exam1-2007 - Name _________________________ Exam 1, 2007:...

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Unformatted text preview: Name _________________________ Exam 1, 2007: CHEM/BCMB 4190/6190/8189 1 Exam 1: CHEM/BCMB 4190/6190/8189 (100 points) Tuesday, 11 September, 2007 1 ). In the example (right), the effect of a 90 ° ( ¡ /2) pulse on the bulk magnetization vector M , applied along the “x” axis (90 ° x), is shown. For a-d below, show the effect of the indicated pulses on the indicated magnetization vectors by drawing the resulting vectors on the coordinate axes. (8 points) a . c . b . d . 2 ). For a single 59 Co nucleus (see Table 1.1 , second-to-last page of exam) in a 400 MHz (9.4 T) magnetic field, calculate the energy difference between the lowest energy and the highest energy nuclear Zeeman levels. ( 5 points ) The spin quantum number (I) for 59 Co is 7/2. In general, there are 2I+1 values for the magnetic quantum number, m (m = I, I-1, I-2…….-I) and 2(I+1) nuclear Zeeman levels or spin states. The energy for a given spin state, E, is: For I = 7/2, therefore, there are eight energy levels, shown in the energy diagram (right), along with the corresponding values for m. The energy difference between the highest and lowest energy states (m = -7/2, m = +7/2, respectively) is: 90 ° x M y z x y z x 270 °-x y z x y z x y z x y z x 90 ° y y z x y z x 720 °-y y z x y z x 90 ° x E = ¡ μ z B = ¡ m ¢ h B E = + 7/ 2 ¡ h B E = + 5/ 2 ¡ h B E = + 3/ 2 ¡ h B E = + 1/ 2 ¡ h B E = ¢ 1/ 2 ¡ h B E = ¢ 3/ 2 ¡ h B E = ¢ 5/ 2 ¡ h B E = ¢ 7/ 2 ¡ h B m = ¡ 7/ 2 m = ¡ 5/ 2 m = ¡ 3/ 2 m = ¡ 1/ 2 m = + 1/ 2 m = + 3/ 2 m = + 5/ 2 m = + 7/ 2 E ¡ E = ( + 7 2 ¢ h B ) £ ( £ 7 2 ¢ h B ) = 7 ¢ h B = 7 ¤ (6.3015 ¤ 10 7 /T/s) ¤ (6.626 ¤ 10 £ 34 Js) ¤ (9.4T) 2 ¥ ¡ E = 4.37 ¤ 10 £ 25 J Name _________________________ Exam 1, 2007: CHEM/BCMB 4190/6190/8189 2 3 ). Fill in the blanks to make the following statements true: a). In NMR, the applied ( B 1 ) field is _____ perpendicular ____ to the static ( B ) field. ( 2 points ) b). At equilibrium, for an ensemble of spin ¡ nuclei in a magnetic field the number of alpha ( ¡ ) spins ( N ¡ ) = X, and the number of beta ( ¢ ) spins ( N ¢ ) = Y. Immediately following a 90° pulse along the y axis, the number of ¡ spins = ____ (X+Y)/2 ___ and the number of ¢ spins = ___ (X+Y)/2 ____. ( 2 points ) c). The number of Zeeman energy levels/states for 10 B is _____ 7 ______. ( 2 points ) d). For a spin ¡ nucleus at equilibrium in a magnetic field, the alpha ( ¡ ) state has a magnetic quantum number of _______ 1/2 ______, its energy is ______ less ______ than that of the beta ( ¢ ) state, and the z component of its associated magnetic moment, μ z , is aligned ____ parallel ____ to the magnetic field. ( 3 points ) e). You record 1D NMR spectrum of your compound using 10 scans. You realize that the signal-to-noise is too low. If you want to record a second spectrum with twice the signal-to- noise as the first, you will need to use ______ 40 ______ scans. ( 2 points ) 4 ). The frequency difference ( £¤ ) between two...
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This note was uploaded on 11/13/2011 for the course CHEM 4190 taught by Professor Staff during the Fall '08 term at University of Georgia Athens.

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exam1-2007 - Name _________________________ Exam 1, 2007:...

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