Bonding in Solids Section 4 (M&T Chapter 7) Structure and Energetics of Metallic and Ionic Solids • We have discussed bonding in molecules with three models: – Lewis – Valence Bond – MO Theory • These models not suitable for describing bonding in solids (metals, ionic compounds) • The structures of many solids (e.g. NaCl(s), Fe(s)) are best described by a lattice model, in which atoms (ions) of the lattice are placed in highly ordered arrangements (crystal lattices) • Arrangement yields maximum net attractive force with other ions/atoms in the lattice and minimum repulsive forces Close-Packing • Treat the atoms of a metal (or ions of an ionic substance) as spheres (e.g. marbles). • Fill a tray until its surface is covered with a layer of marbles, get a picture that looks something like this: • In this model, all spheres are touching the surface of six other spheres (except those on the edges/corners) – a hexagonal arrangement • This arrangement is one layer of what is called a “closepacked” arrangement • See that there are also six spaces around each sphere • If we were to pile another layer on top of this, it would appear as follows… Close-Packing Close-Packing Around each sphere, there are a total of six spaces Put next layer of spheres into three of them When the next closeclose-packed level is added, only three of the spaces surrounding each sphere can be occupied A C AC A C • So far, the levels are different (the spheres are not located in the same space directly above the first level), so we have an “AB-type” arrangement for the two layers • Note the spaces (hollows) that are now present – two types: – “A” – located above a sphere – “C” – located above a space 1 Close-Packing Close-Packing If we were to add another closeclose-packed level on top, we could do so in two ways: • In the upper figure (ABA-type) the top level was created by putting spheres into the spaces labeled “A” of the second layer • The top level here is thus the same as the first (bottom) level (the spheres in this level are located directly above those of the first level) – an “ABAB…” arrangement Put 3rd layer into A hollows C A C A C Put 3rd layer into C hollows ABA arrangement Close-Packing Close-Packed Arrangements •In the lower figure (ABC-type), the top layer was constructed by placing spheres in the spaces labeled “C” of the second layer • These two arrangements shown below, are represented as ball-and-stick arrangements, and not meant to imply that spheres are not touching • Which is which? The coordination number (CN) in each of these lattices is 12. •This results in a level that is different than either of the first two levels (an “ABC…” arrangement) Top layer different than bottom ABC arrangement Close-Packed Arrangements • Close-Packed Arrangements The unit cells are called: a) hexagonal close-packed (hcp) – describing an “ABABAB…” arrangement b) cubic close-packed (ccp) – this describes an “ABCABC…” ordering hcp ccp • The unit cells are shown below as spherical models – Cubic close-packed (ccp) (face-centered cubic) – Hexagonal close-packed (hcp) (this is also known as a “facecentered cubic” arrangement) 2 Interstitial Holes Crystal Structures • The spaces between spheres in the close-packing arrangements are called interstitial holes or interstitial spaces • Two types: – Tetrahedral (four point cavities) – Octahedral (six point cavities - larger) Crystalline: long-range order Amorphous: no long-range pattern Unit Cell Holes in Crystals Unit Cells in the Cubic Crystal System Non-Close Packed Arrangements Number of Atoms in a Unit Cell • In many solids, the percentage of occupied space is less than 74% - these solids assume non-close packed arrangements • Arrangements below depict the simplest model that incorporates all the information of the lattice (a unit cell) 52% of space used 68% of space used What are the coordination numbers in these lattices? Face centered unit cell contains 8 corners @ 1/8 atom + 6 faces @ ½ atom = 4 atoms Simple cubic Body-centered cubic 3 Counting Cell Occupancy 2 atoms 4 atoms 1 atom X-Ray Diffraction : Crystallography circle = bcc diamond = hcp + = ccp (fcc) Data quoted for T = 298 K Diffraction of X-Rays Polymorphism • The lattice structure adopted by an element may change as the temperature and/or the pressure is changed • A substance that can exist in more than one crystalline form is said to be polymorphic • Next slide – a phase diagram for Fe Constructive interference of the x-rays only when the pathlength in the crystal is a whole multiple of the wavelength Bragg Equation: nλ λ = 2d sin θ, n = 1, 2, 3, … • Phase diagram – lines (phase boundaries) separate different phases of a substance • On a line, have both phases present in equilibrium 4 Phase Diagram for Iron Phase Diagram for Iron Look at Fe at 1000 K, 1 atm (α α-Fe). Increase the pressure. What happens? (γγ-Fe) Non close-packed Fe at 200K, 1 atm – increase pressure to 100 atm. What happens? At ambient temperature (~ 300 K) and pressure (1 atm), iron adopts a body-centered cubic (bcc) ordering (α α-Fe) Phase Diagram for Iron • Sometimes, we can heat an element, changing its packing, and rapidly cool it to retain the higher temperature structure (quenching) – this allows higher temperature structures to be studied at ambient temperatures Alloys and Intermetallic Compounds • Everyday examples of alloys: (solute(s), solvent) – Solder (Sn, Pb); low melting point – Stainless steel (Cr, C, Fe); anti-corrosion, strength – Sterling silver (Cu, Ag); doesn’t tarnish Alloys and Intermetallic Compounds • Alloys are intimate mixtures (solid solutions) of metals/elements, or a distinct compound consisting of these elements, with physical properties that differ from those of the elements that make it up • The aim of alloying is to endow the mixture with desirable properties that are inherent to the metals of the mixture (e.g. hardness, ductility) Alloys • Substitutional alloys: created from metals having similar rMet (radii), similar coordination number • Metals are mixed in the molten state (high temperature) and allowed to cool gradually • The solute atoms occupy sites in the lattice that would normally be occupied by solvent atoms • Sterling silver is created from 92.5% Ag, 7.5% Cu – both adopt ccp lattices – rMet(Ag) = 140 pm rMet = metallic radius: half the distance between nearest neighbor atoms in a solid-state metallic – rMet(Cu) = 124 pm lattice 5 Substitutional Alloys Alloys • Interstitial alloys: solute atoms occupy the interstitial spaces (cavities) of a host lattice (e.g. carbon steel) – Low C steel: 0.03 – 0.25% C (steel sheeting) – Medium C steel: 0.25 – 0.70% (bolts, screws, etc.) – High C steel: 0.8 – 1.5% (cutting, drilling tools) Intermetallic Compounds • Some melts (combinations of metals in the molten state) will solidify in arrays that are different than either of the components that make up the mixture – these compounds are called intermetallic compounds • Brass (Cu, Zn) is an intermetallic compound γ-brass has a formula of Cu5Zn8 Diamond • In diamond, C-atoms take on a fcc-type arrangement • Each C-atom is fourcoordinate (tetrahedral) and so possesses a complete octet of electrons through covalent bonds • Also adopted by Si, Ge, Sn and Pb (other C-group elements) Alloys and Intermetallic Compounds • Alloys are intimate mixtures (solid solutions) of metals/elements, or a distinct compound consisting of these elements, with physical properties that differ from those of the elements that make it up • The aim of alloying is to endow the mixture with desirable properties that are inherent to the metals of the mixture (e.g. hardness, ductility) Graphite • The most thermodynamically stable form of carbon (diamond is metastable), graphite consists of layers of carbon sheets that are built from fused, 6-membered carbon rings • The bonding that exists within a layer is covalent (and delocalized), but between planes, dispersion forces (non-covalent) hold the rings together (weak intermolecular force) 6 Unit Cells Unit Cell Contributions • The smallest collection of spheres that describes a lattice (when it is repeated) is the unit cell • Some of the atoms in a unit cell are shared with other cells, and so they do not belong entirely to one cell Corner site Example, the atoms at the corners of the cell shown on the right are shared with seven other cells Each contributes 1/8 of an atom to the cell shown, and is called a “corner site” Face-centered site Face-centered site A face-centered cubic unit cell Ionic Solids Particles: • Corner sites each contribute 1/8 of a sphere to a cell • Edge sites contribute ¼ sphere to the cell • Face sites contribute ½ sphere to the cell • Sites contained in the cell (e.g. the center site of a bcc cell) contribute 1 each How many spheres (atoms/ions) occupy - a simple cubic unit cell? - a body-centered cubic cell? - a face-centered cubic cell? Holes in Crystals Cations (+) – small size Anions (-) – large size UNIT CELL dimensions usually determined by the arrangement of the large ANIONS. (2R)2+(2R)2 = (2R+2r)2 • • r = 0.414R Cations fit into the “holes” within the anion structure Sodium Chloride Radius Ratios • A rough guide for predicting structures of salts (cations and anions). Use rcation/ ranion, or r+/r-) <0.15 Predicted Coordination Number of Cation 2 Predicted Coordination Geometry of Cation Linear 0.15-0.22 3 Trigonal Planar 0.22-0.41 4 Tetrahedral 0.41-0.73 6 Octahedral >0.73 8 Cubic Value of r+/r- rNa+/RCl- = 102 pm/ 181 pm = 0.56, expect Na+ to occupy octahedral holes. 7 Holes in Crystals Four types of holes: trigonal < tetrahedral < octahedral < cubic too small for any cation Final arrangement depends on relative sizes of cations and anions plus the required stoichiometry This is an example of a binary solid (two elements involved) – what is its formula? Dimensions of Cubic Lattices • • • Binary Lattices For cubic unit cells, it is possible to determine cell lengths (edge) and dimensions of lattice spaces For example, a simple cubic cell will have an edge length of 2r, where r is the radius of a sphere It can be shown that the corner-to-corner length indicated in the picture will be √3× ×a In this diagram, let a = 2r Rock Salt Lattice (NaCl) • • • • • √ 3× ×a Each sphere has a radius of r Each edge length (length of the edge of the unit cell) is then 2r (spheres in contact) Can also be shown that - a sphere of radius up to 0.73r can fit in the center -the atoms occupy 52% of the available space of the unit cell Interpenetrating fcc lattices of Na+ ions and Cl- ions (rem: fcc = ccp) Cl- ions are much larger then the Na+ ions (Cl-: 181 pm; Na+: 102 pm). Na+ ions are occupying octahedral holes in the unit cell shown Each Na+, Cl- ion is octahedral (six coordinate) NaCl-type lattice structures exist for many ionic compounds (NaF, NaBr, NaI, NaH, LiX, KX, RbX (X = halide), AgF, AgCl, AgBr, MgO, CaO, SrO, BaO, MnO, CoO, NiO, MgS, CaS, SrS, BaS) a √ 2× ×a Simple cubic unit cell CsCl Lattice • Eight coordinate ions (cations and anions) – body centered cubic • Interpenetrating simple cubic-type lattice • Adopted by CsBr, CsI, TlCl, TlBr CaF2 (Fluorite) Lattice • Eight-coordinate cations (Ca2+, grey spheres) • Four-coordinate anions (F-, blue spheres) • Six cations are face-positioned, shared between adjacent cells. • This lattice type adopted for group II metal fluorides, BaCl2, and f-block metal dioxides • Exchanging the cations and anions in this structure would yield an antifluorite lattice – M2X stoichiometry. Adopted by some group I metal oxides and sulfides (e.g. Na2S). 8 Zinc blende, ZnS (diamond type) lattice • Similar to the fluorite lattice, with removal of half of the anions (so MX2 to MX stoichiometry). • Looks something like the structure shown for diamond – each atom is in a tetrahedral environment How many Zn, S atoms exist in this structure? Wurtzite (ZnS) lattice • Wurtzite formed by high temperature transition from zinc blende • Hexagonal prism unit cell with all ions tetrahedrally sited • How many Zn2+, S2- ions exist in this structure? Zn: grey S: blue β-cristobalite (SiO2) lattice • Again, much like diamond structure, but with oxygen ions between the tetrahedral Si ions. • Si-O-Si bond angle in figure is 180o, while in practice, it is found to be 147o (bonding in SiO2 is not purely electrostatic). Rutile (TiO2) structure • Oxygen ions (white) are trigonal planar while titanium centers (black) are octahedral. • Four oxygen ions are faceoriented, while two are contained in the cell In a pure ionic model, electrostatic attraction would be the only factor that would be expected to hold an ionic lattice together Perovskite (CaTiO3) lattice • A double oxide (oxygen atoms are coordinated to both Ca2+ and Ti4+) • Ca2+ ion is at center of cube unit cell • Ti4+ ions at corners of the cube (eight of these) • O2- ions at each edge of the cube (twelve of these) CdCl2, CdI2 Lattices • Common for MX2 structures to crystallize in this structure • Can observe the layers as ABAB (layered lattices) • I- ions (gold) are arranged in a hcp format with the Cd2+ ions (white) occupying octahedral holes. • In CdCl2, the arrangement is ccp • Attractive forces that exist between these planes is weak (dispersion forces), and so fracture of a crystal of this kind usually produces cleavage planes CdI2 lattice 9 Energy Changes in the Formation of Ionic Crystals Energy Changes in the Formation of Ionic Crystals Born-Fajans-Haber Cycle Na(s) + ½ Cl2(g) → NaCl(s) electron affinity ionization energy ∆Hreaction = ∆Hf(NaCl(s)) bond energy Lattice Energy (Ionic Bond Strength) sublimation Energy Between Two Point Charges Lattice Energies • We have already looked at bond dissociation enthalpies (energy required to break bonds in homonuclear and heteronuclear diatomics) • Energy is also required to break apart ionic lattices, due to the large amount of electrostatic forces that exist between the ions in the lattice • Coulombic forces (attractions, repulsions) • Born forces (electron-electron, nucleus-nucleus) • Consider what happens if we bring two point charges from an infinite separation to form an ion pair: Mz+(g) + Xz-(g) MX(g) • We can calculate the change in internal energy (∆ ∆U) as: ∆U = • • • • Z+Z− r0 e2 4π ε 0 Z-, Z+ are the charges of the ions in electron units e is the charge of an electron (1.602 x 10-19 C) εo is the permittivity of a vacuum (8.854 x 10-12 C2/J.m) r0 is internuclear separation • Because oppositely charged ions are attracted to one another, energy is released in this process • Consider the attractions and repulsions that exist in a rock salt lattice (between oppositely-charged and like-charged ions) ∆U = Z+Z− r0 e2 4π ε 0 10 A Summary of Attractions and Repulsions • The attractions experienced by the Na+ ion are summarized as follows: • 6 Xz- ions, each at a distance d (the ions at the face sites) • 12 Mz+ ions, each at a distance (√ √2)d (the ions at the edge sites) • 8 Xz- ions each at a distance (√ √3)d (the ions at the corner sites) • 6 Mz+ ions each at a distance of (√ √4)d (imagine the next set of pink spheres in figure beyond the face gray spheres) • We must factor these attractions and repulsions into the expression: ∆U = 6 2 12 2 8 Z+ + Z+ Z− − Z + ... (6 Z + Z − ) − 4πε o d 2 3 4 e2 • Convergent series, which yields a number for each lattice type that is called the Madelung constant, M Madelung Constants Lattice Energy (almost there…) • If we sum the interactions (attractive and repulsive) between the ions of this lattice, we get a convergent term (for this lattice, the value converges to a value of ~1.7476). This value is obtained regardless of the actual charges on the ions. • Madelung constants (M) are unique for each coordination environment (i.e., for each type of crystal lattice). Table 7-2, M. & T. MZ + Z − e r0 4π ε 0 2 ∆U = Lattice Energies • The internal energy change for the formation of one mole of an ionic lattice in this arrangement is then calculated as: ∆U = NMZ + Z − e 2 r0 4π ε 0 • N = Avogadro’s number (6.022 x 1023 mol-1) • but what about Born forces? (nuclear-nuclear, electron-electron forces) Lattice Energies • If we consider electrostatic and Born forces, we arrive at the Born-Mayer equation (evaluated at equilibrium internuclear separation, ro) Correction for This equation will enable us to predict lattice energies (called the calculated lattice energy Madelung Constants Born forces NMZ + Z − e 2 ρ 1 − ∆U (0 K ) = r0 4π ε 0 r0 • Lattice energy can be defined as the internal energy change associated with the formation of one mole of the solid from its constituent gas phase ions at 0 (zero) Kelvin. Thus, at 0K, the lattice energy corresponds to the process: Mn+(g) + nX-(g) MXn(s) Lattice energies may be estimated by assuming an electrostatic model (ions are point charges) – a good approximation in some cases. In others, not so good. • ρ corrects for repulsions at short distances. Typically, a value of 30 pm is used for ρ. •Lattice energy can also be defined as the energy required to pull apart an ionic lattice into its gas-phase ions, as defined in M&T, in which case, it is a positive energy 11 Sample Calculation: Lattice Energy • Calculate the lattice energy for NaCl (rNa-Cl = 283 pm) ∆U (0 K ) = NMZ + Z − e ρ 1 − r0 4π ε 0 r0 2 (6.022 × 1023 )(1.7476)(+ 1)(− 1)(1.602 × 10 −19 C ) 2 30 pm ∆U (0 K ) = . 1 − 4π (8.854 ×10 −12 C 2 J −1.m −1 )(283 × 10 −12 m) 283 pm = −766871J / mol Lattice Energies • We see there is only a minor discrepancy between the value obtained with the Born-Mayer equation (-767 kJ/mol) and the Born-Fajans-Haber thermodynamic cycle (-787 kJ/mol) • For NaCl, there’s only a ~2% difference between the calculated and experimental energies (an ionic model provides a good approximation of NaCl) = −767kJ / mol Appendix B rNa+ = 116 pm rCl- = 167 pm Lattice Energies • • • Since the calculated values agree so well (2% error), we see the electrostatic model is a reasonably good assumption for the type of bonding which exists in a NaCl(s) lattice Not true for layered structures like CdI2(s) – recall the forces that exist in this structure We also see that for silver halides, the calculated and experimental energies differ greatly, in the order AgF<AgCl<AgBr<AgI. The bonding with larger halide ions has more covalent character, and thus an ionic approximation does not hold CHEM 245 Interactive 3-D Crystal Structures: http://www.chemtube3d.com/solidstate/_table.htm F Cl Br I Radius Ratios • • • These guidelines often yield incorrect predictions – example: LiBr (r+/r- = 0.38; tetrahedral) Predict only one coordination geometry for a given combination of ions (not helpful for polymorphic samples) Examples: – What is the coordination number of Ti in rutile? (rTi3+ = 75 pm; rO2- = 124 pm) – What is C.N. of Ca in fluorite? (rCa2+ = 126 pm; rF- = 117 pm) – What is C.N. of Zn in zinc blende? (rZn2+ = 77pm; rS2- = 170 pm) Electrical Conductivity in Metals, Semiconductors, and Insulators • Electrical conductivity is a property displayed by metals and some inorganic and organic materials • Loosely defined as the ability of a substance to permit movement of electrons throughout its volume • On a molecular level, electrons can be passed around (atomto-atom) by being promoted into empty orbitals on other atoms • For solid structures a modification of Molecular Orbital Theory called Band Theory is used to explain conductivity 12 Electrical Conductivity and Resistivity Electrical Conductivity and Resistivity • Resistivity (ρ ρ) measures a substance’s electrical resistance for a wire of uniform cross-section R= ρl a = (resistivity / Ωm )(length / m ) (cross − sec tional _ area / m ) 2 • Resistance is measured in ohms (Ω Ω) • Conductivity is 1/resistivity. Units of conductivity are Ω-1m-1 or S/m (S = Siemens; S = Ω-1) “electrical resistivity” Ωm • The resistivity of a metal increases (conductivity decreases) with increasing temperature Electrical Conductivity and Resistivity MO Theory Approach to Band Theory • • The resistivity of a semiconductor decreases (conductivity increases) with increasing temperature Resistivity with temperature for a semiconductor Band Theory of Metals In order for electrons to be able to move through a material, they must jump from an occupied molecular orbital to an unoccupied orbital. In a metal, this should be easily accomplished (metals are highly conductive). Lithium (and other alkali metals) have a half-filled valence s-orbital (occupied). In the infinite solid, there will be a half-full band. Electrons can move into an unoccupied MO with minimal energy cost (small applied potential) unoccupied orbitals produce a “conduction band” density of states is greatest in the middle of the band Band that is created from occupied orbitals is called “valence band” Consider a line of H-atoms that interact through their valence s-orbitals. As more H-atoms interact, more MO’s are created In the infinite structure, there is a continuum of energy states - a “band” (non-quantized) Valence orbitals overlap to create a “valence band” Band Theory of Metals • For the metal, Be (2s2), how does electron movement occur? (valence band is full) • The energy separation of 2s and 2p orbitals in Be is small enough that the conduction band (band that derives from unoccupied orbitals) overlaps the valence band electron movement can occur by an electron jumping from the valence band into an energyenergy-matched unoccupied conduction band orbital 13 Semiconductors and Band Gaps • In many materials (e.g. diamond), there is an energy gap (band gap, Eg) between the valence band and the conduction band • For C, the energy separation that exists between the 2s and 2p valence orbitals is not small enough for valence band-conduction band overlap to occur in the bulk material (e.g. diamond) Band Theory • For semiconductors, thermal energy will enable electrons to move into the empty conduction band, creating “holes” in the valence band (orbitals that were occupied by electrons). The mobile electrons (and holes) give rise to electrical conductivity. • Pure materials that are electrically conductive called intrinsic semiconductors conduction conduction – Metals have either partially filled valence bands or overlapping valence and conduction bands (e.g. Na) – Insulators have fully occupied valence bands that are separated from the conduction band by a significant energy gap (e.g. diamond) – Semiconductors have fully occupied bands that are separated from the valence band by a small energy gap (e.g. Si) Semiconductors band gap overlapping valence and conduction bands valence valence Extrinsic Semiconductors • Certain semiconductors exhibit enhanced electrical conductivity when small quantities of another element are present in the semiconductor lattice (called doping). • The band structure of silicon involves a band gap of approximately 106 kJ/mol (1.11 eV) • Introduction of either gallium or arsenic in very small quantities creates an extrinsic semiconductor, with a band gap of only about 10 kJ/mol (0.10 eV) Si adopts a diamond lattice Semiconductors Semiconductors s2 p2 Ga Ga 14 n-Type Semiconductors Band Theory • Arsenic-doped silicon contains atoms having an additional valence electron (As is gr. 5A, Si Gr. 4A). As in a Si lattice use four of its electrons in bonding (one left over) band gap overlapping valence and conduction bands conduction • Even a small number of As atoms in the Si matrix creates a “donor band” with an energy just below the energy of the Si conduction band (~10 kJ/mol below) conduction valence valence • Thermal energy can excite electrons from the donor band into the Si conduction band, yielding mobile charge carriers Si is n-doped by introducing P or As atoms The charge carriers in the Si band structure are electrons: “n-type” semiconductor p-Type Semiconductors • Ga has one less valence electron than Si • Introduction of a small number of Ga atoms (Gr. 3A) into the Si lattice structure creates a hole (nothing for silicon’s fourth electron to bond to) • The energy of these holes creates a band just above the valence band energy (by ~ 10 kJ/mol), an “acceptor band” • Electrons can occupy this new band, leaving holes in the Si valence band Semiconductors • Semiconductors can be inorganic or organic • Inorganic semiconductors consist of main group elements (Si, Ge, Ga, As, In,…) • Organic semiconductors consist of conjugated carbon structures, typically oligomers or polymers H * n * * n * * S n * * N N n * * n * The mobile holes in Si band structure yield conductivity – a “p-type” type” semiconductor Semiconductor Devices • Thermal population of unoccupied states makes these materials conductive; at 0 K, electrons occupy lowest possible energies The Fermi energy is energy at which an electron is equally likely to be in occupied and unoccupied bands p-type semiconductor at 0 K (left) and at 298 K (right) Semiconductor Devices • The Fermi level serves to set the relative energies of the ptype and n-type interfaces, and is the energy at which an e- is equally likely to be in the valence or conduction band. • In an intrinsic semiconductor, this lies in the middle of the bandgap of the host semiconductor • Doping: – lowers the energy of the Fermi level to the region between the acceptor band and the top of the host’s valence band – raises the energy of the Fermi level to between the donor band and the bottom edge of the host’s conduction band n-type semiconductor at 0 K (left) and at 298 K (right) intrinsic p-type n-type 15 Semiconductor Devices • When n- and p-type semiconductors contact, mobile electrons in the n-type layer near the interface can migrate into the p-type layer, resulting in recombination • The extra electrons in the p-type layer raise its energy and new holes in the ntype layer lower its energy. Charge movement ceases nearly immediately, as the p-type layer accumulates negative charge and the n-type layer positive charge • Recombination results in the formation of a depletion zone at the junction. This depletion zone is accompanied by its own potential Diodes + power supply - When this kind of device is connected to a DC power source, a connection can be made in two ways: 1. Negative terminal to n-type layer, positive terminal to p-type layer (forward bias): as the potential at the negative terminal is made more negative (and the positive terminal more positive), the potential difference of the depletion zone can be overcome and a current flows (e-’s repelled by – terminal; holes repelled by + terminal) Diodes Diodes - power supply + 2. Negative terminal to p-type layer; positive terminal to ntype layer (reverse bias): electrons and holes are pulled away from the depletion zone. Charge cannot move across the junction and there is essentially zero current Photovoltaic Cells • In the absence of an applied potential, electrons can be made to jump from valence to conduction bands by absorbing radiation (e.g. sunlight) and through external connections, can be used to power electronics (solar calculators, solar panels) p,np,n-junction • Diodes are semiconductor devices which permit current to flow in one direction but not the other (current rectifying), made through combinations of p-type and n-type semiconductors • Such devices are useful in electronics (e.g. in common circuitry) Photoswitches • Under reverse bias conditions, if the bandgap is small enough, visible radiation may be sufficient to promote electrons from the valence band to the conduction band (thus current flows in the presence of light – a photoswitch) • Used in sensors (photodetectors for UV, visible, infrared radiation), automatic lights, etc. 16 Light Emitting Diodes (LEDs) • Under forward bias conditions, electrons move from the ntype layer (conduction band) into the p-type layer (valence band) • This movement results in recombination (releases energy). When electrons fall into the holes of the p-doped layer, if the energy change is of the right magnitude, visible light will be emitted (luminescence) • The color of the radiation emitted will depend on the bandgap (Eg), so by varying the bandgap (by controlling the composition of the semiconductor material), different colors can be produced Microscopic to Macroscopic (Atomic Properties to Bulk Properties) atoms, molecules bulk solids •discrete orbitals, energy states •bands, non-quantized Quantum Dots < 10 nm Energy levels depend strongly on size Quantum Dots Photoluminescence of colloidal ZnSe Quantum Dots 17
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