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For visible electromagnetic radiation to be diffracted,

the spacing between lines in a 2d grating must be ofthe same order as wave length range of light.

X ray is suitable. (Atomic spacing)

Electrons at high speeds directed to a metal target:part of their KE is converted into X rays.

X rays emitted by target has continuous range of

wavelengths called as white radiation. Characteristic x-rays: Emitted from heavy elements

when electrons transition between the lower atomicenergy levels. (at high Voltage)

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K- alpha radiation high intensity hence used in

diffraction studies.

X rays directed at the crystal interacts with the

electrons of the atoms.

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Electrons oscillate and form a source ofelectromagnetic radiation.

Emitted waves have same frequency as that of Xrays- in all directions

Emission in a particular direction is a combined effectof oscillations of electrons of all the atoms.

Emissions in phase reinforce one another in onlyspecific directions, depending on the direction ofincident X rays, their wavelength and the spacingbetween the atoms in crystal.

Destructive interference results in other directions.

The waves emitted will have same frequency asincident rays.

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Diffraction of X rays by planes of atoms:

Constructive and non constructive interference

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BRAGGS LAW Easiest way to visualize the diffraction effects

produced by 3-d grating (A diffraction grating is anoptical component with a regular pattern, which splitsand diffracts light into several beams travelling in

different directions) Braggs Law: Make the phases of the beamscoincide when the incident angle equals andreflecting angle.

A beam of x rays of wavelength is directed towards

the crystal at an angle to the atomic planes. The 2nd beam continues to the next layer that is

scattered by atom B. The 2nd beam must travel the extra distance AB +

BC if the two beams are to continue travelingadjacent and parallel.

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This extra distance must be an integral (n) multiple ofthe wavelength () for the phases of the two beams to

be the same. n= AB+BC, where isthe wavelength of the incident X-ray beam, d

hkl

is thedistance between atomic layers in a crystal, is angle

of incidence.

n = AB + BC

but AB = dhklsinand AB =BC, results as n= 2dhklsin(the location of the surfacedoes not change derivation)

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The reflection of x rays is by the atomic planes.

for n=2 and for the plane (100) and for n=1 and the

plane (200) ??

Interplanar spacing of (100) plane is twice that of for(200) plane; similarly for n=3 and for the plane (300)and so on..

Take n=1 and for all reflections for all parallel planes

as (100), (200) if there is no (200) plane with atomson it, then what is designated as (200) reflectionactually refers to the second order reflection from(100) plane.

For cubic crystals, the relation between inter planar

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Q. A diffraction pattern of a cubic crystal of lattice

parameter a= 3.16 Ao is obtained with amonochromatic x-ray beam of wavelength 1.54Ao. Thefirst four lines of this pattern were observed to havethe following values. Determine the inter planarspacing and the miller indices of the reflecting planes.(n=1)

Solution:

Step 1: Evaluate d (inter planar spacing)

for all lines

Step 2: Find (h2+k2+l2) =(a/d)2 values.

Step 3: Arrive the possible hklvalues.

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Line d A0 (h2+k2+l2) =(a/d)2 hkl

1 20.3 2 110

2 29.2 4 200

3 36.7 6 211

4 43.6 8 220

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Experimental technique to determine crystal

structures.

The K-alpha radiation is allowed to incident onthousands of randomly oriented crystals in powder

form.

A Debye- Scherrer camera;consists of a cylindricalcassette with a strip of photographic film positionedround the circular periphery.

If the sample consists of some tens of randomlyorientated single crystals, the diffracted beams areseen to lie on the surface of several cones. The conesmay emerge in all directions, forwards and backwards.

A sam le of some hundreds of cr stals i.e. a

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A circle of film is used to record the diffraction patternas shown. Each cone intersects the film givingdiffraction lines.

The reflected radiation thus lies on the surface of the

cone whose apex is at the point of contact of theincident radiation with the specimen. And since the inter planar spacing d being the same

for all the members of the family of crystal planes,they all reflect at the same Bragg angle hence

lying on the same cone. For n=1, many combinations of d and to satisfy

Braggs Law.

Each value of n represents a cone, hence powderrepresents many cones.

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For every set of crystal planes, by chance, one or

more crystals will be in the correct orientation to givethe correct Bragg angle to satisfy Bragg's equation.

Every crystal plane is thus capable of diffraction. Eachdiffraction line is made up of a large number of smallspots, each from a separate crystal.

Each spot is so small as to give the appearance of a

continuous line. If the crystal is not ground finelyenough, the diffraction lines appear speckled.

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The distance between the two arc spread out film is

termed S and obtained in asS=4R where is theBragg angle expressed in radians.

Ease of conversion, S is measured in mm, indegrees and camera radius to be 57.3 mm (since 1rad=57.3o)

http://www.matter.org.uk/diffraction/x-

ray/powder_method.htm http://www.matter.org.uk/diffraction/x-

ray/experimental_powder_diffraction.htm

http://www.matter.org.uk/diffraction/x-ray/powder_method.htmhttp://www.matter.org.uk/diffraction/x-ray/powder_method.htmhttp://www.matter.org.uk/diffraction/x-ray/experimental_powder_diffraction.htmhttp://www.matter.org.uk/diffraction/x-ray/experimental_powder_diffraction.htmhttp://www.matter.org.uk/diffraction/x-ray/experimental_powder_diffraction.htmhttp://www.matter.org.uk/diffraction/x-ray/experimental_powder_diffraction.htmhttp://www.matter.org.uk/diffraction/x-ray/experimental_powder_diffraction.htmhttp://www.matter.org.uk/diffraction/x-ray/powder_method.htmhttp://www.matter.org.uk/diffraction/x-ray/powder_method.htmhttp://www.matter.org.uk/diffraction/x-ray/powder_method.htm

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Complex structure determination is time consuming Steps to be followed are :

1. Determining the macroscopic symmetry of the crystal

2. The space lattice and its dimensions3. and finally the atomic arrangement with in the unit cell.

Measurement of density of crystal and chemicalcomposition also assists the process.

If the crystal is mono atomic, the space lattice togetherwith the lattice parameters is a complete descriptionof the crystal structure.

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From the above rules the ratio of (h2+K2+l2)values for allowed reflections from differentcrystals :

SC 1:2:3:4:5:6:8.

BCC 1:2:3:4:5:6:7:8 FCC 3:4:8:11:12:16:19:20:24:27 DC 3:8:11:16

A simple comparison of the observed ratios ofsin2values with the above is then sufficient toidentify the crystal

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3.18 Using a diffractometer and a radiation of wavelength

1.54A0, only one reflection from an FCC material isobserved when 2 is 1210. Which plane are we talkingabout? What are the indices of this reflection? What is theinter planar spacing ? Show that next higher indicesreflection cannot occur.

Solution:

Given = 1.54 A0

For FCC, 2 = 1210

From the formula, 2*d*sin()=(1)* from where d can becalculated. d= 0.8846 A0

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Also d= a/(h2+k2+l2)1/2, from where a(lattice

parameter) can be calculated.

The next plane will be half the distance from the first

plane d200= 0.8846 /2

From the above value of d, sin() can be evaluated bysin()= /2*d200

3.28 Explain why there is no end-centered cubic spacelattice?

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2 Multiplying factor

Round off hkl

38.7

45.4

65.7

78.8

83.0

99.6

112.5

117.0

138.1

164.2

3.32 Crystal Structure? Lattice parameter (a)? If possible the element?

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Multiplyingfactor

Round off hkl

0.12

0.238

0.357

0.475

0.593

0.711

0.83

3.34 Crystal Structure? Lattice parameter (a)?? What is the minimum

number of reflections you need to ascertain this structure and why?

Identify the element if possible.

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Schematic diagram of an x-ray diffractometer; T= x-ray source, S=

specimen, C= detector and O the axis around which the specimen and

the detector rotate.

Diffraction Techniques:

Diffractometer is aapparatus uesd to

determine the angles at

which diffraction occurs

for powdered specimens.

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4.60 The metal niobium has a BCC crystal structure. If the angle of

diffraction for the (211) set of planes occurs at 75.990 (first order

reflection) when a mono chromatic x-ray radiation having a wavelength

of 0.1659 nm is used, compute

a) The inter planar spacing for this set of planes

b) The atomic radius for the niobium atom

Solution:

Step1: Calculate d(211)Step 2: Determine the lattice parameter, a

Step 3: Calculate Radius [R=a* (3)1/2 / 4]

4.61

For which set of planes will a first order diffraction peak occur at adiffraction angle of 44.530 for FCC nickel when monochromatic

radiation having a wavelength of 0.1542 nm is used? (R=(0.1246 nm)

Solution:

Compute the inter planar spacing

Evaluate decide the plane based on [square (h,k,l)].