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Main topic: DP: diffraction pattern Reciprocal lattice Bragger’s law Structure factor Amplitude of electron Main topic: DP: diffraction pattern Reciprocal lattice Bragger’s law Structure factor Amplitude of electron beam scattered by a unit cell

How can Diffraction occur Using bragger’s law and reciprocal lattice How can Diffraction occur Using bragger’s law and reciprocal lattice

Something about X-RAY When x ray hit the electrons, it makes electrons to vibrate Something about X-RAY When x ray hit the electrons, it makes electrons to vibrate in the same frequency. those electrons become the new source of X-ray. We call beams which given by electrons SCATTERING BEAM.

Bragger’s law Bragger’s law

d Incident beam Interference occur only when the optical path difference is nl. Bragger’s d Incident beam Interference occur only when the optical path difference is nl. Bragger’s law nl=2 dsinø Scattering beam

Reciprocal lattice Reciprocal lattice

a*, b*, c* are base vectors of reciprocal lattice Through math derivation, we have a*, b*, c* are base vectors of reciprocal lattice Through math derivation, we have In the reciprocal lattice, A point represents sets of parallel (hkl) g is the vector from origin to a single point. |g|=1/dhkl

constructive interference will only occur when constructive interference will only occur when

What affects the intensity Using amplitude equation and structure factor What affects the intensity Using amplitude equation and structure factor

Ri is the vector which defines each atoms in the lattice Multiply ri and Ri is the vector which defines each atoms in the lattice Multiply ri and k We now have Structure factor. The value of structure factor is linked with crystallology It can predict each plane’s DP intensity remember a×a*=1 Ignore the value before ∑

Examples Important structures Examples Important structures

① we need to choose xi, yi, zi：It’s important to know how many atoms ① we need to choose xi, yi, zi：It’s important to know how many atoms are in one lattice, different structure differs. (x, y, z) are the indices of those atoms. ② put h, k, l in the equation BCC: 2 atoms in one lattice (x, y, z)=(0, 0, 0) & (1/2, 1/2) FCC: 4 atoms in one lattice (x, y, z)=(0, 0, 0), (1/2, 0), (1/2. 0, 1/2), (0, 1/2) HCP: 2 atoms (x, y, z)=(0, 0, 0), (1/3, 2/3, 1/2)

BCC: 2 atoms in one lattice (x, y, z)=(0, 0, 0) & (1/2, 1/2) BCC: 2 atoms in one lattice (x, y, z)=(0, 0, 0) & (1/2, 1/2) FCC: 4 atoms in one lattice (x, y, z)=(0, 0, 0), (1/2, 0), (1/2. 0, 1/2) , (0, 1/2) HCP: 2 atoms (x, y, z)=(0, 0, 0), (1/3, 2/3, 1/2)

Examples Extended Examples Extended

Na atoms on an fcc site Cl related to it by the basis vector Na atoms on an fcc site Cl related to it by the basis vector [1/2, 1/2]

Ga located on the fcc lattice As related to it by the basis vector Ga located on the fcc lattice As related to it by the basis vector [1/4, 1/4]

Reference 1. Wiiliams, D. B. , &Barry Carter, C. , (1996). transmission electron microscopy. Reference 1. Wiiliams, D. B. , &Barry Carter, C. , (1996). transmission electron microscopy. new york: Springer ScienceþBusiness Media

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