Solid state phys~ch2.ppt

1、Ch2. Reciprocal Lattice,Diffraction of waves by crystals Scattered wave amplitude Brillouin zones Fourier analysis of the basis Quasicrystals,Diffraction of waves by crystals,Brief review about the wave diffraction: A diffraction will take place when: (1) The wavelength is similar to the scale of th

2、e barrier. (2) The Diffracted waves will be of the same wavelength, phase difference, and the vibration orientation.,Brief review about the wave diffraction,The ordinary waves used are visible light, electro-magnetic wave, whose wavelengths are 500nm, 1m or more. In case of using crystals as diffrac

3、tion barriers, we need an incident beam with the wavelength no more than 10nm.,About the X-ray,X-ray was discovered in 1895 by a German scientist Roentgen. The wavelength of the x-ray covers 0.01 to 10nm just in the range of typical crystal lattice constant. The hard and soft x-ray,About the x-ray d

4、iffraction,In 1912, M.von.Laue E =h; E =h k2=k 2 Derivation of diffraction condition in RL k = G k+G = k (k+G)2 = k2 2kG+G2 = 0; G and -G: RL vectors Diffraction condition in RL: 2kG=G2,Derivation Bragg law from diffraction condition in RL,The reflection plane is (hkl), we have d(hkl) = 2/ G hkl By

5、diffraction condition in RL we have 2kG=G2 and 2Gk sin = G2 and k=2/ 2k sin = G 2(2/)sin=2/d(hkl) 2d sin=n ,Laue equation,Making a scalar product of a1,a2,a3 and k or G: a1 G=2v1,a2 G =2v2,a3 G =2v3 k or G lying at the cones about direction ai(i=1,2,3) To satisfy the three equation, k or G should li

6、e at co-intersection of three cones.,Ewald ball,k,k,G,2,Draw a line in the direction of incident x-ray, an origin is chosen so that the k terminates at any RL point. Draw a sphere of radius k=2/ .,A diffraction beam would be formed if the sphere intercepts any point of RL points satisfying k=k+nG :n

7、th degree diffraction.,Brillouin Zone,Brillouin zone: the statement in electron energy band theory and the elementary excitations. The definition of Brillouin zone: A WS cell in RL. Drawing a Brillouin zone: Take a RL point as the origin, and then connect ,Brillouin Zone,The perpendicular bisecting

8、planes of G form the boundaries of Brillouin zone . Geometrical interpretation of Brillouin zone: By diffraction condition 2kG=G2, we have k ( G)=(G)2. (7) It means that an x-ray beam in the crystal will be diffracted if its k satisfies (7) by its magnitude and direction.,How to form Brillouin Zone:

9、 2D and 1D,In 2D case, take a unit vector a3 normal to the plane defined by a1 and a2 ;calculate the b1,2,3 by the definition;then neglect bi which does not locate in the plane. In 1D case, take 2 unit vectors a2,3 normal to the line and to each other, calculate the b1,2,3 by the definition then neg

10、lect bi which does not locate on the line.,Examples of Brillouin zone,1D lattice,-k=-/a,k= /a,reciprocal lattice, 2D square and oblique lattice,Reciprocal lattice to sc lattice,Primitive translation vectors for DL and RL: a1=ax, a2=ay, a3=az; b1=2/a x, b2=2/a y, b3=2/a z sc Volume of the cells for D

11、L and RL Vc=a3, Vc*= (2 )3/a3 per 1 lattice point Boundaries of Brillouin zone b1=(/a)x, b2=(/a)y, b3= (/a)z The Brillouin zone is a cubic.,Reciprocal lattice to bcc lattice,Primitive translation vectors for DL and RL: a1=a(-x+y+z), a2=a(x-y+z),a3=a(x+y-z) b1=2/a(y+z), b2=2/a(z+x), b3=2/a(x+y)fcc Vo

12、lume of the cells for DL and RL Vc= a3, Vc*= 2(2 )3/a3 per 1 lattice point Boundaries of Brillouin zone (2/a)(y+ z)x, (2/a) (z+x), (2/a)(x+y),Brillouin zone of bcc lattice,A Wigner-Seitz cell of fcc lattice in DL,Reciprocal lattice to fcc lattice,Primitive translation vectors for DL and RL: a1= a(y+

13、z), a2= a(z+x),a3= a(x+y) b1=2/a(-x+y+z), b2=2/a(x-y+z), b3=2/a (x+y-z)bcc Volume of the cells for DL and RL Vc= a3, Vc*= 4(2 )3/a3 per 1 lattice pont Boundaries of Brillouin zone 2/a(x y z) and 2/a(x ), 2/a ( y), 2/a ( z),Brillouin zone of fcc lattice,A Wigner-Seitz cell of bcc lattice in DL,Fourie

14、r analysis of the basis,The scattered beam amplitude:,SG : structure factor defined as the integral over a single cell,Atomic factor,In the definition of atomic factor, we have,where r - rj , and nj(rj-r) refers to the contribution of the atom to the electron density at r .,Definition of atomic fact

15、or fj,atomic factor fj,nj() is an atomic property, fj depending on the electron distribution in a cell fj atomic factor: describing atomic property,About the atomic factor,Combine the definition of Sg and fj:,For jth atom in a cell, rj=xja1+yja2+zja3, thus G rj=(v1b1 + v2b2 + v3b3)(xja1+yja2+zja3) =

16、2(v1 xj+ v2 yj+ v3zj),About the structure factor,Rewrite structure factor SG:,the intensity of amplitude is SG* SG, SG may not be real. the intensity of amplitude deciding by G and rj N1(cell)xS1(basis)=N2(cell)xS2(basis),Structure factor to the bcc lattice,The atoms in bcc lattice: x1=y1=z1=0 and x

17、2=y2=z2= S(v1v2v3)=f1+exp-i(v1+v2+v3) conclusion: S=0when v1+v2+v3=odd integer S=2f when v1+v2+v3=even integer,Diffraction pattern of bcc lattice explained by structure factor to the,1,2,3,1st,2nd,3rd,Phase difference between 1 and 3 is 2,Structure factor to the fcc lattice,The atoms in fcc lattice:

18、 000, 0, 0, 0 S(v1v2v3)=f1+exp-i(v2+v3) + +exp-i(v1+v3) +exp-i(v1+v2) conclusion: S=0 when v1,v2,v3 partly odd or even integer S=4f when v1,v2,v3 total odd or even integer,X-ray diffraction pattern of KCl and KBrl,A typical fcc structure with a sc pattern lattice constant of a/2 and almost fjK=fjCl, only even integer indices planes,A typical fcc pattern: no partly even or odd indices planes,Atomic form factor,The scattering power of the