5 questions, 3 points each, 15 points total possible. 26 Fe Cu Ni Co Pd Ag Ru 101.
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1 Physical Chemistry II Lab CHEM 4644 spring 2017 final exam KEY 5 questions, 3 points each, 15 points total possible h = J s c = m/s 1 GHz = 10 9 s -1. B= h 8π 2 I ν= 1 2 π k μ 6 P Chem Lab II CHEM 4644 spring 2017, average=9.6 5 frequency score <= number shown 1 H Li Na K Rb Cs Fr Be Mg Ca Sr Ba Ra Sc Y La Ac Ce Th Ti Zr Hf Rf Pr Pa V Nb Ta Db Nd U Cr Mo W Sg Pm Np Mn Tc Re Bh Sm Pu Fe Ru Os Hs Eu Am Co Rh Ir Mt Gd Cm Ni Pd Pt Ds Tb Bk Cu Ag Au Rg Dy Cf Zn Cd Hg Cn Ho Es B Al Ga In Tl Nh Er Fm C Si Ge Sn Pb Fl Tm Md N P As Sb Bi Mc Yb No O S Se Te Po Lv Lu Lr F Cl Br I At Ts He Ne Ar Kr Xe Rn Og 294
2 1. fluorescence and phosphorescence Z. Khadraoui et al, Optical Materials, 47, , Consider the TbPO 4 emission spectrum. Excitation wavelength was 370 nm. The atomic state labels can be interpreted like this: the state label " 5 D 4 " means the spin multiplicity (2S+1) is 5, the orbital angular momentum quantum number L is 2, and the total angular momentum quantum number J is 4. a. The most intense peak, at 543 nm, is due to an f-f orbital transition in Tb. Give one reason this is forbidden. Answer: Spin changed (2S+1 from 5 to 7), which is forbidden. A second reason that was not discussed in lab is that the orbital angular momentum quantum number l should change, so f to f is forbidden b. I = I o e t / τ. Following 370 nm excitation, emission intensity decays as in Figure 8. Estimate the lifetime, tau. Note that the scale of the y axis is Figure 8 is not linear; it is a natural logarithm scale. Answer: 1/tau = - slope; tau = 1.7 ms / ln(20) = 0.57 ms.
3 2. anharmonicity of a vibrating bond The following energies of a vibrating nitrogen molecule were given by Feliciano Guistino, Materials Modelling using Density Functional Theory, Oxford U. Press: Oxford, (Note: 1 mev=8.065 cm -1.) Recall that for an anharmonic oscillator, to a good approximation, E n = (n+½)hν - (n+½) 2 x e hν. n E n (mev) a. How do the data show that the vibration is anharmonic, not harmonic? b. Use some of the data to calculate the harmonic vibrational energy hν (or frequency ν) and the anharmonicity constant x e (or x e hν). Show your work. a. The spacing between levels decreases with increasing energy. That is characteristic of anharmonicity. ( ) = ( ) = ( ) = b. Use any two energies to solve algebraically for hν and x e. For example, E 3 = 3.5 hν x e hν and E 0 = 0.5 hν x e hν so E E 0 = ( ) hν = hν = - (21 / 4) hν hν = /21 = mev = 2485 cm -1 or ν = X 1.602X10-19 J/eV / ( J.s) = s -1 Also, 0.5 E E 0 = ( ) x e hν Check: = -(21/4) x e hν x e hν = -( ) (4/21) = 9.4 4/21 = 1.79 mev = 14.4cm -1 or x e = 1.79 / = (unitless) n E n (mev) calculated 308.1/ *1.79 = *1.79 = * 1.79 = *1.79 = *1.79 =
4 3. rovib Pure gas-phase rotational spectra of isotopically pure carbon disulfide were measured by Christian Schroter, et al., (Science, 333, , 2011) using a new method: mass-correlated rotational alignment spectroscopy. Their spectrum of 32 S 12 C 32 S, at right, is a Raman spectrum. The change in rotational quantum number is ΔJ=±2, and only even J is allowed; that is why lines are 8B apart. In the table below are Schroter's measured rotation constants for three isotopes. isotope mass abundance Rotation constant (GHz) 32 S 12 C 32 S S 13 C 32 S S 12 C 34 S a. Why does the heaviest isotope have the smallest rotation constant? b. Why does substitution of carbon-12 with carbon-13 change the rotation constant so little? c. What is 3.27 GHz in cm -1? a. Increasing sulfur mass increases the moment of inertia. The rotation constant is inversely proportional to the moment of inertia, so increasing sulfur mass decreases the rotation constant. Aside: Expect proportionality: (32/34) = b. Because the carbon atom is on the axis of rotation its mass does not contribute to the moment of inertia. c s -1 / ( cm/s) = cm -1.
5 4. EFNMR A pulse sequence for measuring T 1 with an Earth's-Field NMR is shown. One-sentence answers may suffice. a. What is the purpose of the polarizing pulse? b. What is the purpose of the "90 o " pulse? c. How is "T 1 " related to the time, t, shown in the pulse sequence? a. The polarizing pulse enhances magnetization of the sample. b. The 90-degree pulse tips the magnetization vector 90 o onto the y axis, where it can be measured. c. Following the polarizing pulse, the magnetization intensity decays as exp(-t/t 1 ). The greater t, the smaller the magnetization that will remain when the 90 o pulse is applied. 5. CO 2 and CS 2 At right is an infrared absorption spectrum of gaseous carbon dioxide published by P. E. Martin and E. F. Barker in Physical Review in The authors stated that a second, extremely intense, absorption appeared at 4.3 μm. a. Convert μm to cm -1. b. Which vibration is responsible for the peak at μm: symmetric stretch, antisymmetric stretch, or bend? c. Is the mode also allowed in a Raman spectrum? a. 1/( m 100 cm/m) = 668 cm -1 b. bend c. No. CO 2 modes are either Raman active or IR active, not both.
Speed of light c = m/s. x n e a x d x = 1. 2 n+1 a n π a. He Li Ne Na Ar K Ni 58.
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