At equilibrium, the bulk magnetization (M) is oriented along the z axis. Although all of the individual nuclei retain magnetization in the xy plane, their μx and μy components are randomly distributed and thus perfectly cancel, with the result that Mx = My = 0.
The μz components of the various nuclei are also randomly distributed, but in this case the various alignments do not perfectly cancel. There is a slight difference in energies between the different mI states. For 1H, I = ½ and thus mI = +½ or -½. The mI = +½ state is slightly lower in energy than the mI = -½ state, consequently there are slightly more nuclei with mI = +½ than mI = -½.
The random distribution of nuclei between states of different energies is described by the Boltzmann distribution. In this distribution, the difference in energies between two states is the critical value. The energy for the interaction of a nuclear magnetic moment (μ) with the static magnetic field (B) is the dot product of the two vectors:
E = - μ ⋅ B = - μz B = - ℏ γ mI B
The difference between two adjacent energy levels is
ΔE = - ℏ γ ΔmI B = ℏ γ B = ℎ ν
Note that the difference between higher and lower energy states depends upon ΔmI = (-½) - (½) = -1. Take note that the energy difference is related to the Larmor frequency (ν).
The difference in energy for NMR processes is very small. For example, if B = 10.0 T, then ΔE = 0.170 J/mol, which is a million times less than the energy of a covalent bond. More to the point, it is also much less than the energy available from the thermal motion of molecules at room temperature (about 2.5 kJ/mol). Thus there are almost the same number of nuclei in the +½ and -½ states.
The Boltzmann distribution is described by the following equation, in which n0 and n1 represent the number of nuclei in the mI = +½ and -½ states, respectively. R = 8.3145 J K-1 mol-1 is the gas constant and T is the absolute temperature.
n1/n0 = exp(-ΔE/RT) = exp(-ℏγB/RT) = exp(-ℎν/RT)
For the example of 1H is a 10.0 T field, n1/n0 = 0.999931 at 298 K. Although for every one million nuclei there is only an excess of 69 in the lower energy state. Although a small excess, it is sufficient to give a detectable net positive Mz at equilibrium.
Let's suppose the bulk magnetization is not initially at equilibrium. How quickly does M relax to its equilibrium state?
In 1946 Felix Bloch proposed that the relaxation of M to its equilibrium state follows first-order kinetics. Moreover, he described the relaxation of Mz and of Mx and My as occurring on two different time scales.
The relaxation of Mz to its equilibrium value of Meq occurs with a time constant T1, which is called the spin-lattice relaxation time. Relaxation of Mz involves energy transfer between the spin system and the surroundings. Recall from above that E = - μz B and Mz is the sum of the μz of all the nuclei in the system.
The relaxation of Mx and My to their equilibrium values of zero occurs with a time constant T2, which is called the spin-spin relaxation time. Relaxation of magnetization in the xy plane does not involve net energy transfer with the surroundings. Instead this relaxation is an entropic process.
The differential equations describing the way the bulk magnetization changes with time are called the Bloch equations, which contain terms for the first-order rate law for relaxation and terms for the precession.
d Mz/d t = -(Mz - Meq)/T1
d Mx/d t = γ B My - Mx / T2
d My/d t = - γ B Mx - My / T2
The solution of the Bloch equations is illustrated in the simulation shown below. The bulk magnetization is artificially positioned along the y axis at the outset of the experiment. The resulting signal is called a free induction decay (FID).
Run the simulation for various combinations of f, T1, and T2 and carefully observe the behavior of the bulk magnetization and the signal. On the basis of your observations, answer the following questions.
As with other simulations in this series, simulation occurs much more slowly than real precession. The simulation has been slowed to permit visualization of the process.
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