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Lecture VII Why Solids? most elements are solid at room temperature atoms in ~fixed position “simple” case - crystalline solid Crystal Structure Why study crystal structures? description of solid comparison with other similar materials classification correlation with physical properties Metals, insulators & semiconductors? Pure metals: resistivity increases rapidly with increasing temperature. Diamond Resistivity (Ωm) At low temperatures all materials are insulators or metals. 1020- 1010Germanium 100 Copper 10-100 100 200 Temperature (K) 300 Semiconductors: resistivity decreases rapidly with increasing temperature. Semiconductors have resistivities intermediate between metals and insulators at room temperature. Metals and insulators Measured resistivities range over more than 30 orders of magnitude Material Resistivity (Ωm) (295K) Resistivity (Ωm) (4K) 10-5 10-12 Copper 2 10-6 10-10 SemiConductors Ge (pure) 5 102 1012 Insulators Diamond 1014 1014 1020 1020 Potassium “Pure” Metals Polytetrafluoroethylene (P.T.F.E) Early ideas • Crystals are solid - but solids are not necessarily crystalline • Crystals have symmetry (Kepler) and long range order • Spheres and small shapes can be packed to produce regular shapes (Hooke, Hauy) ? Kepler wondered why snowflakes have 6 corners, never 5 or 7. By considering the packing of polygons in 2 dimensions, demonstrate why pentagons and heptagons shouldn’t occur. Empty space not allowed CRYSTAL TYPES Three types of solids, classified according to atomic arrangement: (a) crystalline and (b) amorphous materials are illustrated by microscopic views of the atoms, whereas (c) polycrystalline structure is illustrated by a more macroscopic view of adjacent single-crystalline regions, such as (a). Crystal structure Amorphous structure quartz Definitions 1. The unit cell “The smallest repeat unit of a crystal structure, in 3D, which shows the full symmetry of the structure” The unit cell is a box with: • 3 sides - a, b, c • 3 angles - , , 14 possible crystal structures (Bravais lattices) 3D crystal lattice cubic a=b=c == tetragonal a=bc = = = 90o monoclinic abc = = 90o 90o orthorhombic abc = = = 90o trigonal (rhombohedral) a=b=c = = 90o hexagonal triclinic a=bc abc = = 90o; = 120o 90o Chemical bonding Types: Ionic bonding Covalent bonding Metallic bonding Van der Walls bonding + - + - Bonding in Solids Melting point (K) Molecular crystals Metals Ionic crystals W(3683) C(<3500) BN(3270) 3000 Mo(2883) Pt(2034) Fe(1808) 2000 1000 organic crystals 0 Covalent crystals Cu(1336) Al(933) Pb(600) Na(371) Hg(234) SiO2(2001) LiF(1143) KCl(1063) Si(1683) Ge(1240) Electrons in metals P. Drude: 1900 kinetic gas theory of electrons, classical Maxwell-Boltzmann distribution independent electrons free electrons scattering from ion cores (relaxation time approx.) A. Sommerfeld: 1928 Fermi-Dirac statistics F. Bloch’s theorem: 1928 Bloch electrons L.D. Landau: 1957 Interacting electrons (Fermi liquid theory) Metallic bond Atoms in group IA-IIB let electrons to roam in a crystal. Free electrons glue the crystal Attract eAttract Repel Repel Na+ Na+ Attract Attract e- Additional binding due to interaction of partially filled d – electron shells takes place in transitional metals: IIIB - VIIIB Core and Valence Electrons Most metals are formed from atoms with partially filled atomic orbitals. e.g. Na, and Cu which have the electronic structure Na 1s2 2s2 2p6 3s1 Cu 1s2 2s2 2p6 3s23p63d104s1 Insulators are formed from atoms with closed (totally filled) shells e.g. Solid inert gases He 1s2 Ne 1s2 2s2 2p6 Or form close shells by covalent bonding i.e. Diamond Note orbital filling in Cu does not follow normal rule Simple picture. Metal have CORE electrons that are bound to the nuclei, and VALENCE electrons that can move through the metal. Why are metals good conductors? Consider a metallic Sodium crystal to comprise of a lattice of Na+ ions, containing the 10 electrons which occupy the 1s, 2s and 2p shells, while the 3s valence electrons move throughout the crystal. The valence electrons form a very dense ‘electron gas’. + _ + + + _ _ _ + _ + +_ + _ _ _ + + _+ + _ _ + _ + + + + + + _ + Na+ ions: Nucleus plus 10 core electrons + _ +_ + + We might expect the negatively charged electrons to interact very strongly with the lattice of negative ions and with each other. In fact the valence electrons interact weakly with each other & electrons in a perfect lattice are not scattered by the positive ions. Free classical electrons:Assumptions We will first consider a gas of free classical electrons subject to external electric and magnetic fields. Expressions obtained will be useful when considering real conductors (i) FREE ELECTRONS: The valance electrons are not affected by the electron-ion interaction. That is their dynamical behaviour is as if they are not acted on by any forces internal to the conductor. (ii) NON-INTERACTING ELECTRONS: The valence electrons from a `gas' of non-interacting electrons. They behave as INDEPENDENT ELECTRONS; they do not show any `collective' behaviour. (iii) ELECTRONS ARE CLASSICAL PARTICLES: (iv) ELECTRONS ARE SCATTERED BY DEFECTS IN THE LATTICE: ‘Collisions’ with defects limit the electrical conductivity. This is considered in the relaxation time approximation. Ohms law and electron drift n free electrons per m3 with charge –e ( e = +1.6x10-19 Coulombs ) L V = E/L = IR (Volts) Area A Resistance R = rL/A (Ohms) Resistivity r = AR/L (Ohm m) E = V/L = rI/A = rj (Volts m-1) Electric field E Conductivity s = 1/r (low magnetic field) Force on electron F Drift velocity vd j = sE (Amps m-2) I = dQ/dt Current density j = I/A (Coulomb s-1) dx Force on electrons F = -eE results in a constant electron drift velocity, vd. Area A Charge in volume element dQ = -enAdx vd j= 1 dQ A dt = en dx dt = env d Electrical Conductivity In the absence of collisions, the average momentum of free electrons subject to an electric field E would be given by dp dt = F = eE Field The rate of change of the momentum due to collisions is dp dt At equilibrium Now dp dt Collisions = p / t p dp dt Field Collisions =0 So p = - etΕ j = -nevd = -nep/me = (ne2tp /me) E So the conductivity is s = j/E = ne2tp /me The electron mobility, m, is defined as the drift velocity per unit applied electric field m = vd / E = etp /me (units m2V-1s-1) Examples of ionic bonding • Metal atoms with 1 electron to lose can form ionic bonds with non-metal atoms which need to gain 1 electron: – Eg. sodium reacts with fluorine to form sodium fluoride: So the formula for sodium fluoride is NaF sodium sodiumion atom (Na+) -) fluoride ion (F fluorine atom (Na) (F) Examples of ionic bonding C sC l C ry stal S tructure • C h lo rid e io n s fo rm sim p le cub es w ith cesiu m io n s in th e cen ter Examples of ionic bonding:NaCl •Each sodium atom is surrounded by its six nearest neighbor chlorine atoms (and vice versa) •Electronically – sodium has one electron in its outer shell: [Ne]3s1 and Chlorine has 7 (out of 8 “available” electron positions filled in its outer shell) [Ne]3s23p5 •Sodium “gives up” one of its electrons to the chlorine atom to fill the shells resulting in [Ne] [Ar] cores with Na+ and Cl- ions •Coulombic attraction with tightly bound electron cores NaCl • Potential energy: U = 1 e 2 4 0 r - Madelung constant, m – integer number for ro, the equilibrium position between the ions: U0 = 4 0 B r m 2 e 1 1 r0 m U0 is the cohesive energy, i.e. the energy per ion to remove the ion out of the crystal. Ionic bonding Repulsive potential 1/rm Total potential Attractive potential -1/r Properties of the ionic crystals • medium cohesive energy (2-4 eV/ atom). – low melting and boiling temp. . • Low electrical conductivity. – (the lack of the free electrons). • Transparent for VIS light – ( energy separation between neighbouring levels > 3 eV) • Easily dissolved in water. – Electrical dipoles of water molecules attract the ions Covalent bonding: molecular orbitals Consider an electron in the ground, 1s, state of a hydrogen atom i.e. (r) = 1 3/2 - r/ ao ao e The Hamiltonian is H = where - 2 2 2m a o is the Bohr Radius - where r = e 2 4 o The expectation value of the electron energy is < E > = (r) H (r)dV This gives <E> = E1s = -13.6eV F(r) E1s V(r) + Hydrogen Molecular Ion Consider the H2+ molecular ion in which one electron experiences the potential of two protons. The Hamiltonian is H = - 2 2 + U( r ) = 2m - 2 2 2m - - e- r p+ R r |r - R| We approximate the electron wavefunctions as ( r ) = A [ ( r ) + (| r - R |)] A[ 1 + 2 ] and ( r ) = B [ ( r ) (| r - R |)] B[ 1 2 ] p+ Bonding and anti-bonding states 1 .2 1 .0 (r) 0 .8 0 .6 0 .4 0 .2 Expectation values of the energy are: E = E1s – (R) for (r) (r) - 0 .2 - 0 .4 - 0 .6 - 0 .8 - 1 .0 V(r) - 1 .2 E = E1s + (R) for 0 .0 (r) - 1 .4 -6 -4 0 2 4 6 2 4 6 r 1 .2 (R) - a positive function -2 (r) 1 .0 0 .8 2 0 .6 Two atoms: original 1s state leads to two allowed electron states in molecule. 0 .4 0 .2 0 .0 - 0 .2 - 0 .4 - 0 .6 - 0 .8 Find for N atoms in a solid have N allowed energy states - 1 .0 - 1 .2 - 1 .4 -6 -4 -2 0 r covalent bonding – H2 molecule Anti-bonding 2s bonding Anti-bonding 1s bonding • 8 energy(eV) 6 4 parallel spin 2 0 -2 antiparallel spin -4 -6 R0 0.1 0.2 0.3 nuclear separation (nm) system energy (H2) 0.4 Covalent bonding Crystals: C, Si, Ge Covalent bond is formed by two electrons, one from each atom, localised in the region between the atoms (spins of electrons are anti-parallel ) Example: Carbon 1S2 2S2 2p2 C C 3D Diamond: tetrahedron, cohesive energy 7.3eV 2D Covalent Bonding in Silicon •Silicon [Ne]3s23p2 has four electrons in its outermost shell •Outer electrons are shared with the surrounding nearest neighbor atoms in a silicon crystalline lattice •Sharing results from quantum mechanical bonding – same QM state except for paired, opposite spins (+/- ½ ħ) Covalent bond Atoms in group III, IV,V,&VI tend to form covalent bond Filling factor T. :0.34 F.C.C :0.74 ionic – covalent mixed diamond lattice zinc blend crystals (ZnS, GaAs) As Properties of the covalent crystals • Strong, localized bonding. • High cohesive energy (4-7 eV/atom). – High melting and boiling temperature. • Low conductivity. The Hall Effect Ex, jx Ey Bz vd = vx An electric field Ex causes a current jx to flow. A magnetic field Bz produces a Lorentz force in the y-direction on the electrons. Electrons accumulate on one face and positive charge on the other producing a field Ey . F = -e (E + v B). In equilibrium jy = 0 so Fy = -e (Ey - vxBz) = 0 Therefore Ey = +vxBz jx = -nevx so The Hall resistivity is rH For a general vx. vx+ve or -ve Ey = -jxBz/ne = Ey/jx = -B/ne The Hall coefficient is RH = Ey/jxBz = -1/ne j E The Hall Effect Bz The Hall coefficient RH = Ey/jxBz = -1/ne j=jx vd = vx Ey The Hall angle is given by tan = Ey/Ex = rH/r Ex For many metals RH is quiet well described by this expression which is useful for obtaining the electron density, in some cases. However, the value of n obtained differs from the number of valence electrons in most cases and in some cases the Hall coefficient of ordinary metals, like Pb and Zn, is positive seeming to indicate conduction by positive particles! This is totally inexplicable within the free electron model. The (Quantum)Free Electron model: Assumptions (i) FREE ELECTRONS: The valance electrons are not affected by the electron-ion interaction. That is their dynamical behaviour is as if they are not acted on by any forces internal to the conductor. (ii) NON-INTERACTING ELECTRONS: The valence electron from a `gas' of non-interacting electrons. That is they behave as INDEPENDENT ELECTRONS that do not show any `collective' behaviour. (iii) ELECTRONS ARE FERMIONS: The electrons obey Fermi-Dirac statistics. (iv) ‘Collisions’ with imperfections in the lattice limit the electrical conductivity. This is considered in the relaxation time approximation. Free electron approximation U(r) U(r) Neglect periodic potential & scattering (Pauli) Reasonable for “simple metals” (Alkali Li,Na,K,Cs,Rb) Eigenstates & energies 2 2 d = i U 2m dt k (r, t ) = 0 e i( t k r ) 2 2 Ek = k 2m k = 2(n x / L x , n y / L y , nz / L z ) Ek Unit volume in k-space: 1/(23) |k| Density of states G(E) - this is the number of allowed states within a band: g(E)dE=2g(k)dk with g(k) equal to the density of states within the k-space. 2 is due to two possible spin values. 2 g (k ) g (E ) = dE / dk g(k) is equal to the number of states within the space between two spheras of radii k and k+dk, which is equal to the number of states per unit volume (1/(2)3) multiplied by the volume between the spheras (4k2dk). Thus : g ( k ) dk = 1 ( 2 ) 2 3 4 k dk g ( k ) = k 2 2 2 g (k ) = k 2 2 2 g (k ) g (E ) = 2 = dE / dk For: 2 E = k k 2 1 2 dE / dk 2 2m g (E ) = k 2 2 1 dE / dk = k 2 2 2m 2 2 k = mk 2 2 g(E) E = m 2 3 2 mE Statistics & DOS 2k T B 1 Fermi-Dirac statistics: T=0.1m 1 fFD (E) = e E m kT 0.5 1 2 1 m=EF Density of states: g(E) dk m g ( E ) = 2 g (k ) = 2 3 2mE dE E Occupation of states D(E) 2kBT |kF| Fermi Surface EF E Bound States in atoms F6 F7 F8 F9 0 Electrons in isolated 0 atoms occupy discrete allowed energy levels -1 E0, E1, E2 etc. . Increasing Binding E1 Energy E0 -2 The potential energy of an electron a distance r -3 from a positively charge nucleus of charge q is -4 V (r) = qe 2 4 o r V(r) E2 -5 -8 -6 -4 -2 0 r r 2 4 6 8 Bound and “free” states in solids The 1D potential energy of an electron due to an 0 0 array of nuclei of charge q separated by a distance -1 R is 2 qe V (r) = -2 n 4 o r nR Where n = 0, +/-1, +/-2 etc. -3 V(r) E2 E1 E0 V(r) Solid -3 This is shown as the black line in the figure. -4 V(r) lower in solid (work function). -5 -8 -6 -4 -2 Naive picture: lowest binding energy states can + + become free to move Nuclear positions throughout crystal 00 r2 r +r 4 + R 6 8 + Energy Levels and Bands In solids the electron states of tightly bound (high binding energy) electrons are very similar to those of the isolated atoms. Lower binding electron states become bands of allowed states. We will find that only partial filled band conduct Band of allowed energy states. E + position + + + Electron level similar to that of an isolated atom + Energy band theory 2 atoms 6 atoms Solid state N~1023 atoms/cm3 Metal – energy band theory Insulator -energy band theory diamond semiconductors Intrinsic conductivity ln(s) 1/T s s = s 0se E g / 2 kT Extrinsic conductivity – n – type semiconductor ln(s) s d = s 0d e 1/T E d / kT Extrinsic conductivity – p – type semiconductor Conductivity vs temperature ln(s) s s = s 0se E g / 2 kT s d = s 0d e 1/T E d / kT