Diffraction

Question 1

How diffraction pattern spacing relates to the slit separation

Answer 1

x = slit spacing
1/x is proportional to spacing between diffraction spots

The orientation is retained e.g. horizontal slits make horizontal spots

Question 2

Indexing diffraction spots

Answer 2

in 3D:

e.g. spot labelled 111 will be from (111) family of planes

Question 3

How the angle between non-orthogonal axes changes through diffraction

Answer 3

The angle between the diffraction spots is now the complement of the angle between the original axes

Question 4

Magnification with a lens

Answer 4

1/f = 1/u + 1/v

f = focal length
u = distance between object and lens
v = distance between lens and image

Question 5

What limits the size of the smallest feature you can image

Answer 5

The wavelength of light used but be approximately equal to the size of the smallest feature

Question 6

Abbe’s theorem

Answer 6

At least 2 beams must enter a lens to image a grating

Question 7

Methods of generating x-rays

Answer 7

1. Heat a filament so that electrons are thermionically emitted
2. Accelerate the electrons through a large potential difference
3. Focus the accelerated electrons on a target
4. As the electrons hit the target, they decelerate. They lose energy by heating the target, emitting a photon or by exciting atoms in the target. Excited atoms will then emit a characteristic photon.
   - – The target must be cooled as it heats up rapidly —

Question 8

Continuous spectrum

Answer 8

When electrons hit the target, they emit photons as they decelerate. The only wavelengths not present in the continuous spectrum are those that correspond to a greater energy than the electrons were accelerated to.

If you want to use the continuous spectrum, use an element with a high atomic number as intensity is proportional to atomic number.

Question 9

Characteristic peaks

Answer 9

Bombarding electrons ‘knock’ electrons out of atoms in the target or excite these atoms.
Electrons in these atoms return to lower energy levels and emit photons corresponding to the energy difference between their original and final energy levels.

Characteristic peaks have a greater intensity than the continuous spectrum

Question 10

Why Kalpha1 peak has a greater intensity than the Kalpha2 peak

Answer 10

There are double the electrons in LIII than in LII

Question 11

Foils used to absorb certain wavelengths

Answer 11

Use thin foils with absorption edges
Where the absorption is high, there’s a large probability of photons and electrons combining to create a photo-electron.
Where absorption decreases rapidly is where the energy is too low to create a photo-electron

Question 12

Bragg reflection

Answer 12

Arise from constructive interference between X-rays scattered by atoms within a crystal

Question 13

Powder diffraction

Answer 13

Powder of many crystals in compacted into a polycrystalline block
Each grain is in a random orientation
Produce Debye-Scherrer cones since 2theta can be in any orientation now
On film, the cones are circles made of many individual diffraction spots
Small d-spacing can lead to back reflections as 2theta is so large

Question 14

Debye-Scherrer camera

Answer 14

Strips of film in a circle about the sample.

Question 15

Systematic absences

Answer 15

Occur when planes of atoms reflects beams that are in antiphase with each other –> destructive interference

Question 16

Atomic scattering factor (f)

Answer 16

Wave scattering from an individual atom

For a direct beam, f is maximal and is equal to the number of electrons per ion/atom in the sample.
(atomic number for an atom)

As the angle of scatter increases, f decreases

Question 17

Structure factor (F)

Answer 17

Wave resulting from scattering from multiple atoms

The sum of all the atomic scattering factors taking into account their magnitudes and relative phase factors

F gives the amplitude of the resultant scattered wave

Question 18

Intensity of a scattered wave

Answer 18

Intensity is proportional to the square of the magnitude of the structure factor i.e. structure factor * its complex conjugate

Question 19

Using the structure factor in non-primitive systems

Answer 19

Split the structure factor into the contribution from the lattice and the motif

Multiply the structure factor of the lattice (has no atomic scattering factor) by the structure factor of the motif (with atomic scattering factors)

Question 20

Why intensity is proportional to multiplicity for powder refraction

Answer 20

Number of individual crystals within a polycrystalline block that are in the correct orientation to cause diffraction is proportional to the multiplicity of the family of diffraction planes.
So, intensity is proportional to multiplicity

IN NON-CUBIC SYSTEMS, multiplicity decrease because e.g. a=b not equal to c

Question 21

How to index planes from diffraction pattern

Answer 21

2x = 2theta

Columns:

1. 2x
2. x
3. sin(x)
4. (sin(x))^2
5. (sin(x))^2 / (sin(x1))^2 (may need to be multiplied by a factor to find 6.)
6. h^2 + k^2 + l^2
7. hkl

Question 22

How to find motif from diffraction pattern

Answer 22

Calculate structure factors for some peaks based on possible motifs.
Compare to the observed intensities

Question 23

How to calculate lattice parameters

Answer 23

Most accurate value from a single measurement is to use the highest diffraction angle

Question 24

Systematic absences for hexagonal

Answer 24

h-k = a multiple of 3