x-ray diffraction in crystals

$: X-Ray Diffraction in Crystals$
X-Ray Diffraction in Crystals

Introduction, Braggs law, Miller indices

University of MumbaiDepartment of Chemistry

M.Sc. Chemistry (Inorganic)Semester III, Chem 323, Unit I

Faculty: Dr. Shilpee Sachar

X-RAY – Some Historical facts

• X-rays were discovered in 1895by the German physicistWilhelm Conrad Röntgen andwere so named because theirnature was unknown at thetime.

• He was awarded the Nobelprize for physics in 1901.

Wilhelm Conrad Röntgen(1845-1923)

https://commons.wikimedia.org/wiki/Wilhelm_Conrad_R%C3%B6ntgenSS/Chem 323/X-ray Diffraction

https://commons.wikimedia.org/wiki/Wilhelm_Conrad_R%C3%B6ntgen

X-RAY Properties

• X rays are invisible, highly penetrating electromagnetic radiation of much shorter

wavelength (higher frequency) than visible light.

• Wavelength range: ~ 10-8 m to 10-11 m

• Corresponding frequency range: ~3 × 1016 Hz to 3 × 1019 Hz

http://courses.washington.edu/mengr599/notes/2016-sustainableenergy-luscombe.pdf

SS/Chem 323/X-ray Diffraction

http://courses.washington.edu/mengr599/notes/2016-sustainableenergy-luscombe.pdf

x-ray ≈ 10-10 ≈ 1A° E ~ 104 ev

X-RAY Energy

• Electromagnetic radiation described as having packets of energy, orphotons. The energy of the photon is related to its frequency by thefollowing formula:

hE

=Wavelength , = Frequency , c = Velocity of light

c

hcE

hcE


Production of X-RAYS

• Visible light photons and X-ray photons are both produced bythe movement of electrons in atoms. Electrons occupydifferent energy levels, or orbitals, around an atom's nucleus.

• When an electron drops to a lower orbital, it needs to releasesome energy; it releases the extra energy in the form of aphoton. The energy level of the photon depends on how farthe electron dropped between orbitals.

• X-rays are produced when high energetic electrons interactwith matter


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Evacuated glass bulb

Anode

Cathode

X rays can be produced in a highly evacuated glass bulb, called an X-ray tube, that contains essentially two electrodes—an anode made ofplatinum, tungsten, or another heavy metal of high melting point,and a cathode. When a high voltage is applied between theelectrodes, streams of electrons (cathode rays) are accelerated fromthe cathode to the anode and produce X rays as they strike theanode.

X-RAY TUBE

http://mateosradiologia.blogspot.com/2012/


http://mateosradiologia.blogspot.com/2012/

Monochromatic and Broad Spectrum of X-rays

• X-rays can be created by bombarding a metal target with highenergy (> ) electrons.

• Some of these electrons excite electrons from core states in themetal, which then recombine, producing highly monochromatic X-rays. These are referred to as characteristic X-ray lines.

• Other electrons, which are decelerated by the periodic potentialof the metal, produce a broad spectrum of X-ray frequencies.

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Generation of X-rays (K-Shell Knockout)

An electron in a higher orbital immediately falls to the lower energy level,releasing its extra energy in the form of a photon.

• Consider an atom as consisting of a centralnucleus surrounded by electrons lying in variousshells.

• If one of the electrons bombarding the targethas sufficient kinetic energy, it can knock anelectron out of the K shell, leaving the atom inan excited, high-energy state.

• One of the outer electrons immediately fallsinto the vacancy in the K shell, emitting energyin the process, and the atom is once again in itsnormal energy state.

• The energy emitted is in the form of radiationof a definite wavelength and is, in fact,characteristic K radiation.-X-raysSS/Chem 323/X-ray Diffraction

• The K-shell vacancy may be filled by an electron from any oneof the outer shells, thus giving rise to a series of K lines; Kaand Kb lines

• It is possible to fill a K-shell vacancy either from the L or Mshell, so that one atom of the target may be emitting Karadiation while its neighbor is emitting Kb

• however, it is more probable that a K-shell vacancy will befilled by an L electron than by an M electron, and the result isthat the Ka line is stronger than the Kb line.

• It is impossible to excite one K line without exciting all theothers. L characteristic lines originate in a similar way: anelectron is knocked out of the L shell and the vacancy is filledby an electron from some outer shell.

• These lines fall into several sets, referred to as K, L, M, etc., inthe order of increasing wavelength, all the lines togetherforming the characteristic spectrum of the metal used as thetarget.

Generation of X-rays (K-Shell Knockout)

http://www.unm.edu/~ejpete/pdf/EJP_2-25-2011.pdfSS/Chem 323/X-ray Diffraction

http://www.unm.edu/~ejpete/pdf/EJP_2-25-2011.pdf

For a molybdenum target the K lines have wavelengths of about 0.7A, the L lines about 5A, and the M lines still higher wavelengths.

Ordinarily only the K lines are useful in x-ray diffraction, the longer-wavelength lines being too easily absorbed.

There are several lines in the K set, but only the three strongest are observed in normal diffraction work. These are the Ka1, and Ka2, and Kb1, for molybdenum their wavelengths are:

Ka1= 0.70926A,Ka2= 0.71354A,Kb1= 0.63225A

Generation of X-Rays

http://www.soest.hawaii.edu/HIGP/Faculty/sksharma/GG711/GG711Spectroscopy01Overview.pdf


http://www.soest.hawaii.edu/HIGP/Faculty/sksharma/GG711/GG711Spectroscopy01Overview.pdf

Absorption of X-Rays

• The atoms that make up your body tissueabsorb visible light photons very well.The energy level of the photon fits withvarious energy differences betweenelectron positions.

• Radio waves don't have enough energy tomove electrons between orbitals in largeratoms, so they pass through most stuff.X-ray photons also pass through mostthings, but for the opposite reason: Theyhave too much energy.

...something you won't

see very often (VisibleLight)

X-ray

https://tg24.sky.it/scienze/2017/11/07/scoperta-raggi-x


https://tg24.sky.it/scienze/2017/11/07/scoperta-raggi-x

• A larger atom is more likely to absorb an X-ray photon in thisway, because larger atoms have greater energy differencesbetween orbitals -- the energy level more closely matches theenergy of the photon. Smaller atoms, where the electronorbitals are separated by relatively low jumps in energy, are lesslikely to absorb X-ray photons.

• The soft tissue in your body is composed of smaller atoms, andso does not absorb X-ray photons particularly well. The calciumatoms that make up your bones are much larger, so they arebetter at absorbing X-ray photons.

Absorption of X-rays


Diffraction

• The phenomenon of bending andspreading of waves when they meet anobstruction, is often termed asDiffraction.

• Diffraction occurs with electromagneticwaves, such as light and radio waves,and also in sound waves and waterwaves.

• The most conceptually simple exampleof diffraction is double-slit diffraction,that’s why firstly we remember lightdiffraction.

http://leung.uwaterloo.ca/MNS/102/Lect_2013/Lect_22B.pdf


http://leung.uwaterloo.ca/MNS/102/Lect_2013/Lect_22B.pdf

Light Diffraction

• Light diffraction is caused by light bending around the edge ofan object. The interference pattern of bright and dark linesfrom the diffraction experiment can only be explained by theadditive nature of waves; wave peaks can add together to makea brighter light, or a peak and a through will cancel each otherout and result in darkness.

Thus Young’s light interferenceexperiment proves that light

has wavelike properties.

https://indico.cern.ch/event/193928/contributions/1472192/attachments/280321/391984/SCeTGo-Scenario-CERN.pdf


https://indico.cern.ch/event/193928/contributions/1472192/attachments/280321/391984/SCeTGo-Scenario-CERN.pdf

Constructive & Destructive Waves• Constructive interference is the

result of synchronized lightwaves that add together toincrease the light intensity.

• Destructive İnterference

results when two out-of-phaselight waves cancel each other out,resulting in darkness.

Light Interference


Diffraction from a particle and solid

Single particle• To understand diffraction we also have to

consider what happens when a waveinteracts with a single particle. Theparticle scatters the incident beamuniformly in all directions

Solid material

• What happens if the beam is incident onsolid material? If we consider acrystalline material, the scatteredbeams may add together in a fewdirections and reinforce each other togive diffracted beams


• A crystal is a periodic structure

• When waves are propagated through such periodic structure, we get a

diffraction pattern which gives information of the crystal structure

• We can use X-rays, Electron waves or Neutron waves for generating the

diffraction pattern

Diffraction of Waves by Crystals

The general principles will be the same for each type of waves.


Diffraction of Waves by Crystals

• The diffraction depends on the crystal structure and on thewavelength.

• The structure of a crystal can be determined by studying thediffraction pattern of a beam of radiation incident on thecrystal.

• Beam diffraction takes place only in certain specific directions,much as light is diffracted by a grating.

• By measuring the directions of thediffraction & the corresponding intensities,one obtains information concerning thecrystal structure responsible for diffraction.


X-RAY CRYSTALLOGRAPHY

• X-ray crystallography or XRD (X-Ray Diffraction) is a technique in

crystallography in which the pattern produced by the diffraction of x-rays

through the closely spaced lattice of atoms in a crystal is recorded and then

analyzed to reveal the nature of that lattice.

• The wavelength of X-rays is typically 1 A°, comparable to the interatomicspacing (distances between atoms or ions) in solids.

• Energy of X-rays :


Crystal Structure Determination

• A crystal behaves as a 3-D diffraction grating for x-rays

• In a diffraction experiment, the spacing of lines on the grating canbe deduced from the separation of the diffraction maxima

• Information about the structure of the lines on the grating can beobtained by measuring the relative intensities of different orders

• Similarly, measurement of the separation of the X-ray diffractionmaxima from a crystal allows us to determine the size of the unitcell and from the intensities of diffracted beams one can obtaininformation about the arrangement of atoms within the cell.


X-Ray Diffraction & Bragg Equation

• English physicists Sir W.H. Bragg and hisson Sir W.L. Bragg developed arelationship in 1913 to explain why thecleavage faces of crystals appear toreflect X-ray beams at certain angles ofincidence (theta, θ).This observation isan example of X-ray wave interference. Sir William Henry Bragg (1862-1942),

William Lawrence Bragg (1890-1971)

o 1915, the father and son were awarded the Nobel prize for physics "fortheir services in the analysis of crystal structure by means of Xrays".


Bragg Equation

• Bragg law identifies the angles of the incident radiation relative to thelattice planes for which diffraction peaks occurs.

• Bragg derived the condition for constructive interference of the X-raysscattered from a set of parallel lattice planes.

• W.L. Bragg considered crystals to be made up of parallel planes ofatoms. Incident waves are reflected specularly from parallel planes ofatoms in the crystal, with each plane is reflecting only a very smallfraction of the radiation, like a lightly silvered mirror.

• In mirror like reflection the angle of incidence is equal to the angle ofreflection.


Diffraction Condition

• The diffracted beams are found to occur when the reflections from planesof atoms interfere constructively.

• We treat elastic scattering, in which the energy of X-ray is not changed onreflection.

• When the X-rays strike a layer of a crystal, some of them will be reflected.We are interested in X-rays that are in-phase with one another. X-rays thatadd together constructively in x-ray diffraction analysis in-phase beforethey are reflected and after they reflected.


The line CE is equivalent to the distance betweenthe two layers (d)

Bragg Equation

• These two x-ray beams travel slightly different distances. Thedifference in the distances traveled is related to the distancebetween the adjacent layers.

• Connecting the two beams with perpendicular lines shows thedifference between the top and the bottom beams.

sinDE d


Bragg Law

• The length DE is the same as EF, so the total distance traveled bythe bottom wave is expressed by:

• Constructive interference of the radiation from successive planesoccurs when the path difference is an integral number ofwavelenghts. This is the Bragg Law.

sinEF d

sinDE d

2 sinDE EF d

2 sinn d


Bragg Equation

where, d is the spacing of the planes and n is the order of diffraction.

• Bragg reflection can only occur for wavelength

• This is why we cannot use visible light. No diffraction occurs when the above condition is not satisfied.

• The diffracted beams (reflections) from any set of lattice planes can onlyoccur at particular angles predicted by the Bragg law.

nd sin2

dn 2


Lattice Planes and Miller IndicesImagine representing a crystal structure on a grid (lattice) which is a 3D

array of points (lattice points). Can imagine dividing the grid into sets of

“planes” in different orientations

• All planes in a set are identical

• The planes are “imaginary”

• The perpendicular distance between pairs of adjacent planes is the d-

spacing

Need to label planes to be able to identify them


Exercise - What is the Miller index of the plane below?

Find intercepts on a,b,c:

Take reciprocals

Multiply up to integers:


Miller IndicesRules for Miller Indices:• Determine the intercepts of the face along

the crystallographic axes, in terms of unit cell dimensions.

• Take the reciprocals• Clear fractions• Reduce to lowest termsFor example, if the x-, y-, and z- intercepts are 2, 1, and 3, the Miller indices are calculated as:

• Take reciprocals: 1/2, 1/1, 1/3• Clear fractions (multiply by 6): 3, 6, 2• Reduce to lowest terms • Thus, the Miller indices are 3,6,2.• If a plane is parallel to an axis, its intercept

is at infinity and its Miller index is zero.• A generic Miller index is denoted by (hkl).


Plane perpendicular to y cuts at , 1,

(0 1 0) plane

General label is (h k l) which intersects at a/h, b/k, c/l

(hkl) is the MILLER INDEX of that plane (round brackets, no commas).

This diagonal cuts at 1, 1,

(1 1 0) plane

An index 0 means that the plane is

parallel to that axis

Miller Indices Plane structures

• Miller indices define the orientation of the plane within the unit cell• The Miller Index defines a set of planes parallel to one another (remember the unit cell

is a subset of the “infinite” crystal)• (002) planes are parallel to (001) planes, and so onSS/Chem 323/X-ray Diffraction

d-spacing formula

For orthogonal crystal systems

(i.e. a=b==90) :-2

2

2

2

2

2

2 c

l

b

k

a

h

d

1

For cubic crystals (special case of

orthogonal) a=b=c :-

2

222

2 a

lkh

d

1

e.g. for (1 0 0)

(2 0 0)

(1 1 0)

.

d = a

d = a/2

d = a/2


• A tetragonal crystal has a=4.7 Å, c=3.4 Å. Calculate the separation of the:

(1 0 0)

(0 0 1)

(1 1 1) planes

• A cubic crystal has a=5.2 Å (=0.52nm). Calculate the d-spacing of the (1 1 0) plane.

]ba[c

l

a

kh

d

12

2

2

22

2

Quick Exercise


2d sin = n

X-rays with wavelength 1.54Å are reflected from planes with d=1.2Å. Calculate the Bragg angle, , for constructive interference.

= 1.54 x 10-10 m, d = 1.2 x 10-10 m, =?

d2

nsin

nsind2

1

n=1 : = 39.9°

n=2 : X (n/2d)>1

Bragg’s Law & Miller Indices


Example of equivalence of the two forms of Bragg’s law:

Calculate for =1.54 Å, cubic crystal, a=5Å

2d sin = n

(1 0 0) reflection, d=5 Å

n=1, =8.86o

n=2, =17.93o

n=3, =27.52o

n=4, =38.02o

n=5, =50.35o

n=6, =67.52o

no reflection for n7

(2 0 0) reflection, d=2.5 Å

n=1, =17.93o

n=2, =38.02o

n=3, =67.52o

no reflection for n4

Bragg’s Law & Miller Indices


1d

ha

kb

lc2

2

2

2

2

2

2

Use Bragg’s law and the d-spacing equation to solve a wide

variety of problems

2d sin = n

or

2dhkl sin =

Combining Bragg and d-spacing equation


X-rays with wavelength 1.54 Å are “reflected” from the

(1 1 0) planes of a cubic crystal with unit cell a = 6 Å. Calculate the Bragg

angle, , for all orders of reflection, n.

Combining Bragg and d-spacing equation

056.06

0112

2

222

2 a

lkh

d

1

18d2

d = 4.24 Å

d = 4.24 Å

d2

nsin 1

n = 1 : = 10.46°

n = 2 : = 21.30°

n = 3 : = 33.01°

n = 4 : = 46.59°

n = 5 : = 65.23°

= (1 1 0)

= (2 2 0)

= (3 3 0)

= (4 4 0)

= (5 5 0)


Experimental arrangements for x-ray diffraction

• Since the pioneering work of Bragg, x-ray diffraction has becomeinto a routine technique for the determination of crystal structure.

• Since Bragg's Law applies to all sets of crystal planes, the lattice canbe deduced from the diffraction pattern, making use of generalexpressions for the spacing of the planes in terms of their Millerindices. For cubic structures

• Note that the smaller the spacing the higher the angle ofdiffraction, i.e. the spacing of peaks in the diffraction pattern isinversely proportional to the spacing of the planes in the lattice.The diffraction pattern will reflect the symmetry properties of thelattice.

2 sind n

2 2 2

ad

h k l


Plane spacing for 7 crystal systems


Reference books

• Lesley E. Smart and Elaine A. Moore, Solid state chemistry – An

introduction, 3rd Ed., Taylor and Francis, (2005).

• R. S. Drago, Physical methods for Chemists, 2nd edition, Saunders college

publishing.

• R. A. Scott and C. M. Lukehart, Applications of physical methods to

inorganic and bioinorganic chemistry, John Wiley & Sons Ltd. (2007).

• Clive Whiston, X-ray Methods, John Wiley & Sons Ltd. (2008).


x-ray diffraction in crystals

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