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1、Atoms and Crystals Atomic structure X-ray diffraction Braggs Law Crystal structureAtomic structurel The electron in the hydrogenic atom is attracted by a central force thatis always directed toward the positive Nucleus.l Spherical coordinates centered at the nucleus are used to describe the position

2、 of the electron. The PE of the electron depends only on r.The solution to the Schrodinger equation (r, ,) can be written as the product of the two functions (r, ,) = R(r)Y ( ,)When considering principal, orbital angular momentum, magnetic quantum numbers, it can be rewritten as n,l ,m (r, ,) = Rn,l

3、 (r)Yl ,m ( ,)lln : princial quantum numberl : orbital angular momentum quantum numberml : magnetic quanum numberror = ¥Molecule+Separated atomsFA = Attractive forceFN = Net forceER = Repulsive PEE = Net PEror00roInteratomic separation, rEoFR = Repulsive forceEA = Attractive PE(a) Force vs r(b)

4、 Potential energy vs r(a) Force vs interatomic separation and (b) Potential energy vs interatomic separation.FN = FA + FR = 0ForceRepulsionAttractionPotential Energy, E(r)RepulsionAttractionCovalent bonding- Directional nature, localization- Non-ductile (brittle)- Poor electrical conductivityCN of D

5、iamond structure: 4 Coordination number (CN): Number of nearest neighbor atoms in solidMetallic bonding- Non-directional nature- Delocalized electrons (collective sharing)- Ductile- Go-packed structure with high CN ectrical conductivityIn metallic bonding the valence electrons from the metal atoms f

6、orm a “cloud of electrons” which fills the space between the metal ions and “glues” the ions together through the coulombic attraction between the electron gas and the positive metal ions.Ioinic bondingl NaCl isof Cl- andNa+ ions arranged alternately so that the oppositely charged ionsarest to each

7、otherand attract each other.l There are also repulsive forces between the like ions. In equilibrium the net force acting on any ion is zero.van der Walls bonding(a) The H2O molecule is polar and has a net permanent dipole moment(b) Attractions between the various dipole moments in water gives rise t

8、o van der Walls bonding(a) A permanently polarized molecule is called an electric dipole moment.(b) Dipoles can attract or repel each other depending on their relative orientations.(c) Suitably oriented dipoles can attract each other to form van der Walls bonds.Induced dipole-induced dipole interact

9、ion and the resulting van der Waals forceFrom Principles of Electronic Materials and Devices, Third Edition, S.O. Kasap (© McGraw-Hill, 2005)r¢ = r +u1a1 +u2a2 +u3a3 : La tice translation vectorsVc = a1ga2 ´a3: Volume of primitive la tice ce l2D and 3D Lattice TypesPrimitive translati

10、on vectors in 3D Lattice TypesBCCa = 1 a(x+y -z); a = 1 a(-x +y +z); a = 1 a(x -y +z)123222FCCa = 1 a(x+y); a= 1 a(y +z); a = 1 a(z +x)123222HCP? ?Hexagonald Packed Structure (hcp)Stacking orderABABABABCABCX-ray diffraction§ For electromagnetic radiation to be diffracted the spacing in the grat

11、ing should be of the same order as the wavelength§ In crystals the typical inter-atomic spacing 2-3 Å so the suitable radiation is X-rays§ Hence, X-rays can be used for the study of crystal structuresKaX-raysBeam of electronsCharacteristic radiationdue to energy transitions in the ato

12、mWhiteKbradiationAn accelerating (/decelerating) charge radiates electromagnetic radiation1.40.60.21.0Wavelength (l)Mo Target impacted by electrons accelerated by a 35 kV potentialIntensityTarget Scattering From loosely bound charges § Here the particle picture of the electron & photon come

13、s in handy( ,)E = hElectron knocked aside11112qE2 = h 2D = 2 - 1 : (1- cos2 )No fixed phase relation between the incident and scattered wavesIncoherent does not contribute to diffraction(Darkens the background of the diffraction patterns)(2 , 2 )sKnocked out electron from inner shellVacuumEnergy lev

14、elsL3L2L1Characteristic x-rays (Fluorescent X-rays)K(1016s later seems like scattering!)NucleusEKEL1EL2EL3Fluorescent X-ray§ A beam of X-rays directed at a crystal interacts with the electrons of the atoms in the crystal§ The electrons oscillate under the influence of the incoX-Rays and be

15、come secondary sources of EM radiation§ The secondary radiation is in all directions§ The waves emitted by the electrons have the same frequencyX-rays Þ coherentas the inco§ The emission will undergo constructive or destructive interference with waves scattered from other atomsSe

16、condary emissionIncoX-raysBraggs LawFourier AnalysisTranslation vector T = u1a1 +u2a2 +u3a3 Electron number density n(r +T) = n(r) Expand n(x) in a Fourier series,n(x) = no +åCp cos(2 px / a)+Sp sin(2 px / a)p>0n(x +a) = no +åCp cos(2 px / a +2 p) + Sp sin(2 px / a +2 p)= no +åCp c

17、os(2 px / a) + Sp sin(2 px / a) = n(x)where 2 p / a is a point in the reciprocal la tice or Fourier space of the crystal.Fourier Analysisn(x) = å npexp(i2 px / a) in 1Dp= nn*since n(x) is a real function- ppnp (cos + i sin ) + n- p (cos + i sin )= (np + n- p ) cos + i(np + n- p ) sin =2Renp cos

18、 - 2 Imnp sin From the extention of Fourier series to 3Dn(r) = å nGGexp(iG × r) in 3DInversion of Fourier seriesInversion of Fourier Seriesanp = aò dxn(x) exp(-i2 px / a)-10a p ' òn= a-1å ndx expi2 ( p '- p)x / a)pp '0if p' ¹ p the value of the integral

19、isa(ei 2 ( p '- p )-1) = 0i2 ( p '- p)òcell= V -1dVn(r)exp(-iG × r)nGcReciprocal Lattice Vectorsa2 ´ a3a3× a1a1× a2b = 2b= 2b= 2;123a ga ×aa ga ×aa ga ×a123123123bi gaj =2ijwhere b is primitive vectors of the reciprocal latticePoints in the reciprocal

20、lattice are mapped by set of vectorsG=hb1 +kb2 +lb3n(r + T)=å nG exp(iGgr) exp(iGgT)GBut exp(iGgT)=1, becauseexp(iGgT) = expi(hb1 +kb2 +lb3 ) × (u1a1= expi2 (hu1 + ku2 + lu3 )+ u2a2 + u3a3 )Diffraction ConditionsF = ò dVn(r) expi(k - k') × r =ò dVn(r) exp(-ik × r) w

21、here k - k' = -k, or k + k = k'F = åò dVnG expi(G - k) × rGk=G2k × G+G2 =0 or 2k × G=G 2since both G and - G are reciprocal vectorsDiffraction ConditionsUsing 2k × G=G2 , G=hb +kb +lb , and123spacing between parallel lattice plane d(hkl)=2 / G2(2 / ) sin = 2 / d

22、 (hkl), or2d(hlk)sin =2dsin =nBrillouin Zones11From 2k × G=G2 , k × (G)=(G)222Reciprocal Lattice to sc Latticea1 =ax$;a2 =ay$;a3 =az$.where x$, y$, z$ are orthogonal vectors of unit lengthb1 =(2 /a)x$;b2 =(2 /a)y$;b3 =(2 /a)z$;Boundaries of 1st Brillouin zone are at the midpoints± 1 b

23、 = ± ( /a)x$;± 1 b = ± ( /a)y$;± 1 b = ± ( /a)z$;123222Reciprocal LatticePrimitive bcc Lattice111$a =a(-x+y+z); a =a(x - y+z);a =a(x+y - z);123222= 1 a3The volume of the primitive cell is V= a × a ´ a1232a2 ´ a3a3× a1a1× a2Using b = 2; b= 2; b= 2123a

24、 ga ×aa ga ×aa ga ×a123123123b1 =(2 /a)(y$ + z$ ); b2 =(2 /a)(x$ + z$ ); b3 =(2 /a)(x$ + y$);Teral reciprocal lattice vector isG=hb1 +kb2 +lb3 = (2 /a)(k + l)x$ + (h + l)y$+(h + k )z$12 shortest G can be found as follows(2 /a)( ± y$ ± z$ );(2 /a)( ± x$ ± z$ );(2 /a

25、)( ± x$ ± y$);See Brillion zone (fcc) for bcc lattice.Reciprocal LatticeQ1) Express the primitive lattice vectors of the fcc structure.Q2) Find the volume of the primitive cell of the fcc structure.Q3) Find the reciprocal lattice vectors of the fcc structure.Q4) Find the shortest G's.R

26、eciprocal lattice of 2D hexagonal materials3$x + a siny; a3 =bz;a1 =ax; a =a cos23Q1) Find the volume of the primitive cell of a 2D material with the above lattice vectors.Q2) Find the reciprocal lattice vectors of a 2D material with the above lattice vectors.Q3) Draw a 2D lattice structure and a Br

27、illion zone in 2D xy plane from the reciprocal vectors obtained from Q1.Q4) Find the numerical numbers indicating the corners (K and K) and midpoints(M) of the 1st Brillion zone for the 2D materials with the following lattice constants.A.B.Graphene : 0.246 nm MoS2 : 0.317 nmScattering amplitude and

28、structure factorF = ò dVn(r) expi(k - k') × r =ò dVn(r) exp(-ik × r)FG =N ò dVn(r) exp(-iG × r) = NSG(when k = G)cellsTotal electron concentration n(r) = å nj (r - rj )j-1The structure factorSG =åò dVnj (r - rj )exp( - iG × r)j=åexp( - iG &#

29、215; rj )ò dVnj ()exp( - iGg)where = r - rjjAtomic form factor f j = ò dVnj ()exp( - iGg) Structure factor of the basis in the formSG =å f j exp( - iG × rj )jAnalysis of structure factorThe usual form of this result for atom j is;rj =xja1 +yja2 +z ja3reflection labelled by h, k,

30、l,G ×rj =(hb1 + kb2 + lb3 ) × (xja1 +yja2 +z ja3 ) = 2 (hxjSG (hkl) = å f j exp-i2 (hxj + ky j + lz j )j+ ky j+ lz j )(Example) Strucutre factor of bcc LatticeS (hkl) = f 1+ exp-i (h + k + l)S = 0 when h + k + l = odd integer ;S = 2 fwhen h + k + l = even integer.(Example) Strucutre f

31、actor of fcc LatticeS (hkl) = f 1+ exp-i (k + l)+ exp-i (h + l) exp-i (h + k )Ray 1Ray 2qqd§ The path difference between ray 1 and ray 2 = 2d sinq§ For constructive interference: nl = 2d sinqqqqDeviation = 2qBRAGGs EQUATION§ nl = 2d sinq§ n is an integer and is the order of the r

32、eflection§ For Cu Ka radiation (l = 1.54 Å) and d110= 2.22 Åd=a1102 d220= 1d1102d=ahklh2 + k 2 + l 2 ad220 =8nsinqq10.3420.7ºFirst order reflection from (110)20.6943.92ºSecond order reflection from (110)Also written as (220)AElectronBAtomic scattering factor (f)CStructure fa

33、ctor (S)Unit cell (uc)AtomScattering by a crystalIntensity of the Scattered electronsSets electron into oscillationScattered beams§The electric field (E) is the main cause for the acceleration of the electronThe moving particle radiates most strongly in a direction perpendicular to its motionTh

34、e radiation will be polarized along the direction of its motion§§(0 , 0 )(0 , 0 )Scattering by an ElectronAScattering by an atom µ Atomic number, (path difference suffered by scattering from each e, l)Scattering by an atom µ Z, (q, l)§ Angle of scattering leads to path diffe

35、rences§ In the forward direction all scattered waves are in phaseSin( )f = Atomic Scattering FactorAmplitude of wave scattered by an atom=Amplitude of wave scattered by an electron§ Unit Cell (UC) is representative of the crystal structure§ Scattered waves from various atoms in the UC

36、 interfere to create the diffraction patternThe wave scattered from the middle plane is out of phase with the ones scattered from top and bottom planesScattering by the Unit cell (uc)CScattering by an AtomBIn complex notationi 2 ( h x¢+k y¢+l z¢)iE = Ae= fe§ If atom B is differen

37、t from atom A the amplitudes must be weighed by the respective atomic scattering factors (f)§ The resultant amplitude of all the waves scattered by all the atoms in the UC gives the scattering factor for the unit cell§ The unit cell scattering factor is called the Structure Factor (S)Scatt

38、ering by an unit cell = f(position of the atoms, atomic scattering factors)nn= å fj =1= å fj =1i2 ( hx¢ +ky¢ +lz¢ )ihkl nSeejjjjFor n atoms in the UCjjStructure factor is independent of the shape and size of the unit cellIf the UC distorts so do the planes in it!I µ S 2

39、S = Structure Factor = Amplitude of wave scattered by all atoms in ucAmplitude of wave scattered by an electron = 2 ( h x¢ + k y¢ + l z¢) Simple Cubic e(odd n) ie(even n) i= -1= +1nsAn= ån= åi2 ( hx¢ +ky¢ +lz )¢eiS hklffejjjjnjjj =1j =1ei 2 (h×0+k×0+

40、l×0)S = f=fe0 = f= fS 22 F is independent of the scattering plane (h k l)ei + e-i=Cos( )2eni = e- niAtom at (0,0,0) and equivalent positioeni = (-1)nStructure factor calculationson C- centred Orthorhombic Bei jei2 ( hx¢j +ky¢j +lz¢j )S = f jf ji2 ( h×1 +k × 1 +l×0)

41、ei 2 ( h×0+k ×0+l×0)S = f+ fe22i 2 ( h+k )f 1+ ei ( h+k ) =+ fe=fe02RealS = 2 f= 4 fS 22e.g. (001), (110), (112); (021), (022), (023)f 1+ ei (h+k ) S =S = 0= 0S 2e.g. (100), (101), (102); (031), (032), (033) S is independent of the l indexAtom at (0,0,0) & (½, ½, 0) and

42、equivalent positionsCei jei2 ( hx¢j +ky¢j +lz¢j )S = f jf ji 2 ( h×1 +k ×1 +l)21ei 2 ( h×0+k ×0+l×0)S =+ ffe22i 2 ( h+k +l )f 1+ ei (h+ k +l ) Real=fe0 + fe=2S = 2 f= 4 fS 22e.g. (110), (200), (211); (220), (022), (310)f 1+ ei ( h+k +l ) S =S = 0= 0S 2e.g. (10

43、0), (001), (111); (210), (032), (133)Atom at (0,0,0) & (½, ½, ½) and equivalent positiBody centred Orthorhombic Face Centred Cubic D(½, ½, 0), (½, 0, ½), (0, ½, ½)ei jei2 ( hx¢j +ky¢j +lz¢j )S = f=fjjéi2 ( l +h ) ùi 2 ( h+k )i

44、 2 ( k +l )f êei2 (0)S =+ e+ e+ e222úëûf 1+ ei ( h+k )f 1+ ei ( h+k )+ ei (k +l ) + ei (l +h ) + ei (k +l ) + ei (l +h) =S =S = 4 f= 16 fS 22(h, k, l) unmixede.g. (111), (200), (220), (333), (420)= 0e.g. (100), (211); (210), (032), (033)S = 0S 2(h, k, l) mixedTwo odd and one even

45、 (e.g. 112); two even and one odd (e.g. 122)RealAtom at (0,0,0) & (½, ½, 0) and equivalent positionsTwo odd and one even (e.g. 112); two even and one odd (e.g. 122)1+ ei (e) + ei (o)1 + ei (o) + ei (e)+ ei (o) = 1+1 -1 -1 = 0+ ei (o) = 1-1+1 -1 = 0CASECASEA :B :S = 0S 2= 0(h, k, l) mix

46、ede.g. (100), (211); (210), (032), (033)All odd (e.g. 111); all even (e.g. 222)1+ ei (e) + ei (e)1+ ei ( e) + ei ( e)+ ei (e) = 1 +1+1+1 = 4+ ei (e) = 1+1+1+1 = 4CASECASEA :B :S = 4 f= 16 fS 22(h, k, l) unmixede.g. (111), (200), (220), (333), (420)Unmixed indicesCASEhklAoooBeeeUnmixed indicesMixed indicesC