Pulsed Nuclear Magnetic Resonance

Experiment NMR

University of Florida — Department of Physics PHY4803L — Advanced Physics Laboratory

References Theory C. P. Schlicter, Principles of Magnetic Res- Recall that the hydrogen nucleus consists of a onance, (Springer, Berlin, 2nd ed. 1978, single proton and no neutrons. The precession 3rd ed. 1990) of a bare proton in a magnetic field is a simple consequence of the proton’s intrinsic angu- E. Fukushima, S. B. W. Roeder, Experi- lar momentum and associated magnetic dipole mental Pulse NMR: A nuts and bolts ap- moment. A classical analog would be a gyro- proach, (Perseus Books, 1981) scope having a bar magnet along its rotational M. Sargent, M.O. Scully, W. Lamb, Laser axis. Having a magnetic moment, the proton Physics, (Addison Wesley, 1974). experiences a torque in a static magnetic field. Having angular momentum, it responds to the A. C. Melissinos, Experiments in Modern torque by precessing about the field direction. Physics, This behavior is called Larmor precession. The model of a proton as a spinning positive Introduction charge predicts a proton magnetic dipole moment µ that is aligned with and proportional To observe nuclear magnetic resonance, the to its spin angular momentum s sample nuclei are first aligned in a strong magnetic field. In this experiment, you will learn µ = γs (1) the techniques used in a pulsed nuclear magnetic resonance apparatus (a) to perturb the where γ, called the gyromagnetic ratio, would nuclei out of alignment with the field and (b) depend on how the mass and charge is dis- to measure the small return signal as the mis- tributed within the proton. Determined by aligned nuclei precess in the field. various magnetic resonance experiments to Because the return signal carries informa- better than 1 ppm, to four figures γ = 2.675× tion about the nuclear environment, the tech- 108 rad/(sec tesla). nique has become widely used for material Note that different nuclei will have different analysis. Initially, you will study the behav- spin angular momentum, different magnetic ior of hydrogen nuclei in glycerin because the moments, and different gyromagnetic ratios. signal is easy to find and interpret. Samples The gyromagnetic ratio given above is only for involving other nuclei and other nuclear envi- the hydrogen nucleus. In NMR, the word pro- ronments can also be investigated. ton generally only refers to the nucleus of the

NMR 1 NMR 2 Advanced Physics Laboratory

1H hydrogen isotope (a bare proton), not pro- where tons in other nuclei. For example, the nucleus ω0 = γB0 (6) of 16O has eight protons, but no nuclear spin or is called the Larmor frequency. magnetic dipole moment. While it is possible NMR signals arise from the collective be- to study NMR in other nuclei, such measure- havior of a large number of nuclei in a macro- ments require different experimental settings scopic sample. The net nuclear angular mo- and can not be performed at the same time mentum S is the sum of the individual spin that protons are under investigation. tot angular momenta Exercise 1 Show that the classical value for ∑ S = s (7) the proton gyromagnetic ratio (the ratio of the tot classical magnetic moment to the classical an- The net dipole moment becomes gular momentum) is e/2m (a) assuming the ∑ proton is a point particle of mass m and charge µtot = µ e moving at constant speed in a circular orbit, ∑ = γs and (b) assuming the proton is a solid sphere with a uniform mass and charge density rotat- = γStot (8) ing at a constant angular velocity. (c) Using the proton mass m = 1.67 × 10−27 kg, and The dipole moment per unit volume is called charge e = 1.60 × 10−19 C, find the ratio of the magnetization and is given the symbol M. the true gyromagnetic ratio of the proton given Depending on experimental conditions M may above to this classical value (called the proton vary at different points within a sample. Here g-factor). (d) Why might different nuclei be we will assume M is uniform throughout the expected to have different gyromagnetic ratios? sample volume V so that M = µtot/V . (How- ever, keep in mind there may be effects as- The proton has two energy eigenstates in sociated with a non-uniform sample magnetization.) In the jargon of NMR, it is more a magnetic field B0: spin-up and spin-down. The eigenenergies are governed by the Hamil- common to refer to the sample magnetization tonian rather than the dipole moment, and we will adopt this language in this writeup. In most H = −µ · B0 (2) instances, the term magnetization and dipole Defining the z-axis along B0 moment can be substituted for one another without confusion. B = B zˆ (3) 0 0 It is important to note that Eqs. 7 and 8 and with µ = γs, the Hamiltonian becomes are vector sums, so if the individual proton moments are randomly oriented—as is often H = −γB0sz (4) the case in the absence of any applied magnetic field—the net angular momentum (and where sz is the z-component of the spin. Since net dipole moment) will sum to zero. In a sz has eigenvaluesh/ ¯ 2 for the spin-up state magnetic field, however, a sample of protons at − H and h/¯ 2 for the spin-down state, has an absolute temperature T will come to ther- eigenenergies mal equilibrium with more spins in the lower energy (spin-up) state. The ratio of the num- hω¯ 0 E = ∓ (5) 2 ber of protons in the higher energy state N−

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z to the lower energy state N+ is given by the Boltzmann factor:

N− − = e ∆E/kT (9) M N+ ωωω 0 where k is Boltzmann’s constant and ∆E = hω¯ 0 is the energy diﬀerence between the spin- up and spin-down states. The excess of protons in spin-up states re- B0 sults in a net equilibrium magnetization M0 in the direction of B0. That is, M0 = M0zˆ, where

M0 = µ (N+ − N−) /V (10) and µ = γh/¯ 2 is the magnitude of the z- Figure 1: The magnetization M precesses component of an individual proton magnetic clockwise about an external field B0. moment. M0 can be calculated from the Boltz- mann factor with (N +N−)/V being the den- + angle with the z-axis. The resulting torque sity of protons in the sample. × τ = µtot B0 causes the angular momentum Exercise 2 (a) The ratio of excess spin-up Stot to change according to the equation of protons to the total number of protons would motion: τ = dStot/dt. be expressed: (N − N−)/(N + N−). Eval- + + dS uate this expression in the weak field or high tot = µ × B (11) dt tot 0 temperature limit: kT ≫ hω¯ 0 to lowest order in the small parameter hω/kT¯ . (b) Calculate By substituting Eq. 8 for Stot and dividing the value of M0 for water at room temperature through by V , this equation can be rewritten in a 3 kG field. (c) Why must you determine the density of hydrogen atoms in the sample, dM = γM × B (12) but not include any protons inside the water’s dt 0 oxygen nuclei? (d) The dipole moment of a simple wire loop of area A carrying a current With B0 = B0zˆ and ω0 = γB0 this equation I is µ = IA. So how big is the dipole moment further simplifies to of a 1 cm3 sample of water in a 2 kG field? dM Express your answer in SI units and also as = ω0M × zˆ (13) the current needed in a 1 cm2 loop that would dt give it the same dipole moment. By components We will later show how the nuclear mag- dMz netization M can be turned away from its = 0 (14) dt equilibrium alignment with the z-axis. For dMx the present discussion we assume the config- = ω0My dt uration shown in Fig. 1 has been achieved dMy and M (and µ and S ) make a non-zero = −ω0Mx tot tot dt

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These are the equations for the clockwise pre- is separate from the receiver coil used to pick cession of M on a cone that maintains a con- up the Larmor precession. In our apparatus stant polar angle with the z-axis. This be- one coil performs both jobs. havior is called Larmor precession and the fre- The oscillating field is produced by apply- quency of the precession is ω0, the Larmor fre- ing an ac voltage to the coil at a frequency quency defined after Eq. 5. ω near the Larmor frequency ω0 causing an As M precesses, its dipole field precesses alternating current to flow in the coil. Near with it. The precession can be detected by the the coil center, a linearly polarized magnetic alternating voltage it induces in a receiver coil field is created which oscillates back and forth surrounding the sample. The voltage arises along the coil axis. Choosing an arbitrary ini- from Faraday’s law V = −dϕ/dt and thus the tial phase and taking the amplitude of this coil axis is oriented perpendicular to B0 to field as 2B1, the linearly oscillating B-field can maximize the change in the magnetic flux ϕ be written through the coil as M precesses. ′ B = 2B cos ωt xˆ (17) To get a precession and an induced coil volt- 1 1 ′ age (signal), M must make a non-zero angle The field has been named B1 and the ampli- with B0, i.e., the z-axis. So we define the tude has been chosen as 2B1 because this lin- transverse magnetization Mr early polarized field will next be decomposed into two counter-rotating, circularly polarized Mr = Mxxˆ + Myyˆ (15) components each with half the amplitude of the linearly polarized field. That is, we can and the longitudinal magnetization add and subtract the term B1 sin ωt yˆ to write B′ = B (cos ωt xˆ − sin ωt yˆ) (18) Mz = Mzzˆ (16) 1 1 +B1 (cos ωt xˆ + sin ωt yˆ) The amplitude of the precession-induced coil Each component now represents a vector of signal is proportional to the magnitude M of r constant amplitude B rotating in the xy- M and is independent of M . Note that M 1 r z r plane at an angular frequency ω. These two is zero or positive, while M can be positive, z circularly polarized components rotate in op- negative, or zero. posite directions and when added together In thermal equilibrium, the magnetization is produce the linearly polarized field of ampli- entirely longitudinal: M = M zˆ or M = M 0 0 z 0 tude 2B . and M = 0. With no transverse magneti- 1 r The first component rotates in the same zation, there is no precession and no induced sense as the Larmor precession and is respon- coil voltage. To observe Larmor precession we sible for tipping the magnetization away from need a way to create transverse magnetiza- the z-axis as will be described next. The section. We will show how an oscillating mag- ond, rotating opposite the Larmor precession, netic field can tip the equilibrium longitudinal has little effect on the protons and can be ne- magnetization away from the z-axis so that glected. Neglecting the counter-rotating com- precessional motion will occur. To this end, ponent is called the rotating-wave approxima- the protons are placed inside a drive coil ori- tion and allows us to write what will be called ented with its axis perpendicular to B . The 0 the B field as coil axis will be taken to define the laboratory 1 x-direction. In some apparatuses the drive coil B1 = B1 (cos ωt xˆ − sin ωt yˆ) (19)

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The equation of motion for the magnetiza- The first three terms give the time derivative tion M is still given by Eq. 12 of v as it would appear in the rotating frame and will be written (dv/dt)′. The final three dM = γM × B (20) terms are equal to the expression ω × v where dt ω is the vector angular velocity of the rotating with the replacement of B by B, the vector 0 frame. ( )′ sum of B and the rotating field B given by dv dv 0 1 = + ω × v (27) Eq. 19. dt dt However, it will be mathematically conve- For a frame rotating clockwise about the z- nient to work in a frame rotating clockwise axis at a frequency ω the transformation is about the z-axis at the frequency ω. In this given by Eqs. 27 with ω = −ωzˆ. frame, the unit vectors are time dependent Eq. 20 then becomes ′ ( )′ xˆ = cos ωt xˆ − sin ωt yˆ (21) dM ′ + ω × M = γM × (B + B ) (28) yˆ = sin ωt xˆ + cos ωt yˆ dt 0 1 ′ zˆ = zˆ or ( )′ ( ) dM ω and the B1 field is constant = γM × B + + B (29) dt 0 γ 1 ′ B1 = B1xˆ (22) With B0 = B0zˆ and ω = −ωzˆ we can define The B0 field—lying along the rotation axis—is an effective field along the z-axis the same in both the laboratory and rotating ′ ′ B = B zˆ (30) frames. 0 0 A time dependent vector v in the laboratory where ω frame could be expressed B′ = B − (31) 0 0 γ ′ v = vxxˆ + vyyˆ + vzzˆ (23) With B1 = B1xˆ , the total effective field in the rotating frame becomes where vx, vy, and vz are functions of time. ′ ′ ′ This same vector in a rotating reference frame B = B0zˆ + B1xˆ (32) would be expressed The “on resonance” condition is ω = ω0 = ′ ′ ′ ′ ′ ′ γB , and is achieved by adjusting either the v = vxxˆ + vyyˆ + vzzˆ (24) 0 static field strength B0 (thereby changing the ′ ′ ′ where vx, vy, and vz are new functions of Larmor frequency ω0) or the frequency ω of time determined by vx, vy and vz and the unit the voltage applied to the coil. On resonance, vector transformations describing the rotating ′ the effective field along the z-axis is zero (B0 = frame. 0) and the net effective field lies entirely along ′ ′ Differentiating Eq. 24 gives the rotating x-axis. (B = B1xˆ ). ′ ′ ′ But on resonance or not, the net effective dv dv ′ dv ′ dv ′ = x xˆ + y yˆ + z zˆ (25) field B′ is a constant, and the equation of mo- dt dt dt dt ′ ′ ′ tion for M in the rotating frame becomes ′ dxˆ ′ dyˆ ′ dzˆ ( )′ +vx + vy + vz dt dt dt dM ′ = γM × B (33) (26) dt

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This equation is analogous to Eq. 12, and its (a) (b) (c) solution is also analogous. M precesses clock- z z z wise about the effective field at a frequency ω′ = γB′ where B′ is the magnitude of B′. ω′ is called the Rabi frequency in analogy with the first treatment of this effect in atomic y' y' y' physics by I.I. Rabi. x' x' x' On or near the resonance condition, B1 ≫ ′ B0, the net effective field lies along or nearly ′ along the rotating x -axis. Then, if B1 is Figure 2: (a) Immediately following a π/2 pulsed on and off just long enough to cause pulse, the individual spins—originally aligned 1/4 of a Rabi cycle, M will Rabi-precess to ′ along the z-axis—are Rabi-precessed to the ro- the rotating y -axis. This kind of pulse is ′ ◦ tating y -axis. (b) Afterward, different indi- called a 90 or π/2 pulse. After the π/2 pulse vidual spins Larmor precess slightly faster or ends, it should be clear that in the labora- slower than the average Larmor frequency and tory frame the magnetization will lie in the get somewhat ahead of or behind the y′-axis. xy-plane and it will be Larmor-precessing flat (c) Ultimately, the spins spread out over the out like one blade of a propeller. Were B1 left entire xy-plane. on for a longer time, M would Rabi-precess further than 90◦ and the Larmor precession cone would start to close around the negative arrow) immediately following a π/2-pulse. The pulse has Rabi-precessed the equilibrium z-axis. If the B1 field is pulsed on just long ′ enough to cause 1/2 of a full precession cycle magnetization (dashed arrow) to the y -axis. (such a pulse is called a 180◦ or π pulse), the At the end of the B1 pulse, the now transverse magnetization becomes inverted and points magnetization induces an observable Larmor opposite the external magnetic field. precession signal. After the π/2-pulse, the relaxation of the transverse and longitudinal magnetization Relaxation from their perturbed values Mr = M0 and Mz = 0 back to their equilibrium values of Relaxation in NMR refers to processes that Mr = 0 and Mz = M0, respectively, can occur restore the equilibrium magnetization after a on quite different time scales. perturbation, such as a B1 pulse, has created a non-equilibrium population of nuclear Free Induction Decay spin states. The breadth of possible relaxation studies and the wealth of information Figures 2b and c show a case where the trans- regarding the nuclear environment and the nu- verse magnetization (and thus the signal) de- clear motion that becomes available through cays away quickly, before there is any signifi- those studies is why NMR has become such an cant change in the longitudinal magnetization. important experimental/analytical technique. As in our apparatus, this is often the case, and The sections to follow focus on a small sub- is a result of the inhomogeneities in the B0 set of those studies which highlight the most field. Spins in different parts of the sample general features of relaxation. experience slightly different B0 field strengths Figure 2a shows the magnetization M (solid and thus precess at slightly different Larmor

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frequencies. As shown in Fig. 2b, the individ- bol T2. The spin echo (described later) is an ual dipole moments begin to spread out in the interesting phenomenon which largely cancels xy-plane. In Figs. 2b and c, the x′- and y′-axes the effect of field inhomogeneities and makes ∗ rotate at the average Larmor frequency, cor- measurements of T2 possible even when T2 is responding to the average B0 field strength. much shorter. Some spins precess at this frequency and stay aligned with the y′-axis. However, those spins in regions of higher field strength precess faster Longitudinal Relaxation ′ than the average and get ahead of the y -axis, As described in the previous section, any and those in lower fields fall behind. The vec- transverse magnetization decays to zero as the tor sum of the individual spins, and therefore individual spins fan out. What happens to the net transverse magnetization, decreases. longitudinal magnetization Mz after it is per- As the spins completely spread out over the turbed from equilibrium? Recall (Eq. 10) that xy-plane as in Fig. 2c, the transverse magne- the value of Mz is directly proportional to the tization, and therefore the signal, decays to difference between the populations of protons zero. The decay of the transverse magnetiza- in the spin-up and spin-down states. Also re- tion is called free induction decay (FID). We call that these states differ in energy by the will also refer to the signal picked up by the amounthω ¯ 0. In a sample, individual spins receiver coil as a FID. make transitions from spin-up to spin-down The relaxation of the transverse magnetiza- and vice versa, absorbing and releasing en- tion Mr from an initial value Mi to its equi- ergy depending on the transition direction. At librium value of zero is often modeled by an equilibrium, although the populations of the exponential two states are unequal (more spins are in the −t/T ∗ lower energy spin-down state), the transition Mr(t) = Mie 2 (34) rates in the two directions are equal and there ∗ where T2 is called the transverse relaxation is no net energy transfer into or out of the spin time. While most real FID signals decay non- degrees of freedom. exponentially, the concept of a time constant If the longitudinal magnetization Mz is not associated with the decay of transverse mag- at its equilibrium value M0, the transition netization is nonetheless useful. rates in the two directions will not be equal, Inhomogeneities in the B0 field are not the and there will be a net energy transfer into or only reason the spins fan out. There are out of the spin degrees of freedom. For iso- also randomly fluctuating magnetic fields from lated spins, spontaneous photon emission and other sources in the sample, e.g., the dipole stimulated emission and absorption from the field of one proton moving past another. These background radiation field would reestablish random fields vary over the sample volume and the Boltzmann populations, though it might thus also cause proton precession rates to vary. take quite some time to come to equilibrium. Were the B0 field perfectly uniform, the ran- In a real sample, the spins are not isolated. dom fields (and other processes) would become While stimulated emission and absorption are the dominant source for the decay of the FID. still the major mechanisms by which transi- The decay of transverse magnetization would tions occur, the radiation field experienced by then occur on a longer time scale, called the the spins includes contributions from the mo- spin-spin relaxation time, and given the sym- tion of the electrons and nuclei in the sample.

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In effect, the spins exchange energy with the π-pulse flips the pancake of spins over, 180◦ electrons and other nuclei causing the transi- about the x′-axis. The three illustrated spins tion rates to be much higher than for isolated end up as shown in Fig. 3c. The proton that spins and the approach to equilibrium is much was ahead is now behind the average and the faster. proton that was behind is now ahead. In ad- An exponential approach to equilibrium is dition to no T2 processes, let’s also temporar- often a satisfactory model. If the initial per- ily assume that the spins don’t move around turbed longitudinal magnetization is taken as within the field during the pulse sequence. Mi, Mz(t) would be expressed A spin in a stronger-than-average field stays in that field, continually precessing at a con- −t/T1 Mz(t) = M0 + (Mi − M0)e (35) stant, faster-than-average rate from the begin- ning of the π/2-pulse to well after the end where T1 is called the longitudinal relaxation of the π-pulse. Similarly, a spin in a lower- time. T1 is also called the spin-lattice relax- than-average field continually precesses at a ation time in reference to NMR in solids where constant, slower-than-average rate. The faster the constituents of the lattice (nuclei and elec- precessing proton in Fig. 3c, now behind, be- trons) provide for the exchange of energy nec- gins to catch up to the average. The slower, essary for the relaxation process. now ahead, begins to lose ground to the average. If it took a time t/2 to get to the Spin Echo configuration in Fig. 3b/c, then after an addi- tional time t/2 (i.e., at the time t), the three The basic spin echo technique, used to mea- spins (in fact, all the spins) would be realigned sure the spin-spin relaxation time T , is to ap- 2 along the −y′-axis as shown in Fig. 3d. The ply a π/2-pulse, wait a bit, and then apply transverse magnetization created by the orig- a π-pulse, as illustrated in Fig. 3. In (a) the inal π/2-pulse would be completely restored magnetization M is shown already rotated to ′ and the amplitude of the FID would be as the y -axis by an initial π/2-pulse. With no large as it was at Fig. 3a. This is the echo field inhomogeneities and no T processes, all 2 FID. Afterward, the spins again fan out and spins would precess at the same rate and there again the FID decays away. would be no decay of the transverse magnetization. If we now allow for field inhomo- The magnetization at t (Fig. 3d) will be geneities, then some spins will precess faster smaller than the original magnetization be- than average and some slower. As in Fig. 2, cause spins don’t only fan out due to the B0 the spins eventually fan out into a uniform field inhomogeneities. As mentioned previ- pancake covering the xy-plane and the FID de- ously, the spin-spin relaxation time T2 arises cays to zero. Figure 3b shows the situation at because there are other random fields in the some time t/2 when the spins are completely sample which make individual spins precess fanned out, though in the figure only three at different rates. For liquids, the diffusion spins are shown. One spin is precessing at the of protons within the sample provides an ad- average Larmor frequency, another is precess- ditional reason the spin echo signal will be ing faster than the average, and the third is smaller than the original FID. During the precessing slower. interval from the initial π/2-pulse to the ap- Figure 3b also shows the application of a pearance of the echo, the spins move within second B1 pulse, this one a π-pulse. The the sample. Consequently, if there is any in-

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(a) (b) (c) (d) z z z z s s s a f y' y' a y' y' f f B1 x' x' x' x' π/2π/2π/2 πππ RF Pulses 0 t/2 t FID

Figure 3: The π/2 − π spin echo sequence.

homogeneity in the B0-field, all spins will not to determine T2 as well as the diffusion con- precess at the same rate for the whole time t stant D (under the right conditions such as a as was assumed, and the echo will not recon- known field gradient and sufficiently long T2). ∗ stitute perfectly. Whereas the T2 decay of the FID depends on The amplitude of the echo pulse, as a func- the field inhomogeneity over the whole length tion of the time t when it is made to appear, of the sample, field inhomogeneity in the diffu- decays according to the relation sion term involves the smaller distances spins [ ( )] t γ2G2Dt3 diffuse during the measurement time t. M(t) = M0 exp − + (36) T2 12 Inversion Recovery The first term in the exponential describes the spin-spin relaxation processes. The sec- We next consider the inversion recovery se- ond term arises from the diffusive motion. G quence used to measure T1 and illustrated in is the B0-field gradient, i.e. the maximum rate Fig. 4. Recall that T1 gives the time scale for of change of B0 with position (units of G are relaxation of longitudinal magnetization to- thus tesla per meter) and D is the diffusion co- ward the equilibrium value M0. The inver- efficient which has units of square meters per sion recovery sequence starts with a π-pulse second. applied to a sample initially aligned in the Equation 36 illustrates how spin echo mea- z-direction. In Fig. 4a, the initial magneti- surements provide information about parame- zation (dashed arrow) is inverted by the π- ters of physical interest by unmasking the ef- pulse, and ends up pointing along the nega- fects of field inhomogeneities responsible for tive z-axis: Mz = −M0. In Fig. 4b, the in- ∗ the quick and relatively uninteresting T2 de- verted M becomes smaller as it relaxes back cay of the FID. Fits to the data can be used toward equilibrium. M continues shrinking,

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(a) (b) (c) τ z z z long c

M τ 1/ ω 0 c ˜ 0 y' y' y' M short τc B x' B x' B x' 1 1 1 ω ω M 0 M Figure 5: The spectral density function changes as the mean time between collisions π π/2 RF τ changes. Pulses c

∗ t in Fig 4b. The FID will decay with the T2 time constant, but its initial size will be pro- FID portional to Mz(t). By varying the time t at which the second π/2 pulse is applied, the behavior of Mz(t) can be measured.

Figure 4: The π − π/2 inversion recovery se- Exercise 3 Make a plot of Mz(t) (assuming quence. exponential decay) and describe what effect the second π/2-pulse will have on M if it is applied ∞ passes through zero, and then increases until at t = 0, t = T1 ln 2, and t = ? it reaches the equilibrium value M0. An exponential decay of the longitudinal Spectral Density Function magnetization from a perfectly inverted value An important concept in NMR relates to the M = −M to the equilibrium value M = M z 0 z 0 intensity of the random electromagnetic fields is modeled ( ) experienced by spins in different frequency − t/T1 intervals. This spectral density function is Mz(t) = M0 1 − 2e (37) strongly related to the motion of the spins and Because the perturbed magnetization has therefore depends indirectly on the tempera- no component in the xy-plane after it is in- ture and (for liquid samples) the viscosity of verted by the π-pulse (nor while it is relax- the sample. At low temperatures and high ing toward equilibrium), there is no trans- viscosity the mean time tc between molecular verse magnetization to produce a FID. To ob- collisions is long and the spectrum is domi- serve the changing Mz(t), the magnetization nated by low frequency fluctuations. At high is tipped over to the xy-plane by applying a temperatures and low viscosities, tc is short second pulse, this time a π/2-pulse, at the de- and the spectrum spreads out to higher fre- sired observation time t. This is illustrated quencies. Figure 5 shows how the frequency

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distribution spreads out as the collision time signal power gen. decreases. splitter attenuator gate amp

Recall that the B1-field must oscillate near 50 Ω the resonant frequency ω0 to cause a Rabi- diode pairs precession of M away from its equilibrium D alignment with the z-axis, and thus effect a B A C change in the relative populations of the two hybrid spin states. So, too, must the random fields junction 50 Ω have intensities in the frequency interval near sample Cs coil ω to stimulate transitions and reestablish the tuning Cp 0 magnet equilibrium longitudinal magnetization M = capacitors z doubly M0. Thus at low temperatures (long tc) where balanced voltage low frequencies dominate the spectrum, T is mixer preamp 1 LO long. While at very high temperatures (short IF RF scope tc) the spectrum is spread so thin that again rf amp the intensity near ω0 is small and T1 is long. At intermediate collision times one expects to diode pair find a minimum in T1. Theory predicts the shortest T1 when tc is of order 1/ω0.

T2 is also affected by these “resonant” field fluctuations but in addition, field fluctuations Figure 6: Schematic of our pulsed NMR ap- along the z-axis at any frequency will cause paratus. the individual spins to fan out; just as they fan out due to field inhomogeneities. It is these 1 fluctuations that can cause T2 to be apprecia- amplitude of 10 dBm (0.707 Vrms, 2.00 Vpp). bly shorter than T1. This value is chosen because the 50-50 split- Besides varying the sample temperature or ter reduces the voltage at its two outputs by viscosity to influence the relaxation times, 3 dBm and 7 dBm is the recommended ampli- adding small amounts of paramagnetic ions to tude for the local oscillator (LO) input of the the sample will change the amplitude of the doubly balance mixer. frequency fluctuations. These ions have large This RF is reduced another 10 dB by the at- magnetic moments, which when whizzing by tenuator to about -3 dBm so that it will be at the protons, cause very large fluctuations. the appropriate level before being fed to the Their importance can be investigated by vary- RF gate. The gating signal for the RF gate ing the concentration of the ions. is not shown in the schematic. It is derived from a TTL circuit in a separate interface box which is driven by timing signals from a Na- Apparatus tional Instruments 6601 timer/counter board inside the computer. The output of the RF A schematic of the experimental apparatus is 1 dBm is a unit of absolute power and is defined√ by shown in Fig. 6. 2 2 the expression 10 log10(V /V0 ) where V0 = 0.05 = The RF signal generator should be set to 0.224 volts is the rms voltage across a R = 50 Ω resistor 2 produce a 16 MHz, continuous RF signal at an that will dissipate energy at a rate of V0 /R = 1 mW.

July 18, 2019 NMR 12 Advanced Physics Laboratory gate then consists of short RF pulses whose of the signal at the mixer output will be the amplitude and frequency are set by the signal difference between the FID frequency (at the generator amplitude and the attenuator set- Larmor frequency ω0) and the signal genera- tings and whose durations are set by the com- tor frequency ω. The B0 field strength will be puter generated gating signal. adjusted so that the “beat” frequency ω − ω0 The gated RF pulses are amplified to about is around 1 kHz. 50 Vrms by the power amplifier. However, the The signal from the mixer is filtered and RF gate is not perfect and even while the com- amplified by a voltage preamp before being puter gating signal is off, a small amount of RF measured with a digital oscilloscope. still passes through. After amplification, this gate-off RF signal has an amplitude around 1 V and is largely eliminated by the two sets Experiment of diode pairs. The magnet is field regulated using a Hall The pulsed RF is delivered to the coil probe attached to one pole face. The instruc- through a hybrid junction which when prop- tor will show you how to bring up the magnet. erly balanced, sends half the incident power Remember to open the cooling water supply to the coil and half to the 50 Ω terminator. and drain line. While it is important to have The receiver circuity (RF amplifier and dou- the water running while the magnet supply is bly balanced mixer) thus sees very little of the on it is even more important to turn off the transmitter power driving the B1-field. water lines when the magnet supply is The sample coil and the two capacitors Cs turned off. If the water is left running for and Cp form a resonance circuit, called a tank more than a few hours while the supply is not circuit. Cs and Cp, called the series and par- running, humidity in the air will condense all allel capacitors, are variable and must both over the internal cooled parts of the supply. If be properly tuned (1) to get the resonance you think the water has been left on for more frequency√ of the tank circuit—approximately than a few hours with the supply off, do not 1/ L(Cp + Cs)—equal to the drive frequency turn the supply back on without first asking of the signal generator and (2) to make the an instructor to check it for condensation. overall impedance of the tank circuit equal to 50 Ω. Satisfying these two conditions en- Tuning the tank circuit sures that the current oscillations in the coil will be maximized (giving the largest possible Make sure the RF power amplifier is ei- amplitude of the rotating B1 field). Matching ther off or its output is temporarily dis- the total impedance to 50 Ω (with no imagi- connected. If the gate signal from the com- nary, i.e., reactive component) is also needed puter stays high (as it may when starting or for proper operation of the hybrid junction. shutting down the computer), the high (50 W) The hybrid junction also routs the FID to output power will be on continuously instead the receiver circuitry. The first component of of only for short bursts and the components the receiver is a low noise RF amplifier pro- downstream will fry in no time at all. tected from large signals by a diode pair to Check the RF coil and remove any sample ground. The amplified RF signal is frequency vial that may have been left in it from the pre- “mixed” with the steady 16 MHz signal from vious experimenters. Place a glycerin sample the other output of the splitter. The frequency in the coil taking care to center the glycerin

July 18, 2019 Pulsed Nuclear Magnetic Resonance NMR 13 near the center of the coil and then place the pacitance in the correct direction. If it goes coil near the center of the magnet. up, you are going the wrong way; change the Set the signal generator for a 10 dBm output parallel capacitance in the opposite direction. (0.7 Vrms, 2 Vpp) and a frequency of 16 MHz. Continue this two-step procedure until the os- This signal goes to the splitter which provides cilloscope signal becomes fairly non-sinusoidal two equal RF voltages of 7 dBm amplitude at and has an amplitude under 50 mV. The per- its outputs. fectly balanced circuit would show a doubled Disconnect the function generator from the frequency around 32 MHz. splitter and connect it instead to the hybrid Check with the instructor if this tuning pro- junction “A” input. Check that the hybrid cess is taking more than 15 minutes or so. junction “D” output is terminated with a 50 Ω When balanced, note how the balance is af- terminator. Disconnect the “B” output and fected when you move the sample coil a few monitor it with an oscilloscope. Be sure to use centimeters farther or closer to the pole faces. a BNC tee and 50 Ω terminator at the oscillo- The balancing is important, but getting it per- scope input. Get a clear stable trace showing fect is not critical. the 16 MHz waveform. The basic idea in tun- When you are finished tuning, return the ing the tank circuit capacitors is that when the function generator signal to the 7 dBm splitter impedance of the branch connected to the “C” input and return the splitter’s “A” and “B” output matches the 50 Ω impedance attached outputs to the “LO” input of the mixer and to the “D” output, the “B” output will be as to the input of the RF amp, respectively. small as possible. When the “C” and “D” outputs are not balanced, more of the “A” input will pass through to the “B” output. Learning the NMR pulse sequencer Adjust the series capacitor knob for a min- Temporarily rout the gating signal from the imum signal on the oscilloscope and note its computer interface to an oscilloscope input amplitude. Next adjust the back-most knob and rout the trigger signal from the computer 2 for the parallel capacitors by about 1/20 of interface to the oscilloscope trigger source. a turn at first (you may need to go in even Launch the NMR pulser program and view- smaller steps as you get close to the correct ing the gating pulses on the oscilloscope, learn setting) and note whether you are turning the how to externally trigger the oscilloscope and knob clockwise or counterclockwise. Either is see the gating pulses. Also learn how to op- OK at first. Because your hand on this knob erate the program to produce gating pulse- seems to significantly affect the tuning, don’t pairs. You can adjust the width of each pulse, worry about how it affects the oscilloscope sig- the separation between them, and the pulse nal, but move your hand away and then again sequence repetition rate. The separation be- adjust the series capacitance for a minimum tween the start of the (first) A-pulse and the on the oscilloscope. If the voltage at the min- start of the (second) B-pulse is labeled Td. Be- imum goes down after adjusting the series ca- tween the start of one A-pulse and the next is pacitance, you are changing the parallel ca- the repeat time Tr. Try turning the A- and B-pulses on and off by clicking on the appro- 2The parallel capacitors are two capacitors in series that are in parallel with the coil (go to ground). The priate buttons. The rising edge of the trigger series capacitor is a single capacitor in series with the output can be set to transition at the end of coil. the A-pulse, the B-pulse or at the temporal

July 18, 2019 NMR 14 Advanced Physics Laboratory position of a spin echo. It can also be off- changes the amplitude to the input (and out- set from these positions with the pretrigger put) of the gate and thus it also changes the control, which should be kept at zero. Play input to the RF amplifier. Then, because the around with the buttons and controls to get RF amplifier has a fixed gain, changing the familiar with their workings. Check that the attenuator will also change the amplitude of gating pulse durations and the separations be- its RF output. If the attenuation is reduced tween them agree with the values as specified too much, the input to the amplifier will be- on the program control panel. Check with an come too big and the output waveform will be- instructor if they don’t. come distorted. If the attenuation is raised too much, the RF amplitude out will become unnecessarily small, requiring unnecessarily long Creating the RF pulses Ta and Tb to get the desired π/2 or π pulses. Now, reconnect the gating signal from the The attenuator should be set so that the am- computer interface to the gating input on the plitude of the output RF pulse is right about RF gate and rout the pulsed RF from the 50 Vrms and is not distorted significantly. output of the RF gate to the input of the Exercise 4 Show that this amplitude would RF power amplifier. Disconnect the output produce 50 W of rms power into a 50 Ω load, from the power amplifier and turn it on. The the power limit for this amplifier. What would power level meter on the front panel should the peak-to-peak RF voltage (as measured on barely move because it measures average out- the oscilloscope waveform) be for this 50 W put power (into a 50 Ω load) and is only on for condition? short pulsed periods. If you ever see the amplifier output meter rise above zero Now further zoom in to see the amplitude while the output is connected to any- of the 16 MHz signal in the region where the thing, turn it off immediately. Some- gate should be turning it off, say, in the re- thing will likely be cooking. gion before the gating turns on. Because the Look at the amplified RF pulses using a 10× gate is not perfect, some small amplitude RF oscilloscope probe. (The pulse amplitude is leaks through to its output even when the RF too big for a direct connection to the oscillo- is supposed to be gated off. This leakage is scope.) Pull off the 10× probe’s clip lead and amplified by the RF amplifier and shows up push the straight tip end into the BNC jack. in its output. How big is this leakage com- The fit is a bit loose. Just rest the probe on pared to the 50 Vrms that it outputs when the the bench top with the tip connected. Look at RF is gated on? the RF pulses straight out of the power am- Connect the RF amplifier output to the plifier. Set the oscilloscope for the 10× probe, crossed diode circuit and move the 10× probe and adjust the time base and volts/division to the BNC connector after the diodes. Again (gain) so you can see the amplitude and du- look at the amplitude of the RF pulses when ration of the RF pulses. Then zoom in time the gate is on and off and describe the changes to see the turn on and the waveform for the in these two regions compared to the direct 16 MHz oscillations inside the RF pulse. output from the RF amplifier. The attenua- While looking at the RF pulses with the os- tion during the gated on period is mostly due cilloscope, adjust the attenuator around the to the 50 Ω resistor to ground in the crossed suggested value of 10 dB. This adjustment diode circuit. The diodes are responsible for

July 18, 2019 Pulsed Nuclear Magnetic Resonance NMR 15 attenuating the leakage RF during the gated magnetic field around that B-field strength off period. Explain how the diodes attenu- until you see a FID. Keep your eye on the ate the leakage nearly perfectly without doing oscilloscope because decaying oscillations at much attenuation during the gated on period. the beat frequency (the FID) will appear at Set the pulser program for A-pulses only resonance. Once you have a signal, finger (turn B off). Make the A-pulse width 12 µs, tighten the coarse control locking screw and and set the repetition time to 250 ms. Rout adjust the fine control to get the FID beat fre- the amplified RF pulses from the power am- quency around 1 kHz. Check that its locking plifier through the series diodes to the hybrid mechanism—a sliding lever at the base of the junction “A” terminal and rout the “B” termi- dial—is also loose (pushed left) during adjust- nal to the diode-protected input of the RF sig- ments. ∗ nal amplifier (not the power amplifier). Con- The FID decays with the time constant T2 , nect the output of the RF signal amplifier to which will vary if the sample is moved around the RF input of the mixer. The mixer’s LO (lo- in the field. Move the sample up/down, cal oscillator) input is taken from one of the back/forth, and left/right trying to find the two RF splitter outputs driven directly by the most uniform field region characterized by the function generator. The mixer’s IF (interme- longest decay. diate frequency) output is then routed to the If the B0 field varies—even by a little—the voltage preamp input. View the output from FID frequency varies by a lot. If your FID the preamp on the oscilloscope. Start with frequency is not very stable, the problem may this preamp set to AC coupling, high-pass fil- lie with the potentiometers connected to the tering to 100 Hz, low-pass filtering to 10 kHz, fine and coarse adjustment knobs. They may and gain to 1000. Set the oscilloscope to a be a little noisy (have a varying resistance) time base around 2 ms/div and the gain to depending on where they are set and whether around 50 mV/div. You are now ready to find they are locked in position. We have found a resonance. less problems when the coarse control knob is in the locked position. The fine control knob Finding the FID can be left in the unlocked position. If the FID frequency is still unstable, try finding a better Now it’s time to adjust the B0 field strength spot for the potentiometers—moving both di- to find the free induction decay. Put the glyc- als to keep the beat frequency around 1 kHz. erin sample into the sample coil (from the top) Remember to unlock both before making ad- taking care to get the 5 mm or so of glycerin justments and to lock down the coarse control centered in the 1 cm coil. You can put a small afterward. piece of tape on the sample if it slides around Another cause of field instability seems to too easily in the coil. be associated with the cooling water. We have Make sure the screw marked LOCK located found that changing a dirty filter in the water below the coarse dial on the magnet power line sometimes improves the magnet stability. supply is loosened. Use Eq. 6 and the pro- If your efforts to get a stable B0 fail and the ton gyromagnetic ratio to determine the ex- beat frequency continues to vary, you can still pected B-field needed to achieve a Larmor frequency of 16 MHz.3 Then slowly vary the factor of 2π difference between an angular frequency ω in units of rad/s and that same frequency f = ω/2π 3 Throughout this experiment be careful with the in Hz.

July 18, 2019 NMR 16 Advanced Physics Laboratory get good data. This is because mostly you will precesses at the Larmor frequency and decays make measurements of the starting amplitude quickly due to ﬁeld inhomogeneities. The pre- of the FID envelope. As long as the envelope cessing Mr(t) induces a pickup in the sample is stable, this measurement is independent of coil at the Larmor frequency with an ampli- the beat frequency. tude proportional to Mr(t). After amplify- If necessary, you can hit the stop button on ing that signal, mixing it, and amplifying it the oscilloscope to hold a good-looking FID again, the signal measured by the oscilloscope from which to make measurements. now oscillates at the beat frequency, (around

Whether or not you are having B0-field sta- 1 kHz), but its amplitude should be propor- bility problems, purposely vary the magnetic tional to Mr(t). That is, the envelope of the field fine control setting to see how the FID decaying oscilloscope signal should be propor- changes as the beat frequency changes. What tional to the magnitude Mr(t) role do the the preamp filter settings play in As the theory is usually concerned with the this regard? Adjust those as well. Report on starting amplitude Mr(0), the starting ampli- your observations and describe the issues asso- tude of the envelope should provide the desired ciated with the preamp settings and the FID measure. You should measure a peak-to-peak beat frequency and the field instabilities. Re- amplitude using the oscilloscope cursors—a turn the preamp filters to the recommended measure that should be proportional to Mr(0). 100 Hz and 10 kHz settings. Typical oscilloscope traces of FIDs and a spin echo are shown in Fig. 7 and demonstrate sev- Measurements eral issues that arise when trying to determine a peak-to-peak amplitudes as the separation In general, pulse sequences should not be re- of two horizontal oscilloscope cursors. Keep peated at intervals faster than 5T1. It takes in mind that the envelope is that of the oscil- at least this long for the sample to return to lations and there will be noise riding on top of thermal equilibrium after being perturbed. those oscillations. You should try to set the

With only A-pulses on, vary Ta to find cursors as if the noise were absent. pulses which rotate the magnetization by π/2, The leftmost trace shows a fairly strong FID π, 3π/2, 2π, and 5π/2. The π/2-pulse is at the where the noise component is small. In this first maximum in the FID signal. The π-pulse case, the cursors have been “eyeballed” a bit is at the first minimum, etc. Use the dura- above and below the trace to try to take into tion of the A-pulses relative to the rotation account where the envelope amplitude would of the magnetization they cause to determine be at t = 0 (t = 0 is at the external trigger the Rabi frequency and then use it to esti- or at the sharp spike arising from the strong mate the amplitude of your B1-field. You can RF drive pulse leaking into the detection sys- assume you are exactly on resonance (where tem). For the weaker FID in the middle, sig- ′ ω = γB1) for the calculation. nificant noise rides on top of the signal. The Applying π/2 A-pulses, adjust the B0-field cursors should again be set to give the am- slightly and notice the change in frequency of plitude of the envelope at t = 0, but because the detected signal. Describe how the signal of the noise, properly positioned cursors now changes as you go through the zero beat fre- may actually cross with some of the trace. The quency. third figure is from a spin echo with a small The true transverse magnetization Mr(t) noisy signal. For a spin echo, one is trying to

July 18, 2019 Pulsed Nuclear Magnetic Resonance NMR 17

Figure 7: The “size” of the FID should be chosen as the peak-to-peak amplitude at the start of a FID, not as the amplitude at the first maximum (first two traces). The size of a spin- echo signal (third trace) should be chosen as the peak-to-peak amplitude at the middle of the signal. When noise is significant, try to imagine what the signal would look like without it and measure these amplitudes as if it were absent. The figures show horizontal cursors positioned to take into account these issues and give a reasonable estimate of the peak-to-peak amplitudes needed.

measure the envelope amplitude at the mid- T1 dle of the signal. For the case shown in the figure, the noise is fairly large and has com- Set the A-pulse to a π-pulse by adjusting Ta. ponents on the same time scale as that of the Then turn the A-pulse off and set the B-pulse signal. The cursor separation has been set to to a π/2-pulse. Then turn the A-pulse back try to match the size of the oscillations at the on. Starting at Td small, as you increase Td, middle of the trace as if the noise were absent. you should see the FID decrease in size until In any case, be careful not to measure am- it is nearly zero as Mz increases from around plitudes when the oscillations are not clearly −M towards zero, and then it will increase bigger than the noise. 0 again as Mz increases to its equilibrium value M0. The point where the signal goes to zero, the crossover, occurs at t = 0.693T1. Mea- sure the amplitude of the FID as a function T ∗ 2 of Td. Check if your decay is exponential by computer fitting the amplitude of the FID as Look at the FID envelope for a π/2-pulse. a function of t = Td in Eq. 35 with M0, Mi Does it appear exponential? Make a rough es- and T1 as fitting parameters. Do not fit to ∗ − timate of T2 . Estimate the dimensions of your Eq. 37, which is Eq. 35 with Mi = M0. The ∗ sample and assuming T2 is entirely due to B0- fitted value for Mi should come out close to field inhomogeneities, estimate the gradient of −M0, but if the A-pulse is not a good π-pulse, the B0-field. For the estimate, you may as- the longitudinal magnetization following this ∗ sume T2 represents the time for protons at one pulse will not be perfectly inverted and Mi will end of the sample to make one more Larmor be smaller (in magnitude) than M0. Fitting precession than those at the other end. the data to a formula with Mi fixed at −M0

July 18, 2019 NMR 18 Advanced Physics Laboratory can then lead to systematic errors in the fitted scope trigger must be set for normal (not auto) T1. so the FIDs will not disappear between pulses. For the fit, you must treat the FID am- A good way to determine how big Tr needs plitude as negative for signals before the to be is to measure the zero-crossing time for crossover and positive for those afterward. the inversion recovery sequence, increasing Tr Near the crossover you may want to throw out until the crossover time stops changing. This data points as it may be difficult to determine also lets you check the Tr > 5T1 rule of thumb. their correct sign. Does the fitted time con- Report on the results of this experiment and stant agree with the crossover time? Deter- the validity of this rule. mine the uncertainty in the fitted time con- Because of the long T2, the spin echo mea- stant. surements in distilled water clearly show the the effect of diffusion. Unfortunately, the

T2 field gradient G is not well determined in the present apparatus. So use your measurements Set up the necessary A- and B-pulse widths and the fitted coefficient of the t3 term (in the and find the spin echo. Measure the FID argument of the exponential) with the known amplitude of the echo as a function of Td. value of D (about 2.3 × 10−9 m2/s at 25◦C) Perform a fit to Eq. 36 with t = 2Td (why to determine G and compare with the cruder t = 2Td, not t = Td?) and determine a value ∗ determination based on T2 . and uncertainty for T2. Perform fits both with Finally, measure T1 using an alternative and without the t3 term in the exponential. method that is sometimes used when T1 is Can you see the effects of diffusion and field long. This method applies a single A-pulse inhomogeneities? Why would the effects of repeatedly at fixed time intervals Tr. This one diffusion be difficult or impossible to measure pulse (with its repeat time Tr) is then used for if T2 is too short? Comment on the relative ∗ both preparation and probing. For an A-pulse sizes of T2 , T2 and T1. What should be their that causes a precession through an angle θ predicted ordering and why? (θ = π/2 is typically used) the steady state amplitude of the magnetization immediately CHECKPOINT: Measurements for following each A-pulse can be shown to be: the glycerin sample and the determi- ∗ − − nations of its T2 , T1, and T2 should be 1 exp( Tr/T1) M = M0 sin θ · (38) complete. 1 − cos θ exp(−Tr/T1) (You might want to try to derive this equation Distilled Water yourself.) Change to the distilled water sample. Be sure To use Eq. 38, measure the size of the FID the liquid is centered in the coil. using a π/2 A-pulse (leave the B-pulse off) as The time constants T1 and T2 for distilled you vary the pulse repeat time Tr. Start off by water will be much longer—on the order of fitting your results to Eq. 38 with M0 and T1 as seconds. This means data acquisition for in- fit parameters but with θ fixed at the nominal version recovery and echo sequences will take value of π/2 (1.57). The reason that fitting much longer. (Recall that Tr needs to be at θ is problematic should become clearer if you least 5T1 to let the spins reestablish equilib- try the fit with θ still fixed but off the nominal rium populations.) For long T1, the oscillo- value (at, say, 1.47 and 1.67). What happens

July 18, 2019 Pulsed Nuclear Magnetic Resonance NMR 19

to T1 for these conditions? What happens to T1 if you let θ be a third ﬁtting parameter? Is the ﬁtted value of θ reasonable? Discuss the results.

Special projects There are many studies that can be performed with this apparatus. Here are three you might consider. 1. Measure the gyromagnetic ratio of the 19F (fluorine) nucleus in the liquid Fluo- rinert (perfluorotributylamine or FC-43) or in the solid teflon (tetrafluoroethy- lene). Only try the solid sample after finding a FID with the liquid. Solids, in general, have short T1 and T2. Thus, in order to see enough beats within the FID, you may need to go with a higher beat frequency and higher band pass filtering. You may also have to increase the amplifier gain to see the weak FID. The Fluorinert FID should show a beating pattern. Learn enough about chemical shifts to relate them to the (at least two) Larmor frequencies that must be present in the sample in order to create a beating FID. (See auxiliary material for chemical shifts in FC-43.) How can you determine the very small difference in the two Larmor frequencies from the beat pattern? Express the difference in ppm of the mean Larmor frequency.

2. Add small amounts of CuSO4 (copper sul- fate) to distilled water samples to determine how the time constants depend on the concentrations of paramagnetic ions. 3. Make mixtures with varying ratios of glycerin and water, which changes the viscosity of the mixture. Measure how the time constants depend on the viscosity.

July 18, 2019