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Objectives By the end of this section you should: understand the concept of diffraction in crystals be able to derive and use Braggs law know how X-rays.

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Presentation on theme: "Objectives By the end of this section you should: understand the concept of diffraction in crystals be able to derive and use Braggs law know how X-rays."— Presentation transcript:

1 Objectives By the end of this section you should: understand the concept of diffraction in crystals be able to derive and use Braggs 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 So separation must be same order as, but greater than, wavelength of light. Thus for diffraction from crystals: Interatomic distances Å so = Å X-rays, electrons, neutrons suitable

4 Diffraction from crystals ?

5 Beam 2 lags beam 1 by XYZ = 2d sin so 2d sin = n Braggs Law

6 We normally set n=1 and adjust Miller indices, to give 2d hkl sin = 2d sin = n 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 m, d = 1.2 x m, =? n=1 : = 39.9° n=2 :X (n /2d)>1

7 Use Braggs law and the d-spacing equation to solve a wide variety of problems 2d sin = n or 2d hkl sin =

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

9 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. Combining Bragg and d-spacing equation d = 4.24 Å

10 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) 2d hkl sin =

11 X-rays and solids X-rays - electromagnetic waves So X-ray photon has E = h X-ray wavelengths vary from Å; those used in crystallography have frequencies 2-6 x Hz 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 Electrons stopped by target; kinetic energy converted to X-rays continuous spectrum of white radiation, with cut- off at short (according to h =½mv 2 ) 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 Two processes lead to two forms of X-ray emission:

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

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 x10 8 ) / (1.54 x ) = 1.95 x Hz E = h = x x 1.95 x = 1.29 x 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 m= 0.05Å

20 Filter As well as characteristic emission spectra, elements have characteristic absorption wavelengths 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 = Å = 2d hkl sin

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 - Braggs 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.


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