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Example 1 For BCC iron, compute (a) the interplanar spacing, and (b) the diffraction angle for the (220) set of planes. The lattice parameter for Fe is 0.2866 nm. Also, assume that monochromatic radiation having a wavelength of 0.1790 nm is used, and the order of reflection is 1. Solution: (a) The value of the interplanar spacing is determined, with a= 0.2866 nm, and h= 2, k = 2, l = 0, and since we are considering the (220) planes. Therefore,   b) The value of ɵ may now be computed, with n=1 since this is a first-order reflection: The diffraction angle is 2ɵ or:

Example 2 For BCC iron, compute (a) the interplanar spacing, and (b) the diffraction angle for the (220) set of planes. The lattice parameter for Fe is 0.2866 nm. Also, assume that monochromatic radiation having a wavelength of 0.1790 nm is used, and the order of reflection is 1. Solution: From the data given in the problem, and realizing that 36.12° = 2θ, the interplanar spacing for the (311) set of planes for rhodium may be computed using the following Equation: In order to compute the atomic radius we must first determine the lattice parameter, a, and then R , since Rh has an FCC crystal structure. Therefore,

Example 3 The metal niobium has a BCC crystal structure. If the angle of diffraction for the (211) set of planes occurs at 75.99o (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1659 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for the niobium atom. Solution: From the data given in the problem, and realizing that 75.99° = 2θ, the interplanar spacing for the (211) set of planes for Nb may be computed using the following Equation: In order to compute the atomic radius we must first determine the lattice parameter, a, and then R since Nb has a BCC crystal structure. Therefore

Example 4 The results of a diffraction experiment using X-Rays with λ = 0.7107 Å (radiation obtained from a molybdenum, Mo, target) show that diffracted peaks occur at the following 2θ angles: Find the crystal structure, the indices of the plane producing each peak, and the lattice parameter of the material.

Example (Solution) First calculate the sin2 θ value for each peak, then divide through by the lowest denominator, 0.0308.

Example (Solution Continued) Then use the 2θ values for any of the peaks to calculate the interplanar spacing & thus the lattice parameter. Picking peak 8: 2θ = 59.42° or θ = 29.71° This is the lattice parameter for body-centered cubic iron. So, for example: Ǻ Ǻ

Example 5 How much should be the energy of X-ray to generate wavelength of 10-10m, (h= 4.1357 × 10-15 eV.s, and c=3×108m)