Chapter 2 - Physical Methods for Characterizing Solids

1 1X-rays were discovered by Wilhelm Rontgen, a German physicist in generate x-rays, three things are needed. a source of electrons a means of accelerating the electrons at high speeds a target material to receive the impact of the electrons and interact with them. Typical cathode element is difference is 20-50 must be water (wavelength) depends on the anode metal, often Cu, Mo, Co. Lines occur because bombarding electrons knock out e-from K shell (n = 1), which are filled by electrons in higher shells. Electrons falling from L shell (n = 2) give rise to Kalines, whereas e-from M shell (n = 3) give the Kblines.

2 (K 1and K 2doublets, etc.)X-ray diffractionX-ray tube2 Monochromatic radiation (single wavelength or a narrow range of wavelengths) is required for X-ray diffraction. Typically, the K line is selected and the K line is filtered out by using a thin metal foil of the adjacent (Z-1) element. (nickel filters K line of copper) A monochromatic beam of X-rays can also be selected by reflecting from a plane of a single observed intensity Iis given by: where is a linear absorption coefficient and tis the path length through which the X-rays are moving. = Diffraction of X-rays Max von Laue used a crystal of copper sulfate as the diffraction grating (Nobel Prize 1914).

3 Crystalline Solids consist of regular arrays of atoms, ion, or molecules with interatomic spacing on the order of 100 pm or 1 . The wavelength of the incident light has to be on the same order as the spacing of the atoms. and Bragg determined crystal structures of NaCl, KCl, ZnS, CaF2, CaCO3, C (diamond). Reflection of X-rays only occurs when the conditions for constructive interference are sourceX-ray detectorDestructive InterferenceConstructive InterferenceDifference in path length = BC + CDBC = CD = dhklsin hklDifference in path length = 2dhklsin hklMust be an integral number of wavelengths, n = 2dhklsin hkl (n = 1, 2, 3.)

4 = 2dhklsin Bragg Equation3 Powder DiffractionA powder may be composed of many small and finely ground crystals, known as crystallites. These crystallites are (assumed to be) randomly oriented to one another. If the powder is placed in the path of a monochromatic X-ray beam, diffraction will occur from the planes in those crystallites that are oriented at the correct angle to fulfill the Bragg condition. The diffracted beams make an angle of 2 with the incident CrystalPolycrystalline PowderIn the Debye-Scherrerphotographic method, a film was wrapped around the inside of a X-ray powder, sealed in a glass capillary tube, diffracts the X-rays (Bragg s law) to produce cones of diffracted beams.

5 These cones intersect a strip of photographic film located in the cylindrical camera to produce a characteristic set of arcs on the film. The film can be removed and examined. Using the radius of the camera and the distance along the film from the center, the Bragg angle 2 and therefore the d-spacing for each reflection can be X-ray Diffractometer-use a scintillation or CCD detector to record the angles and intensities of the diffracted resolution obtained by a diffractometeris better than photographic film, since the sample helps refocus the X-ray is more readily measured and digitally the peaksWhich planes are responsible for each reflection?

6 -need to index the reflections (assign the correct hkl)-difficulty may range from simple to extremely difficultThe planes giving rise to the smallest Bragg angle will have the largest a primitive cubic system, with the 100 planes having the largest separation (the 010 and 001 planes also reflect at this position))(222lkhadhkl++=)()sin(2222lkha ++= )(4)(sin222222lkha++= )sin(2 d=222221alkhd++=Bragg s LawFor a cubic system5 For aprimitive cubicclass,all integral values of theindicesh,k, + k2 + l21001110211132004210521162208300, 2219 Note that the value 7 is missing in the sequence, since there is no possible integral value that h2+ k2 + l2 = 7If the intensity of diffraction of the pattern is plotted against sin2( ), one would obtain equally spaced lines with the 7th, 15th, 23rd, etc.

7 Is easy to identify a primitive cubic system and (by inspection) assign indices to each of the cubic unit cell dimension acan be obtained from anyof the indexed the experimental error is relatively constant, the reflection with the largest Bragg angle is chosen to minimize error in the lattice parameter calculation (or perform least squares on all reflections).)(4)(sin222222lkha++= 2 = IntensityDiffractometer equipped with a copper X-ray sources emit X-rays with a wavelength, = . 6 Absences due to lattice centeringThe pattern of observed lines for the two other cubic crystalline systems, body-centered and face-centered is different from primitive.

8 The differences arise because the centering leads to destructive interference for some reflections and the missing reflections are known as systematic ais the cell dimension, the planes have a spacing planes of two face centered unit reflection from the 200 plane is exactly out of phase with the 100 reflection and destructive interference occurs and no 100 reflection is , in order for a reflection to be observed for fcc, the hklindices must be all odd or all similar analysis of the bccsystem would find that the sum of the reflections must be even (h+k+l= 2n)-these absences apply to all crystal systems, not just list of h2+ k2 + l2for cubic crystalsForbidden numbersPrimitive, PFace Centered, FBody Centered, ICorresponding hkl1100221103311144420052106621178882209 221, 3001010310111131112121222213320141432115 161616400If the observed sin2( ) follows in a ratio of 1, 2,3,4,5,6, 8.

9 , then the unit cell is likely primitive common factor is 2/4a27 Indexing Example 2 ( )IntensityDiffractometer equipped with a copper X-ray sources emit X-rays with a wavelength, = . Peak #2 ( ) ( )d ( ) Bragg s Law, = 2dhklsin Look for a common factor, Z, which can be divided into each value to give an integer quotient. Let Z= in this #2 ( ) ( )d ( )1/d2(1/d2) that in this example, we obtain all integers and all the quotients are integers that indicate a primitive unit #2 ( ) ( )d ( )1/d2(1/d2) * of Miller indicesThe (300) and (221) peaks fall at the same location, both give h2+ k2+ l2= 9.

10 ()sin(2222lkha++= The lattice parameter, a, is Systematic absencesSymmetry elementAffected reflection Condition for reflection to be presentPrimitive latticePhklnoneBody-centered latticeIhklh+ k+ l= evenFace-centered lattice Ahklk+ l= evenBhklh+ l= evenChklh+ k= evenFace-centered latticeFhklh k lall odd or eventwofold screw, 21along ah00h = evenfourfold screw, 42alongah00h = evensixfold screw, 63alongc00lldivisible by 3threefold screw, 31, 32alongc00lldivisible by 3sixfold screw, 62, 64alongah00hdivisible by 4fourfold screw, 41, 43alongah00hdivisible by 4sixfold screw, 61, 65alongc00lldivisible by 6 Glide plane perpendicular to bTranslation a/2 (aglide)h0lh= evenTranslation c/2 (cglide)h0ll= evenb/2 + c/2 (nglide)h0lh+ l= evenb/4 + c/4 (dglide))