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Lecture 18 Hydrogen’s wave functions and energies (c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has been developed and made available online by work supported jointly by University of Illinois, the National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the sponsoring agencies. The energy expression The nuclear charge Z me e E=2 2 2 2 32p e 0 n 2 Discrete energies are negative 4 This explains the experiment. Note that the angular momentum quantum numbers do not enter the energy expression. Homework challenge #4 The special theory of relativity states that a mass cannot travel faster than the speed of light. By assuming that the energy of the ground-state hydrogenic atom is equal to the negative of the classical kinetic energy (cf. the virial theorem) of the electron and using the above speed limit, can we find an upper limit of the atomic number Z? Does this explain the fact that there are only 120 or so atom types in nature and not so many more? The energy levels When an electron is given an energy greater than that required to excited into the highest state, it escapes from the Coulomb force of the nucleus – ionization into an unbound, continuum state Z me e E=2 2 2 2 32p e 0 n 2 4 + H®H +e - There are an infinite number of bound states with a negative energy Atomic orbitals AO has the form: Y nlm sm = N nl Rnl (r)Ylm (f ,q )s sm l s l s With three orbital quantum numbers: n, l, ml. n = 1, 2, 3,… Principal quantum number l = 0, 1, 2,…, n–1 ml = –l, –(l–1),…, (l–1), l Also spin quantum numbers: s = ½, ms = ±½. 1 1 , 1 1 They are orthonormal 2 , 2 2 , 2 functions Shells The orbitals are classified by their principal quantum number n. n = 1, K shell. n = 2, L shell. n = 3, M shell, etc. Because energies are determined by n, the orbitals in the same shell have the same energy. Subshells For each value of n, we classify the orbitals by l. l = 0, s subshell (1 orbital). l = 1, p subshell (3 orbitals because 2l +1-fold degeneracy: ml = – 1,0,1). l = 2, d subshell (5 orbitals). Atomic orbitals Principal quantum number n 3d+2 Angular momentum quantum number l Angular momentum quantum number ml The s orbitals The higher the quantum number n, the more diffuse the orbitals are 6 2.0 1.0 1s 0.8 1.5 2s 5 3s 4 0.6 1.0 0.4 3 2 0.5 1 0.2 2 4 6 8 10 12 14 2 0 2 4 6 8 10 r / a0 12 14 0.5 r / a0 1 4 6 8 r / a0 The higher the quantum number n, the higher the energy and the more (n–1) nodes The orbital has a kink 10 12 14 Homework challenge #5 Given the fact that the electron in the hydrogen atom can exist exactly on the nucleus, why is it that the energy of the atom is not −∞ (stability of matter of the first kind)? Given the fact that the particles in a chemical system interact through two-body Coulomb forces, why is it that the energy of the system grows only asymptotically linearly with the number of particles (not quadratically as the number of particle-particle pairs does) (stability of matter of the second kind)? The p orbitals: radial part The higher the quantum number n, the more diffuse the orbitals are 1.4 5 2p 1.2 4 1.0 3p 3 0.8 2 0.6 1 0.4 5 0.2 10 15 20 25 30 1 5 10 15 20 r / a0 l æ rö Rnl (r) = N nl ç ÷ Lnl ( r )e- r /2n è nø 25 30 2 r / a0 The p orbitals are zero and kinked at the nucleus. The number of nodes is n–2. The pz orbital Angular part 3 l = 1, m = 1, Ylm = sin q eij 8p Radial part 1 æZö n = 2, l = 1, Rnl = ça ÷ 4 6è 0ø 3/2 re- r /4 l = 1, m = 0, Ylm = 3 cosq 4p 3 l = 1, m = -1, Ylm = sin q e- ij 8p The product of n = 2 radial function and l = 1 and ml = 0 angular function gives rise to r cosθ = z. It is a product of a spherical s-type function times z. x r sin cos y r sin sin z r cos The pz orbital e - r /4 r cosq = z z / a0 re- r /4 cosq z / a0 The px and py orbitals Angular part 3 l = 1, m = 1, Ylm = sin q eij 8p Radial part 1 æZö n = 2, l = 1, Rnl = ça ÷ 4 6è 0ø 3/2 re- r /4 l = 1, m = 0, Ylm = 3 cosq 4p 3 l = 1, m = -1, Ylm = sin q e- ij 8p The l = 1 and ml = ±1 angular functions do not lend themselves to such simple interpretation or visualization; they are complex. x r sin cos y r sin sin z r cos The px and py orbitals Radial part 1 æZö n = 2, l = 1, Rnl = ç ÷ 4 6 è a0 ø Angular part 3/2 re - r /4 However, we can take the linear combination of these to make them align with x and y axes. r sinq (eij + e-ij ) = 2r sinq cosj = 2x 3 l = 1, m = 1, Ylm = sin q eij 8p 3 l = 1, m = 0, Ylm = cosq 4p 3 l = 1, m = -1, Ylm = sin q e- ij 8p r sinq (eij - e-ij ) = 2ir sinq sinj = 2iy We are entitled to take any linear combination of degenerate eigenfunctions (with the same n and l) to form another eigenfunction (with the same energy and total angular momentum but no well-defined ml). x r sin cos y r sin sin z r cos The px and py orbitals The d orbitals The linear combination of d+2, d+1, d0, d–1, d–2 can give rise to dxy, dyz, dzx, dx –y , d3z –r . They are degenerate (the same n and l = 2) and have the same energy and same total angular momentum. They no longer have a well-defined ml except for ml = 0 (d3z –r ). 2 2 2 2 2 2 Size of the hydrogen atom Calculate the average radius of the hydrogen atom in the ground state (the electron is in the 1s orbital). 1 -r/a Y ( r,q ,j ) = e 3 p a0 0 ˆ dt = W = ò Y WY ò * ¥ 0 0 p 2p 0 0 ò ò n ax xe ˆ 2 sinq dr dq dj Y*WYr n! dx n 1 a The solution r = ò Y rY d t = ò r Y d t * 2 1 ¥ p 2p -2r /a0 2 = 3 ò ò ò re r sinq drdq df p a0 0 0 0 4 ¥ 3 -2r /a0 3 = 3ò re dr = a0 2 a0 0 Radial distribution functions Because the wave function is the product of radial (R) and angular (Y) parts, the probability density is also the product … 2 1 = òòò Y n,l ,m r 2 sin q dr dq df l 2 2 = òòò Rn,l (r) Yl ,m (f ,q ) r 2 sin q dr dq df l 2 = ò Rn,l (r) r dr 2 2 1 = òò Yl,m (f ,q ) sinq dq df Radial distribution function (probability of finding an electron in the shell of radius r and thickness dr) l Size of the hydrogen atom Calculate the most probable radius of the hydrogen atom in the ground state (the electron is in the 1s orbital). R(r ) = 2 a 3 0 4 e R(r ) 3 2 - r/a0 2 R(r ) r 2 2 1 0 1 2 3 r / a0 4 5 6 The solution 0= ( 2 ¶ R r2 ¶r ) = 4 æç 2re -2r /a0 a è 3 0 2r -2r /a0 ö e ÷ a0 ø 2 r = a0 Most probable point Average radius Most probable radius 4 R(r ) 3 2 2 R(r ) r 2 2 1 0 1 2 3 r / a0 4 5 6 Summary Examining the solutions of the hydrogenic Schrödinger equation, we have learned the quantum-mechanical explanations of chemistry concepts such as discrete energies of the hydrogenic atom ionization and continuum states atomic shell structures s, p, and d-type atomic orbitals atomic size and radial distribution function