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X-RAY PRODUCTION Lecture 1 091409 Wavelength: between 0.1 and 100 Å. 1- Bombard a metal target with a beam of high energy electrons (what we do in the lab). 2- Expose a substance to a primary X-ray beam to produce a secondary fluorescent beam of X-rays usually used for analytical analysis (geochemistry). 3- Use a radioactive source whose decay produces X-rays. 4- Cyclotron, synchrotron, linear accelerator (very high flux of X-rays). Continuous spectra: in a Coolidge tube, most of the electrons are not brought to a full stop by a single collision with the target, and therefore, one gets a continuum of radiation. Cost ~$4800; lifetime: ~1000-20000 hours Continuous X-ray spectrum is called “white radiation” or bremstrahlung. KE = eV = ½ mv2 = hυ = hc/λ If KE is a maximum, then υ = maximum and λ = minimum, called λSWL (short wavelength limit) = hc/eV Ex.: λSWL = (6.626x10-34 J s)(2.998x108 ms-1)/(1.602x10-19 coul)(V) λSWL = 12.4x10-7 J m/(coul)(V) Since 1 J = 1 (coul)(V) and 1 m = 1.0x1010 Å: λSWL = 12.4x10-7 (coul)(V)(m)/(coul)(V) λSWL = (12.4x10-7 m) x (1.0x1010 Å m-1) = 12.4x103 Å Called the Duane-Hunt law: has been used to get highly accurate value of Planck’s constant, h. This lower wavelength limit, λSWL, is dependent only on the accelerating voltage, V, and is independent of the target material. Explanation : The conversion of entire KE of an electron to radiant energy. Therefore, complete deceleration down to zero KE. For longer wavelengths, assume that the bombarding electrons lose different amounts of energy when they strike the target and some get decelerated quickly, while others lose energy in steps, etc. The production of X-rays is very inefficient, with less than 1% conversion to X-rays; the rest is all heat. Therefore, cooling of X-ray tubes is required. The power dissipated at the anode is ~1.5 to 3 kW. (Rotating anodes~12kW) Characteristic line spectra: Can also get line spectra superimposed on white radiation: X-ray line spectra are very simple: with Z > 23 =Vanadium, usually get 2 series of lines: K-series and L-series. Ex: Element Cr Cu Mo Cs Z 24 29 42 55 K-series α (Å) β (Å) 2.290 2.085 1.5418 1.38 0.71073 0.632 0.401 0.355 L-series α (Å) β (Å) 21.714 21.323 2.892 2.683 Each shell is separated into 2n-1 discrete energy levels with respect to the 3 other quantum number (l, ml, and s). For K shell, n = 1; for L, n = 2; and for M, n = 3. Therefore, L has 2(2)-1 = 3 levels and M has 2(3)-1 = 5 levels, etc. The selection rules make some of the transitions impossible: ex.: inner L to inner K shell is not allowed. Therefore, the possible allowed transitions are: 2 different L-shell to K-shell transitions, yielding Kα1 and Kα2; two different M-shell to K-shell transitions, yielding Kβ1 and Kβ2. Another series of transitions from the Mshell to the L-shell would yield the Lα-series of X-ray lines, and the N-shell to the L-shell would yield the Lβ-series of X-ray lines. Minimum accelerating voltage required for line spectra increases with increasing Z. Mo (Z=42) gives no line spectra below 20 kV. W (Z=74) produces no lines at 50 kV, but get K lines at ≥ 70 kV. Limited target materials: conductive, solid, dense, high-melting; therefore, use transition metals Z = 21 - 30 and 39 - 48. Line spectra arise from inner electrons being knocked out of the atom (Auger electrons & Auger spectroscopy) and, when the electrons in higher energy shells drop in to fill the vacancies, then X-rays are produced. Absorption edges: If you take a thin sheet (~0.001 in.) of a metal and bombard it with X-rays of varying wavelengths, you will find the absorption edges of that particular element: As the wavelength increases, there is an increase in absorption until a discontinuity (absorption edge) occurs; this is where the electrons no longer can be knocked out of the inner K-shell (the incoming electrons do not have enough energy). Then the absorption continues to increase until the absorption of the L-electron occurs, etc. Usually use CuKα radiation for studying the structure of organic materials because the radiation penetrates enough and the elements to be determined usually have Z<20, and their fluorescence is soft. Also, as we will see later, we can use Cu to determine the absolute configuration of molecules (Mo radiation does not allow one to do this). MoKα radiation is used mainly for compounds containing “heavy” metals (Z > 20). However, in our experience, we can do very well in determining structures with heavy metals using Cu radiation, as long as we make appropriate absorption corrections (see below). Three different ways to make monochromatic X-rays: 1- Filter: for 1st transition row elements, use a metal Z-1 from that of the metal used to produce the radiation; for 2nd transition row elements, use Z-2 from that of the metal used to produce the radiation. Ex.: Use a Ni filter ~0.1 mm thick in front of a Cu X-ray tube. Since the Ni’s absorption edge lies between CuKα and CuKβ, it will absorb Kβ completely, but will also absorb about 50% of the Kα. For Mo radiation, use Zr filter (it will allow only MoKα to come through. 2- Monochromator: Use large single crystal of some material to monochromatize the X-rays, and slits to separate out the wavelength of interest. Examples of some crystals used: Topaz 2.67-0.24 Å NaCl 5.55-0.49 Å Graphite ? Å 3- Göbel mirrors: mainly used by protein crystallographers. USES OF X-RAYS: 1- Absorption 2- Fluorescence 3- Diffraction 1- Absorption: Can tell if structural steel of Ti-Al alloy has cracks or weaknesses. Can tell how much liquid is in a sealed Al can (beer and soda). Beer’s law holds: I = Ioe-μt where I = transmitted intensity Io = initial intensity linear absorption coefficient (cm-1) t = path length (cm) Ex.: for a solid like CaSO4, we can calculate the bulk absorption of the material by looking up the mass absorption coefficient (μ/ρ) for each element (as example, for Cu radiation), then multiplying by the % composition of each element in the compound; this will give the μ/ρ contribution for each element. After summing these, we will have Σ μ/ρ for the whole compound; then, multiplying by ρ (the density of the material), we will then have μ, the linear absorption coefficient (in cm-1). Element Ca S O μ/ρ(cm2/g) 172 91 13 Weight Fraction μ/ρ x Weight Fraction 40.08/136.15 = 0.2944 50.64 32.07/136.15 = 0.2355 21.43 64.00/136.15 = 0.4701 6.11 Σ μ/ρ = 78.18 cm2/g Therefore, μ = Σ μ/ρ x ρ = 78.18 cm2/g x 2.96 g/cm3 = 231.4 cm-1 μ/ρ is a function of the radiation. If one were to use Mo radiation instead of Cu, the μ/ρ values for these elements are: Ca = 19.00, S = 9.6, O = 1.15 cm2/g. Therefore, Σ μ/ρ = 8.39 cm2/g and μ, the linear absorption coefficient, is 24.8 cm-1, approximately 10 times less than the absorption calculated for Cu radiation. We will have to take this absorption into account during the data reduction process. 2- Fluorescence: Used for non-destructive analytical analyses: paintings, rocks, minerals (geologists), and many other applications. 3- Diffraction: Both powder and single crystal diffraction. THE SOLID STATE: Crystals denote symmetry. Q: Are all xtals from same batch identical? A: No. They generally have same habit (all needles or all plates); however, they may have different shapes and sizes, as well as different habit. Q: Anything in common? A: Yes, same interfacial angles (the angles between the normals to the faces). Definitions: Unit cell: smallest repeat unit that will fill up entire space or lattice. Asymmetric unit: smallest repeat unit required to be able to make the unit cell using all the available symmetries. Lattice: regular array of points in space, either 2-D or 3-D. In 2-D, each cell requires 2 vectors, a & b, to describe it. In 3-D, each cell requires 3 vectors, a, b & c, to describe it. If each point in a space lattice is replaced by an identical atom, or group of ions, or group of atoms (a molecule), then there is obtained a crystal structure. The lattice is an array of points; in a crystal structure, each point is replaced by an atom, or a group of ions or atoms (or a molecule). The positions of atoms in the unit cell are denoted by coordinates given as fractions of the unit cell dimensions. Ex.: A unit cell has sides a = 10.05 Å , b = 12.25 Å, c = 17.69 Å. An atom at 0.1234, 0.4255, and 0.8106 would be at (0.1234x10.05 Å) = 1.240Å along a, (0.4255x12.25 Å) = 5.212Å along b, and (0.8106x17.69 Å) = 14.34 Å along c, relative to the origin at 0.00, 0.00, 0.00 at the corner of the unit cell. (NOTE: ALL the corners in the unit cell are at 0,0,0 relative to some unit cell!!). RAL 072209