DIFFRACTION: INTENSITIES OF DIFFRACTED BEAMS · 2º Semestre de 2017 October 2017 . Outline 1....

DIFFRACTION:

INTENSITIES OF DIFFRACTED BEAMS

Course: Crystallography and diffraction

Professors: Walter José Botta Filho

Guilherme Zepon

1

2º Semestre de 2017

October 2017

Outline

1. Introduction

2. Scattering by an electron

3. Scattering by an atom

4. Scattering by a unit cell

5. Structure Factor Calculation

6. Other factors affecting the relative intensity

in XRD.

7. Intensity of diffraction peaks from

polycrystalline sample

2

1. Introduction

X-ray diffraction in Crystals

3

• The atoms in the unit cell affect the intensities but not the directions of the diffracted beams!!!

• Consider two orthorhombic unit cells with two atoms of the same kind per unit cell, but one base-centered and one is body-centered.

• Suppose that Bragg’s law is satisfied for particular values of λ and θ employed for the (001) planes of both structures.

• It is easy to see that, even satisfying the Bragg’s law, There is no 001 reflection from the body-centered lattice.

Body-centered Base-centered

𝐴𝐵𝐶 = 1λ = 2𝑐𝑠𝑒𝑛𝜃 𝐴𝐵𝐶 = 1λ = 2𝑐𝑠𝑒𝑛𝜃

𝐷𝐸𝐹 =1

2λ = 2𝑐𝑠𝑒𝑛𝜃

Completely out of phase.

Destructive Interference!!!


4

• X-ray beam is an electromagnetic wave.

• The oscillating electric field of an x-ray beam will set any electron it encounters into oscillatory motion about its mean position.

• An accelerating or decelerating electron emits an electromagnetic wave. Thus, an electron which has been set into oscillation by an X-ray beam emits electromagnetic wave during its motion.

•In this sense, an electron is said to scatter x-rays. The scattered beam being simply the beam radiated by the electron under the action of the incident beam.


5

•The scattered beam has the same wavelength and frequency as the incident beam and is said to be coherent with it.

•Although x-rays are scattered in all directions by an electron, the intensity of the scattered beam depends on the angle of scattering.

• The Thompson’s equation shows the intensity of the beam scattered by a single electron of charge e coulombs (C) and mass m (kg), at a distance r (meters) from the electron.

𝐼 = 𝐼0𝜇04𝜋

2 𝑒4

𝑚2𝑟2𝑠𝑒𝑛2𝛼 = 𝐼0

𝐾

𝑟2𝑠𝑒𝑛2𝛼

Thompson Equation

Where:

𝐼0 = 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑏𝑒𝑎𝑚

𝜇0 = 4𝜋𝑥10−7𝑚. 𝑘𝑔. 𝐶−2

𝛼 = 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑖𝑛𝑔 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒

𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛.


6

•Suppose the incident beam is traveling in the direction Ox and encounter an electron at O.

• We must determine the scattered intensity at P in the xy plane.

• An unpolarized incident beam, such as that issuing from an x-ray tube, has its electric vector E in a random direction in the yz plane.

• This beam may be resolved into two plane-polarized components, having electric vectors Ey and Ez.

𝐸2 = 𝐸𝑦2 + 𝐸𝑧

2

𝐸𝑦2 = 𝐸𝑧

2 =1

2𝐸2

•On the average, Ey will be equal to Ez since de direction of E is perfectly random.

•Since the intensity of a wave is proportional to the square of its amplitude, therefore:

𝐼𝑂𝑦 = 𝐼𝑂𝑧=1

2𝐼0


7

•The y component of the incident beam accelerates the electron in the direction Oy. It therefore give rise to a scattered beam whose intensity at P is:

•Similarly, the intensity of the scattered z component is given by:

𝐼𝑝𝑦 = 𝐼𝑂𝑦𝐾

𝑟2𝑠𝑒𝑛2 < 𝑦𝑂𝑃 = 𝐼𝑂𝑦

𝐾

𝑟2

𝜋/2

𝐼𝑝𝑧 = 𝐼𝑂𝑧𝐾

𝑟2𝑠𝑒𝑛2

𝜋

2− 2𝜃 = 𝐼𝑂𝑧

𝐾

𝑟2𝑐𝑜𝑠2 2𝜃

• The total scattered intensity at P is obtained by summing the intensity of these two scattered components.

𝐼𝑃 = 𝐼𝑝𝑦 +𝐼𝑝𝑧

𝐼𝑃 =𝐾

𝑟2(𝐼𝑂𝑦 + 𝐼𝑂𝑧𝑐𝑜𝑠

22𝜃)

𝐼𝑃 = 𝐼0𝐾

𝑟21 + 𝑐𝑜𝑠22𝜃

2

Thompson Equation


8

•Since K = 7,94x 10-30 m², the Thompson equation shows that the intensity of the scattered beam is only a minute fraction of the incident beam.

•The Factor 1+𝑐𝑜𝑠22𝜃

2 is called

polarization factor . This is a rather unfortunate term because this factor enters the equation simply because the incident beam is unpolarized.

•If a monochromator is used, the polarization factor must include and additional term depending on the Bragg angle for the monochromator.

𝐼𝑃 = 𝐼0𝐾

𝑟21 + 𝑐𝑜𝑠22𝜃

2

Thompson Equation


9

• There is another way in which na electron can scatter x-ray, called Compton effect.

• It occurs whenever x-rays encounter loosely bound or free electrons.

• The x-ray can be considered a stream of x-ray quanta or photon, each of energy 𝒉𝒗𝟏.

• When such a photon strikes a loosely bound electron, the collision is an elastic one like that of two billiard balls.

• The electron is knocked aside an the photon deflects through an angle 2θ. Since some of the energy is used to provide kinetic energy for the electron, the energy 𝒉𝒗𝟐of the photon is less than its energy before impact. Thus, the wavelength of the scattered radiation is slightly greater than the wavelength of the incident beam.

• Radiation so scattered is called Compton modified radiation, and it has the important characteristic that its phase has no fixed relation to the phase of the incident beam . For this reason it is also known as incoherent radiation.

• It cannot take part in diffraction because its phase is only randomly related to the incident beam and cannot therefore produce any interference effect.

• It has the undesirable effect of increasing the background in diffraction patterns.

Compton effect

∆λ = λ2 − λ1 = 0.0486 𝑠𝑒𝑛2𝜃


10

•When an X-ray encounters an atom, each electron in it scatters part of the radiation coherently in accordance with the Thompson equation.

•The net effect is that coherent scattering by an atom is due to the electrons contained in that atom.

• Consider the waves scattered in the forward direction by electrons A and B. These waves are exactly in phase on a wave front such as XX’, because each wave has traveled the same distance before and after scattering.

• However, the other scattered waves shown in the figure have a path difference equal to (CB-AD) and are thus somewhat out of phase along a wave front such as YY’.

•Partial interference occur between the waves scattered by A and B, with the result that the net amplitude of the wave scattered in this direction is less than that of the wave scattered by the same electron in the forward direction.

X-ray scattering by an atom


11

• Atomic scattering factor (f): It is the quantity used to describe the “efficiency” of scattering of a given atom in a given direction. It is defined as a ratio of amplitudes:

• For any atom scattering in the forward direction its atomic scattering factor is equal to its atomic number (𝑓 = 𝑍).

•As 𝜽 increases, the wave scattered by individual electrons become more and more out of phase and 𝒇 decreases.

• The atomic scattering factor depends also on the wavelength of the incident beam: at fixed value of 𝜽, 𝒇 will be smaller the shorter the wavelength.

𝑓 =𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑣𝑒 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑏𝑦 𝑎𝑛 𝑎𝑡𝑜𝑚

𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑣𝑒 𝑠𝑐𝑎𝑡𝑡𝑒𝑟𝑒𝑑 𝑏𝑦 𝑜𝑛𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛


12

• Atomic scattering factor (f) - Apendix Cullity


13

• Atomic scattering factor (f) - Apendix Cullity


14

• The coherently scattered radiation from all the atoms undergoes reinforcement in certain directions and cancellation in certain directions, thus producing beams. Diffraction is, essentially, reinforced coherent scattering.

• To arrive at an expression for the intensity of a diffracted beam, the coherent scattering must be considered, not from an isolated atom but from all the atoms making up the crystal;

• The directions of these beams are fixed by Bragg’s law, which is, in a sense, a negative law:

“If Bragg’s law is not satisfied, no diffracted beam can occur; however, Bragg’s law may be satisfied for a certain set of atomic planes an yet no diffraction may occur.”

• Assuming that Bragg’s law is satisfied, the goal is to find the intensity of the beam diffracted by a crystal as a function of atom position.

Let’s solve this problem !!!!!


15

• Consider an orthogonal unit cell.

• Take atom A as the origin and orient the incident beam (𝑺𝟎) so that 𝒉𝟎𝟎 diffraction occurs. Therefore:

𝛿2′1′ = 𝑀𝐶𝑁 = 2𝑑ℎ00𝑠𝑒𝑛𝜃 = λ

From the definition of Miller indices,

𝑑ℎ00 = AC =𝑎

ℎ

How is this reflection affected by x-rays scattered in the same direction by atom b, located at a distance x from A?

Obs: Note that only this direction need be considered since only in this direction is Bragg law satisfied for ℎ00 reflection.

x y

z


16

• Clearly, the path difference between ray 3’ and 1’ will be less than λ; by simple proportion it is found to be:

𝛿3′1′ = 𝑅𝐵𝑆 =𝐴𝐵

𝐴𝐶λ =𝑥

𝑎/ℎλ

• Phase differences may be expressed in angular measure as well as in wavelength: two rays differing in path length by one whole wavelength, are said to differ in phase by 360° (or 2𝜋 radians). If the path difference is 𝜹, then the phase difference From the definition of Miller indices,∅ in radians is given by:

∅ =𝛿

λ2𝜋

Obs: The use of angular measure is convenient because it makes the expression of phase differences independent of the wavelength, whereas the use of wave length is meaningless unless the wavelength is specified.

x y

z


17

𝛿3′1′ =𝑥

𝑎/ℎλ

•Then, the phase difference between the wave scattered by atom 𝑩 and the scattered by atom 𝑨 at the origin is given by

∅3′1′ =𝛿3′1′λ2𝜋 =2𝜋ℎ𝑥

𝑎

• If the position of the atom 𝑩 is specified by its fraction coordinate ( 𝑢 = 𝑥/𝑎 ), then the phase difference becomes

∅3′1′ = 2𝜋ℎ𝑢

x y

z


18

•This reasoning may be extended to three dimensions, in which atom 𝑩 has actual

coordinate 𝒙 𝒚 𝒛 or fractional coordinates 𝒙

𝒂

𝒚

𝒃

𝒛

𝒄

equal to 𝒖 𝒗 𝒘, respectively.

• The following relationship applies between the phase difference between the wave scattered by atom B an that scattered by atom 𝑨 at the origin, for the 𝒉𝒌𝒍 reflection.

∅ = 2𝜋(ℎ𝑢 + 𝑘𝑣 + 𝑙𝑤)

• This relation is general and applicable to a unit cell of any shape.

• These two phases may differ, not only in phase, but also in amplitude if atom 𝑩 and the atom at origin are of different kinds !!!!!!

x y

z


19

•The problem of scattering from a unit cell is resolved by adding waves of different phase and amplitude in order to find the resultant wave. Waves scattered by all the atoms of the unit cell, including the one at origin, must be added.

•The figure aside shows the resultant wave 3 when the waves 1 and 2 with the same

frequency 𝑣 (therefore the same λ) but with different amplitude 𝐴 and phase ∅ are added.

• The equation of the two waves can be written as:

𝐸1 = 𝐴1𝑠𝑒𝑛(2𝜋𝑣𝑡 − ∅1)

And

𝐸2 = 𝐴2𝑠𝑒𝑛(2𝜋𝑣𝑡 − ∅1)

•The resultant wave 3 is also a sine wave, but of different amplitude and phase

Vector addition of waves


20

•Using complex numbers to describe a wave vector and to find an analytical expression for our problem.

•Draw the wave vector in the complex plane. The amplitude of the wave is given by A, the length of the vector, and the phase o the wave is ∅ , the angle between the vector and the axis of real numbers.

• The analytical expression for the wave is the complex number

𝐴𝑐𝑜𝑠∅ + 𝑖𝐴𝑠𝑒𝑛∅

Comparing the power-series expansions of 𝑒𝑖𝑥,cos 𝑥, and 𝑠𝑒𝑛 𝑥 and yields

𝑒𝑖𝑥 = cos 𝑥 + 𝑖𝑠𝑒𝑛 𝑥

or

𝐴𝑒𝑖∅ = 𝐴 cos∅ + 𝐴𝑖𝑠𝑒𝑛 ∅


21

•The intensity of a wave is proportional to the square of its amplitude. Thus

𝐴𝑒𝑖∅2= 𝐴𝑒𝑖∅𝐴𝑒−𝑖∅ = 𝐴2

• Adding the scattered waves from each of the atoms in the unit cell requires addition of complex numbers representing the amplitude an phase of each wave..

• The amplitude of each wave is given by the appropriate value of 𝒇.

• The phase of each wave is given in terms of the 𝒉𝒌𝒍 reflection considered and the 𝒖𝒗𝒘 coordinates of the atom by

∅ = 2𝜋(ℎ𝑢 + 𝑘𝑣 + 𝑙𝑤)

• Thus, any scattered wave can be expressed in the complex exponential form:

𝐴𝑒𝑖∅ = 𝑓𝑒2𝜋𝑖(ℎ𝑢+𝑘𝑣+𝑙𝑤)


22

•The resultant wave scattered by all the atoms of the unit cell is called Structure Factor (F).

• 𝑭 describes how the atoms arrangement (given by uvw for each atom) affects the scattered beam.

• 𝑭 is obtained by simply adding together all the waves scattered the individual atoms.

• If a unit cell contain atoms 1,2,3,...N, with fractional coordinates 𝑢1𝑣1𝑤1, 𝑢2𝑣2𝑤2, 𝑢3𝑣3𝑤3... 𝑢𝑁𝑣𝑁𝑤𝑁, then the structure factor for the hkl reflection is given by

𝐹 = 𝑓1𝑒2𝜋𝑖(ℎ𝑢1+𝑘𝑣1+𝑙𝑤1) + 𝑓2𝑒

2𝜋𝑖(ℎ𝑢2+𝑘𝑣2+𝑙𝑤2)+𝑓3𝑒2𝜋𝑖(ℎ𝑢3+𝑘𝑣3+𝑙𝑤3) +⋯+𝑓𝑁 𝑒

2𝜋𝑖(ℎ𝑢𝑁+𝑘𝑣𝑁+𝑙𝑤𝑁)

• This equation may be written as

𝐹 = 𝑓𝑛𝑒2𝜋𝑖(ℎ𝑢𝑛+𝑘𝑣𝑛+𝑙𝑤𝑛)

𝑁

1


23


𝑁

1

•𝑭 is a complex number and it express both the amplitude and phase of the resultant wave.

• Its absolute value 𝑭 gives the amplitude of the resultant wave in terms of the amplitude of the wave scattered by a single electron. So, 𝑭 is defined as a ratio of amplitude:

𝑭 =𝒂𝒎𝒑𝒍𝒊𝒕𝒖𝒅𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒂𝒗𝒆 𝒔𝒄𝒂𝒕𝒕𝒆𝒓𝒆𝒅 𝒃𝒚 𝒂𝒍𝒍 𝒕𝒉𝒆 𝒂𝒕𝒐𝒎𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍

𝒂𝒎𝒑𝒍𝒊𝒕𝒖𝒅𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒂𝒗𝒆 𝒔𝒄𝒂𝒕𝒕𝒆𝒓𝒆𝒅 𝒃𝒚 𝒐𝒏𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏

• The intensity of the beam diffracted by all the atoms of the unit cell in a direction predict by Bragg’s law is proportional simply to 𝑭 𝟐. This value is obtained by multiplying 𝑭 by its complex conjugate 𝑭∗


24

Some useful relations

• 𝒆𝝅𝒊 = 𝒆𝟑𝝅𝒊 = 𝒆𝟓𝝅𝒊 = −𝟏

• 𝒆𝟐𝝅𝒊 = 𝒆𝟒𝝅𝒊 = 𝒆𝟔𝝅𝒊 = +𝟏

• 𝒆𝒏𝝅𝒊 = (−𝟏)𝒏 𝒘𝒉𝒆𝒓𝒆 𝒏 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓

• 𝒆𝒏𝝅𝒊 = 𝒆−𝒏𝝅𝒊, 𝒘𝒉𝒆𝒓𝒆 𝒏 𝒊𝒔 𝒂𝒏𝒚 𝒊𝒏𝒕𝒆𝒈𝒆𝒓


25

BCC structure (Fe, Cr, etc)

o Two atoms of same kind (same 𝒇) located at the lattice points 𝟎𝟎𝟎 𝒆 𝟏

𝟐

𝟏

𝟐

𝟏

𝟐.


𝑁

1

o Replacing 𝒖𝒏, 𝒗𝒏 𝒆 𝒘𝒏, we find

𝐹 = 𝑓𝑒2𝜋𝑖(ℎ0+0𝑘+0𝑙) + 𝑓𝑒2𝜋𝑖(12ℎ+12𝑘+12𝑙)

o Rearrenging

𝐹 = 𝑓 + 𝑓𝑒𝜋𝑖(ℎ+𝑘+𝑙) = 𝑓 1 + 𝑒𝜋𝑖(ℎ+𝑘+𝑙)

𝑭 = 𝟐𝒇 when 𝐡 + 𝐤 + 𝐥 is even.

𝑭 = 𝟎 when 𝐡 + 𝐤 + 𝐥 is odd.

Systematically absent

reflections


26

FCC structure (Al, Ni, etc)

o Four atoms of same kind (same 𝒇) located at the lattice points 𝟎𝟎𝟎 𝒆 𝟏

𝟐

𝟏

𝟐𝟎 ,𝟏

𝟐𝟎𝟏

𝟐, 𝟎𝟏

𝟐

𝟏

𝟐.


𝑁

1

o Replacing 𝒖𝒏, 𝒗𝒏 𝒆 𝒘𝒏, we find

𝐹 = 𝑓𝑒2𝜋𝑖(ℎ0+0𝑘+0𝑙) + 𝑓𝑒2𝜋𝑖(1

2ℎ+1

2𝑘+0𝑙) + 𝑓𝑒2𝜋𝑖(

1

2ℎ+0𝑘+

1

2𝑙) + 𝑓𝑒2𝜋𝑖(0ℎ+

1

2𝑘+1

2𝑙)

o Rearranging

𝑭 = 𝒇 𝟏 + 𝒆𝝅𝒊(𝒉+𝒌) + 𝒆𝝅𝒊(𝒉+𝒍) + 𝒆𝝅𝒊(𝒌+𝒍)

𝑭 = 𝟒𝒇 for umixed indices.

𝑭 = 𝟎 for mixed indices.

Systematically absent

reflections


27

“The structure factor is independent of the shape and size of the unit cell”

• This means that any body-centered cell will have missing reflections for those planes which have (h+k+l) equal to and odd number.


28

B2 structure (NiAl,AlCo, AgZn, etc)

o Ni (𝑓𝑁𝑖) located at the lattice points 000 and Al (𝑓𝐴𝑙) located at 𝟏

𝟐

𝟏

𝟐

𝟏

𝟐.


𝑁

1

o Replacing 𝒇𝒏 𝒂𝒏𝒅 𝒖𝒏, 𝒗𝒏 𝒆 𝒘𝒏, we find

𝐹 = 𝑓𝑁𝑖𝑓𝑒2𝜋𝑖(ℎ0+0𝑘+0𝑙) + 𝑓𝐴𝑙𝑒

2𝜋𝑖(12ℎ+12𝑘+12𝑙)

o Rearranging

𝐹 = 𝑓𝑁𝑖 + 𝑓𝐴𝑙𝑒𝜋𝑖(ℎ+𝑘+𝑙)

𝑭 = 𝒇𝑵𝒊 + 𝒇𝑨𝒍 when 𝐡 + 𝐤 + 𝐥 is even.

𝑭 = 𝒇𝑵𝒊 − 𝒇𝑨𝒍when 𝐡 + 𝐤 + 𝐥 is odd

No Systematically

absent reflections


29

L12 structure (Ni3Al, Cu3Au, etc)

o Al (𝑓𝐴𝑙)located at the lattice points 𝟎𝟎𝟎 and Ni (𝑓𝑁𝑖) at 𝟏

𝟐

𝟏

𝟐𝟎 ,𝟏

𝟐𝟎𝟏

𝟐, 𝟎𝟏

𝟐

𝟏

𝟐.


𝑁

1

o Replacing 𝒇𝒏 𝒂𝒏𝒅 𝒖𝒏, 𝒗𝒏 𝒆 𝒘𝒏, we find

𝐹 = 𝑓𝐴𝑙𝑒2𝜋𝑖(ℎ0+0𝑘+0𝑙) + 𝑓𝑁𝑖𝑒

2𝜋𝑖(1

2ℎ+1

2𝑘+0𝑙) + 𝑓𝑁𝑖𝑒

2𝜋𝑖(1

2ℎ+0𝑘+

1

2𝑙) + 𝑓𝑁𝑖𝑒

2𝜋𝑖(0ℎ+1

2𝑘+1

2𝑙)

o Rearranging

𝑭 = 𝒇𝑨𝒍 + 𝒇𝑵𝒊 𝒆𝝅𝒊(𝒉+𝒌) + 𝒆𝝅𝒊(𝒉+𝒍) + 𝒆𝝅𝒊(𝒌+𝒍)

𝑭 = 𝒇𝑨𝒍 + 𝟑𝒇𝑵𝒊 for umixed indices.

𝑭 = 𝒇𝑨𝒍 − 𝒇𝑵𝒊for mixed indices.

No Systematically

absent reflections

6. Other factors affecting the relative intensity in XRD

30

Six Factors affect the relative intensity of the diffraction lines on a powder pattern:

1. Polarization factor

2. Structure factor

3. Multiplicity Factor

4. Lorentz Factor

5. Absorption Factor

6. Temperature Factor


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Multiplicity Factor

• Consider a cubic lattice.

• In a powder specimen, some of the crystals will be so oriented that 100 diffraction occurs. Other crystal may be oriented is such a way that 010 and 001 diffraction occur. However, 𝑑100 = 𝑑010 = 𝑑001, so the diffracted beams form part of the same diffraction cone.

• The same cubic crystal has four sets of planes of form {111} which have the same spacing but different orientation, namely, (111), (111 ), (111) and (11 1), whereas there are only three sets of the form {100}.

•Therefore, the probability that {111} will be correct oriented for diffraction is 4/3 the probability that {100} will be oriented. So, it follows that the intensity of the {111} reflection will be 4/3 that of the {100} reflection, other things being equal.

•The relative proportion of ℎ𝑘𝑙 planes contributing to the same reflection enters the intensity equation as the quantity 𝒑, the multiplicity factor, which may be defined as the number of permutations of positions and sign of ±ℎ,±𝑘, ±𝑙 for planes having the same d and 𝑭𝟐.


32

Multiplicity Factor

• Parallel planes with different Miller indices, such as (100) and (1 00), are counted separately, yielding numbers which are double given in the last slide.

•The value of 𝒑 depends on the crystal system


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Lorentz Factor

• It is a set of geometricall factors which influence the intensity of the diffracted beam.

The Lorentz-polarization factor equation is

𝐿𝑜𝑟𝑒𝑛𝑡𝑧 − 𝑝𝑜𝑙𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 = 1 + 𝑐𝑜𝑠22𝜃

𝑠𝑒𝑛22𝜃. 𝑐𝑜𝑠𝜃

• The overall effect of these geometrical factors is to decrease the intensity of reflections at intermediate angles compared to those in forward and backward directions.

• See Section 4.9 of B.D Cullity and S.R. Stock !!!


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Absorption Factor

• The intensity of the diffracted beam depends on the absorption which takes place in the specimen itself.

• In a diffractometer measurement, the absorption factor A is independent of θ.

• For small θ the specimen area irradiated is large but the effective depth of x-ray penetration is small.

• For large θ the irradiated area is small, but the penetration is depth is relatively large.

•The net effect is that the effective irradiated volume is constant and independent of θ.

•Absorption occur in any case, and the larger the absorption coefficient of the specimen, the lower the intensity of the diffracted beams, other things being equal.

𝑑𝐼𝐷 = 𝑎𝑏𝑙𝐼0𝑒−𝜇 𝐴𝐵+𝐵𝐶 𝑑𝑥


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Temperature Factor

Increased thermal vibration of the atoms, as the result of an increase in temperature, has three main effects:

• The unit cell expands, causing changes in periodicity 𝑑 and therefore in the 2𝜃 position.

• The intensity of the diffraction lines decreases

• The intensity of the background scattering between lines increases.

In intensity calculations, the temperature effect is included by introducing the temperature factor 𝒆−𝟐𝑴, which is a number by which the calculated intensity is to be multiplied to allow for thermal vibration of the atoms.

Temperature factor of Iron at

20°C as function of (𝒔𝒆𝒏𝜽

𝝀)

Effect of thermal

vibration of the aoms on

a powder pattern (very

schematic)

7. Intensity of diffraction peaks from polycrystalline sample

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Now we can gather together the factors discussed so far into an equation for the relative intensity of powder diffraction lines!!!!

𝑰 = 𝑭 𝟐𝒑𝟏 + 𝒄𝒐𝒔𝟐𝟐𝜽

𝒔𝒆𝒏𝟐𝜽𝒄𝒐𝒔𝜽𝒆−𝟐𝑴

Obs: In diffractometer measurements, the absorption factor is independent

of 𝜽 and so does not enter into the calculation of relative intensities.


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𝑰 = 𝑭 𝟐𝒑𝟏 + 𝒄𝒐𝒔𝟐𝟐𝜽

𝒔𝒆𝒏𝟐𝜽𝒄𝒐𝒔𝜽𝒆−𝟐𝑴


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The international Table brings the general and special reflection conditions for each space group.


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I432 and a = 4Å


40

I432 and a = 4Å


41

I432 and a = 4Å


42

I432 and a = 4Å

DIFFRACTION: INTENSITIES OF DIFFRACTED BEAMS · 2º Semestre de 2017 October 2017 . Outline 1....

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Transcript of DIFFRACTION: INTENSITIES OF DIFFRACTED BEAMS · 2º Semestre de 2017 October 2017 . Outline 1....