Last week of Course q Today -- Atom Mole s, S s, cule olids q S s with m e ctrons fille according to thePauli e tate any le d xclusion principle q Ne tim C que s of quantumm chnanics xt e onse nce e q Me tals, insulators, se iconductors, supe m rconductors, lase . . rs, Final Exam Monday March 3 Review session Sunday March 2 Extra office hours (Wed.,Sun.,Mon) Any students who cannot take the 214 final and the conflict should email Professor Tai C. Chiang [[e-mail]] ASAP, and no later than noon, Wednesday. Do not email Professor Clegg (as it says on the web site). He will just forward your email Building Atoms, Molecules and solids +e n=3 n=2 n=1 r U(r) a r 6 +e +e 4 r 3 2 even 1 Overview q Atom C ic onfigurations S s in atom with m e ctrons fille according to thePauli e tate s any le d xclusion principle q q Mole cular Wave functions: origins of covale bonds nt q Exam : H + H H2 ple q Ele ctron e rgy bands in S ne olids S s in atom with m e ctrons fille according to thePauli e tate s any le d xclusion principle q Review session Sunday Extra office hours (TBA) Re at from pe Le 12 ct. Let's start building more complicated atoms to study the Periodic Table. For atoms with many electrons (e.g., carbon: 6, iron: 26, etc.) - what energies do they have? <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> Fromspe of com x atom Wolfgang Pauli (1925) de d a ne rule ctra ple s, duce w : &quot;<a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> &quot; &quot;No two e ctrons* can bein thesam quantumstate i.e in a give atomthe cannot have le e , . n y thesam se of quantumnum rs n, l, m, m&quot; e t be l s -- i.e e ry atom orbital with n,l,m can hold 2 e ctrons: ( ) ., ve ic le l q The fore e ctrons do not pileup in thelowe e rgy state i.e the(1,0,0) orbital. re , le st ne , , q The aredistribute am y d ong thehighe e rgy le ls according to thePauli Principle r ne ve . q Particle that obe thePauli Principlearecalle &quot;fe ions&quot; s y d rm *Note Morege rally, no two ide : ne ntical fe ions (any particlewith spin of 3 e rm /2, /2, tc.) can bein thesam quantumstate e . Filling the atomic orbitals according to the Pauli Principle n 4 3 s l=0 p 1 d 2 f 3 g 4 Energy En = - 13.6 eV 2 Z n2 2 is valid only for one electron in the Coulomb potential of Z protons. The energy levels shift as more electrons are added, due to electron-electron interactions. Nevertheless, this hydrogenic diagram helps us keep track of where the added electrons go. Example: Na Z = 11 1s22s22p63s1 l 0 1 2 labe l s p d f #orbitals (2l+1) 1 3 5 7 1 3 Z = atomic number = number of protons Act 1: <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> 1. Which of thefollowing state (n,l,m s) is/areNOT allowe s l,m d? (a). (2, 1, 1, -1/2) (b). (4, 0, 0, 1/2) (c). (3, 2, 3, -1/2) (d). (5, 2, 2, 1/2) (e (4, 4, 2, -1/2) ). 2. Which of thefollowing atom e ctron configurations violate thePauli ic le s <a href="/keyword/exclusion-principle/" >exclusion principle</a> ? (a). 1s2, 2s2, 2p6, 3d10 (b). 1s2, 2s2, 2p6, 3d4 (c). 1s2, 2s2, 2p8, 3d8 (d). 1s1, 2s2, 2p6, 3d5 (e 1s2, 2s2, 2p3, 3d11 ). Act 1: <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> - Solution 1. Which of thefollowing state (n,l,m s) is/areNOT allowe s l,m d? (a). (2, 1, 1, -1/2) (b). (4, 0, 0, 1/2) (c). (3, 2, 3, -1/2) (d). (5, 2, 2, 1/2) (e (4, 4, 2, -1/2) ). m = -l, -(l-1), ... (l-1), l n&gt;l 2. Which of thefollowing atom e ctron configurations violate thePauli ic le s <a href="/keyword/exclusion-principle/" >exclusion principle</a> ? (a). 1s2, 2s2, 2p6, 3d10 (b). 1s2, 2s2, 2p6, 3d4 (c). 1s2, 2s2, 2p8, 3d8 (d). 1s1, 2s2, 2p6, 3d5 (e 1s2, 2s2, 2p3, 3d11 ). Act 1: <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> - Solution 1. Which of thefollowing state (n,l,m s) is/areNOT allowe s l,m d? (a). (2, 1, 1, -1/2) (b). (4, 0, 0, 1/2) (c). (3, 2, 3, -1/2) (d). (5, 2, 2, 1/2) (e (4, 4, 2, -1/2) ). m = -l, -(l-1), ... (l-1), l n&gt;l 2. Which of thefollowing atom e ctron configurations violate thePauli ic le s <a href="/keyword/exclusion-principle/" >exclusion principle</a> ? (a). 1s2, 2s2, 2p6, 3d10 (b). 1s2, 2s2, 2p6, 3d4 (c). 1s2, 2s2, 2p8, 3d8 (d). 1s1, 2s2, 2p6, 3d5 (e 1s2, 2s2, 2p3, 3d11 ). 2(2l+1) = 6 allowe e ctrons d le 2(2l+1) = 10 allowe e ctrons d le Filling Procedure for Atomic Orbitals --example: Bromine Dueto e ctron-e ctron inte le le ractions, thehydroge le ls fail to giveus thecorre filling orde n ve ct r as wego highe in thepe r riodic table . Theactual filling orde is give in thetablebe r n low. Ele ctrons areadde by proce ding along d e thearrows shown. Brom is an e m nt with Z = 35. Find its e ctronic configuration (e 1s2 2s2 2p6 ...). ine le e le .g. As you le d in che istry, thevarious be arne m haviors of all thee m nts (and all them cule le e ole s m up fromthe ) is all dueto theway thee ctrons organizethe se s, according to ade m le m lve quantumm chanics. e Example Problem 1 What is the electronic structure of lithium (3 electrons)? That is, what quantum numbers do the electrons have? S olution: The guiding principle is to find the lowest energy. This involves (for atoms without too many electrons) putting the electrons into the smallest possible n state, because energy depends only on n (to a good approximation). Electron #1: (n, , m , ms) = (1,0,0,1/2) Electron #2: (n, , m , ms) = (1,0,0,-1/2) As you saw in Act 1, the first two electrons have n = 1. This forces them to have = 0 and m = 0. All electrons have s = 1 /2, so it is not listed. ms is always +1/2 or -1/2. The first two electrons can have n = 1, but the third must Electron #3: have n = 2. = 0 has lower energy than = 1, because of 1 (n, , m , ms) = (2,0,0, /2) effects of those n=1 electrons. m doesn't affect the energy, s (symmetry) so either value is OK. Whenever an atom has a single electron in a higher energy state (n value) than the others, that electron is not tightly bound, and the atom can easily lose it. This kind of atom is chemically very reactive. All of the alkali metals (group IA) have this electronic configuration. Supplement: Example Problem 2 A hydrogen atom is in the n = 3, = 2, m = -2 state. To what states can the electron fall when it emits a photon? Which are the strongest (most likely) transitions? Selection Rule for S olution: Final state quantum numbers: n = 1, 2 = 0, 1 m = any electric-dipole transitions: l= 1 Remember the &quot;selection rules&quot;: n is negative here* (conservation of energy) 0 (conservation of angular momentum) ml any value Selection Rule on m: Final states with strongest m=0, 1 transition: =1 photon is circularly polarized photon is linearly polarized n=2 m = -1 = 1 for strongest transition (dipole allowed). = 1 requires n 2. m = -1, 0, or 1. m = 1, only possibility consistent with initial m . *In a different atom, it may be possible to have a n = 0 transition, and still conserve energy (e.g., go from 3s 3p). Act 2: <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> Part 2 ThePauli e xclusion principleapplie to all fe ions in all situations (not just to s rm e ctrons in atom C le s). onside e ctrons in a 2-dim nsional infinitesquarewe r le e ll pote ntial. 1. How m e ctrons can bein thefirst e d state i.ene lowe afte the any le xcite s, xt st r ground state s? (a). 1 (b). 2 (c). 3 (d). 4 (e 5 ). 2. I f the are4 e ctrons in thewe what is thee rgy of them e rge one re le ll, ne ost ne tic (ignoring e inte -e ractions, and assum thetotal e rgy is as low as possible ing ne )? (a). (h2/8m 2) x 2 L (b). (h2/8m 2) x 5 L (c). (h2/8m 2) x 10 L Act 2: <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> Part 2 - Solution ThePauli e xclusion principleapplie to all fe ions in all situations (not just to s rm e ctrons in atom C le s). onside e ctrons in a 2-dim nsional infinitesquarewe r le e ll pote ntial. 1. How m e ctrons can bein thefirst e d state any le xcite s? (a). 1 (b). 2 (c). 3 (d). 4 (e 5 ). Theground statehas quantumnum rs (1, 1), and can hold 2 e ctrons, i.e be le ., (1,1,1/2) &amp; (1,1,-1/2) Thefirst e d statehas two de ne single le xcite ge rate -e ctron state (2,1) and s: (1,2), e of which can have2 e ctrons: (2,1,1/2), (2,1,-1/2), (1,2,1/2), ach le (1,2,-1/2) 2. I f the are4 e ctrons in thewe what is thee rgy of them e rge one re le ll, ne ost ne tic (ignoring e inte -e ractions, and assum thetotal e rgy is as low as possible ing ne )? (a). (h2/8m 2) x 2 L (b). (h2/8m 2) x 5 L (c). (h2/8m 2) x 10 L Act 2: <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> Part 2 - Solution ThePauli e xclusion principleapplie to all fe ions in all situations (not just to s rm e ctrons in atom C le s). onside e ctrons in a 2-dim nsional infinitesquarewe r le e ll pote ntial. 1. How m e ctrons can bein thefirst e d state any le xcite ? (a). 1 (b). 2 (c). 3 (d). 4 (e 5 ). Theground statehas quantumnum rs (1, 1), and can hold 2 e ctrons, i.e be le ., (1,1,1/2) &amp; (1,1,-1/2) Thefirst e d statehas two de ne single le xcite ge rate -e ctron state (2,1) and s: (1,2), e of which can have2 e ctrons: (2,1,1/2), (2,1,-1/2), (1,2,1/2), ach le (1,2,-1/2) 2. I f the are4 e ctrons in thewe what is thee rgy of them e rge one re le ll, ne ost ne tic (ignoring e inte -e ractions, and assum thetotal e rgy is as low as possible ing ne )? le xt xcite (a). (h2/8m 2) x 2 Theground statehas 2 e ctrons. Thene two arein thefirst e d L state thee rgy of which is: , ne (b). (h2/8m 2) x 5 L (h2/8m x (12 + 22) = 5(h2/8m L2) L2) (c). (h2/8m 2) x 10 L Bonding between atoms -- How can two neutral objects bind together? H + H H2 +e Continuum of free electron states. Let's represent the atom in space by its Coulomb potential centered on the proton (+e): r n=3 n=2 n=1 e 2 U( r ) = - r In this picture the potential energy of the two protons in an H2 molecule look something like this: +e +e r e 2 e 2 U( r ) = - - r - r1 r - r2 The energy levels for this potential are more complicated, so we consider a simpler potential that we already know a lot about. Particle in a box -- Finite square well potential (analogue to Molecular binding) Continuum of free electron states. Bound states If this were the `atomic' potential, then this would be the `molecular' potential: Again, we don't know exactly what the energy levels are, although in 1-D we could solve the equation exactly if we had to. For now we settle for a qualitative understanding... Just consider the ground state: Let's say that the lowest energy level is about equal to that of an infinite well: L = 1 nm 1.505 eV nm 2 E = 0.4 eV (2L)2 For convenience, we are going to plot the electronic wavefunction with the energy level as a baseline: A `Molecular' Wavefunctions and Energies `Atomic' Wavefunctions: L = 1 nm A 1.505 eV nm 2 E = 0.4 eV (2L)2 `Molecular' Wavefunctions: 2 `atomic' states 2 `molecular' states even odd When the wells are far apart, `atomic' functions don't overlap. The single electron can be in either well with equal probability, and E = 0.4 eV. `Molecular' Wavefunctions and Energies Wells far apart: d L = 1 nm even 1.505 eV nm 2 E = 0.4 eV (2L)2 odd 1.505 eV nm 2 E = 0.4 eV 2 (2L) (&quot;Degenerate&quot; states) Wells closer together: d even `Atomic' states are beginning to overlap and distort. 2 even and 2 odd are not the same (note center point). Energies for these two states are not equal. (The degeneracy is broken.) odd Act 3: Symmetric vs. Antisymmetric states 1. Which statehas thelowe e rgy? r ne (a). even (b). odd even d odd 2. What will happe to thee rgy of even as thetwo we com toge r (i.e as n ne lls e the ., d is re d)? duce (a). E incre s ase (b). E de ase cre s (c). E stays thesam e Act 3: Symmetric vs. Antisymmetric states - Solution 1. Which statehas thelowe e rgy? r ne (a). even (b). odd even d The&quot;curvature argum nt can behard to &quot; e odd apply (e it doe work insidethe .g., sn't barrie unle you ke p track of sign of r, ss e curvature Be r to count thenum r of ). tte be ze ro-crossings. even has none while , odd has one . 2. What will happe to thee rgy of even as thetwo we com toge r (i.e as n ne lls e the ., d is re d)? duce (a). E incre s ase (b). E de ase cre s (c). E stays thesam e Act 3: Symmetric vs. Antisymmetric states - Solution 1. Which statehas thelowe e rgy? r ne (a). even (b). odd even d The&quot;curvature argum nt can behard to &quot; e odd apply (e it doe work insidethe .g., sn't barrie unle you ke p track of sign of r, ss e curvature Be r to count thenum r of ). tte be ze ro-crossings. even has none while , odd has one . 2. What will happe to thee rgy of even as thetwo we com toge r (i.e as n ne lls e the ., d is re d)? duce (a). E incre s ase (b). E de ase cre s (c). E stays thesam e As d be e ve sm thecurvaturein even is re d, com s ry all duce re ducing thee rgy. What doe this m an for thetwo ne s e &quot;atom s&quot;? Energies as a function of distance between wells When the wells just touch (be ing onewe wecan solvefor the com ll) e rgie e ne s asily 2L = 2 nm even 1.505 eV nm 2 E1 = 0.1 eV (4L)2 (n = 1 state) odd 1.505 eV nm 2 2 E2 2 = 0.4 eV (4L)2 (n = 2 state) As thewe arebrought thee n statealways has lowe kine e rgy (sm r lls ve r tic ne alle curvature . Theodd statestays at about thesam e rgy (incre dueto large ) e ne ase r curvatureand de asedueto thinne barrie cre r r). Eodd 0.4 eV Eeven 0.1 eV odd = 0.4 0.1 eV = 0.3 eV even = splitting between even and odd states. d (mainly ground state lowering) <a href="/keyword/exclusion-principle/" >exclusion principle</a> and bonding Up to now wehaveconside d only thee rgy of thestate whe the 's a re ne s n re singlee ctron. le Eodd 0.4 eV Eeven 0.1 eV odd even d ns re ore le What happe if the is m than onee ctron? 2 e ctrons: Lowe e rgy for both in lowe state spin up and down le st ne st 3 e ctrons: Onee ctron m go in highe e rgy (odd) state le le ust r ne 4 e ctrons: Both e n and odd state are&quot;fille le ve s d&quot; &gt; 4 e ctrons: Must start filling thehighe state of thewe le r s lls Molecular Wavefunctions and Energies -- back to Coulomb Potential +e Atomic ground state: (1s) r A n=1 e 2 U( r ) = - r ( r ) e - r / a0 Molecular states: +e +e r +e +e r even odd Bonding state Antibonding state Remember: These are single-electron states. To understand the bonding, we must see how electrons fill these states following the <a href="/keyword/pauli-exclusion-principle/" ><a href="/keyword/pauli-exclusion/" >pauli exclusion</a> principle</a> Remember the particle in potential with two square wells Think of positive charges in the wells + even d + bonding anti bonding odd Look where the electron density is: The repu...