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4th Set of Notes Chp. 11 and 14 Notes on Rotational Motion and Hydrogen Atom Spherical Polar Coordinates (r, goes from 0 to 2 goes from 0 to Rotational Motion (Particle on a Sphere) (Rigid Rotor) Rotor is a dumbbell with masses ma and mb at the ends, held together by a massless bar equal to the bond length “R” Solve for the energy levels of this system. Use the general method as always: 1) Write down the kinetic and potential parts of the Hamiltonian 2) Find solutions to the eigenvalue equation H= E What would the Hamiltonian be in Cartesian coordinates? Let mass ma have coordinates xa,ya,za, and mass mb have coordinates xb, yb, and zb. This proves very difficult if not impossible to solve. Not making use of the symmetry of the situation!!! Make reference to the center of mass of the system. Separate into a problem of translation and rotation which you separate out the translation and rotation parts. Transform eqn into a mass, a reduced mass , where = mamb/(ma+mb) rotating at a fixed distance from the origin of the center of mass. This is essentially then a particle on a _________. The Hamiltonian in Schrodingers eqn for the rigid rotor or particle on a sphere then becomes: two separable eqns. H = H1 + H2 H1 = -h/ 2 /2mA+mB) (.......................................................................................... H2 =-h/ 2 /2 ( 1/r2 /r (r2 /r) + 1/(r2sin) / (sin /) + 1/(r2sin2) 2/ 2) H1 is the hamiltonian for a free particle translating in space. And H2 is rotation of a particle on a sphere. But this is rigid rotor so r is ________________ and /r is equal to _____________. And R2 = I, where I is called the moment of inertia. Get: ............-h/ 2 /2 = 1/R2 [1/(sin) / (sin /) + 1/(sin2) 2/ 2)] Use this SIMPLIFIED HAMILTONIAN to find the allowed rotational energy levels via? Use trick of plugging in functions which depend only on and . ie. Plugging in, and manipulating the eqn. and get sin21/(sin) / (sin /) + 2IE/h/ 2 ] = -1/ 2/2 If this is to hold for all values of and then both sides must equal a constant. Call it 1. sin21/(sin) / (sin /) + 2IE/h/ 2 ] 2. -1/ 2/2 Looking 1st at number 2, this is a differential eqn whose solution is: () = Aexp(im), / = A i m exp(im), 22 = A i2 m2 exp(im) so -1/22-i2 m2 = m2 = Thus = m2 If the wavefunction is to be “single-valued” then m = 0,+/-1,+/-2,.... How? Consider Euler’s relationship For the other part, more manipulations with the substitution that x = cos are required, but eventually we reach a differential equation of the form: (1-x2) 2/x2 - 2x /x + (2IE/h/ 2 - /(1-x2)) = 0 ...........................(C) (1-x2) ”- 2x ’ + (l(l+1) - m2/(1-x2)) = 0 .........................................(D) “ means 2nd derivative with respect to . ‘ means 1st derivative with respect to This second equation (D) is the Associated Legendre equation. Solutions to this eqn. exist only for the conditions that l is a positive integer or zero, and that /m/ < l. Solutions are the associated Legendre Polynomials l m lm Comparing (C) and (D) above we find that our eqn for the dependence of the rigid rotor is just the same as the Associated Legendre Eqn. with the stipulation that: ........................ 2IE/h/ 2 = l (l+1) Thus the energy eigenvalues for the rigid rotor are: ..............................E = l (l+1) ............... where (l = 0,1,2,3, ...... ) ............................. (E) .................... or .... E = J (J+1) .............. where (J = 0,1,2,3, ....... ) Often m is called ml since the value of m depends on the value of l. In summary the energy eigenfunctions for the Rigid Rotor (the solutions to the wavefunctions satisfying Schrodingers Eqn) are: where and A is found by? The normalized product of and is called te Spherical Harmonics (see Table 12.3). Should be able to write out the wavefunctions (eigenfunctions) such as 2,1 for every value of ..l.. there are ___________________ values of ml. Also since E = El = ......................................................... where l = 0,1,2,3,... and I = mr2 What can be said about 1,0, 1,1, 1,-1? So the degeneracy of the lth state is? Associated with ROTATION is ANGULAR MOMENTUM. Angular momentum, J, is a vector quantity, meaning it has both direction and magnitude. Classically the magnitude of the J, is related to Erot by Erot = J2/(2I), where I is the moment of inertia or Erot = L2/2I ...........................(F) Comparing Eqns. (E) and (F) leads to the conclusion that the magnitude of the total angular momentum .................................. L2 = l(l+1)h/ 2 ............. and in fact L2 rot = l(l+1)h/ 2 rot The angular momentum vector quantity is given by J = r x p ................. where r and p are the radial direction of the particle and the linear momentum as defined in the fig. above. In general for vectors a and b, the cross product a x b is given by: The magnitude of the yth component is given by: (axbz - azby). The magnitude of the xth component is given by (aybz - byaz) So for angular momentum L = r x p. In cartesian coordinates the magnitude of the components in the x, y, and z, directions become: lx = ly = lz = In terms of quantum mechanics, this becomes Also putting it in terms of the symmetry of the problem (spherical polar coordinates) gives: lx ly lz See if the eigenfunctions to the rigid rotor will also give eigenvalues for the zth component of the angular momentum operator. Also lz = lz Find that Lz = hm / l .................................................. where ml = 0, +/-1, +/-2, ..... The zth component of the angular momentum can only take on discrete values: 2l+1values. Since ml is restricted by l. Thus the projection of L on the z axis is allowed to be only certain discrete (quantized) values). The plane of rotation of the particle can only be oriented in certain discrete ways. Quantum Mechanics says that a rotating body may not take up any arbitrary orientation with respect to some axis, only certain discrete orientations are allowed Furthermore: by uncertainty principle, if lz is known, lx and y can not be precisely known. A better model is a cone model where lx and ly are not known. So far angular momentum due to orbit of a particle. (Orbit of electron about nucleus!) Also have spin angular momentum - motion of electron about its own axis - internal angular momentum of the electron is called its ________________. Use quantum number s intead of l, and ms instead of ml, for zth component. Magnitude of spin angular momentum S is: Magnitude of the zth component of the sz is: For an electron only one value of s is allowed:......... s = ½ electron i s aspin ½ particle. Thus the magnitude of the angular momentum is: neutrons and protons are also spin ½ particles, more massive, but spin slower. ½ integral spin particles are called ______________ integral spin particles are called ________________ Matter is an assembly of fermions held together by forces conveyed by bosons. Elementary particls are all fermions. Electronic Energy: Ideal Model is the Hydrogen Atom Hydrogenic Atoms - Atoms which have a single electron H, He+, Li2+, etc. Spectra of H emission revealed line spectra Lyman Balmer Paschen We can solve Schrodingers equation for this case if we consider the hydrogen atom as a system of two interacting point particles. What is potential? Coulombic attraction? Charge of nucleus is +Ze, Charge of electron is -e Z is atomic number So potential energy is: V = -Ze2/r or V = -Ze2 /(4or) (in SI units) r = distance between the electron and the nucleus Write Schrodingers Eqn. Cartesian x1, y1, z1 coordinates of nucleus, m1 is the mass of the nucleus ............... x2, y2, z2 coordinates of electrons -h/2/2m1 (....................................................) + -h/2/2m2 (......................................................) + V ET V= -ze2/ [(x2-x1)2 + (y2-y1)2 + (z2-z1)2]1/2 Since interested in motion of the electron about the nucleus, try to separate this into two eqns., one representing the translational motion of the molecule as a whole, ......and the other - involves the relative motion of the nucleus and the electron Introduce new variables x, y, z, which are the Cartesian coordinates of the center of mass system, and r, and , and the polar coordinates of the relative motions. Then for the relative motion coordinates x = x2 - x1 = r sincos y = y2 - y1 = r sinsin z= Have two parts which depend only on x, y, and z and another which depends on r, and Each part must be equal to a constant or in this case zero. Resulting Eqns are: ................................................................................................. = 0 ...........................(A) and ............................................................................................. = 0 ................................... (B) Eqn A is the eqn for the motion of a ______________ ____________________, the translation of the ______________ _____________ in space. Eqn,. B is identical with the wave eqn of a single partile of mass under the influence of a potential function V(r,Actually, the potential depends only on ______ as it is centrosymmetric. Try separating Eqn. B Further: ...................(r,) = R(r)() Introducing this into eqn. B and dividing by Rmultiplying by r2sin gives: sin2R d/dr(r2dR/dr) + 1/d2/d2 + sin/ d/d(sin d/d) + 2/h/2 (E-V(r)) r2sin2 =0 Note that this is starting to separate into three parts: one depending only on r, another depending only on , and one on . Now there is a part depending only on so as before: 1/2 d2/d2 = must be equal to a constant, and looks like RIGID ROTOR so = -ml2 substituting in this -ml2 for the dependence and dividing by sin2 gives: 1/R d/dr (.................................................................................................................................. ................................................ = 0. Now the 2nd and third parts of this equation are independent of r and the remaining part is independent of so again equate each to a constant. Set part = - and r part = +. Also then multiplying the part by and the r part by R/r2 then it all falls out wonderfully as: 1/sin d/d(sin d/d) - ml/sin2 + = 0 ..................................... (C) 1/r2 d/dr(r2dR/dr) - /r2R + 2h/2(E-V(r))R = 0 .................................... (D) Solved the first part before Found that = l (l+l) and was the Associated Legendre Polynomial. Also Remember that the dependence was = (1/2)½ exp(iml) and the product Yl ml which is known as the _______________ ________________ which are listed in TABLE ______. restrictions are: l = 0, 1, 2, … ml = -l, -l+1, …..., +l So just have the Solution of the R Eqn. to contend with. Note we haven’t put in V(r) yet so this is applicable to any system of two particles which interact with one another in a way expressible by V(r). Put in V(r) = -Ze2/r (in cgs) or -Ze2/(40 r) (in SI) Consider E to be negative - bound states; the total energy is insufficient to ionize Now the R eqn. is easier to solve with the following substitutions: a2 = - 2m/h/2 E,...... l = m Ze2/(h/2 ), ...... = 2r, .......... and S() = R(r) then the wave eqn. becomes: 1/2 d/d(2 dS/d) + [-1/4 - l(l+ 1)/2 + ] S = 0 ...........0 < < infinity Recognize that this is differential eqn. whose solution are the Associated Laguerre polynomials. Rn l is the Associated ______________ _______________ Note that the functions depend on the value of the quantum numbers n and l. The quantum number n has the restriction such that n = 1, 2, 3, 4, ........... And the quantum number l has the restriction such that l = 0, 1, 2, .........n-1. NOTE that the value of l is restricted by the value of n. This is how the Associated Laguerre Polynomials work. The total wavefunction then is: n,l,ml = Rn l Yl m Also remember that ml is restricted by the values of l such that ml = -l, ..., -1, 0, 1, .....+l and there are 2l + l values of ml for every value of l. Whydrogen are the energy eigenfunctions and are the ONE ELECTRON OR HYDROGENIC ATOMIC ORBITALS An orbital is a one electron eigenfunction. Why is it called one-electron? Atomic Orbital - one electron energy wavefunction describing the distribution of an electron in an atom. How does it describe the distribution? n l m = Rn l Yl m is the form of all hydrogenic wavefunctions. The three quantum #s n, l, and ml define which hydrogenic orbital or eigenstate the electron is occupying. l and ml came from the _____________ solution. Remember l describes the magnitude of the angular momentum by L = l l + 1)½ h/; so the orbital angular momentum of the electron is given by this. Also the zth component of the angular momentum of the electron is given by: lz = ml h/ What about the ENEGY OF THE HYDROGEN ATOM? The energy eigenvalues are determined in the solution of the differential eqn. above and the energy is: En = -hcR/n2; ............... hCR = 2e4/(32202 h/2) (in SI) **** Note that the energy depends only on the quantum number n: n = 1,2,3, ......... for hydrogenic atoms. Orbital Energy Diagram (depends only on n) Summarizing then: The eigenfucntions (eigenstates, one electron atomic orbitals) are defined by the quantum #s n - principal quantum # - tells about the size and energy of the electron in the atom ......... allowed values of n are n = l - azimuthal or angular momentum quantum # -tells about the ____________of the __________ _________. ..........allowed values of l are l = ml - zth component of the ______________ __________ or also called the magnetic quantum #; ...... since it tells about the magnitude of the ____ _________ of the ___________ __________. ...... allowed values are ml = n, l, ml is the product of angular spherical harmonics and radial associated Laguerre Polymials. The electron also has an intrinsic momentum, and INTRINSIC angular momentum - spin angular momentum. Full specification of the electronic state of a hydrogenic atom then requires the use of another .........quantum #. Since the spin angular momentum quantum number s is fixed at ½, use ms .........instead. ms - specifies the orientation of the spin angular momentum ....... Allowed values for ms are ms = SO n l ml ms is the Definitions: 1) All orbitals of a given n are said to be in the same _________ or ________. Those orbitals in the same shell have the same _______________ and similar ___________. n = 1, 2, 3, ... ......K, L, M, N, ...... 2) All atomic orbitals with the same value of n, but different l values are said to be in different _____________ or _____________ of a given shell. l = 0, 1, 2, 3, 4, ......... n-1 ......s, p, d, f, g, Letters run alphabetically with the omission of j. For n =2, l = 1, or 21 this is called the __________ subshell or sublevel. 3) ml = -l, ........, 0, ............ +l For the case above, 21, the allowed values of ml are ? ........................ and 21 turns into: when all three of the values of ml are considered. Thus when ml is included we are looking at the 2p orbitals. An electron occupying the 2p orbital is called a 2p electron. Since l can go from 0 to n-1, there are ____ subshells of a shell. # of orbitals is 2l + 1 so for n = 1 l = _____ and ml = ________________ so there are ____ orbitals. Example: Case when n = 3? Overall, the number of orbitals in a shell with principal quantum number n is ______. Thus in a hydrogenic atom each shell is n2 __________________. Remember we also have spin angular momentum in addition to the orbital angular momentum Talked about this before. It is the motion of the electron about its own axis - __________ angular momentum Use quantum number s instead of l, and ms instead of ml, for the zth component. Magnitude of spin angular momentum, S, is: and the magnitude of the zth component, sz is: for a single electron only one value of s is allowed s=1/2 and ms = ? In each orbital can only have at most 2 __, one with ms = ½ and other with ms = -1/2 ENERGY DIAGRAM The orbital occupied in the ground state is ______________________ or _____________________________ which is the 1s level or state. Writing out the wavefunction Table Since wavefunction is largest at r=0, expect probabiltiy to be largest there as well! Decays exponentially form maximum at r=0. The ground state of the hydrogen atom n=1, ml = 0 has the electron closest to the nucleus. The successive excited states where n=2, n=3, ……. show the electron to be spread out over more space away from the nucleus. . Way to represent the probability density of electron is 2 shown by density of shading generated by computer. Fig The probability of finding the electron at a distance r from the nucleus is proportional to the volume of a spherical shell of radius r around the nucleus. The volume is directly proportional to r2 . THUS the probability of finding the electron at a distance r is proportional to r2 2 . Plots of this versus r/ao are given where ao is called the Bohr radius (radius of Bohr’s n=1 orbital = 0.0529A. Fig A second way is a boundary surface. Surface whose shape and size represent capturing about 90% of the electron probability. For 1s, the surface is a ______________ Fig Example: For the 1s orbital of the hydrogen atom, calculate the probability of finding the electron within a distance ao from the nucleus. Radial nodes # of radial nodes = # of places where the radial wavefunction passes through zero or where the probability is equal to zero. How many radial nodes are there??? See figure 1s has _________ nodes. 2s has ________ nodes. etc. These can be solved for by setting the radial part of the wavefunction to zero. R10 The wavefunction or more precisely, *d gives the probability of finding an electron in any region. For 1s * is proportional to exp(-2r/a0) Also talk about probability of finding electron on spherical shell of thickness dr ar radius r. Interested in spherical shell of volume 4r2dr which is d so the Probability = 4r2 2 dr Since r2 increases with radius from zero at the nucleus and 2 decreases to 0 at infinity. P on the other hand is zero both at _______________ and at________________. Pmax (how do you get it) gives the most probable radius. So far only considered S one-electron atomic orbitals p orbitals are different p has nonzero angular momentum L = l(l+1)½ h/ l is equal to______ for electron occupying orbitals in p subshell. All orbitals with l > 0 have zero amplitude at the nucleus so no probability of finding electron there. Fact that electron in p subshells have nonzero angular momentum changes the shape! Classically there is a centrifugal force that flings the electrons away from the nucleus. 2p subshell There are three 2p orbitals ml distinguishes Angular Nodes x,y plane is a nodal plane of the orbital The 2pz orbital is 2 1 0 2px, and 2py orbitals are real linear combinations of the _____ and ________ d subshell and d orbitals n = 3 .......... l is 0,1,or ____ In the shell there is one 3s, three 3ps, and five 3d orbitals have Spectroscopic Transitions For a transition - electron changes its eigenstate or orbital. Goes from state with quantums ni, li, ml .... to a state with nf lf, ml E = Enf - Eni = Ephoton = h = h c/ Can electron move to any old atomic orbital? Angular momentum must be conserved. Since energy given by a photon or taken away by a photon. The angular momentum must change to compensate. Go from l=2 to l=0 subshells? What is the angular momentum of a photon? Give Selection Rules: Grotrian Diagram Many Electron Atoms For the hydrogenic atoms there was one electron surrounding the nucleus with an effective charge Z. For many-electron atoms must consider not only the interaction between the nucleus and electrons, but also the interaction of the electrons with themselves What would Schrodinger’s equation look like H= -h/2/2me where ri is the distance for the ith electron from the nucleus ri j is the distance between the ith and jth electrons Z is the atomic number .... and ...... me is the mass of the electron Consider He, Li+ Be2+ etc. Z = 2,3,4 etc H If we omit the term which is the interaction between the electrons (ri j) terms, this eqn. is separable into N three-dimensional equations (one for each electron) (CALLED THE ORBITAL APPROX) To this degree of approx. the wavefunction for the atom may be built up out of singleelectron wavefunctions - That is a solution to the equation for the many electron atom with Omission of electron interaction terms, And Z -> Zeff where the individual orbitals resemble the hydrogenic orbitals, but with the nuclear charge modified to account for the “screening effect”? Within the orbital approx. the electronic structure can be expressed by reporting its atomic electronic configuration which is a list of the occupied hydrogenic atomic orbitals. Remember in General Chemistry we did this. Used Aufbau Principle, or Building-up Principle (Add electrons in succession into atomic hydrogenic orbitals) Need also the Pauli Exclusion Principle No more than two electrons may occupy any given orbital, and if two do occupy one orbital, their spins must be paired. No spin angular momentum when the spin is paired. The spin of one electron cancels the spin of the other. Also another way of stating Pauli, no two electrons in an atom may have the same ________ __________ ____________. For the hydrogen atom then - with one electron- that electron goes into the 1s orbital n = 1 l = ............., ml Atomic Electronic Configuration is: 1s1 What about Li? In building up - build up based on energy of electronic orbital? Do electrons in a 2p and 2s have the same energy? Shielding Effects Many-electron atoms experience Coulombic repulsion forma all the other electrons present. Other electrons “effectively” reduce the attraction of the outer electrons to the nucleus. Zeff = Z - where is the shielding constant Zeff is different for p and s electrons! Why? Figure “s” electrons have higher probability of being closer to the nucleus. - s experiences less shielding since it penetrates further. Principal Quantum # muost important generally! Combining penetration and shielding shows that energies of subshells with same principal quantum #s is s,p,d Many Electron vs Hydrogenic Consider Li again (Building up) Aufbau principle says that the order of occupation is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 6p etc. Get this order from the periodic table - where are the main exceptions? Abbreviated Notation [Noble Gas Symbol] rest of config Example: Write the Atomic Electronic Configs for the following many-electron atoms C P Fe Cr Cr2+ Ag O2Valence Electrons come off lst when forming cation. Valence electrons in the outermost shell of an atom int its ground state - responsible for chemical bonds. Core Electrons - non valence electrons - not as reactive. What about electron orbitals? Electronic configuration gives an idea about the arrangement of electrons in? Atomic electronic orbital diagram gives more information! There is more information as the arrangement of the electrons in their orbitals - Again based on Energy! Hunds Rule: An atom in its ground state (lowest energy state) adopts a configuration with the greatest # of unpaired electrons. Spin Correlation - electrons with same spin stay well apart from each other - SO less repulsion Repulsive term is smaller, it makes positeve contribution to the electronic energy - so energy is more negative when spins are parallel. Consider atomic orbital diagram for Carbon .... C What about N Periodicity 1) Size Size increase going up or down a group (family, column) size decrease going left to right across the period (row) 2) Ionization Energy lst Ionization Energy - energy required to remove and electron Na -> Na+ + 1eTrend See Fig 9.21 Overhead (kink in curve at Be and N Why? 2nd Ionization energy? Does this orbital approximation work well enough for all predictions? Can we get better approximations to the true wavefunctions - orbitals? Well remember the problem is the term V= ri j is the separation of electrons i j this term is difficult to deal with (can’t deal with it analytically) must use approcimations Hartree Fock Procedure If we know what the rough structure of the atom is, (Use orbital approx as the start - that is approximate by hydrogenic orbitals) then a Schrodinger eqn can be written for this electron by ascribing a potenetial energy to it arises from the nuclear attraction and now the average repulsion from other electrons in their approximate orbitals. For instance take AluminumL orbital approximation suggests configuration of with hydrogenic atomic orbitals consider 3p .......... Solve for 3p in approx Schrodinger with numerical methods where Vee depends on the wavefunction of the other electrons and is the average term. Once 3p is solved move on to 3s now with Vee modified with the new 3p This is solved giving a new value for 3s. Move on to 2p, 2s etc. and iterate ............ These solutions are said to be self-consistent. The Electron-Electron interaction causes several terms (with different energies) to arise from a given atomic electronic interaction. This means that the Atomic Electronic Configurations lack some information. So in addition to Atomic Electronic Configurations, USE TERM SYMBOLS 1) Electron - Electron Interaction .... .A) Configuration interaction - spins of electrons may interact and cause splittings ..... B) Spin Orbit Coupling (interaction) between the spin and orbital angular momenta of an .......... electron. ... ......... ml & ms values tell about the relative orientation of individual angular momenta w respect to each other. Different Orientations of the spin and orbital angular momenta correspond to different couplings and different overall energies