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Materials IA > Course C - Diffraction > Flashcards

Course C - Diffraction Flashcards

1
Q

what is a wave

A

the movement of energy but not matter from one place to another through an oscillation

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2
Q

what is wave interference

A

when two waves meet at a point they interfere, the displacement at any point is equal to the sum of the displacements of the individual waves

two special cases:
- constructive interference: Δ = nλ, phase diff. = 2pi, max amplitude wave formed

  • destructive interference, Δ = (n+1/2)λ, phase diff. = pi, no wave
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3
Q

what occurs in single slit diffraction

A
  • when a planar wave passes through a slit, the wave spreads out and we get circular wavefronts
  • this is because slit gives secondary source of waves (infinite wavelets (Huygens))
  • these interfere to form sinc function on a screen
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4
Q

what occurs in double slit diffraction

A
  • now there are two spherical wavefronts which interfere continuously forming an interference pattern on a detector
  • we obtain equally spaced maxima spreading out with decreasing intensity from centre
    nλ = d sinθ
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5
Q

what occurs with diffraction from a 1D diffraction grating

A
  • same idea as for double slit
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6
Q

what occurs with diffraction from a 2D diffraction grating

A
  • assume grating of vertical dimension a and horizontal dimension b
  • the horizontal slits are responsible for forming vertical maxima
  • the vertical slits are responsible for forming horizontal maxima
  • we end up with 2D spacing of spots/maxima
  • the maxima separation vertically = 1/a
  • maxima separation horizontally = 1/b
  • similar things occur for non-rectangular shaped gratings
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7
Q

give the thin lens equation for optical microscopes

A

1/f = 1/u + 1/v
f = distance from lens to back focal plane, focal length
u = distance from object to lens
v = distance from lens to image

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8
Q

state Abbe’s theorem

A

“To resolve a diffraction grating of slit separation d then at least two beams should enter the lens, the image is effectively formed by the interference of the diffracted beams and the direct beam at the image plane”

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9
Q

approximately, what is the smallest feature which can be resolved by an optical microscope

A

λ

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10
Q

explain the production of X-rays

A
  • large PD placed over a filament, heats up, electrons emitted by thermionic emission
  • V large PD placed between the filament and a target metal, cathode, this causes the electrons to accelerate towards the target metal
  • they collide with the metal and two mechanisms occur:
    1) the deceleration of electrons through multiple inelastic collisions produces a continuous spectrum of X-rays, Bremsstrahlung radiation
    2) some of the incident electrons will remove electrons from shells around an atom, this causes higher energy electrons to drop down to take their place and in the process they release a photon of X-ray frequency, this produces a characteristic spectrum
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11
Q

how/why do we filter the spectrum of X-rays produced

A
  • we often don’t want the different λ’s of the continuous or characteristic spectrum

To avoid this we use a filtering material:
-these are materials that absorb different λ C-rays different amounts and usually have a sharp ‘drop-off’
- we want this drop off absorption edge to be just below the λ we want to keep

  • for Cu k-alpha we use Ni
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12
Q

how can X-rays be detected

A

Historically: Photographic film
Now GM counters or semiconductors or scintillator materials

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13
Q

how can we derive the Bragg equation

A
  • model the crystal as containing planes of atoms and consider the X-rays as being reflected by these planes
  • think about two rays entering a material and being reflected off of two adjacent planes
  • calculate the phase difference using sinθ and the interplanar spacing, this gives

nλ = 2d(hkl)sinθ

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14
Q

what do we mean when we say diffraction from the (200) plane of single primitive crystal

A
  • clearly no X-rays can be diffracted from (200) in a primitive crystal as there are no atoms to diffract from
  • what it actually is, is the second order diffraction from the (100) planes, i.e. where n=2
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15
Q

what is the only information we need about a single crystal to be able to calculate Bragg angles for a single crystal

A

lattice parameter and wavelength of light

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16
Q

How are Bragg angles generally measured for single crystals in a diffractometer

A
  • generally when changing θ to observe different diffractions, it is easier to have a fixed source and move both the sample and detector

NOTE: the sample and detector must be rotated by different amounts, a 1:2 gear ratio must be used

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17
Q

what must we be careful of when measuring diffraction from planes in a single crystal (angles of sample)

A
  • if the normal of the (100) plane is in the plane of the source and the detector, then only {100} like planes can be detected
  • if it is not e.g. (111) then it cannot be measured using a simple diffractometer setup
  • either the sample must be rotated through an angle or the detector must be out of the plane
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18
Q

what occurs in powder diffraction

A
  • if rather than a single crystal we have many small crystals all oriented randomly (powder)
  • because of the random orientations, some grains will have the correct orientation to diffract from a certain set of planes
  • this means for a single through beam and no rotation of the sample, there will be reflections from many different planes
  • given the only restriction is the angle on reflection, not the plane, we end up with cones of max. intensity, Debye-Scherrer cones
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19
Q

how can we observe the Debye- Scherrer cones (3)

A

Using a flat sheet of photographic paper:
- we obtain concentric circles
- the through beam must be stopped using a beam stop
- the setup does limit observable angles

Debye-Scherrer camera:
- ribbon of photographic film is wrapped around the sample
- resulting film shows entrance and exit, distance between corresponds to 180 degrees
- curved lines can be viewed on the ribbon and distances measured to obtain data on the diffraction angle

Electric diffractometer:
- measure intensity on single path through 180 degrees
- results plotted electronically

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20
Q

considering pure bcc, what are the systematic absences and why

A
  • (100), there is a plane of atoms between each (100) plane so this gives destructive interference and hence that plane doesn’t show up
  • (111) is absent for the same reason
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21
Q

considering pure fcc, what are the systematic absences and why

A
  • (100) is absent as there is a plane of atoms between each (100) plane, this gives destructive interference with the (100) planes themselves and hence there’s no peak
  • (110) is absent for the same reason
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22
Q

are there any systematic absences in cubic p

A

no

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23
Q

explain the scattering factor, f, give some of its features

A

“the scattering factor, f, defines the wave scattering from an individual atom”

  • the greatest intensity of X-rays after passing through an atom is at θ = 0, this is because whilst every electron scatters electrons, the θ=0 direction will be where there’s only constructive interference, giving the greatest intensity
  • I decreases (gaussian like) with θ as more destructive interference occurs at greater θ
  • the scattering amplitude, I at θ = 0 is the scattering factor,f
  • f is proportional to Z
24
Q

define the structure factor, F

A

“The wave resulting from scattering from multiple atoms”

25
Q

how do we calculate the structure factor, F

A
  • we must consider the path difference of the waves that diffract from the different atomic positions
  • we assume that we have a primitive lattice and d(hkl) corresponds to a phase difference of 2pi
  • hence the phase difference from a position (x1,y1,z1) in fractional coordinates is
    φn = 2π (hxn + kyn + lzn)

we can combine our scattering factors to get a structure factor by considering the scattering factor for an atom at to be
fn(vector) = fn (cosφn + isinφn) = fn exp(iφn)

then

F(hkl)(vector) = Σfn exp(iφn)

26
Q

what is a physical interpretation of the structure factor, F

A

|F(hkl)| = amplitude of resultant wave
so
Intensity prop to |F(hkl)|^2

27
Q

how can the structure factor be used to determine the systematic absences (or lack of) for a cubic P lattice

A

lattice = atoms @ (0,0,0), only 1 type of atom, X
so
F(hkl) = Σ fx exp(iφn)

and
φn = 0

so
F(hkl) = fx

no systematic absences

28
Q

how can the structure factor be used to determine the systematic absences for a cubic F lattice

A

atoms @ (0,0,0), (1/2, 1/2, 0), (0, 1/2, 1/2), (1/2, 0, 1/2)
all of the same type, X

F(hkl) = fx (exp(0) + exp(iπ(h+k)) + exp(iπ(k+l)) +exp(iπ(h+l)) )

F(hkl) = fx (1+ (-1)^(h+k) + (-1)^(k+l) + (-1)^(h+l))

so, F(hkl) = 0 when h,k,l are mixed odd and even

this means systematic absences occur when h,k,l are mixed odd and even

29
Q

how can the structure factor be used to determine the systematic absences for a cubic I lattice

A

atoms @ (0,0,0) (1/2,1/2,1/2), all of same type X

F(hkl) = fx (exp(0) + exp(iπ(h+k+l))
= fx (1+(-1)^(h+k+l))

so F(hkl) = 0 where h+k+l is odd

so systematic absences occur where h+k+l is odd

30
Q

give a summary of what is systematically absent for cubic P, F, I lattices (THIS IS ALSO IN DATA BOOK)

A

Absent for cubic P = none
Absent for cubic F = h,k,l mixed odd and even
Absent for cubic I = h+k+l odd

31
Q

what is the condition on the structure factor for a plane, (hkl), to be absent

A

F(hkl) = 0

32
Q

give a simplified (factorised) equation for the structure factor

A

F(hkl) = [Σ(on m lattice points in cell) exp(2πi(hxm + kym + lzm))] [Σ(on n atoms in motif) fn exp(2πi(hxm + kym + lzm))]

33
Q

give the three factors (and hence overall proportionality) that contribute to the intensity of a diffraction peak for a certain plane (hkl)

A

Geometric factors:
- at small and very large angles, the portion of the diffracted intensity from each Debye-Scherrer cone will be greater
- close to 90 degrees, it will be lower

Multiplicity, M(hkl):
- A crystal will only diffract if its in the correct orientation
- but different planes have different multiplicities so in powder diffraction they can be more or less likely to diffract
- the more likely they are to diffract, the greater the intensity of the peak in powder diffraction

F(hkl):
- intensity prop to. |F(hkl)|^2

34
Q

give the equation for calculating multiplicities

A

M(hkl) = 3! 2^A / n!

A = number of non-zero indices in (hkl)
n = number of repeated indices in (hkl)

35
Q

what is the general method for indexing a diffraction pattern, give the equations which show why this works

A

we know for cubic materials
d(hkl) = a/sqrt(N)
N = h^2+k^2+l^2

we know Bragg’s law
λ = 2d(hkl)sinθ

so
(sin^2(θ))/ N = λ^2 / 4a^2 = const.

so
Nn/N1 = sin^2(θn)/sin^2(θ1)

  • so the best thing to do when indexing peaks is draw a table with columns
    2θ, θ, sinθ, sin^2(θ), sin^2(θ)/sin^2(θ1), N, (hkl)

N can be found by taking multiples of sin^2(θ)/sin^2(θ1) to give integers

(hkl) must be worked out from N such that h^2+k^2+l^2 = N

36
Q

what is the formula for calculating lattice parameter from 2θ values and wavelength and N

A

rearrange Bragg’s law to get
a = λsqrt(N) / 2 sin(θ)

37
Q

why should we always use the greatest 2θ value for calculating a lattice parameter

A

we can implicitly differentiate the equation
a = sqrt(N) d(hkl) to give

δd/d = -δθ / tan(θ)

so uncertainty is least in d(hkl) (and hence lattice parameter) where θ is greatest

38
Q

define reciporical space

A

officially the Fourier transform of a real lattice

really just a mathematical construct

39
Q

what is a plane in reciprocal space

A

a point

40
Q

what is a set of planes in reciprocal space, what is the spacing

A

a set of points

the spacing between the points is the reciprocal of the true interplanar spacing

41
Q

what is the ‘official’ way to form a reciprocal lattice from a real lattice

A
  • draw out the real lattice
  • consider all planes in the real lattice which aren’t systematically absent
  • draw these planes as points from an origin where the distance to the point is the reciprocal of the true interplanar spacing in real space and the direction is in the direction of the normal of the real plane
42
Q

what is the quick way to draw a real cubic lattice in reciprocal space

A

draw a square array of indexed points for each c value

e.g. c = 0, has points with labels of form (hk0) representing all planes of form (hk0)

then remove from each square array of points the points which have indices of planes that are systematically absent for the real lattice

43
Q

what do real space cubic P,I,F lattices form in reciprocal space

A

cubic P forms cubic P

cubic I forms cubic F

cubic F forms cubic I

44
Q

how can we represent Bragg’s Law in reciprocal space/ how does it work (Ewald Sphere)

A

1) draw out reciprocal lattice
2) draw a vector to represent incoming x-ray beam, point towards the origin (point of (000)), and make the vector have length 1/λ, it must also be incident at the same angle as it normally is

3) draw a circle centered on the START of that vector such that the circle passes through (000)

4) any point that it passes through represents a plane that would be visible if that wavelength of light was incident to the crystal at that angle

NOTE: this also works in 3D with spheres

45
Q

if we only know the wavelength of some incident radiation and want to find the angle at which a certain plane will diffract, how can we do this using an Ewald sphere

A
  • draw a vector incident to (000) and draw the circle as normal
  • rotate the circle around until it passes through the suitable plane
  • then measure the angle between the incident vector and the line joining (000) and (0 bar1 0)
46
Q

what will be the form of the Ewald sphere/circle if white radiation is used

A
  • in this case there is a large range of wavelengths
  • this represents a range of different radii of our circle
  • hence two circles can be drawn and the area between them can be shaded to work out which planes can diffract
47
Q

what is the purpose of electron microscopy

A

we know that the smallest distinguishable thing in microscopy is approximately equal to the wavelength of the radiation used

  • electrons have a very small wavelength (think of de Broglie) so smaller objects can be distinguished
48
Q

explain the structure of a TEM

A
  • electrons produced using an electron gun (thermionic emission from tungsten wire)
  • a Wehnelt sits just below the filament at about -200V, acts as cathode
  • there’s an anode at 1-30KV below which accelerates the electrons
  • EM lenses can cause the electrons to spiral down the column, the current in the electromagnets can be changed to change focal length
  • samples must be V thin to prevent electrons interacting too much with matter
49
Q

what are the two key things about constructing an Ewald sphere for electron microscopy

A
  • the wavelength is much smaller so the radius of the circle drawn is much much larger
  • the crystals studied are usually very thin
  • this means the reciprocal lattice points are no longer points but instead stretched into rods
50
Q

what do the differences in the Ewald sphere in electron microscopy mean for the number/ sort of planes that are satisfied by the condition

A
  • as we have a V large Ewald sphere and elongated rods, the diffraction condition of many different planes can be satisfied simultaneously
  • this is how electron diffraction patterns arise
51
Q

what can electron diffraction patterns be interpreted as

A
  • sections of reciprocal space
  • the central spots of from many rods being intersected by the same Ewald Sphere
  • the spots further out are still from intersections with the Ewald sphere but from points on c=1 or greater
52
Q

how can lattice parameters be determined from TEM diffraction patterns

A

assume tan(2θ) = 2θ = R/L
where
R is the distance from the through beam spot to the detected diffracted maximum
L is the distance from sample to screen

use Bragg’s law to give

d(hkl) = λL/R

53
Q

give a brief overview of SEM and the different types of electrons and what they show

A

structure of SEM is similar to TEM but no thin sample neeeded

  • the electrons interact with the sample and transfer their energy to the atoms/electrons in the sample
  • most of the energy is dissipated as heat
  • some electrons escape as secondary electrons, back-scattered electrons or X-rays

secondary electrons = detected by fixed detector as function of beam position
back-scattered electrons = good for compositional contrast
X-rays = good for compositional analysis

54
Q

how should we construct a reciporical lattice for a non-cubic structure

A

Be VERY careful:
- draw out some unit cells
- set an origin
- consider the major planes e.g. (100,010) and their interplanar distances and normals then sketch accordingly

OR
use the equations for
d* = 1/d
and
θ* = 180° - θ

55
Q

what can we say about the angles between points in reciprocal space compared to the angles between the planes in real space

A

θ* = 180° - θ

Materials IA (9 decks)

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