Understanding X-ray Diffraction and Bragg's Law in Crystallography

1. Objectives By the end of this section you should: • understand the concept of diffraction in crystals • be able to derive and use Bragg’s law • know how X-rays are produced • know the typical emission spectrum for X-rays, the source of white radiation and the K and K lines • know about Compton scattering

2. Diffraction - an optical grating Path difference XY between diffracted beams 1 and 2: sin = XY/a  XY = a sin  For 1 and 2 to be in phase and give constructive interference, XY = , 2, 3, 4…..n so a sin  = n where n is the order of diffraction

3. Consequences: maximum value of  for diffraction sin = 1  a =  Realistically, sin <1  a >  So separation must be same order as, but greater than, wavelength of light. Thus for diffraction from crystals: Interatomic distances 0.1 - 2 Å so  = 0.1 - 2 Å X-rays, electrons, neutrons suitable

4. Diffraction from crystals ?

5. Beam 2 lags beam 1 by XYZ = 2d sin  so 2d sin  = nBragg’s Law

6. e.g. X-rays with wavelength 1.54Å are reflected from planes with d=1.2Å. Calculate the Bragg angle, , for constructive interference.  = 1.54 x 10-10 m, d = 1.2 x 10-10 m, =? n=1 :  = 39.9° n=2 : X (n/2d)>1 2d sin  = n We normally set n=1 and adjust Miller indices, to give 2dhkl sin  = 

7. Use Bragg’s law and the d-spacing equation to solve a wide variety of problems 2d sin  = n or 2dhkl sin  = 

8. Example of equivalence of the two forms of Bragg’s law: Calculate  for =1.54 Å, cubic crystal, a=5Å 2d sin  = n (1 0 0) reflection, d=5 Å n=1, =8.86o n=2, =17.93o n=3, =27.52o n=4, =38.02o n=5, =50.35o n=6, =67.52o no reflection for n7 (2 0 0) reflection, d=2.5 Å n=1, =17.93o n=2, =38.02o n=3, =67.52o no reflection for n4

9. Combining Bragg and d-spacing equation X-rays with wavelength 1.54 Å are “reflected” from the (1 1 0) planes of a cubic crystal with unit cell a = 6 Å. Calculate the Bragg angle, , for all orders of reflection, n. d = 4.24 Å

10. d = 4.24 Å n = 1 :  = 10.46° n = 2 :  = 21.30° n = 3 :  = 33.01° n = 4 :  = 46.59° n = 5 :  = 65.23° = (1 1 0) = (2 2 0) = (3 3 0) = (4 4 0) = (5 5 0) 2dhkl sin  = 

11. X-rays and solids X-rays - electromagnetic waves So X-ray photon has E = h X-ray wavelengths vary from .01 - 10Å; those used in crystallography have frequencies 2-6 x 1018Hz Q. To what wavelength range does this frequency range correspond? c =  max = 1.5 Å min = 0.5 Å

12. In the classical treatment, X-rays interact with electrons in an atom, causing them to oscillate with the X-ray beam. The electron then acts as a source of an electric field with the same frequency  Electrons scatter X-rays with no frequency shift

13. Production of X-rays

14. Two processes lead to two forms of X-ray emission:  Electrons stopped by target; kinetic energy converted to X-rays continuous spectrum of “white” radiation, with cut-off at short  (according to h=½mv2) Wavelength not characteristic of target  Incident electrons displace inner shell electrons, intershell electron transitions from outer shell to inner shell vacancy. “line” spectra Wavelength characteristic of target

15. Each element has a characteristic wavelength. For copper, the  are: CuK1 = 1.540 Å CuK2 = 1.544 Å CuK = 1.39 Å Typical emission spectrum

16. Many intershell transitions can occur - the common transitions encountered are: 2p (L) - 1s (K), known as the K line 3p (M) - 1s (K), known as the K line (in fact K is a close doublet, associated with the two spin states of 2p electrons)

17. Copper K X-rays have a wavelength of 1.54 Å and are produced when an electron falls from the L shell to a vacant site in the K shell of a copper atom. Calculate the difference in the energy levels between the K and L shells of copper. E = h  = c/  = (3 x108) / (1.54 x 10-10) = 1.95 x 1018 Hz E = h = 6.626 x 10-34x 1.95 x 1018 = 1.29 x 10-15 J ~ 8 keV

18. Some radiation is also scattered, resulting in a loss of energy [and hence, E=h, shorter frequency and, c= , longer wavelength. The change in frequency/wavelength depends on the angle of scattering. This effect is known as Compton scattering It is a quantum effect - remember classically there should be no frequency shift

19. Calculate the maximum wavelength shift predicted from the Compton scattering equation. = 4.85 x 10-12m = 0.05Å

20. Filter As well as characteristic emission spectra, elements have characteristic absorptionwavelengths e.g. copper

21. We want to choose an element which absorbs K [and high energy/low  white radiation] but transmits K e.g. Ni K absorption edge = 1.45 Å As a general rule use an element whose Z is one or two less than that of the emitting atom

22. Monochromator Choose a crystal (quartz, germanium etc.) with a strong reflection from one set of lattice planes, then orient the crystal at the Bragg angle for K1  = 1.540 Å = 2dhklsin

23. Example: A monochromator is made using the (111) planes of germanium, which is cubic, a=5.66Å. Calculate the angle at which it must be oriented to give CuK1 radiation d=3.27Å =2d sin = 13.62°

24. Summary • Crystals diffract radiation of a similar order of wavelength to the interatomic spacings • We model this diffraction by considering the “reflection” of radiation from planes - Bragg’s Law • X-rays are produced by intershell transitions - e.g. L-K (K) and M-K (K) • The interaction of X-rays with matter produces a small wavelength shift (Compton scattering) • Filters can be used to eliminate K radiation; monochromators are used to select K1 radiation.