Please use these examples to answer the questions below:

N 1:th = Z fn821ri(hun+kvn+lwn) 1 1. Simple Cubic (SC) There is one atom at the origin. 000 F2 = 12 X-ray diffraction intensity is same for all (hkl) planes. This calculation is not limited to the cubic. u, V, w are position of atom in the unit of lattice constant. u =v=w = 16 at the body center. Instead of taking one atom at the origin, 000, one can assume that 1/8 of atoms are located at 8 corners. 1/8 (000,100, 010, 001,110,101, 011, and 111). Conrm that the same structure factor will be obtained. Fhkl fne2ni(hun +kun +lwn) Similar to SC, BCC structure can be 2. Body Centered cubic (BCC) considered to have one There are two atoms in the unit atom in the Body Center cell. position and 1/8 atom 000 and 1/21/21/2. each at each of 8 F = fe2xi(0) + fe2ni(h/2 + k/2 +1/2) corners. = fl 1 + exi(h+k+1)] F = 2f for h+k+1= even F2 =412 F= 0 for h+k+l = odd F2 = 0 Peaks such as (001), (111) will be absent. Peaks are observed for (110), (200), (211), (220), ...1} Draw an FCC unit cell and give the coordinates (u,v,w) describing the positions of all the atoms in such a unit cell. 2} Calculate the structure factor Fhk. for such a FCC unit cell for xray scattering from a given set of crystallographic planes {hkl}