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Part 1. the Electronic Structure of Atoms Hydrogen-Like Atoms

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Part 1. the Electronic Structure of Atoms Hydrogen-Like Atoms Part 1. The Electronic Structure of Atoms The wave function of an atom contains all the information about the properties of the atom. The wave functions for the hydrogen atom and other one-electron atoms (such as He+ and Li2+) can be calculated exactly by solving Schrödinger’s equation. Approximate methods must be used to calculate the wave functions for atoms containing two or more electrons. These numerical calculations can be very accurate, but require complicated computer calculations. So we start with (brief review from Chem 331): Hydrogen-Like Atoms What are they? He+ (Z = 2) , Li2+ (Z = 3), Be3+ (Z = 4), etc. Ions with one nucleus and one electron (two bound particles), atomic number Z > 1. Also a few rare, but interesting, “exotics”, such as muonium (more on that in the tutorial!). Why are they important? Schrödinger’s equation can be solved exactly for hydrogen-like atoms, using the reduced mass concept to simplify the two-body problem to an effective one-body problem. This greatly expands the number of chemical systems that can be treated analytically by quantum mechanical principles. (For molecules and multi-electron atoms, cumbersome and less accurate numerical methods must be used to solve Schrödinger’s equation.) For hydrogen-like atoms, the electrostatic potential energy and the reduced mass in Schrödinger’s equation for the hydrogen atom e2 V (r) = − 40r m m = e p me + mp are changed to (notice the factor of Z) Ze 2 V (r) = − 40r m m = e N me + mN me is the electron mass and mN is the nucleus mass. Schrodinger Equation for Hydrogen-Like Atoms H (r,,) = E (r,,) with Hamiltonian operator h2 Ze 2 H = − 2 − 2 8 40r The Laplacian operator 2 can be separated into radial and angle-dependent terms ˆ2 2 1 2 1 L = r − r 2 r r r 2 2 The angle-dependent terms can be represented by the operator Lˆ2 for the square of the angular momentum. ˆ2 ˆ2 ˆ2 ˆ2 L = Lx + Ly + Lz 1 1 2 = 2 sin + 2 2 sin sin Partial separation of variables leads to solutions of Schrödinger’s equation of the form nm (r,,) = Rn (r)Ym (,) Rnℓ(r) is called the hydrogen-like radial wave function and Yℓm(θ,) are spherical harmonics. As for the hydrogen atom, principal quantum number n = 1, 2, 3, 4, … orbital angular momentum quantum number ℓ = 0, 1, 2, …, n − 1 (ℓ < n) magnetic quantum number m = 0, ±1, ±2, … , ± ℓ A Few Solutions to Schrödinger’s Equation for Hydrogen-Like Atoms: 2 2 (the Bohr radius is a0 = ε0h /πμe ) 3/ 2 1 Z n = 1, ℓ = 0, m = 0 100 = exp(−Zr / a0 ) a0 3/ 2 1 Z Zr = 2 − exp(−Zr / 2a ) n = 2, ℓ = 0, m = 0 200 0 32 a0 a0 3/ 2 1 Z Zr = cos exp(−Zr / 2a ) n = 2, ℓ = 1, m = 0 210 0 32 a0 a0 3/ 2 1 Z Zr = sin exp(−Zr / 2a )exp(i) n = 2, ℓ = 1, m = ±1 211 0 64 a0 a0 Okay, but how can a chemist or physicist use these wave functions? 1. Energies of Hydrogen-Like Atoms For an isolated hydrogen atom (no externally applied magnetic fields or other fields), the energy depends only on the principal quantum number n. The same result obtained for hydrogen-like atoms: e4Z 2 En = − 2 2 2 (no externally applied fields) 8 0 h n Why is the energy negative and proportional to Z2? The energy equation can be simplified by lumping the constants together to form a single parameter. If the energy is expressed in cm−1, the unit favored by spectroscopists, this factor called the Rydberg constant. Photon energies hv are inversely proportional to wavelengths hv = hc/ so energies are converted from units of Joules to wavenumbers by dividing by hc (and don’t forget to convert from m−1 to cm−1!) ~ E e4 Z 2 E = n = − n 2 3 2 hc 8 0 h cn ~ Using energies En in wavenumbers defined by ~ Z 2 E = − n n2 the Rydberg constant (in wavenumbers) is e4 = 2 3 8 0 ch The Rydberg constant depends on fundamental physical constants and the mass of the nucleus. 5 −1 For the hydrogen atom, H = 1.096 775 856 10 cm 5 −1 For the deuterium atom, D = 1.097 074 275 10 cm 5 −1 In the limit me/mN → , = 1.097 373 153 10 cm Notice the extraordinary precision of Rydberg constants. This is possible because frequencies can be measured very precisely and accurately. 2. Ionization Energies: AZ-1 → AZ + e– The ionization energy of a hydrogen-like atom is defined as the energy required to take the atom from its ground state (ni = 1) to the dissociated state nf = where the electron and nucleus are infinitely far apart. In wavenumbers: 2 2 ~ ~ ~ Z Z 2 E = E − E = − − − = Z IONIZATION f i 2 12 ~ -1 For the hydrogen atom (Z = 1) EIONIZATION = H = 10967.75 856 cm = 2.178722 10−19 J = 13.605698 eV For hydrogen-like atoms, ignoring small changes in the reduces mass, the ionization energies are ~ 2 2 EIONIZATION Z H = 13.606Z eV 13.606 22 = 54.424 eV for He+, 13.606 32 = 122.45 eV for Li++, etc. 3. Electronic Spectra of Hydrogen-Like Atoms The selection rules for transitions between atomic energy levels are Δn unrestricted Δℓ = ±1 Δm = ±1, 0 Transitions from any initial ni value to any final nf are therefore “allowed”. For emission spectra: nf < ni For absorption spectra, nf > ni The photon energy ΔE = hv = Ef − Ei in wavenumbers is ~ ~ 1 1 E − E = Z 2 − f i n 2 n 2 i f Hydrogen Atom Spectra (in Wavenumbers) Electronic Transitions for Hydrogen Atoms (in nm) 4. Electron Density – Important for Atomic Structure For a hydrogen-like atom, the probability of finding the electron in the volume element dτ = dxdydz is (nℓm)* nℓm dτ. (Recall that the asterisk denotes the complex conjugate of nℓm, where all values of i in nℓm are converted to −i.) Transforming to more-convenient spherical r, θ, coordinates with dτ = r2sinθdrdθd gives 2 electron density = r sinθnℓm(r,θ,)* nℓm(r,θ,) Finding the maximum value of the electron density gives the most probable locations of the electron. The locations of zero electron density are called nodes. Nodes are required to make the wave functions orthogonal: ( nm ) * nm d = 0 if n n or or m m Electron Densities for n = 1, 2, 3 and ℓ = 0 (s), 1 (p), 2(d) (graphical display) Stereo plots of probability electron probability densities See: D. T. Cromer, Stereo plots of hydrogen-like electron densities, Journal of Chemical Education, 45, 626−631 (1968) Online Views of Hydrogen Orbitals www.falstad.com/qmatom • shows real wave functions (used by chemists) • also complex wave functions • phase information can be color coded • views can be rotated in three dimensions • radial probability distribution can be plotted • time evolution also simulated Plotting probability distributions How? Monte Carlo Method (“Algorithm”) Compute values of the probability distribution function (pdf) at a large number of randomly selected points in space. Plot points in proportion to the value of the pdf. Step 1 Calculate the maximum value of the probability distribution function: (*nlmnlm)max Step 2 Use a random number generator (0 < RAND < 1) to calculate random values of r, , on the intervals 0 to rmax, 0 to , 0 to 2 r = RAND rmax = RAND = RAND 2 The radial extent rmax is chosen so that *nlmnlm 0 at r = rmax. Step 3 At the randomly selected point r, , calculate *(r,,) (r,,) p = [ *(r,,) (r,,)]max Step 4 Calculate a fourth random number q = RAND between 0 and 1. Plot the point at r, , if p > q. repeat (go to Step 2) … Example: Probability Density Plot for the Hydrogen 2s Orbital hydrogen 2s orbital: n = 2 l = 0 m = 0 1 1 r (r) = 2 − e−r / 2a0 200 3/ 2 4 2 a0 a0 maximum value at r = 0: 1 1 [ 200 (r)]max = 200 (0) = 3/ 2 2 4 2 a0 At randomly selected point r, , in three-dimensions, calculate 2 r −r / a0 2 − e * 200 (r) 200(r) a0 p = * = [ 200 (r) 200(r)]max 4 Plot a point at r, , if p > random number q between 0 and 1. Repeat many times: at r = 0.223567a0, calculate p = 0.630876 and q = 0.389954. Plot this point. at r = 2.07378, calculate p = 0.000684 and q = 0.440932. Do not plot this point. Etc. 20 “Monte10 Carlo” Probability Density Plots 0 a 0 / 1s 100*(r)100(r) z -10 20 -20 10 200*(r)200(r) -20 -10 0 10 20 x / a0 0 a 0 2s view along y / axis z -10 10,000 points 200(r) > 0 red -20 200(r) < 0 blue -20 -10 0 10 20 x / a0 20 10 0 a 3s 300*(r)300(r) 0 / z -10 -20 -20 -10 0 10 20 x / a0 Other “Monte Carlo” Applications 1. Quantum Computing Conventional “silicon-based” computers use strings of binary bits 01100101011101000 … (two possible values, 0 or 1) A string of n binary bits exists as one of 2n possible values. But quantum computer bits (qbits) are superpositions of 0 and 1. A qbit can take an infinite number of possible values, making quantum computing, in theory, very fast and very powerful. Still under development. 2. Monte Carlo Error Analysis Use random numbers to simulate random experimental errors. Important for the design of experiments.
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