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CH107 Special Topics Part A: The Bohr Model of the Hydrogen Atom (First steps in Quantization of the Atom) Part B: Waves and Wave Equations (the Electron as a Wave Form) Part C: Particle (such as an Electron) in a Box (Square Well) and Similar Situations 1 A. The Bohr Model of the Hydrogen Atom n3 n2 n1 n3 Photon 2 Setting a Goal for Part A • You will learn how Bohr imposed Planck’s hypothesis on a classical description of the Rutherford model of the hydrogen atom and used this model to explain the emission and absorption spectra of hydrogen (Z = 1) and other single electron atomic species (Z = 2, 3, etc). 3 Objective for Part A • Describe how Bohr applied the quantum hypothesis of Planck and classical physics to build a model for the hydrogen atom (and other single electron species) and how this can be used to explain and predict atomic line spectra. 4 Setting the Scene At the turn of the 19th/20th centuries, classical physics (Newtonian mechanics and Maxwellian wave theory) were unable explain a number of observations relating to atomic phenomena: • Line spectra of atoms (absorption or emission) • Black body radiation • The photoelectric effect • The stability of the Rutherford (nuclear) atom 5 The Beginnings of Bohr’s Model In 1913, Niels Bohr, a student of Ernest Rutherford, put forward the idea of superimposing the quantum principal of Max Planck (1901) on the nuclear model of the H atom. He believed that exchange of energy in quanta (E = hn) could explain the lines in the absorption and emission spectra of H and other elements. 6 Basic Assumptions (Postulates) of Bohr’s Original Model of the H Atom 1. 2. 3. 4. The electron moves in a circular path around the nucleus. The energy of the electron can assume only certain quantized values. Only orbits of angular momentum equal to integral multiples of h/2p are allowed (meru = nħ, which is equation (1); n = 1, 2, 3, …) The atom can absorb or emit electromagnetic radiation (hn) only when the electron transfers between stable orbits. 7 First Steps: Balance of Forces The centripetal force of circular motion balances the electrostatic force of attraction: e- (me) r + (Ze+) u meu2 Ze2 = r 4p r 2 0 or 2 meu = Ze2 4p 0 r (2) 0 is known as the permitivity of vacuum 8 Determination of Potential Energy (PE or V) and Kinetic Energy (KE) The PE arises from the attractive electrostatic force between the electron and the nucleus. In general, V(r) = _ r F(r) dr 0 _ 2 Ze For the Bohr H atom, V(r) = 4p0 r The KE exists by virtue of the electron's (circular) motion. 1 meu2 KE = 2 9 Determination of Total Energy and Calculation of u 2 Hence the total energy is Ze 1 2 _ meu 2 4p0 r 2 Ze 1 _ m u2 (since = = e 2 4p0 r meu2 ) (3) Imposition of the quantum restriction meru = nh (1) and substituting for r in equation (2) gives 2 Ze u = 2h0n (4) 10 Determination of Total Energy Substitution of the value of u from equation (4) into equation (2) gives En = _ e4me Z2 802h2 n2 (5) This is the Bohr equation for quantized electron energy levels: the integer n, introduced in equation (1) with the quantum hypothesis, defines the energy for each stable orbit, known as energy levels. 11 Determination of Total Energy Continued The expression inside the brackets of equation (5) is a constant, equivalent to 2.18 x 10-18 J. This is a convenient unit of electronic energy and is called the rydber g. Equation (5) thus reduces to Z2 (rydberg) _ En = n2 (6) where n = 1, 2, 3,... and for H, Z = 1 12 Energy Level Diagram for the Bohr H Atom n En 0.00 J _ 1 rydberg ) 16 _ ( 1 rydberg ) 9 4 -1.36 x 10-19 J ( 3 -2.42 x 10-19 J 2 -5.45 x 10-19 J ( 1 _ -2.18 x 10-18 J ( 1.00 rydberg) _ 1 rydberg ) 4 13 Determination of Orbit Radius Elimination of u between equations (1) and (4), and solving for r, gives rn = 0h2 pmee2 n2 (6) Z This equation describes the radii of the allowed orbits corresponding to different values of n, and with energies given by equation (5). As before, Z = 1 for the H atom and n = 1, 2, 3,... 14 Determination of Orbit Radius Continued All the parameters in equation (6), apart from Z and n2, are constants, which is equivalent to 5.29 x 10-11 m (0.529 A). This is known as the Bohr r adius (a0), and is a convenient measure of electronic distance in atoms. Thus equation (6) reduces to n2a rn = 0 (7) Z 15 Electronic Radius Diagram of Bohr H Atom Each electronic radius corresponds to an energy level with the same quantum number r 3 =9a0 r 1 = a0 n =3 r 2 =4a0 n =2 n =1 r 4 =16a0 n =4 16 The Line Spectra of Hydrogen The major triumph of the Bohr model of the H atom was its ability to explain and predict the wavelength of the lines in the absorption and emission spectra of H, for the first time. Bohr postulated that the H spectra are obtained from transitions of the electron between stable energy levels, by absorbing or emitting a quantum of radiation. 17 Electronic Transitions and Spectra in the Bohr H atom Electron relaxation n3 n2 Emission n1 n3 Electron promotion Photon Absorption 18 Qualitative Explanation of Emission Spectra of H in Different Regions of the Electromagnetic Spectrum 7 6 5 4 Brackett series (mid infrared) Energy 3 2 Paschen series (near infrared) Balmer series (visible) n = 1 Lyman series (ultraviolet) 19 Quantitative Explanation of Line Spectra of Hydrogen and Hydrogen-Like Species When H or H-like species (a one-electron ion) undergoes a transition between two energy _ hn levels Ei and Ef, E = + ni E nf hn nf Emission Z2e4me _ Z2e4me hn = 802h2 nf 2 802h2 ni 2 ni > nf hn = hn ni Absorption Z2e4me _ Z2e4me 802h2 ni 2 802h2 nf 2 nf > ni 20 …….Continued From the previous slide, n = Z2e4me 2 3 80 h n = Z2e4me 802h3 1 _ 1 (Emission) nf 2 ni 2 (8) 1 _ 1 (Absorption) (9) 2 2 ni nf 4 The constant e me is equivalent to Rydberg's 802h3 empirical constant (3.29 x 1015 s-1), hence 1 _ 1 (Emission) (10) n = (3.29 x 1015 s-1) Z2 2 2 nf ni n = (3.29 x 1015 s-1) Z2 1 _ 1 2 2 (Absorption) (11) ni nf 21 Conclusion For the H atom (Z = 1), the predicted emission spectrum associated with nf = 1 corresponds to the Lyman series of lines in the ultraviolet region. That associated with nf = 2 corresponds to the Balmer series in the visible region, and so on. Likewise, the lines in all the absorption spectra can be predicted by Bohr’s equations. 22 Aftermath – the Successes and Failures of Bohr’s Model • For the first time, Bohr was able to give a theoretical explanation of the stability of the Rutherford H atom, and of the line spectra of hydrogen and other single electron species (e.g. He+, Li2+, etc). • However, Bohr’s theory failed totally with two-and many-electron atoms, even after several drastic modifications. Also, the imposition of quantization on an otherwise classical description was uneasy. • Clearly a new theory was needed! 23 B. Waves and Wave Equations 24 Setting a Goal for Part B • You will learn how to express equations for wave motions in both sine/cosine terms and second order derivative terms. • You will learn how de Broglie’s matter wave hypothesis can be incorporated into a wave equation to give Schrödinger-type equations. • You will learn qualitatively how the Schrödinger equation can be solved for the H atom and what the solutions mean. 25 Objective for Part B •Describe wave forms in general and matter (or particle) waves in particular, and how the Schrödinger equation for a 1dimensional particle can be constructed. •Describe how the Schrödinger equation can be applied to the H atom, and the meaning of the sensible solutions to this equation. 26 The Basic Wave Form Wave amplitude y(x) Basic wave properties c = n (1) A = amplitude = wavelength = n = frequency c = velocity = wavenumber E = hn = hc/ = hc (2) 27 Basic Wave Equations For a travelling wave, _ 2pn t (3) p 2 x y (x ) = Asin For a standing wave, p 2 x sin ( ) A yx = (4) 28 Characteristics of a Travelling Wave 29 Characteristics of a Standing Wave 30 Wave Form Related to Vibration and Circular Motion y x Looking along x axis The wave form appears as a vibration or oscillation This can be resolved into two opposing circular motions 31 The Dual Nature of Matter – de Broglie’s Matter Waves • We have seen that Bohr’s model of the H atom could not be used on multi-electron atoms. Also, the theory was an uncomfortable mixture of classical and modern ideas. • These (and other) problems forced scientists to look for alternative theories. • The most important of the new theories was that of Louis de Broglie, who suggested all matter had wave-like character. 32 de Broglie’s Matter Waves • De Broglie suggested that the wave-like character of matter could be expressed by the equation (5), for any object of mass m, moving with velocity v. = h (5) mv • Since kinetic energy (Ek = 1/2mv2) can be written as 2 Ek = (mv) 2m • De Broglie’s matter wave expression can thus be written h = (6) 2mEk 33 de Broglie’s Matter Waves, Continued • Since h is very small, the de Broglie wavelength will be too small to measure for high mass, fast objects, but not for very light objects. Thus the wave character is significant only for atomic particles such as electrons, neutrons and protons. • De Broglie’s equation (5) can be derived from (1) equations representing the energy of photons (from Einstein – E = mc2 – and Planck – E = hc/) and also (2) equations representing the electron in the Bohr H atom as a standing wave (mevr = nh/2p; n = 2pr) 34 The Electron in an Atom as a Standing Wave An important suggestion of de Broglie was that the electron in the Bohr H atom could be considered as a circular standing wave 35 Differential Form of Wave Equations • Consider a one-dimensional standing wave. If we suppose that the value y(x) of the wave form at any point x to be the wave function y(x), then we have, according to equation (4) y(x) = A sin 2px or B cos 2px (7) • Of particular interest is the curvature of the wave function; the way that the gradient of the gradient of the plot of y versus x varies. This is the second derivative of y with respect to x. 36 Differential Form of Wave Equations, Continued Thus 2 d2y(x) = _ A 2p sin 2px dx2 2y d (x) _ or = dx2 (8) 2p 2 y (x) (9) Equation (9) is a second order differential equation whose solutions are of the form given by equation (7). 37 Differential Wave Equation for a OneDimensional de Broglie Particle Wave We now consider the differential wave equation for a one-dimensional particle with both kinetic energy (Ek) and potential energy (V(x)). h d2y 2pmv 2 y _ (x) If = , then = 2 mv dx h Manipulating the above equation to get a kinetic energy term, since Ek = (mv)2/2m 2 d2y _ h 8p2m dx2 = h2 d2y 8p2m dx2 = _ (mv)2 y(x) 2m Ek y(x) or, (10) 38 The Schrödinger Equation for a Particle Moving in One Dimension • Equation (10) shows the relationship between the second derivative of a wave function and the kinetic energy of the particle it represents. • If external forces are present (e.g. due to the presence of fixed charges, as in an atom), then a potential energy term V(x) must be added. • Since E(total) = Ek + V(x), substituting for Ek in Equation (10) gives 2 2 y h d _ V(x) = E y(x) + (11) 2 2 8p m dx This is the Schrödinger equation for a particle moving in one dimension. 39 The Schrödinger Equation for the Hydrogen Atom Erwin Schrödinger (1926) was the first to act upon de Broglie’s idea of the electron in a hydrogen atom behaving as a standing wave. The resulting equation (12) is analogous to equation (11); It represents the wave form in three dimensions and is thus a second-order partial differential equation. 2 _ h 2 8p 2 m x or 2 + _ h2 2 y 2 + 2 2 z 2 y + + V(x,y,z) y (x,y,z) = Ey (x,y,z) (12) Vy = Ey 2m 40 The Schrödinger Equation, Continued • The general solution of equations like equation (12) had been determined in the 19th century (by Laguerre and Legendre). • The equations are more easily solved if expressed in terms of spherical polar coordinates (r,q,f), rather than in cartesian coordinates (x,y,z), in which case, 2 1 r2 r r2 r in equation (12) becomes q sin + q r2sinq q 1 2 + 2 2q f2 r sin 1 41 Erwin Schrödinger Students: I hope you are staying awake while the professor talks about my work! Love, Erwin 42 Spherical Polar Coordinates z The electron position with respect to the nucleus in spherical polar coordinates is r,q ,f (in cartesian coordinates this is x,y,z) x P q r 0 f q is the angle between OP the z axis. f (the azimuthal angle) is y the angle between the projection of OP onto the xy plane. x = rsinqcosf y = rsinqsinf z = rcosq r 2 = x2 + y2 + z2 43 Solutions of the Schrödinger Equation for the Hydrogen Atom • The number of solutions to the Schrödinger equation is infinite. • By assuming certain properties of y (the wave function) - boundary conditions relevant to the physical nature of the H atom - only solutions meaningful to the H atom are selected. • These sensible solutions for y (originally called specific quantum states, now orbitals) can be expressed as the product of a radial function [R(r)] and an angular function [Y(q,f)], both of which include integers, known as quantum numbers; n, l and m (or ml ). 44 Solutions of the Schrödinger Equation for the Hydrogen Atom, Continued Y(r,q,f) = Rnl(r)Ylm(q,f) (13) • The radial function R is a polynomial in r of degree n – 1 (highest power r(n-1), called a Laguerre polynomial) multiplied by an exponential function of the type e(-r/na0) or e(-/n), where a0 is the Bohr radius. • The angular function Y consists of products of polynomials in sinq and cosq (called Legendre polynomials) multiplied by a complex exponential function of the type e(imf). 45 Solutions of the Schrödinger Equation for the Hydrogen Atom, Continued • The principal quantum number is n (like the Bohr quantum number = 1, 2, 3,…), whereas the other two quantum numbers both depend on n. • l = 0 to n - 1 (in integral values). • m = -l through 0 to +l (again in integral values). • The energies of the specific quantum states (or orbitals) depend only on n for the H atom (but not for many-electron atoms) and are numerically the same as those for the Bohr H atom. 46 Orbitals • Orbitals where l = 0 are called ‘s orbitals’; those with l = 1 are known as ‘p orbitals’; and those with l = 2 are known as ‘d orbitals’. • When n = 1, l = m = 0 only; there is only one 1s orbital. • When n = 2, l can be 0 again (one 2s orbital), but l can also be 1, in which case m = -1, 0 or +1 (corresponding to three p orbitals). • When n = 3, l can be 0 (one 3s orbital) and 1 (three 3p orbitals) again, but can also be 2, whence m can be –2, -1, 0, +1 or +2 (corresponding to five d orbitals). 47 Energy Levels of the H atom 48 The 1s Wave Function of H and Corresponding Pictorial Representation The 1s wave function is the solution of the Schrödinger equation when n = 1; l = 0; m = 0. It represents the orbital with the lowest energy. Y100 = 2 a0-3/2 e-r/a0 2 1 p R10 is a function Y is a constant for of r0 and of e-r/a0 00 all cases where l = m = 0 (s orbitals), indicating spherical symmetry The 3D boundary surface of the 1s orbital 49 The 2pz Wave Function of H and Corresponding Pictorial Representation The 2pz wave function is the solution of the Schrödinger equation when n = 2; l = 1; m = 0. It represents the p orbital pointing along the z axis. Y210 = 1 2 6 R21 a function of r, as well as of e-r/2a0 a0-5/2 r e-r/2a 0 2 3 p cosq Y10 is a function of q only, indicating directionality along the z axis The 3D boundary surface of the 2pz orbital 50 C. Particle in a One-Dimensional Box oo oo V(x) 0 0 x a 51 Setting a Goal for Part C • You will learn how the Schrödinger equation can be applied to one of the simplest problems; a particle in a onedimensional box or energy well. • You will learn how to calculate the energies of various quantum states associated with this system. • You will learn how extend these ideas to three dimensions. 52 Objective for Part C • Describe how the Schrödinger equation can be applied to a particle in a onedimensional box (and similar situations) and how the energies of specific quantum states can be calculated. 53 Particle in a One-Dimensional Box • The simplest model to which the Schrödinger equation can be applied is the particle (such as a ‘1-D electron’) in a one-dimensional box or potential energy well. • The potential energy of the particle is 0 when it is in the box and beyond the boundaries of the box; clearly the particle is totally confined to the box. • All its energy will thus be kinetic energy. 54 Defining the Problem V(x) = 0 (0 < x < L) V(x) = oo (x < 0 or x > L) oo oo 0 L V(x) 0 x 55 Setting up the Schrödinger Equation The Schrödinger equation for this model is derived from equation (11) in Part B, with V(x) = 0 for 0 < x < L _ h2 d2y 8p2m dx2 = Ey (1) Separating variables in the Schrödinger equation gives equation (2) d2y dx2 2 _ 8p mE y = h2 -k2y (2) 56 Solution of the Schrödinger Equation and use of the First Boundary Condition The solution to this equation is y(x) = A sin(kx) or B cos(kx), but after imposition of the boundar y condit ion that y (x) = 0 at x = 0, the solution must be y(x) = A sin(kx) Substituting this into equation 1 gives _ 2 A k sin(kx) = _ 8p2mE A sin(kx) h2 (3) 57 Evaluation of the Constant k and Use of the 2nd Boundary Condition From equation (3), k = 8p2mE 1 2 and hence h2 y(x) = A sin 8p2mE 1 2 x (4) h2 Imposition of the 2nd boundary condit ion, that y(x) = 0 at x = L, implies equation (5) 8 p2 mE h2 1 2 L = np (5) , since sin(x) is zero only when x = np 58 Determination of the Energy Levels This boundary condition is satisfied if E is restricted to values that satisfy equation (5): that is if En is the value of energy that satisfies equation (5) for given allowed value of n, then En = n2h2 8mL2 (6) (n = 1, 2, 3,....) This gives rise to the set of energy levels opposite, showing the corresponding wave forms The spacing of energy levels; E = En-1 _En = h2 8 mL 2 (2n + 1) 59 Determination of the Constant A From equations (4) and (5), it can be seen that y(x) = A sin npx (7) L A is called the nor malization const ant . To evaluate A, we refer to the well-behaved nat ur e of y that the integral of y2 over all space (here between x = 0 and x = L) must be 1, since the particle is somewhere in the box. a a i.e npx dx y2dx = A2sin2 =1 L 0 0 L 2 or A2 2 = 1 from which, A = L npx 2 ( n = 1, 2, 3,....) (8) Hence y(x) = sin L L 60 A Particle in a Three-Dimensional Box • The arguments in the previous slides can be extended to a particle confined in a 3D box of lengths Lx, Ly and Lz. • Within the box, V(x,y,z) = 0; outside the cube it is • A quantum number is needed for each dimension and the Schrödinger equation includes derivatives with respect to each coordinate. • The allowed energies for the particle are given by h2 nx2 ny2 nz2 En xnynz = 2+ 2+ Ly Lz2 8m Lx (9) 61 Calculation of Energies of a Particle in a 3-D Box Comparative energies of a particle confined to a box of sides 2L, L, L. Enxnyn = z h2 nx2 8m (2L)2 + ny2 L2 + nz2 L2 For the ground state, nx = ny = nz = 1 Hence E111 = h2 8mL2 1 +1 + 1 4 = 9h2 32mL2 62 ….Continued For the excited states where one of nx, ny or nz is 2, the others are 1, we have E211 = h2 8mL E121 = E112 = 2 4 + 1 + 1 4 h2 8mL 1 2 4 = + 4 + 1 3h2 12h2 8mL2 32mL2 = 21h2 32mL2 The last two are degenerate (of the same energy). If all the box sides had been L, then all three of the above excited state energies would be degenerate. Note that there are many other excited states; E311, E321, etc. 63 Calculation of Energy Spacing in Different Situations Consider two situations: (1) an electron in a one-dimensional box of length 1.0 A. (2) an electron in a cube of lengths 10 cm on an edge. Calculate the energy difference between the ground state and the first excited state. (1) E = 3h2 8mL 2 = 3 (6.626 x 10-34 Js) 2 8 (9.11 x 10-31 kg)(1.0 x 10-10 m) 2 = 1.807 x 10-17 J ( 10,880 kJ/mol) 64 …Continued (2) For the cube, where Lx = Ly = Lz = L = 10 cm (0.1 m) E111 (ground state) = E211 = E121 = E112 = h2 8mL2 h 12 + 12 + 12 E = 2 8mL2 22 + 12 + 12 3 h2 8mL2 (First excited state) E = 3 (6.626 x 10-34 Js) 2 8 (9.11 x 10-31 kg)(0.1 m)2 = 1.807 x 10-35 J This is equivalent to 1.088 x 10-14 kJ/mol. The energy levels in general are so close together that they appear continuous; quantum effects are minimal, quite unlike the case with the electron in a 1-dimensional box of atom-size length 1.0 A. 65