Exercises 16.3a, 16.5a, 16.13a, 16.14a, 16.21a, 16.25a.

Transcription

1 SPECTROSCOPY Readings in Atkins: Justification 13.1, Figure 16.1, Chapter 16: Sections 16.4 (diatomics only), 16.5 (omit a, b, d, e), 16.6, 16.9, 16.10, (omit b), (omit c). Exercises 16.3a, 16.5a, 16.13a, 16.14a, 16.21a, 16.25a. Reduced Mass Consider a system of two particles with masses m 1 and m 2 and a potential energy which depends only on the separation of the particles. The classical energy is given by E = 1 2 m 1 ṙ m 2 ṙ V ( r 2 r 1 ) the dots signifying derivative wrt time. Introduce two new variables, the particle separation r and the position of the center of mass R: r = r 2 r 1, R = m 1r 1 + m 2 r 2 m where m = m 1 + m 2. In terms of the new coordinates and r 1 = R + m 2 m r, r 2 = R m 1 m r E = 1 2 mṙ µṙ2 + V (r) where r = r and µ is the reduced mass µ m 1m 2 m 1 + m 2 1

2 or 1 µ = 1 m m 2 Note that, if m 2, then µ m 1. The term containing Ṙ represents the kinetic energy of a single hypothetical particle of mass m located at the center of mass R. The remaining terms represent the relative motion of the two particles. It has the appearance of a single particle of effective mass µ moving in the potential field V (r). E rel = 1 2 µṙ2 + V (r) = p2 2µ + V (r) We can thus write the Schrödinger equation for the relative motion { } h2 2µ 2 + V (r) ψ(r) = Eψ(r) When we treated the hydrogen atom, it was assumed that the nuclear mass was infinite. In that case we can set µ = m, the mass of an electron. The Rydberg constant for infinite nuclear mass was calculated to be R = 2π2 me 4 h 3 c = 109, 737 cm 1 If instead, we use the reduced mass of the electron-proton system µ = mm m + M m m This changes the Rydberg constant for hydrogen to R H 109, 677 cm 1 2

3 in perfect agreement with experiment. In 1931, H. C. Urey evaporated four liters of hydrogen down to one milliliter and measured the spectrum of the residue. The result was a set of lines displaced slightly from the hydrogen spectrum. This amounted to the discovery of deuterium, or heavy hydrogen, for which Urey was awarded in 1934 Nobel Prize in chemistry. Estimating the mass of the deuteron, 2 H 1, as twice that of the proton, gives R D 109,707 cm 1 Another interesting illustration of reduced mass is positronium, a short-lived combination of an electron and a positron, the electron s antiparticle. The electron and positron mutually annihilate with a half-life of approximately 10 7 sec and positronium decays into gamma rays. The reduced mass of positronium is µ = m m m + m = m 2 half the mass of the electron. Thus the ionization energy equals 6.80 ev, half that of hydrogen atom. Positron emission tomography (PET) provides a sensitive scanning technique for functioning living tissue, notably the brain. A compound containing a positron-emitting radionuclide, for example, 11 C, 13 N, 15 O or 18 F, is injected into the body. The emitted positrons attach to electrons to form short-lived positronium, and the annihilation radiation is monitored. 3

4 Sculpture of Positronium Annihilation Vibration of Diatomic Molecules Diatomic molecule with nuclear masses M A, M B µ = M AM B M A + M B V (R) = E elec (R) from solution of electronic Schrödinger equation. Schrödinger equation for vibration { h2 d 2 } 2µ dr + V (R) χ(r) = Eχ(R) 2 4

5 Vibrational Energies of Diatomic Molecule Expand V (R) in Taylor series about R = R e V (R) = V (R e ) + (R R e )V (R e ) (R R e) 2 V (R e ) +... Harmonic oscillator approximation V (R) 1 2 k(r R e) 2 Choose energy origin so V (R e ) = 0. At minimum, V (R e ) = 0. Best match to a parabola when force constant is found from k d2 V (R) dr 2 R=Re 5

6 Ground state vibrational energy, for quantum number v = 0 E 0 = hω = h Dissociation energy from ground vibrational state In wavenumber units k µ D 0 = D e 1 2 hω hcd 0 = hcd e 1 2 ν cm 1 An improved approximation which account for anharmonicity, deviation from harmonic oscillator. Now have finite number of vibrational energy levels and possibility of dissociation of the molecule at high enough energy. Morse potential: V (R) = hcd e { 1 e a(r R e) } 2 a = ( µω 2 2hcD e ) 1/2 Note V (R e ) = 0, minimum of potential well. Energy levels of Morse oscillator In wavenumber units E v = (v ) hω (v )2 hωx e hce v = (v ) ν (v )2 x e ν 6

7 Note higher vibrational energy levels are spaced closer together, as in real molecule. Vibrational transitions of diatomic molecules occur in the infrared. A molecule will absorb or emit radiation only if it has a non-zero dipole moment. Thus HCl is infrared active while H 2 and Cl 2 are not. Vibration of Polyatomic Molecules A molecule with N atoms has a total of 3N degrees of freedom for its nuclear motions, since each nucleus can be displaced in three independent directions. Three of these degrees of freedom correspond to translational motion of the center of mass. For a nonlinear molecule, three more degrees of freedom determine the orientation of the molecule in space, and thus its rotational motion. This leaves 3N 6 vibrational modes. For a linear molecule, there are just two rotational degrees of freedom, which leaves 3N 5 vibrational modes. For example, the nonlinear molecule H 2 O has three vibrational modes while the linear molecule CO 2 has four vibrational modes. The vibrations consist of coordinated motions of several atoms in such a way as to keep the center of mass stationary and nonrotating. These are called the normal modes. each normal mode has a characteristic resonance frequency ν i, which is usually determined experimentally. To a reasonable approximation, each normal mode behaves as an independent harmonic oscillator of frequency ν i. The normal modes of H 2 O and CO 2 are pictured below. 7

8 Normal Modes of H 2 O and CO 2 A normal mode will be infrared active only if it involves a change in the dipole moment. All three modes of H 2 O are active. The symmetric stretch of CO 2 is inactive because the two C-O bonds, each of which is polar, exactly compensate. Note that the bending mode of CO 2 is doubly degenerate. Bending of adjacent bonds in a molecule generally involves less energy than bond stretching, thus bending modes have lower wavenumbers that stretching modes. Rotation of Diatomic Molecules 8

9 The rigid rotor model assumes that the internuclear distance R is a constant. This is not a bad approximation since the amplitude of vibration is generally of the order of 1% of R. The Schrödinger equation for nuclear motion then involves the three-dimensional angular momentum operator, written Ĵ rather than ˆL when it refers to molecular rotation. The solutions to this equation are already known and we can write Ĵ 2 2µR 2 Y JM(θ, φ) = E J Y JM (θ,φ) J = 0, 1, 2... M = 0,±1... ± J where Y JM (θ, φ) are spherical harmonics in terms of the quantum numbers J and M, rather than l and m. Since the eigenvalues of Ĵ 2 are J(J + 1) h 2, the rotational energy levels are E J = h2 J(J + 1) 2I The moment of inertia is given by I = µr 2 = M A R 2 A + M B R 2 B where R A and R B are the distances from nuclei A and B, respectively, to the center of mass. In wavenumber units, the rotational energy is expressed hce J = BJ(J + 1) cm 1 where B is the rotational constant. The rotational energy-level diagram is shown below. Each level is (2J + 1)-fold degenerate. 9

10 Rotational Energies hce J = BJ(J + 1) Again, only polar molecules can absorb or emit radiation in the course of rotational transitions. The radiation is in the microwave or infrared region. The selection rule for rotational transitions is J = ±1. 10

11 Molecular Parameters from Spectroscopy Table of spectroscopic constants for hydrogen halides: ν/cm 1 B/cm 1 H 19 F H 35 Cl H 81 Br H 127 I The force constant can be found from the vibrational constant. Equating the energy quantities hω = hc ν, we find k ω = 2πc ν = µ Thus with k = (2πc ν) 2 µ µ = m Am B m A + m B = M AM B M A + M B u where u = kg, the atomic mass unit. M A and M B are the atomic weights of atoms A and B, on the scale M( 12 C) = 12. Putting in numerical factors k = 58.9 ( ν/cm 1 ) 2 M A M B M A + M B N/m This gives 958.6, 512.4, and N/m for HF, HCl, HBr and HI, respectively. These values do not take account of anharmonicity. 11

12 The internuclear distance R is determined by the rotational constant. By definition of the rotational constant, and with hcb = h2 2I B = h 4πcI I = µr 2 = m Am B m A + m B R 2 = M AM B M A + M B ur 2 kg m 2 Solving for R / MA M B R = B M A + M B pm For the hydrogen halides, HF, HCl, HBr, HI, we calculate R = 92.0, 127.9, 142.0, pm, respectively. Rotation of Nonlinear Molecules A nonlinear molecule has three moments of inertia about three principal axes, designated I a, I b and I c. The classical rotational energy can be written E = J2 a 2I a + J2 b 2I b + J2 c 2I c where J a, J b, J c are the components of angular momentum about the principal axes. For a spherical rotor, such as CH 4 or 12

13 SF 6, the three moments of inertia are equal to the same value I. The energy simplifies to J 2 /2I and the quantum-mechanical Hamiltonian is given by The eigenvalues are Ĥ = Ĵ2 2I E J = h2 J(J + 1) J = 0, 1, I just as for a linear molecule. But the levels of a spherical rotor have degeneracies of (2J + 1) 2 rather than (2J + 1). A symmetric rotor has two equal moments of inertia, say I c = I b I a. The molecules NH 3, CH 3 Cl and C 6 H 6 are examples. The Hamiltonian takes the form Ĥ = Ĵ2 a 2I a + Ĵ2 b + Ĵ2 c 2I b ( = Ĵ ) Ĵa 2 2I b 2I a 2I b Since it its possible to have simultaneous eigenstates of Ĵ 2 and one of its components Ĵa, the energies of a symmetric rotor have the form ( J(J + 1) 1 E JK = + 1 ) K 2 2I b 2I a 2I b J = 0, 1, 2... K = 0, ±1, ±2... ± J There is, in addition, the (2J + 1)-fold M degeneracy. 13