1. X Ray Diffraction
© D Hoult 2009

3. Some x ray are

4. Some x ray are

5. Some x ray are reflected or

6. Some x ray are reflected or diffracted or

7. Some x ray are reflected or diffracted or scattered by the first layer of atoms

10. Other parts of the incident beam are diffracted by the next layer of atoms (and lower layers)

13. path difference =

14. path difference = 2 d sin q

15. path difference = 2 d sin q
If this path difference is equal to

16. path difference = 2 d sin q
If this path difference is equal to nl then the two sets of waves will interfere constructively

17. When x rays are diffracted (reflected, scattered) by the atoms in a crystal we will find maxima at angles given by

18. When x rays are diffracted (reflected, scattered) by the atoms in a crystal we will find maxima at angles given by
n l = 2 d sin q

19. When x rays are diffracted (reflected, scattered) by the atoms in a crystal we will find maxima at angles given by
n l = 2 d sin q
This is known as the Bragg equation

25. What is the distance between these nuclei ?

26. Mass of 1 mol of Na Cl =

27. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl =

28. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3

29. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3
Volume of 1 mol of Na Cl =

30. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3
Volume of 1 mol of Na Cl = 2.66 × 10-5 m3

31. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3
Volume of 1 mol of Na Cl = 2.66 × 10-5 m3
This volume contains

32. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3
Volume of 1 mol of Na Cl = 2.66 × 10-5 m3
This volume contains 6.02 × 1023 Na Cl pairs

33. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3
Volume of 1 mol of Na Cl = 2.66 × 10-5 m3
This volume contains 6.02 × 1023 Na Cl pairs
Volume associated with each pair is

34. Mass of 1 mol of Na Cl = 58.5 g
Density of Na Cl = 2.2 g cm-3
Volume of 1 mol of Na Cl = 2.66 × 10-5 m3
This volume contains 6.02 × 1023 Na Cl pairs
Volume associated with each pair is 4.42 × m3

36. Volume associated with each ion is 2.21 × 10-29 m3

37. Therefore the length of the side of one of these cubes is

38. Therefore the length of the side of one of these cubes is the cube root of this volume

39. d = 2.8 × m

40. So the separation between layers of ions in the crystal is also d = 2
So the separation between layers of ions in the crystal is also d = 2.8 × m

41. Separation between layers of ions = 2.8 × 10-10 m

42. Separation between layers of ions = 2.8 × 10-10 m
Calculate the first two angles at which diffraction maxima would be observed using x-rays of wavelength

43. Separation between layers of ions = 2.8 × 10-10 m
Calculate the first two angles at which diffraction maxima would be observed using x-rays of wavelength l = 1.5 × m

44. Separation between layers of ions = 2.8 × 10-10 m
Calculate the first two angles at which diffraction maxima would be observed using x-rays of wavelength l = 1.5 × m

45. Separation between layers of ions = 2.8 × 10-10 m
Calculate the first two angles at which diffraction maxima would be observed using x-rays of wavelength l = 1.5 × m
about 15° and

46. Separation between layers of ions = 2.8 × 10-10 m
Calculate the first two angles at which diffraction maxima would be observed using x-rays of wavelength l = 1.5 × m
about 15° and
about 32°

47. Typical results of an x ray diffraction measurement