X Ray Diffraction © D Hoult 2009
Some x ray are
Some x ray are
Some x ray are reflected or
Some x ray are reflected or diffracted or
Some x ray are reflected or diffracted or scattered by the first layer of atoms
Other parts of the incident beam are diffracted by the next layer of atoms (and lower layers)
path difference =
path difference = 2 d sin q
path difference = 2 d sin q If this path difference is equal to
path difference = 2 d sin q If this path difference is equal to nl then the two sets of waves will interfere constructively
When x rays are diffracted (reflected, scattered) by the atoms in a crystal we will find maxima at angles given by
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
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
What is the distance between these nuclei ?
Mass of 1 mol of Na Cl =
Mass of 1 mol of Na Cl = 58.5 g Density of Na Cl =
Mass of 1 mol of Na Cl = 58.5 g Density of Na Cl = 2.2 g cm-3
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 =
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
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
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
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
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 × 10-29 m3
Volume associated with each ion is 2.21 × 10-29 m3
Therefore the length of the side of one of these cubes is
Therefore the length of the side of one of these cubes is the cube root of this volume
d = 2.8 × 10-10 m
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 × 10-10 m
Separation between layers of ions = 2.8 × 10-10 m
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
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 × 10-10 m
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 × 10-10 m
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 × 10-10 m about 15° and
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 × 10-10 m about 15° and about 32°
Typical results of an x ray diffraction measurement