Indexing System for Crystal Plane-Solid State Physics-Lecture Slides, Slides of Solid State Physics

Download Indexing System for Crystal Plane-Solid State Physics-Lecture Slides and more Slides Solid State Physics in PDF only on Docsity!

 There are two parts of indexing system

1. Crystal Direction

2. Crystal Planes

 The orientation of a plane in a lattice is specified by Miller indices  They are defined as follows:  We find intercept of the plane with the axes along the primitive translation vectors a 1 , a 2 and a 3.  Let’s these intercepts be x, y, and z, so that x is fractional multiple of a 1 , y is a fractional multiple of a 2 and z is a fractional multiple of a 3. Therefore we can measure x, y, and z in units a 1 , a 2 and a 3 respectively.  We have then a triplet of integers ( x y z)  Then we invert it (1/x 1/y 1/ z) and reduce this set to a similar one having the smallest integers by multiplying by a common factor.  This set is called Miller indices of the plane ( hkl). For example, if the plane intercepts x, y, and z in points 1, 3, and 1, the index of this plane will be (313).

 Its name is static structure factor or simply structure factor

 In condensed matter physics and crystallography, structure factor is a mathematical description of how a material scatters incident radiation

 The structure factor is a particularly useful tool in the interpretation of interference patterns obtained in X- ray, electron and neutron diffraction experiments.

 The static structure factor is measured without resolving the energy of scattered photons/electrons/neutrons

 Energy-resolved measurements yield the dynamic structure factor

 When the diffraction condition ∆k=G is satisfied, the scattering amplitude is determined by a crystal of N cells may be written as:

 FG = = NSG

 Where the quantity SG is called structure factor and is defined as an integral over a single cell with r = o at one corner and n(r) is the electron concentration.

6/20/2012 10

 Draw a vector AO of length 1/λ in the direction of the incident x-ray beam which terminates at the origin O of the reciprocal lattice.

 Taking A as the center, draw a sphere of radius AO which may intersect some point B of the reciprocal lattice

 Let the coordinates of pt. B be (h’,k’,l’) which may have a highest common factor n

 The coordinates are of the type (nh, nk, nl), where h, k, l are not the common factor

 Vector OB is the reciprocal vector

docsity.com

 Therefore, it must be normal to the plane (h’k’l’) or (hkl) and have length equal to 1/dhkl or n/dhkl

 Thus

 From the geometry, let us such a plane AE,

 if <EAO = θ, is the angle b/w incident ray and normal

 Then, from ∆AOB, we have

 OB = 2OE = 2OASinθ = (2Sinθ)/λ

 (2Sinθ)/λ = n/dhkl

 2dSinθ = nλ

 Which is Bragg’s law

docsity.com