EXAM NMR (8N090) November , am

Transcription

1 EXA NR (8N9) Novembe 5 9, am Remaks: 1. The exam consists of 8 questions, each with 3 pats.. Each question yields the same amount of points. 3. You ae allowed to use the fomula sheet which has been handed out and a (gaphical) calculato. 4. Please use at least 3 sheets of pape, because the exam will be coected by diffeent teaches: 1 sheet fo questions 1 en, 1 sheet fo questions 3 en 4, 1 sheet (o moe) fo questions 5-8. page 1/1

2 Fo all questions it is assumed that a static magnetic field B has been applied along the +z-axis. 1. Basic pinciples a. Daw the enegy diagam fo a spin I 3/ in an extenal magnetic field B and give an expession fo the enegy fo each enegy level. 3 E m -3/, E γ h 3/ B 1 m -1/, E γ h 1/ B 1 m 1/, E γ h 1/ B 3 m 3/, E γ h 3/ B b. Conside the enegy diagam of question 1a. Between which enegy levels can tansitions take place? Tansition between enegy levels is possible fo m ± 1. c. What is the atio between paallel and anti-paallel poton ( 1 H) spin populations at 1.5 Tesla and 37 C (31 K)? γ( 1 H) ad T -1 s -1, h Planck s constant J s, k Boltzmann s constant J/K γh E E 1/ E1/ B h h π, B 1.5 T E J p p E1/ ( 1/ 1/ ) 1 kt E E E / e kt kt E e e 1/ kt 1/ e Basic pinciples a. What is the effective magnetic field in the otating fame that otates with fequency ω ω? (No f-field has been applied yet.) page /1

3 & ω ( ) ot γ B + Thus: γ ω B + new effective B field in the otating fame γ Effective B field becomes zeo when coodinate system otates at the Lamo fequency ω ω γb b. Give an expession fo the effective magnetic field in the otating fame that otates with v fequency ω B t B1 cos( ω t) e has been applied. ω, when an f-field f ( ) x Thee is no effective B field along the z-axis (see question a). The fequency of the f-field is ω, thus the B 1 field lies still in otating fame that otates with fequency ω ω. The phase of the f-field is zeo, so the B 1 field lies along the x-axis: v ot B B e eff 1 x c. Suppose the magnetization is at equilibium ( (,,1) ). What should be the length of the f-pulse of question b, if we want to otate the magnetization by 9 degees (towads the y-axis)? 1 ( t) 1 cosω1 B, ϕ, Ω (,,1 ) (, sinω t, t) π 9 pulse: ω t ( t) (, 1,) 9 pulse length: 1 t o 9 π ω 1 In the following questions it is assumed that f-pulses ae always on-esonance. 3. Relaxation a. Why is nuclea spin elaxation not a spontaneous pocess? The lifetimes of excited nuclea spins ae extemely long because of the low tansition enegies associated with nuclea esonance. The time equied fo spontaneous emission in page 3/1

4 NR is so long (oughly equivalent to the age of the univese!) that this has no effect on the spin populations, so stimulated emission must be opeative fo elaxation to occu. b. In the figue below the spectal density function J(ω) is shown fo thee molecules: one with slow tumbling, one with fast tumbling and one with an intemediate tumbling ate. Which of the thee molecules has the shotest T 1 elaxation time fo the given Lamo fequency (see figue)? Explain why. ω (Lamo fequency) The molecule that has slow tumbling will have the shotest T 1. The spectal density function epesents the fequency distibution of the fluctuating magnetic field and is popotional to the pobability of finding a component of the motion at a given fequency. The molecule with slow tumbling has the highest pobability of finding a fluctuating magnetic field at the given Lamo fequency and theefoe T 1 elaxation will be most efficient. c. Give an expession fo z as a function of time afte the application of a 9 degee fpulse. R t ( t) e ( ) R1t R t ( ) + e ( ) + ( e ) z z 4. Pulse sequences a. Suppose you measue a spectum of wate with a T elaxation time of 3 ms (eg in muscle tissue) and this poduces a peak with a full width at half height (FWHH) of Hz. Is this what you expect? If not, what could be the cause? page 4/1

5 R 1 1 Intinsic linewidth: FWHH π T π.3 π 1.6 Hz The eal linewidth is almost a facto boade. This could be due to local field inhomogeneities. Real linewidth: FWHH γ B * T π π T b. Deive expessions fo the magnetization at time points 1,, 3 and 4 in the spin-echo expeiment indicated in the figue below. At time point you stat with equilibium,,1. magnetization ( ) excitation pulse 9 y efocusing pulse 18 x τ τ : (,,1) 1: ( 1,, ) equilibium magnetization along x-axis cosωτ sinωτ 1 R : τ Rτ e sinωτ cosωτ e (cosωτ,sinω τ,) 1 T elaxation + Lamo pecession (T1 elaxation ignoed) Rτ 3: e (cosωτ, sinωτ,) flip aound x-axis 4: e R cosωτ sinωτ cosωτ τ Rτ e sinωτ cosωτ sinω τ 1 page 5/1

6 e e R τ R τ e R τ ( 1) ( cos ω τ + sin ω τ, sinω τ cosω τ cosω τ sinω τ, ) ( 1,, ) Spin-echo expeiment efocuses Lamo pecession thus also efocusing of B -field inhomogeneities! (B -field inhomogeneities lead to diffeent fequencies depending on the position of the spin) c. You use the spin-echo expeiment to measue the T elaxation time of a small molecule. You epeat the expeiment a numbe of times and fo each expeiment you incease the echo time τ. The T elaxation time that you detemine is 3 ms, but it should have been 5 ms. What is the poblem? The poblem is diffusion. When the echo time becomes lage, the assumption that the B - field that the spin expeiences (and theefoe the fequency) is the same befoe and afte the 18 degee pulse does no longe hold, because the molecule will have diffused to a diffeent position with a diffeent B -field due to B -field inhomogeneities. Theefoe, the signal is not completely efocused, leading to a moe apid decline of the signal as a function of the echo time and an undeestimation of the T elaxation time. The undeestimation will depend on the diffusion ate of the molecule (will be lage fo smalle molecules) and the magnitude of the B -field inhomogeneities. 5. Chemical shift, J-coupling and exchange a. Why do the CH and CH 3 potons in ethanol (see stuctual fomula below) have a diffeent chemical shift? Which potons have the highest fequency? Give you easoning. Because the CH and CH 3 potons have a diffeent chemical envionment and theefoe a diffeent shielding fom the applied static magnetic field by the suounding electons. The electonegative oxygen shifts the electon density away fom the CH potons, leading to educed electonic shielding and thus a highe Lamo fequency. page 6/1

7 b. Conside a coupled 3-spin system AX with chemical shift positions given by δ A 5. ppm, δ 1.5 ppm and δ X 3. ppm elative to a caie fequency (B ) of 1 Hz. Sketch the NR spectum fo this compound when the J-coupling constant between nucleus A and (J A ) 4 Hz and J X Hz. Assume equal T 1 and T elaxation times fo all esonances. 4 Hz 4/1 ppm.4 ppm, Hz. ppm A X 4 Hz Hz 4 Hz Hz δ (ppm) c. Conside two molecules, A and B, which ae involved in an exchange eaction: A k AB k BA B The exchange between molecules A en B is slow on the chemical shift time scale and theefoe we obseve one peak fo molecule A and one sepaate peak fo molecule B. What is the effect on the longitudinal magnetization of molecule A ( ) when we educe the magnetization of B to zeo? Explain you answe using an expession fo. Longitudinal effects of exchange: d dt d dt R R 1 A 1 B ( ) ( ) k + k AB BA T 1A ( ) ( ) + k k AB longitudinal exchange elaxation BA T 1B k + k Reduce the magnetization of B to zeo: ( ), ( t) d ( ) Thus: dt T 1A k AB So, will decease with k AB as a esult of the exchange eaction. At the same time, will incease (i.e. ecove) with 1/T 1A due to T 1 elaxation. The final effect depends on the AB AB + k k BA BA page 7/1

8 magnitude of k AB elative to 1/T 1A. When the exchange is fast and T 1 elaxation is slow, the decease in will be lage compaed to a situation whee the exchange is slow and T 1 elaxation is fast. 6. Signal acquisition and pocessing a. Daw the fequency spectum fo a spin with fequency ω in case you did only detect the signal along the x-axis (eal signal). You will find a peak at both ω and -ω. b. What is the poblem in question 6a and how can you fix this? Positive and negative fequencies cannot be distinguished when you do a Fouie tansfomation on a eal signal. You can fix this by measuing a complex signal. This is called quadatue detection and can be pefomed by using two coils (one along x and one along y) o by electonic manipulation. c. Suppose you measue a 1 H NR spectum with a spectal width of ppm and 56 complex points at a field stength of 7 Tesla. What is the digital esolution ( ν) of the spectum? γ( 1 H) ad T -1 s -1 SW Digital esolution: ν N 7 Tesla (1/π) γ 7 98 Hz SW ppm Hz ν SW N Hz 7. D NR a. When do you obseve a coss peak in a D EXSY spectum and when do you obseve a coss peak in a D COSY spectum? page 8/1

9 The appeaance of coss peaks in a D EXSY spectum is an unambiguous indication that magnetization between two esonances has been exchanged (though chemical exchange o coss elaxation). In a D COSY spectum, you obseve coss peaks between spins that shae a mutual J coupling. b. What ae axial peaks? Axial peaks ae signals that ae not fequency labeled in ω 1. Axial peaks ae caused by T 1 elaxation, but also by expeimental impefection. Fo example, spins that ae not excited by the fist 9 pulse, have no evolution in t 1 and thus ae not fequency labeled in ω 1. c. Why is the digital esolution usually lowe ( ν lage) fo the ω 1 diection of a D NR spectum compaed to the ω diection? Because fo each point in the ω 1 diection you need to epeat the whole expeiment, leading to vey long total expeiment times fo many points in the ω 1 diection. Theefoe, the numbe of points (N) in the ω 1 diection is usually smalle than fo the ω diection, esulting SW in a lage ν ( ν ). N 8. agnetic esonance imaging a. What is the coespondence between D NR spectoscopy en D R imaging? Fo both D NR spectoscopy en D R imaging, one of the dimensions needs to be detected indiectly. Fo both techniques, this is done by having the magnetization acquie a cetain phase. Fo D NR spectoscopy the phase develops due to Lamo pecession and depends on the Lamo fequency and the length of t 1 (the latte is vaied fo diffeent expeiments). Fo D R imaging the phase develops due to the phase encoding gadient and depends on the position of the spin and the amplitude of the phase encoding gadient (the latte is vaied fo diffeent expeiments). b. Wate and fat have a diffeent chemical shift. The diffeence in chemical shift between wate en fat is about 3.5 ppm. This causes a so-called wate-fat shift in RI images: in the RI image the fat does not appea at its actual location but it is shifted to a diffeent page 9/1

10 location. Calculate the wate-fat shift in the x-diection fo a 7 T scanne and a ead-out gadient G x of 1 mt/m. γ( 1 H) ad T -1 s -1 x + γ G x wate: ω ( ) ωwate x wate x + γ G x fat: ω ( ) ω fat x fat At fixed fequency ω(x): x x wate x ( ωwate ω fat ) ( γ Gx ) ( δ wate δ fat ) γ B ) ( γ Gx ) ( ) ( ).45 1 fat 3 m c. The y-diection of the RI image is acquied using phase-encoding. Fo the phase encoding, the expeiment is epeated a numbe of times with diffeent amplitudes fo the phase-encoding gadient. Will you also obseve a wate-fat shift in the y-diection? Give you easoning. The phase developed due to the phase encoding depends on (1) the Lamo fequency and () the position of the spin along the phase-encoding diection and the amplitude of the phaseencoding gadient. E.g. fo wate: ϕ ω( y ) δ ( ω + γ G x ) δ wate x wate. Howeve, because the length/duation of the gadient (δ) is constant, the phase accumulation due to Lamo pecession is also constant and the phase diffeences between the diffeent expeiments ae not influenced by diffeences in Lamo fequency. Theefoe, thee is no wate-fat shift in the y-diection. page 1/1