Lecture -93-3The reciprocal lattice
Since vectors in real space have dimensions of
length, those in reciprocal space have dimensions of
length-1 .The inverted distance between plans is
determines the position of a point in the reciprocal
lattice,so it is considered from the point of view of
mathematics vector ,that we call reciprocal lattice
vector (𝑮hkl).
‫𝑮ן‬hkl‫𝐴=ן‬/𝑑 hkl(where 𝐴=1 or 2𝜋 )
or 𝑑 hkl = 2𝜋/I𝑮hklI
If 𝒂, 𝒃, and 𝒄are the fundamental translation vectors
of a real lattice, then the fundamental reciprocal
vectors𝒂*, 𝒃*, and𝒄* are defined as following:
𝒃×𝒄
𝒂*=2𝜋 𝒂.𝒃×𝒄
𝒃* =2𝜋
𝒄* =2𝜋
𝒄×𝒂
𝒂.𝒃×𝒄
𝒂×𝒃
𝒂.𝒃×𝒄
Note 1: The denominator is the volume of the unit
cell, which does not depend on the order
Note 2:𝒂* is not necessarily parallel to 𝒂 but is
required to be perpendicular to both 𝒃 and 𝒄.
Note 3: 𝟐𝝅may be neglected from above Eqs.
We can write zeros for the various dot products:
𝒂*.𝒃 =𝒂*. 𝒄 = 𝒃*. 𝒄= 𝒃*. 𝒂=𝒄*.𝒂= 𝒄*. 𝒃=0
Whereas
𝒂*. 𝒂 = 𝒃*. 𝒃= 𝒄*. 𝒄=2𝜋
These consequences are illustrated in Fig.21 for a
2D oblique real lattice.
b
𝒃*
𝑎
𝒂*
Fig.21
The inverse of the transformation defined in Eq.
𝒂* =2𝜋
𝒃×𝒄
is made in the same fashion as the first.
𝒂.𝒃×𝒄
We can easily confirm that
𝒂=2𝜋
𝒃∗×𝒄∗
𝒂∗.𝒃∗×𝒄∗
and so forth.
Note 4: The 𝑎*, 𝑏,*, 𝑐*are orthogonal if 𝑎, 𝑏, 𝑐 vectors
are also orthogonal (see Fig.22)
𝑐
𝑐*𝑏
𝒃*
𝑎𝑎*
Fig.22
The point in the real lattice express as
𝑻=𝑛1𝒂+𝑛2𝒃+𝑛3𝒄
And in reciprocal space in the set of translations that
we called reciprocal lattice vectors :
𝑮hkl= ℎ𝒂*+ 𝑘𝒃*+ 𝑙𝒄*
Where ℎ, 𝑘, 𝑙 are integers
Note:when𝑻. 𝑮hkl=(𝑛1𝒂+𝑛2𝒃+𝑛3𝒄). (ℎ𝒂*+ 𝑘𝒃*+ 𝑙𝒄*)
=2𝜋(integer)
Then 𝑮hklsatisfied the diffraction condition
Problems:
1-Find the primitivereciprocal vectors of the Sc lattice
2- Find the primitivereciprocal vectors of the Bcc lattice
3- Prove that the reciprocal vector𝑮hklis perpendicular
always to the plane (ℎ𝑘𝑙)
4- Show that the spacing between two parallel successive
plane (hkl) is 𝑑 hkl = 2𝜋/I𝑮hklI
5- Show that 𝑑 hkl = 2𝜋/I𝑮hklI is equivalent to 𝑑=
In Sc lattice
𝑎
ℎ2+ 𝑘 2+ 𝑙2