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This 33 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 795V at North Carolina State University taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/224007/ch-795v-north-carolina-state-university in Chemistry at North Carolina State University.
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Date Created: 10/15/15
Chemistry 795T Lecture 5 The Schrodinger equation for hydrogen The electronic structure of atoms Absorption spectra of atoms Manyelectron atoms NC State University The Solar Spectrum There are gaps in the solar emission called Frauenhofer lines The gaps arise from specific atoms in the sun that absorb radiation Experimental observation of hydrogen atom Hydrogen atom emission is quantized It occurs at discrete wavelengths and therefore at discrete energies The Balmer series results from four visible lines at 410 nm 434 nm 496 nm and 656 nm The relationship between these lines was shown to follow the Rydberg relation HT T Atomic spectra Atomic spectra consist of series of narrow lines Empirically it has been shown that the wavenumber of the spectral lines can be fit by 1 2 2 quot1 quot2 n2gtn1 where R is the Rydberg constant and n1 and n2 are integers Electronic SII IIGIIII B 0 Hydrogen The Schrodinger equation for hydrogen Separation of variables Radial and angular parts Hydrogen atom wavefunctions Expectation values 1 Spectroscopy of atomic hydrogen Sellriitlinger equation for hydrogen The kinetic eneruv onerator The Schrodinger equation in three dimensions is 1 2 72 5U V 5 E5 2 The operator deIsquared is V2 82 82 82 8X2 8y2 822 The procedure uses a spherical polar coordinate system Instead of x y and z the coordiantes are 6 1 and r SGIII lIiIIQBI Bllll lillll tor IIVIII IIQBII The form of the potential The Coulomb potential between the electron and the proton is V Ze247tsor The hamiltonian for both the proton and electron IS 712 2 7712 2 H 2mN VN Z meVe V 1 Separation of nuclear and electronic variables results in an electronic equation in the centerof mass coordinates H h22uV2 Ze247t801 1 u1me1mN It would be impossible to solve the equation with all three variables simultaneously Instead a procedure known as separation of variables is used The steps are 1 Substitute in LPr6l RrY6l 2 Divide both sides by RrY6l 3 Multiply both sides by 2M r The wavefunction solutions of the angular equation are spherical harmonics Yzm 94 These functions describe the angular distribution of atomic orbitals and are the wavefunctions for the rigid rotor of polyatomic molecules The degeneracy of a given orbital is 2l1 and the angular momentum of the electron is Val 1 The effective Il l llli l I BSIIII 0 IE S lllli ll 0f the angular nart The solutions for the angular part result in a term in potential energy equal to V 2ll12ur2 This term contains the contributions to the energy from angular terms Together with the Coulomb potential the effective potential energy is Veg Z6247T801 l l2ll12ur2 The radial equation f l IWIII39IIQBII Making the above approximations we have an radial hamiltonian energy operator 1 2 V2R 292 T 2p 47tsor 2er2 The solutions have the form RM NM pl e39P2 Lngr Where p 2ZrnaO and a0 47c80h2m62 NM is the normalization constant Lngr is an associated Laguerre polynomial The Bohr radius The quantity a0 47cso r3912me2 is known as the Bohr radius The Bohr radius is 210 0529 A Since it emerges from the calculation of the wave functions and energies of the hydrogen atom it is a fundamental unit In socalled atomic units the unit of length is the Bohr radius 80 1 A is approximately 2 Bohr radii You should do a dimensional unit analysis and verify that 210 has units of length The normalization constant depends on n and l 12 mi 1 M 2nn 13 The associated Laguerre polynomials are n1OL1x 1 These are given 1 n 2 0 L2 X 2K2 X for completeness n2 1L x 3 3 n3 o L x 33 3x12x2 n3 1ij 44 x 5 n3 2 Lgx Each of the radial equation solutions is a polynomial multiplying an exponential The normalization is obtained from the integral I RzRnrzdr 1 0 The volume element here is rzdr which is the r part of the spherical coordinate volume element rzsinedrdedcl A MAPLE worksheet attached to this lecture illustrates the normalization of the first three radial functions The worksheet includes plots of the functions When examining a plot keep in mind that you can plot the wave function or the square of the wave function We often plot the square of the wave function because the integral of the square of the wave function gives the probability Energy levels If IWIII IIQBII atom The energy levels of the hydrogen atom are specified by the principal quantum number n 64 E I42 21 2 12 3271 80 17 All states with the same quantum number n have the same energy g All states of negative energy are bound states states of positive energy are unbound and are part of the continuum IE Rydberg GDIISlaIII 1 The energy levels calculated using the Schrodinger equation permit calculation of the Rydberg constant One major issue is units Spectroscopists often use units of wavenumber or cm39l At first this seems odd but hv hck hcv where i3 is the value of the transition in wavenumbers 4 R1 He in cm391 70 32712802 2 All of the orbitals of a given value of n for a shell n 1 2 3 4 correspond to shells K L M N Orbitals with the same value of n and different values of l form subshells i O 1 2 correspond to subshells s p d 7 Using the quantum numbers that emerge from solution of the Schrodinger equation the subshells can be described as orbitals llvdrogenic orbitals 39 s orbitals are spherically symmetrical The 1s wavefunction decays exponentially from a maximum value Of171780312 at the nucleus g p orbitals have zero amplitude at rO and the electron possesses an angular momentum of hll 1 The orbital with m O has zero angular momentum about the z axis The angular variation is cosG which can be written as zr leading to the name pZ orbital Hydrogen 1 S radial waveiunction 20 The 13 orbital has no nodes and decays 15 exponentially g R13 21a032ep2 10 1 13 n1andi0are the quantum numbers for this orbital 05 The Radial Ilistrihution in Hydrogen 28 and 2 orbitals 06 L1 2 L112 P Exnectation values The expectation or average value of an observable ltrgt is given by the general formula ltrgt I Werr 0 As written the above integral describes the expectation value of the mean value of the radius r Integration over the angular part gives 1 because the spherical harmonics are normalized The volume element can be written dr r2dr The mean value is ltrgt 5Ur5Ur2dr 5U5Ur3dr 0 0 M PlE worksheet on Expectation Average values There is a MAPLE worksheet attached to this lecture that illustrates the use of expectation values The example of the position ltrgt is given for the 1s 2s and 3s radial wave functions The expectation value or average value of r gives the average distance of an electron from the nucleus in a particular orbital Since the 2s orbital has one radial node and the 3s orbital has two radial nodes the average distance of the electron from the nucleus is shown to increase as ltrgt for 3s gt ltrgt for 23 gt ltrgt for 1s SIIBGII OSGDIW of atomic IIVIII DQBII Spectra reported in wavenumbers 39 Rydberg fit all of the series of hydrogen spectra with a single equation w Absorption or emission of a photon of frequency v occurs in resonance with an energy change AE hv Bohr frequency condition 39 Solutions of Schrodinger equation result in further selection rules 39 A transition requires a transfer from one state with its quantum numbers n1 l1 m to another state n2 l2 m2 7 Not all transitions are possible there are selection rules Al r 1 m 0 i1 39 These rules demand conservation of angular momentum Since a photon carries an intrinsic angular momentum of 1 Manyelectron atoms The orbital annroximation The orbital approximation to the total many electron wavefunction Pr1 r2 is to rewrite it as a product of oneelectron wave functions wr1wr2 so that L1101 r2 The configuration of an atom is the list of occupied orbitals The Pauli exclusion principle states that the spins must be paired if two electrons are to occupy one orbital and no more than two electrons may occupy an orbital 4 In manyelectron atoms the s p d etc orbitals of each shell are not degenerate s An electron distance r from the nucleus experiences the nuclear charge Z shielded by all of the other electrons The effective charge is Zeff Zeff Z o where G is the shielding parameter i An s electron has a greater penetration through inner shells than a p electron of the same shell because the p electron has a node at the nucleus The Aufbau nrincinle Aufbau means buildingup in German I The configurations of atoms are built up by population of the hydrogenic orbitals Imagine a bare nucleus with charge Z and then add Z electrons to the orbitals in the following order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s etc Hund s rule an atom in its ground state adopts a configuration with the greatest number of unpaired spins l0 nrohlem 0f multinle BIBBII IIIIS The central difficulty with application of the Schrodinger equation is the presence of electronelectron interaction terms in the potential energy I J Ze2 e2 V i147580riij2147580r1j electron nuclear electron electron No analytical solutions HartreeFock procedure Find solutions that optimize the electron in each orbital in the presence of the field of all of the other orbitals quotMIMEFlick III IIGBIIIII B As an example for He we can write the two electron wave function as a product of orbitals L11r1r2 r1 r2 The probability distribution 1r2r2dr2 for electron 2 corresponds to a charge density in classical physics Therefore we can say that the effective potential felt by electron 1 is mm r2riu r2dr2 The SBIfGDIISiSlBIII field method The effective or average potential can be used in a one electron hamiltonian operator for electron 1 Aeff H1 r1 Vi Vf vo The Schrodinger equation is solved for electron 1 A eff H1 MI1 81 r1 Start with a trial function 1r2 and solve for 1r1 Using 1r1 calculate an effective potential for 2 and solve for 1r2 Continue until convergence is reached Atomic term SVIIIIIIIIS The letter indicates the total orbital angular momentum quantum number of all electrons The left superscript gives the multiplicity 281 where S s1 s2 s3 The right subscript gives the total angular momentumJLS LS 1 LS The term symbol is ZSHL J IIIIIIII39S I IIIBS llBlBl lllillB the term symbol of the l llllll state Each state is designated by a term symbol corresponds to a wave function that is an eigenfunction 0sz and 82 with unique energy The state with the larges value of S is the most stable For states with the same value of S the state with the largest value of L is the most stable If the states have the same value of L and S subshell less than half filled smallestJ is most stable subshell more than half filled the state with the largest value OH is the most stable
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