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3.16  X-Ray Diffraction: Determination of Crystal Structures  •  91

              Figure 3.24
              Diffraction pattern
              for powdered lead.
              (Courtesy of Wesley
              L. Holman.)






                                    One of the primary uses of x-ray diffractometry is for the determination of crystal
                                 structure. The unit cell size and geometry may be resolved from the angular positions
                                 of the diffraction peaks, whereas the arrangement of atoms within the unit cell is associ-
                                 ated with the relative intensities of these peaks.
                                    X-rays, as well as electron and neutron beams, are also used in other types of mate-
                                 rial investigations. For example, crystallographic orientations of single crystals are possi-
                                 ble using x-ray diffraction (or Laue) photographs. The chapter-opening photograph (a)
                                 was generated using an incident x-ray beam that was directed on a magnesium crystal;
                                 each spot (with the exception of the darkest one near the center) resulted from an x-ray
                                 beam that was diffracted by a specific set of crystallographic planes. Other uses of x-rays
                                 include qualitative and quantitative chemical identifications and the determination of
                                 residual stresses and crystal size.



                        EXAMPLE PROBLEM 3.14
                          Interplanar Spacing and Diffraction Angle Computations
                          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. 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 d hkl  is determined using Equation 3.22, with a   0.2866 nm,
                             and h   2, k   2, and l   0 because we are considering the (220) planes. Therefore,

                                                          a
                                               d hkl =
                                                            2
                                                       2
                                                    2h + k + l  2
                                                         0.2866 nm
                                                   =                  = 0.1013 nm
                                                         2
                                                               2
                                                    2(2) + (2) + (0) 2
                          (b)  The value of u may now be computed using Equation 3.21, with n   1 because this is a first-
                             order reflection:
                                                        nl    (1)(0.1790 nm)
                                                  sin u =   =              = 0.884
                                                       2d hkl  (2)(0.1013 nm)
                                                          -1
                                                    u =  sin  (0.884) = 62.13
                             The diffraction angle is 2u, or
                                                     2u = (2)(62.13 ) = 124.26
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