exam2-2007 - Name _ Exam 2: CHEM/BCMB 4190/6190/8189 (106...

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Name _________________________ Exam 2, 2007: CHEM/BCMB 4190/6190/8189 1 Exam 2: CHEM/BCMB 4190/6190/8189 (106 points) Tuesday, 2 October, 2007 1 ). In the example (right), the effect of a 90 ° ( ± /2) pulse applied along the “x” axis (90 ° x) is shown for a bulk magnetization vector ( M 0 ) at equilibrium (on the ‘ z ’ axis). For ‘b’, ‘c’, ‘d’ and ‘f’ below, show the effects of the indicated pulses by drawing the missing (originating or resulting) vectors on the coordinate axes. For ‘a’ and ‘e’, fill in the blank with the correct pulse that will promote the indicated movement of the bulk magnetization vector (there may be more than one correct answer for some of these). (12 points) a . d . b . e . c . f . 90 ° x M 0 y z x y z x 90 ° - x y z x y z x 180 ° y , 90 ° - x y z x y z x this vector is in the x - z plane 2 ± y y z x y z x y z x y z x 90 ° - x , 90 ° - y 90 ° y y z x y z x this vector is in the x - z plane y z x y z x 135 ° x
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Name _________________________ Exam 2, 2007: CHEM/BCMB 4190/6190/8189 2 2 ). The inversion-recovery method was used to estimate the relaxation time constant ( T 1 ) for the H A hydrogen of the compound shown (right). The signals from H A at various delays ( ± ) are shown. Estimate T 1 for H A . Shown your work or otherwise explain your reasoning. ( 5 points ) In the inversion-recovery sequence, a 180 ° pulse on magnetization initially at equilibrium (+z) inverts the spin populations to give -z magnetization, with M Z = -M 0 . The 180 pulse is followed by a variable delay ( ± ), which is then followed by a 90 pulse to create the observable transverse magnetization. We know that, following a 180 pulse, the return of bulk magnetization along z to equilibrium (M Z = M 0 ) is described by the following first order equation: For the relaxation data shown, the magnitudes of the signals at the various delay times are not shown. However, the value of the delay that produces no observable signal (M Z = 0) is reasonably easy to estimate from the data (we’ll call this delay zero ). For the data shown, zero is somewhere between 0.7 and 0.8 s, we’ll call it 0.725 s. We can then rearrange the equation to calculate T 1 : H A H X R-O CH2-R ² M z = M 0 (1 ± 2 e ± t / T 1 ) M z = M 0 (1 ± 2 e ± ² zero / T 1 ) 0 = M 0 (1 ± 2 e ± zero / T 1 ) ± M 0 = ± M 0 2 e ± zero / T 1 1 2 = e ± zero / T 1 ln 1 2 = ± zero / T 1 T 1 = zero ln 2 Thus , T 1 ³ 0.725 ln 2 ³ 1.05 s
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Name _________________________ Exam 2, 2007: CHEM/BCMB 4190/6190/8189 3 3 ). The pulse sequence shown (A, right), called the Carr-Purcell spin-echo experiment can be used to measure the transverse relaxation time constant T 2 . Also shown (B, right) is data from an experiment to determine T 2 . a.
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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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exam2-2007 - Name _ Exam 2: CHEM/BCMB 4190/6190/8189 (106...

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