# NMR Spectroscopy

## Equilibrium State

In an NMR experiment, the sample is placed in a strong magnetic field ( B ) that is oriented along the z axis. At equilibrium, the bulk magnetization ( M ) of the sample is also 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 (among the allowed mz states), but in this case the various alignments do not perfectly cancel. There is a slight difference in energies between the different mz states. For 1H, I = ½ and therefore mz = +½ or -½. The mz = +½ state is slightly lower in energy than the mz = -½ state, consequently there are slightly more nuclei with mz = +½ than mz = -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 B is the dot product of the two vectors:

E   =   - μB   =   - μz B = - ℏ γ mz B

The difference between two adjacent energy levels is

ΔE   =   - ℏ γ Δmz B   =   ℏ γ B   =   ℏ ν

Note that the difference between higher and lower energy states depends upon Δmz = (-½) - (½) = -1. Take note that the energy difference is related to the Larmor frequency ( ν = ΔE / ℏ ).

The difference in energy for NMR processes is extremely small. For example, if B = 10.0 T, then for a proton ΔE = 0.170 J/mol, which is a million times less than the energy of a covalent bond. Moreover, 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 n and n represent the number of nuclei in the mz = +½ and -½ states, respectively. R = 8.3145 J K-1 mol-1 is the gas constant and T is the absolute temperature.

n   n
=   exp( -ΔE / RT ) = exp( -ℏ γ B / RT ) = exp( -ℏ ν / RT )

For the example of 1H is a 10.0 T field, n / n = 0.999931 at 298 K. 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.

## Bloch Equations

Let's suppose the bulk magnetization is not initially at equilibrium. How quickly does M relax to the 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 to occur with a different time constant than the relaxation of Mx and My .

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 contributions from 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 how the bulk magnetization changes with time are called the Bloch equations, which contain terms for the first-order rate law for relaxation and for precession about the z axis.

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

## Free Induction Decay

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 simulation then shows how the bulk magnetization evolves over time, with precession and relaxation both contributing to the evolution of M. The signal from the x detector is displayed on the graph. The signal showing the relaxation of the bulk magnetization to its equilibrium value is called a free induction decay (FID).

The simulation, showing the bulk magnetization as a green arrow, is in the rotating frame. Recall that f is the difference between the Larmor frequency ( ν = γ B ) and the spectrometer frequency ( νo ) and can be positive or negative.

Run the simulation for various combinations of f, T1, and T2 . Carefully observe the behavior of the bulk magnetization and the signal. On the basis of your observations, answer the following questions.

• How does f affect the simulation? How does it affect the FID?
• How does T1 affect the simulation? How does it affect the FID?
• How does T2 affect the simulation? How does it affect the FID?
• Is the magnitude of the bulk magnetization (M) a constant or does it vary during the experiment? Explain this behavior.

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f (Hz)
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