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Phys 446: Solid State Physics / Optical Properties

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Phys 446: Solid State Physics / Optical Properties Solid State Physics Lecture 2 (Ch. 2.1-2.3, 2.6-2.7) Last week: Phys 446: • Crystals, • Crystal Lattice, Solid State Physics / Optical • Reciprocal Lattice, and Properties • Types of bonds in crystals Today: Fall 2007 • Diffraction from crystals • Importance of the reciprocal lattice concept Lecture 2 Andrei Sirenko, NJIT 1 Lecture 2 Andrei Sirenko, NJIT 2 Crystal Lattice Reciprocal Lattice Lecture 2 Andrei Sirenko, NJIT 3 Lecture 2 Andrei Sirenko, NJIT 4 Diffraction of waves by crystal lattice The Bragg Law • Most methods for determining the atomic structure of crystals are Conditions for a sharp peak in the based on scattering of particles/radiation. intensity of the scattered radiation: • X-rays is one of the types of the radiation which can be used 1) the x-rays should be specularly • Other types include electrons and neutrons reflected by the atoms in one plane • The wavelength of the radiation should be comparable to a typical 2) the reflected rays from the interatomic distance of a few Å (1 Å =10-10 m) successive planes interfere constructively hc hc λ(Å) = 12398/E(eV) ⇒ The path difference between the two x-rays: 2d·sinθ ⇒ E = hν = ⇒ λ = λ E few keV is suitable energy the Bragg formula: 2d·sinθ = mλ for λ = 1 Å The model used to get the Bragg law are greatly oversimplified • X-rays are scattered mostly by electronic shells of atoms in a solid. (but it works!). Nuclei are too heavy to respond. – It says nothing about intensity and width of x-ray diffraction peaks • Reflectivity of x-rays ~10-3-10-5 ⇒ deep penetration into the solid ⇒ x-rays serve as a bulk probe – neglects differences in scattering from different atoms – assumes single atom in every lattice point – neglects distribution of charge around atoms Lecture 2 Andrei Sirenko, NJIT 5 Lecture 2 Andrei Sirenko, NJIT 6 The Bragg Law and Diffraction grating Meaning of d for 2D Compare Bragg Law 2d·sinθ = mλ 2d·sinθ = mλ X-ray Diffraction d Lecture 2 Andrei Sirenko, NJIT 7 Lecture 2 Andrei Sirenko, NJIT 8 Meaning of d for 3D X-rays are EM waves http://www.desy.de/~luebbert/CrystalCalc_Cubic.html 2d·sinθ = mλ d 111 n d = hkl 222 Intercepts: a,a,a hkl Reciprocals: a/a, a/a, a/a ++ = 1, 1, 1 222 Miller index for this plane : (1 1 1) abc na⋅ da111 =≈3.13 A for Si with =5.431 A Lecture 23 Andrei Sirenko, NJIT 9 Lecture 2 Andrei Sirenko, NJIT 10 X-rays and X-ray tube X-rays and Synchrotrons Bragg Law 2d·sinθ = mλ Bragg Law 2d·sinθ = mλ for m=1 2d > λ Synchrotron radiation X-ray tube Natural Synchrotron Radiation Stars and Galaxies Accelerating electron emits light Electronic transitions Lecture 2 Andrei Sirenko, NJIT 11 Lecture 2 Andrei Sirenko, NJIT 12 Synchrotron Diffraction condition and reciprocal lattice Radiation produced by Von Laue approach: relativistic – crystal is composed of identical electrons in atoms placed at the lattice sites T accelerators – each atom can reradiate the incident (since 1947) radiation in all directions. – Sharp peaks are observed only in the directions for which the x-rays scattered from all lattice points interfere constructively. k k' Consider two scatterers separated by a lattice vector T. k = k'= Incident x-rays: wavelength λ, wavevector k; |k| = k = 2π/λ; k k' Assume elastic scattering: scattered x-rays have same energy (same λ) ⇒ wavevector k' has the same magnitude |k'| = k = 2π/λ •Synchrotron Radiation from a storage ring Condition of constructive interference: ( k ' − k ) ⋅ T = m λ or ()k'−k ⋅T = 2πm is the most bright manmade source of white light •Useful for materials studies from Far Infrared and UV to X-ray Define ∆k = k' - k - scattering wave vector Then ∆k = G , where G is defined as such a vector for which G·T = 2πm Lecture 2 Andrei Sirenko, NJIT 13 Lecture 2 Andrei Sirenko, NJIT 14 We obtained the diffraction (Laue) condition: ∆k = G where G·T = 2πm We got ∆k = k' – k = G ⇒ |k'|2 = |k|2 + |G|2 +2k·G ⇒ G2 +2k·G = 0 Vectors G which satisfy this relation form a reciprocal lattice 2k·G = G2 – another expression for diffraction condition A reciprocal lattice is defined with reference to a particular Bravais lattice, which is determined by a set of lattice vectors T. Now, show that the reciprocal lattice vector G = hb1 + kb2 + lb3 is orthogonal to the plane represented by Miller indices (hkl) Constricting the reciprocal lattice from the direct lattice: Let a1, a2, a3 - primitive vectors of the direct lattice; T = n1a1 + n2a2 + n3a3 plane (hkl) intercepts axes at points x, y, and z given in units a , a and a Then reciprocal lattice can be generated using the primitive vectors 1 2 3 By the definition of the Miller indices: where V = a1·(a2×a3) is the volume of the unit cell Then vector G = m1b1 + m2b2 + m3b3 We have bi·aj = δij define plane by two non-collinear vectors u and v lying within this plane: Therefore, G·T = (m1b1 + m2b2 + m3b3)·(n1a1 + n2a2 + n3a3) = 2π(m1n1+ m2n2+ m3n3) = 2πm The set of reciprocal lattice vectors determines the possible scattering wave prove that G is orthogonal to u and v: vectors for diffraction analogously show Lecture 2 Andrei Sirenko, NJIT 15 Lecture 2 Andrei Sirenko, NJIT 16 Now, prove that the distance between two adjacent parallel planes of Ewald Construction for Diffraction the direct lattice is d = 2π/G. Condition and reciprocal space The interplanar distance is given by the projection of the one of the vectors xa1, ya2, za3, to the direction normal to the (hkl) plane, which is the direction of the unit vector G/G ⇒ The reciprocal vector G(hkl) is associated with the crystal planes (hkl) and is normal to these planes. The separation between these planes is 2π/G 2k·G = G2 ⇒ 2|k|Gsinθ = G2 ⇒ 2·2πsinθ /λ = 2π/d ⇒ 2dsinθ = λ ∆k θ 2dsinθ = mλ - get Bragg law k k' Lecture 2 Andrei Sirenko, NJIT 17 Lecture 2 Andrei Sirenko, NJIT 18 Reciprocal Space: Accessible Area for Diffraction Summary Various statements of the Bragg condition: 2d·sinθ = mλ ; ∆k = G ;2k·G = G2 Reciprocal lattice is defined by primitive vectors: A reciprocal lattice vector has the form G = hb1 + kb2 + lb3 It is normal to (hkl) planes of direct lattice Only waves whose wave vector drawn from the origin terminates on a surface of the Brillouin zone can be diffracted by the crystal First BZ of bcc lattice First BZ of fcc lattice Lecture 2 Andrei Sirenko, NJIT 19 Lecture 2 Andrei Sirenko, NJIT 20 Solid State Physics Diffraction process: Lecture 2 (continued) (Ch. 2.4-2.5, 2.9-2.12) 1) Scattering by individual atoms 2) Mutual interference between scattered rays Scattering from atom i(k⋅r−ωt) 2π Consider single electron. Plane wave u = Ae k = k = λ A i(kR−ωt) Atomic and structure factors Scattered field: u'= f e f – scattering length of electron e R e Experimental techniques R – radial distance A Neutron and electron diffraction Two electrons: u'= f eikR []1+ ei∆k⋅r e R A or, more generally u'= f eikR []ei∆k⋅r1 + ei∆k⋅r2 e R similar to single electron with A ikR i∆k⋅rl many electrons: u'= f e e i∆k⋅r e ∑ f = f e l Lecture 2 Andrei Sirenko, NJIT 21 Lecture 2R Andrei lSirenko, NJITe ∑ 22 l 2 Atomic scattering factor (dimensionless) is determined by intensity: 2 i∆k⋅rl I ~ f = fe ∑e electronic distribution. l If n(r) is spherically symmetric, then 2 this is for coherent scatterers. If random then I ~ Nfe r 2 1 2 0 Scattering length of electron: 2 sin()∆k ⋅r fe = []()1+ cos 2θ / 2 re f = 4πr n(r) dr a ∫ 0 ∆k ⋅r 2 1 e −15 classical electron radius re = ≈ 2.8×10 m 2 2 4πε 0 mc in forward scattering ∆k = 0 so f = 4π r n(r)dr = Z a ∫ Z - total number of electrons i∆k⋅r i∆k⋅r 3 In atom, f e l → f n(r)e l d r e ∑ e ∫ l n(r) – electron density Atomic factor for forward scattering is equal to the atomic number f = n(r)ei∆k⋅rl d 3r (all rays are in phase, hence interfere constructively) a ∫ - atomic scattering factor (form factor) Lecture 2 Andrei Sirenko, NJIT 23 Lecture 2 Andrei Sirenko, NJIT 24 Scattering from crystal i∆k⋅r i∆k⋅R Since ∆k = G, c crystal scattering factor: l l iG⋅Rl i2πm fcr = ∑e = ∑ fale the lattice factor becomes S = ∑e = ∑e = N l l l l th Rl - position of l atom, fal - corresponding atomic factor 2 iGs⋅ j Then scattering intensity I ~ |fcr| where fFNNfe=⋅= rewrite fcr = FS⋅ cr∑ aj j i∆k⋅s j - structure factor of the basis, where F = f e ∑ aj summation over the atoms in unit cell G = Ghkl = hb1 + kb2 + lb3 if sj = uja1 + vja2 + wja3 j i(u ja1 +v ja2 +w ja3 )(hb1 +kb2 +lb3 ) 2πi(hu j +kv j +lw j ) i∆k⋅Rc - lattice factor, summation over all Then F = f e = f e and S = e l ∑ aj ∑ aj ∑ unit cells in the crystal j j l structure factor c Where Rl = Rl + s j Lecture 2 Andrei Sirenko, NJIT 25 Lecture 2 Andrei Sirenko, NJIT 26 Example: structure factor of bcc lattice (identical atoms) 2(πihu++ kv lw ) structure factor jjj Ffe= ∑ aj j Two atoms per unit cell: s1 = (0,0,0); s2 = a(1/2,1/2,1/2) πi(h+k +l) F = fa [1+ e ] ⇒ F=2fa if h+k+l is even, and F=0 if h+k+l is odd Diffraction is absent for planes with odd sum of Miller indices For allowed reflections in fcc lattice h,k,and l are all even or all odd 4 atoms in the basis.
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