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Unformatted text preview: Ch 24a Winter Term 2009‐10 Introduction to Biophysical Chemistry Problem Set # 1 Answers Key Distributed: 15 January 2010 Due: 25 January 2010, 1 p.m. 1. (10 points) In the treatment of the interaction of an electron with a light wave discussed in class, we obtain the following result for the out‐of‐phase polarizability α90: α90 = (2 e2/m){ (ωω’)/[(ωo2 –ω2)2 + 4ω’2ω2]}, where ω’ is the damping frequency η/2m. Show that the half‐width at half maximum height of α90 is approximately ω’. Note that near resonance, ω ~ ωo, as ω’ is typically very small compared to ωo. For example, if ω = (ωo2 + ω’2)½ ± ω’, then ω2 = ωo2 ± 2ωoω’. Thr result obtained indicates that dissipative loss determines the width of the resonance absorption, or the spectral line, classically. α90 = (2 e2/m){ (ωω΄)/[(ωo2 –ω2)2 + 4ω΄2ω2]} Maximum occurs at ω = ωo where α90 = (2 e2/m){ (ωoω΄)/[ 4ω΄2ωo2]} = (e2/m)/2 ωoω΄ Half maximum is then (e2/m)/4 ωoω΄. Locate the value of ω = ω½ such that α90 = (2 e2/m){ (ω½ω΄)/[(ωo2 –ω½2)2 + 4ω΄2ω½2]} = (e2/m)/4 ωoω΄ or ( 2) (4 ω½ωo(ω΄)2 = (ωo2 –ω½2)2 + 4ω΄2ω½2 or 8 ω½ωo(ω΄)2  4ω΄2ω½2 = (ωo2 –ω½2)2 Since the halfwidth at halfmaximum height is small compared with ωo, we have ω½ ~ ωo, so or or Note that (ωo –ω½) is just the position of the halfmaximum on the low frequency side, and (ωo +ω½) is the position of the halfmaximum on the high frequency side. Now (ωo –ω½)(ωo + ω½) ≈ ± 2 ω΄ωo could be approximated by (ωo –ω½)(ωo + ωo) ≈ ± 2 ω΄ωo . So or i.e., ω½ is close to the resonance frequency, 8 ωo2 ω΄2  4ω΄2ωo2 = 4ω΄2ωo2 ≈ (ωo2 –ω½2)2 (ωo2 –ω½2) ≈ ± 2 ω΄ωo (ωo –ω½)( (ωo + ω½) ≈ ± 2 ω΄ωo (ωo –ω½) (2ωo) ≈ ± 2 ω΄ωo (ωo –ω½) ≈ ± ω΄ , the desired result. 2. (15 points) James P. Allen, Chapter 10, problem 10.10. (2 points for parts (a) and (b); 3 points for parts (c) and (d); and 5 points for part (e)) Six electrons in a onedimensional box of length L = 10 nm. 4 3 2 1 n 0 L 0 L Right: Distribution of 6 electrons in the ground state. Quantum number of highest occupied state is n = 3; quantum number of lowest unoccupied state is n = 4. Left Distribution of electrons in first excited state 3. (25 points) (a) For a particle in a one‐dimensional box of length L, calculate the amplitude square of the transition dipole from the ground state (n = 1) to the first excited 2 state (n = 2), i.e. μ12 . (10 points) 2 (b) Repeat the same calculation for the corresponding transition, i.e. μ01 , for a harmonic oscillator with force constant k and mass m. Be sure that the wavefunctions you use are normalized. (15 points) The wavefunctions for a particle in a one‐dimensional box are given by Φn = (2/L)½ sin(nπx/L), where n = 1, 2, 3, ….. The wavefunctions for the ground (n = 0) and first‐excited (n = 1) states of the harmonic oscillator are, respectively, Φ0(x) = exp‐(βx2/2) and Φ1(x) = 2 (β)½ x exp‐(βx2/2), where β = 2π(mk)½/h. ___________________________________________________________________ 4. (10 points) The iron porphyrin of a heme protein can usually be converted to the ferrous form from the ferric state by a one‐electron titration. This reduction can be followed spectrophotometrically. As is often the case, the absorption spectrum of the ferrous species differs somewhat from that of the ferric protein. (a) What are required properties of the two spectra in order for there to be at least one isobestic point, i.e., a wavelength at which the absorbance of the system is invariant during the titration? (5 points) At any wavelength λ, A(λ)/l = ε1(λ) c1 + ε2(λ)c2 = ε1(λ) c1 + ε2(λ)(co – c1) = [ε1(λ) ‐ ε2(λ)] c1 + ε2(λ)co where A is the absorbance of the sample; c1 and c2 are the concentrations of the two species under a particular condition; and ε1(λ) and ε2(λ) denote the extinction coefficients of the two species at wavelength λ. For an isobestic point, [ε1(λ) ‐ ε2(λ)] = 0, since absorbance is constant for varying c1 or c2. So the extinction coefficients of the two species must be the same at the isobestic point. (b) Discuss what you would expect if, during the titration, there happens to be a trace of oxygen in the system that converts part of the ferrous protein to the oxy‐ form, i.e., oxygen bound species, which has its own characteristic absorption spectrum. (5 points) Now there will be three species, ferric heme, ferrous heme, and the dioxygen adduct, and the isobestic point will disappear. 5. (20 points) (a) Show that for a diatomic molecule AB, the kinetic energy associated with the vibration of the two atoms about the center of mass is ½ μ(dR/dt)2, where μ is the reduced mass (mA mB)/ (mA + mB) and R is the distance between the two atoms. (10 points) Let xA and xB denote the distance of the two atoms from the center of mass of the diatomic molecule AB. Then the kinetic energy associated with the vibration is ½ mA (dxA/dt)2 + ½ mB (dxB/dt)2 By virtue of the definition of the center of mass, mA R = (mA + mB) xB and mB R = (mA + mB) xA where R = xA + xB, the internuclear separation of the molecule. Therefore the kinetic energy associated with the vibration is ½ mA [mB2/(mA + mB)2](dR/dt)2 + ½ mB [mA2/(mA + mB)2](dR/dt)2 = ½ [(mAmB2 + mBmA2)/(mA + mB)2](dR/dt)2 = ½ [mAmB/(mA + mB)](dR/dt)2 (b) If the vibrational frequency of 12C16O occurs at 2143 cm−1, where would you expect the corresponding frequencies for the 12C18O, 13C16O, and 13C18O isotopomers? (10 points) C16O : μ = (12.000)(15.995)/(12.000 + 15.995) = 6.8562 ; (μ)½ = 2.618; v C O = 2143 cm‐1 12
12 16 C18O : μ = (12.000)(17.999)/(12.000 + 17.999) = 7.2000 (μ)½ = 2.683; v C O = 2143 x (2.618/2.683) cm‐1 = 2091 cm‐1 13C16O : μ = (13.003)(15.995)/(13.003 + 15.995) = 7.1723 (μ)½ = 2.678; v C O = 2143 x (2.618/2.678) cm‐1 = 2095 cm‐1 13C18O : μ = (13.003)(17.999)/(13.003 + 17.999) = 7.5492 (μ)½ = 2.748; v C O = 2143 x (2.618/2.748) cm‐1 = 2042 cm‐1 12
12 18 13 16 13 18 6. (10 points) The following stretching vibrations occur in motions of proteins: ← → ← → ← → ← → ← → C=O, O−H, C=N, C=C, and N−H. Which do you predict will contribute mainly to infrared absorption and which mainly to Raman scattering? C=O stretch, O‐H stretch, C=N stretch, and N‐H stretch will contribute mainly to infrared absorption. C=C stretch mainly to Raman scattering. 7. (10 points) Describe the normal modes of vibration of SO2 (bent) and CS2 (linear). Which are Raman active and which are infrared active? ____________________________________________________________________ 8. (15 points) When acetylene, HCCH, is irradiated with the 435.83 nm line of a mercury lamp, a Raman line is observed at 476.85 nm. (a) Determine the fundamental frequency of vibration associated with this Raman line. (3 points) (b) Is this Raman line a Stokes or anti‐Stokes line, and why? (2 points) (c) Propose a possible assignment of mode for this vibration. (2 points) (d) How would you use isotopically substituted acetylene to confirm the assignment you have proposed in (c)? (5 points) (e) Predict the relative position you would expect the Raman line for the isotopically substituted acetylene to appear. (3 points) 9. (25 points) Blue copper proteins, such as plastocyanin from spinach, azurin from Pseudomonas aeruginosa, etc., are copper proteins characterized by a single copper atom coordinated by two histidine residues and a cysteine residue in a trigonal planar structure, and a variable remote axial ligand (R). The blue color of these proteins is due to a strong cysteine thiolate S → Cu(II) ligand‐to‐metal charge transfer (LMCT) absorption near 600 nm, giving these proteins their characteristic blue color. In the ground electronic state, the electronic structure of the Cu(II)‐S bond could be described by the sharing of a pair of electrons in a molecular orbital formed from the Cu dx2‐y2 atomic orbital and a sulfur (S) orbital. The simplest wave function that one could write to describe this molecular orbital would be a linear combination of the Cu dx2‐y2 atomic orbital and a S orbital, namely ΨMO1 (1) = a1 φA(1) + b1 φB (1) where a1 and b1 are coefficients in the LCAO MO (a molecular model made up of a linear combination of atomic orbitals centered on A and B, respectively). If A corresponds to the thiolate S, and B the Cu(II), then a1 >> b1. That is, in the ground electronic state, the molecular orbital has mostly S character. . (a) Show that a12 + b12 ≈ 1, if the wave function for the above molecular orbital is written from normalized atomic orbitals, and if we may neglect the small overlap that exists between the two atomic orbitals. (5 points) = ∫ ΨMO1 (1)* ΨMO1 (1) dτ = ∫ [a1* φA(1)* + b1* φB (1)*][ a1 φA(1) + b1 φB (1)] dτ ∫ [a1* φA(1)* + b1* φB (1)*][ a1 φA(1) + b1 φB (1)] dτ
+ b1* b1 ∫φB (1)* φB (1)] dτ = a1* a1 ∫φA(1)* φA(1) dτ + a1* b1 ∫φA(1)* φB (1)] dτ + b1* a1 ∫φB (1)* φA(1)* dτ ≈ a1* a1 + b1* b1 = 1 if (1) the atomic orbitals φA(1) and φB (1) are normalized to begin with, i.e., if ∫φA(1)* φA(1) dτ =1 and ∫φB (1)* φB (1)] dτ = 1; and (2) the overlap integrals ∫φA(1)* φB (1)] dτ and ∫φB (1)* φA(1)* dτ are sufficiently small that we may neglect them. (b) A similar molecular orbital could be written for the first excited state: ΨMO2 (1) = a2 φA(1) + b2 φB (1) . If the two molecular orbitals ΨMO1 (1) and ΨMO2 (1) must be orthogonal, show that │b2 │ >> │a2 │. (5 points) = ∫ ΨMO2 (1)* ΨMO1 (1) dτ ∫ [a2* φA(1)* + b2* φB (1)*][ a1 φA(1) + b1 φB (1)] dτ
+ b2* b1 ∫φB (1)* φB (1)] dτ = 0 = a2* a1 ∫φA(1)* φA(1) dτ + a2* b1 ∫φA(1)* φB (1)] dτ + b2* a1 ∫φB (1)* φA(1)* dτ (the two MOs are orthogonal!) Again, if (1) the atomic orbitals φA(1) and φB (1) are normalized, and we may assume that their overlap is small, then a2* a1 (1) + a2* b1 (0) + b2* a1 (0) + b2* b1 (1) ≈ 0 or a2* a1 + b2* b1 ≈ 0, a2* / b2* = − b1 / a1 or a2* a1 = − b2* b1 or So if │b1 / a1 │ << 1, then │a2* / b2* │ << 1. Note that if a1 and b1 are of the same sign, then a2 and b2 will be of the opposite sign. (c) The 600 nm absorption observed for the blue copper center has been assigned to an electronic transition from MO1 to MO2 as described above. Since a1 >> b1 and a2 << b2, the absorption is frequently referred to as a cysteine thiolate S → Cu(II) LMCT transition. Obtain an expression for the transition moment of this transition to confirm the validity of this assertion. (10 points). Transition moment for the electronic transition from MO1 to MO2 = μ12 = −│e │∫ ΨMO2 (1)* rA1 ΨMO1 (1) dτ = −│e │ ∫ [a2* φA(1)*
+ b2* φB (1)*] rA1 [ a1 φA(1) + b1 φB (1)] dτ (where underlie designate vector quantity) = −│e │[ a2* a1 ∫φA(1)* rA1 φA(1) dτ + a2* b1 ∫φA(1)* rA1 φB (1)] dτ + b2* a1 ∫φB (1)* rA1 φA(1)* dτ + b2* b1 ∫φB (1)* rA1 φB (1)] dτ ] First integral is zero because φA(1) is an atomic wavefunction, which has spherical symmetry. Second and third integral is small, because small overlap between φB (1) and φA(1) wavefunctions, which are centered on the two different atoms. So the transition moment for the transition = μ12 = −│e │ b2* b1 ∫φB (1)* rA1 φB (1)] dτ = −│e │ b2* b1 ∫φB (1)* (R + rB1 )φB (1)] dτ noting that rA1 = R + rB1 (see diagram below), = −│e │ [ b2* b1 ∫φB (1)* R φB (1)] dτ + b2* b1 ∫φB (1)* rB1 φB (1)] dτ ] The first integral is just −│e │(b2* b1) R, and the second integral vanishes because φB(1) is also an atomic wavefunction and has spherical symmetry. Therefore μ12 = −│e │(b2* b1) R (d) Given the result derived in (c) above, what is the direction of the transition dipole moment for the chargetransfer transition? (5 points) The transition moment is aligned along R, that is, along the bond between the thiolate S and the Cu(II) ion. That is why the transition is a thiolate → Cu(II) transition (LMCT). ________________________________________________________________________ The end ...
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 Winter '08
 Richards,J
 Electron, pH

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