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 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 Bragg’s Law

6. e. g. X-rays with wavelength 1. 54Å are reflected from planes with d=1
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
2d sin  = n
We normally set n=1 and adjust Miller indices, to give
2dhkl sin  = 

7. 2dhkl sin  =  2d sin  = n or
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. X-rays with wavelength 1.54 Å are “reflected” from the
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°
= (1 1 0)
= (2 2 0)
= (3 3 0)
= (4 4 0)
= (5 5 0)
2dhkl sin  = 

11. X-rays and solids c =  max = 1.5 Å min = 0.5 Å
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 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 = Å
CuK2 = 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
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 = x x 1.95 x 1018
= 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  = 1.540 Å = 2dhklsin
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 = 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.