has a direct relation to the diffraction condition:
Wigner-Seitz cell
Therefore, the Brillouin Zone exhibits all wavevectors, k, which
can be Bragg-reflected by a crystalk � (1/2 G) = (1/2 G)2
The First Brillouin ZoneAn example: Rectangular Lattice
� The Zone we have drawn above using the Wigner-Seitz method is called the first Brillouin zone. The zone boundaries are k = +/- π/a (to make the total length to a side 2π/a in reciprocal space).
� The 1st Brillouin zone is the smallest volume entirely enclosed by the planes that are perpendicular bisectors of the reciprocal lattice vectors drawn from the origin.
� Usually, we don�t consider higher zones when we look at diffraction. However, they are of use in energy-band theory
Reciprocal lattice to SC lattice� The primitive translation vectors of any simple cubic
lattice are:
� Using the definition of reciprocal lattice vectors:
� We get the following primitive translation vectors of the reciprocal lattice:
321
213
321
132
321
321 2 2 2
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ו×
=ו
×=
ו×
= πππ
a1 = a x a2 = a y a3 = a z
b1 = (2π/a)x b2 = (2π/a)y b3 = (2π/a)z
This is another cubic lattice of length 2π/a
Reciprocal lattice to SC lattice
The boundaries of the first Brillouin zone are the planes normal to the six reciprocal lattice vectors +/- b1, +/- b2, +/-b3 at their midpoints:
+/- (π/a)
The length of each side is 2π/a and the volume is (2π/a)3
2π/a
Reciprocal lattice to BCC lattice� The primitive translation vectors for the BCC lattice are:
� The volume of the primitive cell is ½ a3 (2 pts./unit cell)� So, the primitive translation vectors in reciprocal space
are:
� What lattice is this?
a1 = ½ a (x + y - z)a2 = ½ a (-x+y + z)a3 = ½ a (x - y + z)
b1 = 2π/a (x + y)b2 = 2π/a (y + z)b3 = 2π/a (z + x)
Reciprocal lattice to BCC lattice� This is the FCC lattice!� So, the fourier transform of the BCC lattice is the FCC
lattice (what do you expect for the FCC lattice, then?)� The general reciprocal lattice vector, for integrals ν1,ν2,
and ν3 is then:
� The shortest G vectors are the following 12 vectors, where choices of sign are independent:
G = ν1 b1 + ν2 b2 + ν3 b3= (2π/a)[(ν2 + ν3)x +(ν1 + ν3)y + (ν1 + ν2)z)]
(2π/a)(+/-y +/- z) (2π/a)(+/-x +/-z) (2π/a)(+/-x +/-y)
Reciprocal lattice to BCC lattice� This is the first Brillouin zone of
the BCC lattice (which has the same shape as the Wigner-Seitz cell of the FCC lattice). It has 12 sides (rhombic dodecahedron).
� The volume of this cell in reciprocal space is 2(2π/a)3, but it only contains one reciprocal lattice point.
� The vectors from the origin to the center of each face are:
(π/a)(+/-y +/- z) (π/a)(+/-x +/-z) (π/a)(+/-x +/-y)
Reciprocal lattice to the FCC lattice
� The primitive translation vectors for the FCC lattice are:
� The volume of the primitive cell is 1/4 a3 (4 pts./unit cell)� So, the primitive translation vectors in reciprocal space are:
� This is, of course, the BCC lattice� The volume of this cell is 4(2π/a)3 in reciprocal space
a1 = ½ a (x + y)a2 = ½ a (y + z)a3 = ½ a (z + x)
b1 = (2π/a) (x + y - z)b2 = (2π/a) (-x+y + z)b3 = (2π/a) (x - y + z)
Reciprocal space to the FCC lattice
� This is the first Brillouin zone of the FCC cubic lattice. It has 14 sides bound by:
(2π/a)(+/-x +/-y +/- z) 4π/a
(8 of these vectors)
and
(2π/a)(+/-2x)(2π/a)(+/-2y)(2π/a)(+/-2z)
(6 of these vectors)