X-Ray Diffraction

ME 215

Exp#1

X-Ray Diffraction

• X-rays is a form of electromagnetic radiation having a range of wavelength from 0.01-7 nm (0.01x10-9 to 7x10-9 m)

• Visible light wavelength is 0.56 m (560 nm)• Spacing between atoms in metals range from 2-

3 A (0.2-0.3 nm)• Electromagnetic radiations behave like waves as

well as particles.• Energy of x-ray (KeV) = 1.24/(nm)

Properties of X-rays

• An x-rayphoton can interact with the electrons in the target by one of 4 ways:– No interaction (the x-ray

photons experience no change in energy )

– Completely absorbed (photon energy is transferred to the target electrons. X-ray radiography)

Properties of X-rays

– The x-ray photon is completely absorbed and another x-ray photon of lower energy is produced (x-ray fluorescence)

– The x-ray photon produces an oscillating electric field in the electrons of the target object. The target object generate photons if the same wavelength at many different angles w.r.t. the incident photons. Each interaction is elastic scattering. (This is the main interaction that is important for x-ray diffraction)

12

1 < 2

Interference of Waves

• Constructive interference: mutual reinforcement of the scattered x-rays– Can happen if the

difference in distances traveled by the various x-ray parallel beams are a multiple of wavelength. (d = n*)

• Destructive interference: scattered beams are out of phase and cancel each other. (d = n*/2)

• Diffraction: constructive interference of x-rays

Bragg’s Law

• Bragg’s law: the positions of thte discrete x-ray spots I the diffraction pattern is cuased by the x-ray reflection by equally spaced parallel planes of atoms.

• Bragg’s law is necessary but not sufficient– n = 2*d*sin– n is the diffraction order =1,2,.. : wavelength : Bragg’s angle – d : spacing between planes

• For cubic crystals, d(hkl) = a/√(h2+k2+l2)

Reflection planes in cubic lattice

Example#1

• For a BCC crystal, compute the diffraction angles of the {110} planes assuming d(110) = 0.1181 nm and a monochromatic x-ray of = 0.1541 nm.– n =1

• sin = /(2*d) = 0.1541/(2*0.1181) = arcsin (0.1541/(2*0.1181)) = 40.7• Diffraction angle = 2*= 81.4

– n =2 = arcsin (2*0.1541/(2*0.1181)) = arcsin (1.305) • Hence, 2nd order diffraction is not possible

Example#2

• Determine the diffraction angles when {111} planes of Cu (a = 3.6151 Aº) are in diffraction condition? = 1.660 Aº– d(111) = a/√(1+1+1) = 2.087 Aº– For n=1

=arcsin (/(2*d)) = 23.45º• 2 = 46.9º

– For n=2 =arcsin (2/(2*d)) = 52.7º• 2 = 105.4º

– For n=3 =arcsin (3/(2*d)) = arcsin (1.19) • Not possible

Powder Diffraction

• Powder diffractometer:– An x-ray instrument used to

determine angles of diffraction (2) for a polycrystalline specimen or a powder as a function of diffracted beam intensities. The specimen must be flat and polycrystalline.

Powder Diffraction

• Each peak in the diffraction pattern correspond to a set of crystallographic planes {hkl}

• Diffraction pattern can be used to:– Identify crystalline phases– Determine unit cell size and R– Structure determination by

indexingDiffraction rules:– SC all planes– BCC h+k+l=even– FCC h,k,l unmixed

Example

• Determine the unit cell size for BCC iron from the powder diffraction pattern (=1.542)