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Unformatted text preview: Final Exam Thu, Sep 8, 2011 10:00 11:40 am Chapters 11, 12.1 12.7, 14.1, 14.3 5 Closed Book and Closed Notes No electronic devices except calculator At least one assigned homework problem will appear on the exam. Review session on Wed Sep 7 at 4:00 pm in 179 Chem Practice Final Exam on website. Nuclear Magnetic Resonance (NMR) I Neutrons Protons Spin ( I ) Examples even even 0 12 C 6 , 16 O 8 even odd 1 H, 15 N 7 , 19 F 9 odd even 13 C, 31 P odd odd 1 2 H, 14 N E = = B HO CH 2 CH 3 2 B E h = 2 B E h = (MHz) B = 2 = = h 2 h = h 10 6fold smaller than E for e transition = 800 MHz, = 0.375 m B = 0 800,000,000 Hz 800,002,400 800,004,000 B = 18.8 T E  B o I Time Domain vs. Frequency Domain B o E = B Nuclear Magnetic Moment ( ) vs Time: = I x = cos( t) real axis i y = isin( t) imag axis (i) x,y = x + i y = ei 0t Frequency Domain (MHz) Frequency (MHz) Zeeman Splitting ( E = ) Larmor Precession ( = B ) d (t)/dt = (t) B (t) FT = B o /2 = 800 MHz E = 3 x 1025 J 2 B h 2 B h + = B FID(t) B E = 10 6fold smaller than e transition MHz 800 2 = = FT FT Fourier Transform: A( ) D( ) S( ) = A( ) + iD( ) real imag s(t) = M x + iM y ) / exp(( ) cos( ) ( 2 T t t t M x = M x (t) M y (t) ) / exp(( ) sin( ) ( 2 T t t t M y = Fourier Transform: S( ) = A( ) + iD( ) A( ) D( ) Low Sensitivity of NMR ( ) ( ) J h T s T E 25 1 1 7 10 35 . 5 2 8 . 18 10 75 . 26 = = Energy difference at a magnetic field of 18.8T: Boltzmann population (300K): Absorption is proportional to net concentration of ground state ( N = N N ) NMR Absorption is 10000fold weaker than for IR (need mM or higher)....
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This note was uploaded on 09/16/2011 for the course CHE 107B CHE 107B taught by Professor Ames during the Summer '11 term at UC Davis.
 Summer '11
 AMES

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