Bragg’s Law-Solid State Physics-Lecture Slides, Slides of Solid State Physics

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IN THE NAME OF ALLAH

SODIUM CHLORIDE STRUCTURE

 Each Na+^ ion is octahedrally coordinated to six Cl- ions  The unit cell is fcc with four Cl-^ ions occupying all the four fcc positions  The four Na+^ ions occupying all the four octahedral void position  Interchangeable  Na+^ : 0 0 0 ; 1/2 1/2 0 ; 1/2 0 1/2 ; 0 1/2 1/  Cl-^ : 1/2 1/2 1/2 ; 0 0 1/2 ; 0 1/2 0 ; 1/2 0 0

RELATIONSHIP B/W LATTICE

PARAMETER AND ATOMIC RADIUS

 For sc; a = 2r

 For bcc; √3a = 4r

 For fcc; √2a = 4r

X-RAY DIFFRACTION

 Introduction  When an atomic electron is irradiated by a beam of x-ray, it starts vibrating with a frequency equal to that of incident beam.  An accelerated charge emits radiations, the vibrating electron present inside a crystal become source of secondary radiations having the same frequency as the incident x-ray.  These secondary x-rays spread out all over the directions.  The phenomenon is called “scattering of x-ray by atomic electrons”  If the wavelength of incident radiation is quite large than that of inter-atomic distance of lattice, all the emitted radiations shall be in phase with one another.  If the wavelength of incident radiation is equal to inter-atomic distance of lattices then the emitted radiations are out of phase.  These may be constructive or destructive interference.  The phenomenon consisted on Bragg‟s Law

BRAGG‟S LAW DERIVATION

 Bragg‟s Law in mathematical term is  nλ = 2 dSinθ Where the variable d is the distance between atomic layers in a crystal, and the variable lambda is the wavelength of the incident X-ray beam and n is an integer.

BRAGG‟S LAW CONDITION

CONTINUE…

 Let zA and zC be the perpendiculars drawn from the point z on the incident and reflected point of ray 2 respectively  The path difference b/w ray 1 and 2 is given by  (AB + BC); hence AB and BC are equal to each other and  AB = BC = dSinθ, we get  So, Path difference = 2dSinθ  For constructive interference of ray 1and 2, the path difference must be an integral multiple of wavelength λ, i.e. 2 dSinθ = nλ  The diffraction takes place for those values of d, θ, λ and n, which satisfies the Bragg‟s conditions