2.2 THE NUCLEUS IN A MAGNETIC FIELD 9 Finally, remember that different isotopes of the same element can have different nuclear spins, some of which are detectable Nucleus Multiplicity m values by NMR, others of which are not. -1.0,1 u EXAMPLE 2.3 Predict the nuclear spin I of He, Li and'Li, and indicate which are detectable by NMR o}6-3,-}.-,l D Solution: Assign each nucle described above Nucleus A Z N Group Predicted/ Detectable? All nuclei with /=, have a spherical(symmetrical) distri bution of spinning charge, so the electric and magnetic fields 4221 surrounding such nuclei are spherical, homogeneous, and bLi isotropic in all directions. By contrast, nuclei with 1>have 7Li 3 n/2(n odd) Ye a nonspherical distribution of spinning charge, resulting in nonsymmetrical electric and magnetic fields. This imparts an The actual values of /are /=0 for He, /=I for L. and electric quadrupole( Q)to the nucleus, a property that can 1=2 for 7Li 0 complicate their NMR behavior. As a result, the most com- monly studied nuclei are those with a nuclear spin of Just when we begin to understand that a nuclear spin number of i gives rise to two spin orientations in a magnetic 2. 2 THE NUCLEUS IN A MAGNETIC FIELD even 6. What is the significance of this? The explanation is 2.2.1 More about the nuclear Zeeman Effect actually quite straightforward. Although single atomic parti cles such as protons, neutrons, and electrons can adopt only As we have said, a nucleus with nuclear spin / adopts 2/+I two magnetic spin orientations, complex nuclei can adopt nondegenerate spin orientations in a magnetic field. The states more than two. In fact, the total number(multiplicity) of separate in energy, with the largest positive m value corre- possible spin states (i. e, the different values of m)is deter- sponding to the lowest energy(most stable)state. It is this mined solely by the value of 1: separation of states in a magnetic field that is the essence of he nuclear Zeeman effect Multiplicity = 2/+ I a The energy of the ith spin state(Ei)is directly proportional the value of m; and the magnetic field strength Bo( that is Each of these 2+ I states has its own spin quantum number energy is quantized in units of hBo/2n) m in the range m=-1,-1+l,.., I-1, I(listed in the order of decreasing energy and increasing stability). Thus, for nu- clei with /=, the multiplicity is 2, and these two states are y h Bo i and m=-1 a EXAMPLE 2.4 Calculate the spin state multiplicity for In this equation h(Planck,s constant)and t have their usual each of the nuclei below, and list the value of m for each state meanings, while y is called the magnetogyric ratio, a propor- from highest to lowest energy tionality constant characteristic of the isotope being examined (more on this a little later). The minus sign in the equation B.C, 4N 170 3IP follows from the convention of making a positive m corre spond to a lower (negative)energy. Figure 2.3 graphically D Solution: The / values for these nuclei appear among the depicts the variation of spin state energy as a function of above list of three groups. Using Eq. (2.2), we first calcu magnetic field strength for two different nuclei, one with late the multiplicity of spin states and then their m values. /=I, the other with /=I. Notice that as field strength increases, the difference in energy(AE)between any two spin Nucleus Multiplicity values states also increases proportionally. For a nucleus with /=i this difference is B 4 E
10 MAGNETIC PROPERTIES OF NUCLEI E 0 △E Bo igure 2.3. Nuclear Zeeman effect. (a) A nucleus with /=4. (b)A nucleus with/= 1. The arrow beside each spin state line indicates the orientation of the magnetic moment in a vertical magnetic field =()2 will become significant later. Relative sensitivity is the strength of the NMR signal that is generated by a fixed number of nuclei of a given isotope relative to the signal h obtained from an equal number of H nuclei. If we compare the data on natural abundance and sensitivity, we see why historically the most easily studied nuclei were H, F, and 3P. Indeed, prior to the 1970s these three were the only nuclei And now you realize why values of*i were picked for m routinely studied with commercially available instrumenta (and s too, for that matter). It is so that the difference inenergy tion More recently, however, instruments have become avail- between two neighboring spin states will always be an integer able that can routinely examine a wide variety of other multiple of y h Bo/2兀 ubiquitous elements, including C(which is of immense The magnetogyric ratio y describes how much the spin importance to organic chemists ), SN, 23Na, and 29Si, to men state energies of a given nucleus vary with changes in the tion just a few external magnetic field. Each isotope with nonzero nuclear spin has its own unique value of y, though the magnitude of EXAMPLE 2.5(a)What is the energy difference be- depends on the units selected for Bo. We will use the unit tesla (T)for magnetic field strength so that y has units of radians (b)Of c?o spin states of H in a magnetic field of 5.87T per tesla per second(2T radians in one cycle of 360). As we will see in Chapter 3, modern commercial NMR spectro ters are equipped with magnets that generate fields ranging D Solution: (a) Use Eq (2. 4)and the y values in Table 2. from ca 5 to 16 T For comparison, the earths magnetic field is a mere 6 x l0-5T (Note: Earlier books on NMR used the gauss AE 2t or kilogauss for magnetic field strength; I T= 10G=10 kG. In Table 2.1 are listed many of the common isotopes examined by NMR techniques, together with their nuclear 267512×10° rad T-s-)663×103Js)(5.87T constants. Notice that a bare proton has the largest y value of 2(3.14rad) any nuclear particle, while heavier nuclei, surrounded by many subvalence electrons. tend toward lower values. this =1.66×10-25J
2.2 THE NUCLEUS IN A MAGNETIC FIELD 11 (b)For C, Y=672640 x 106, so AE= 4.18 x 10-263, quency is independent of m, so that all spin orientations of a about one-fourth the difference for H given nucleus precess at the same frequency in a fixed mag netic field 2.2.2 Precession and the Larmor Frequency We now know that nuclei with I# 0. when immersed in a a EXAMPLE 2.6(a)At 5.87 T, what is the precession magnetic field, adopt 21+ I spin orientations, each with a frequency v of a H nucleus? A C nucleus?(b)In what different energy. But before these nuclei can absorb photons. region of the electromagnetic spectrum does radiation of these frequencies occur? they must be oscillating in some sort of uniform periodic motion(Section 1.3). Fortunately, quantum mechanics re- quires that the magnetic moments are actually not statically Q Solution:(a)For H, using the y value from Table 2.1, we aligned exactly parallel or antiparallel to the external mag netic field, as Figure 2.2 implied. Instead, they are forced to remain at a certain angle to Bo, and this causes them to YB2(267.512×10 radT-Is)X5.78T "wobble"around the axis of the field at a fixed frequency 2(3.14rad Why is this so? If you have ever played with a spinning top, you may know =2.50x108s-1=250MHz that it is the spin angular momentum of the top that prevents it from falling over and also causes it to wobble in addition to Similarly, for C, v= 62.9 MHz spinning. This periodic wobbling motion that the top assumes in a gravitational field is called precession. The earth pre After you have completed these calculations the"long cesses on its axis in much the same way, though much more way, using Bo andy values from Table 2. l, try this easier slowly. In an exactly analogous way, the magnetic moment way. In Table 2. 1, the data in the column labeled v are the vector of a nucleus in a magnetic field also precesses with a precession frequencies(in megahertz)of each nucleus in characteristic angular frequency called the Larmor fre- a 1.00-T magnetic field. Simply multiplying these num- quency(o), which is a function solely of y and Bo: bers by the actual field strength(in tesla), directly gives he value of v at any other field strength. Thus, for H, o=y Bo (25) v=(425759MHT)587T)=250MHz The angular Larmor frequency, in units of radians per second. can be transformed into linear frequency v(in reciprocal (b) From Table l I we note that these frequencies fall in seconds or hertz)by division by the rf region (26) l EXAMPLE 2.7 At what magnetic field strength do pro- tons('H nuclei)precess at a frequency of 300 MHz? This precessional motion causes the tip of the magnetic mo- D Solution: Rearranging Eq(2.6), we find ment vectors(either up or down) to trace out a circular path, as shown in Figure 2. 4. Note also that the precession fre- 2TTV (314rad(300×10s-) 267512×10°radT-l NMR have immersed the collection of nuclei in a magnetic field m=+1/2 m=-1/2 each nucleus is precessing with a characteristic frequency. to do is irradiate them with Figure 2.4. Precession of the magnetic moment in each of the two electromagnetic radiation of the appropriate frequency, right? possible spin states of an/=5 nucleus in external magnetic field B ell
12 MAGNETIC PROPERTIES OF NUCLEI 2.3 NUCLEAR ENERGY LEVELS AND sion)is exactly as likely to occur as absorption. In such a case, RELAXATION TIMES no net absorption is possible, a condition called saturation Is there any reason to expect that there will be an excess of 2.3.1 Boltzmann Distribution and Saturation nuclei in the lower spin state? The answer is a qualified yes For any system of energy levels at thermal equilibrium, there In Chapter I we hinted that once a particle absorbs a photon, will always be more particles in the lower state(s)than in the the energy originally associated with the electromagnetic upper state(s). However, there will always be some particles radiation appears somehow in the particle's motion. Where in the upper state(s). What we really need is an equation does the energy go in the case of precessing H nuclei? relating the energy gap(AE) between the states to the relative Because there are only two spin states possible. the energy populations of (numbers of particles in)each of those states goes into a spin flip. That is, the photons energy is absorbed This time, quantum mechanics comes to our rescue in the by a nucleus in the lower energy spin state(m=+ ) and the form of the Boltzmann distribution nucleus is flipped into its higher energy spin state(m=-) his situation is depicted in Figure 2.5. And remember that this spin flip does not change the precessional frequency of he nucleus where P is the population(or fraction of the particles)in each We have already calculated the energy gap between these state, T is the absolute temperature in Kelvin(not to be two spin states [Eg (2.4)1, and this must equal the energy of confused with nonitalicized T for Tesla), and k(the Boltzmann the absorbed photon [Eq(1.3)]. Combining these with Eq constant)has a value of 1.381 x 10-23JK (2.6)gives us ■ EXAMPLE28At25°C(298K) what fraction of ' H Thus e- yhBo-hVpreession-photon=hy photo (2.7) nuclei in 5.87 T field are in the upper and lower states? See Example 2.5 for the value of AE. Thus, as we expected from Chapter l, for resonance to occur, the radiation frequency must exactly match the precessional D Solution: Use Eq(.8)and the results of Example 2.5 But there is a fly in the ointment. Quantum mechanics tells △E s that, for net absorption of radiation to occur, there must be Im=-= exp kT more particles in the lower energy state than in the higher one. 1.66×10-25J If the two populations happen to be equal, Einstein predicted =eA-1381×102)Jk1×298K =0.99996 heoretically that transition from the upper(m=-5)state to the lower(m =+))state(a process called stimulated emis- Figure 2.5. Relative energy of both spin states of an /=i nucleus as a function of the strength of the external magnetic field B
2.3 NUCLEAR ENERGY LEVELS AND RELAXATION TIMES Since there are only two spin states, Pom=-1/2) 0.50001. At 95% of equilibrium =0.49999andP 0.05(Peg-Po). Use Eq. (2,9)an 0.50001 P As you can see from the above example, the difference in -exp populations of the two H spin states is exceedingly small, on the order of 20 ppm. And the difference for other elements is even smaller because of their smaller y values. But do not 0.05 despair. This difference is sufficient to allow an NMR signal to be detected. It is this small difference, however that ac counts in part for the relatively low sensitivity of NMR Taking the natural logarithm (In) of both sides of the spectroscopy compared to other absorption techniques such equation, we find as infrared and ultraviolet spectroscopy. Remember, factors such as a stronger magnetic field (Bo), a larger magnetogyric 3.00= =3.007 ratio, or a lower temperature all contribute to a larger popula tion difference. reduce the likelihood of saturation and lead Since T=0.20s,t=3(0.20s)=0.60s to a more intense NMR signal 2.3.2 Relaxation Processes (b) when the field strength retums to zero. the collection of nuclei will decay exponentially toward the original Our NMR theory is almost complete, but there is one more equal populations of spin states at a rate still governed by thing to consider before we set about designing a spectrome- the same value of TI ter. We indicated previously that at equilibrium in the absence of an extemal magnetic field, all nuclear spin states are Figure 2.6 graphically depicts the situation in Example 2.9 degenerate and, therefore, of equal probability and popula The arrows in the three diagrams below the graph represent the distribution of individual precessing 'H magnetic mo- tion. Then, when immersed in a magnetic field, the spin states ments, either up or down. Initially there are equal numbers of establish a new(Boltzmann)equilibrium distribution with a nuclei in each spin state. But at equilibrium in the magnetic slight excess of nuclei in the lower energy state eld, there is a 20 ppm excess of up spins(exaggerated in the A relevant question is this: How long after immersion in middle diagram). When the magnetic field is turned off,the the external field does it take for a collection of nuclei to collection decays back to the original equal distribution reequilibrate? This process is nor infinitely fast. In fact, the The values of T, range broadly, depending on the particular rate at which the new equilibrium is established is governed type of nucleus, the location of the nucleus(atom)within a byaquantity called the spin-lattice(or longitudinal) relaxa- molecule, the size of the molecule, the physical state of the tion time, TI. The exact relation involves exponential decay: sample(solid or liquid), and the temperature For liquids or solutions, values of 10-2-102 s are typical, though some quadrupolar nuclei have( faster) relaxation times of the order = exp (2.9) of 10-4 s. For crystalline solids, T, values are much longer (Section 2.5). For now. just remember that the larger the value where Pea-Pr is the difference between the equilibrium of T, the longer it takes for a collection of nuclei to reach population of a given state (for example, the m =+3 state)and return to) equilibrium. the population after time t and the subscript zero refers to r There is another reason why the magnitude of Ti is impor 0 (Peg is thus the population at I=oo.) tant. Suppose we have a boltzmann distribution of nuclei precessing in a magnetic field, and we irradiate the collection I EXAMPLE .9(a)Suppose that for a certain set ofh to cause transitions(spin flips)between the lower(n=+ level and the upper(m=-i)states. Because there is initially immersion in a 5.87-T magnetic field will it take for an such a small difference between the populations of the two nitially equal distribution of H spin states to progress 95% states, it will not be long before the populations are equalized of the way toward equilibrium?(b)What would happen if the through the absorption of the photons! This, of course, means magnet were turmed off at this point? the spin system has become saturated and no further net absorption is possible. However, if we tum off the source of D Solution: (a)From Example 2.8, we know that the final rf radiation, the system can relax back to the Boltzmann equilibrium population of the m=+ state will be distribution (at a rate controlled by T) and absorption can