The
nuclei of all elements carry a charge. When the spins of the protons and
neutrons comprising these nuclei are not paired, the overall spin of the
charged nucleus generates a magnetic dipole along the spin axis, and the
intrinsic magnitude of this dipole is a fundamental nuclear property called
the nuclear magnetic moment, µ. The symmetry of the charge distribution
in the nucleus is a function of its internal structure and if this is spherical
(ie analogous to the symmetry of a 1s hydrogen orbital), it is said to
have a corresponding spin angular momentum number of I=1/2, of which examples
are ^{1}H,
^{13}C, ^{15}N, ^{19}F, ^{31}P
etc. Nuclei which have a non-spherical charge distribution (analogous to
e.g. a hydrogen 3d orbital) have higher spin numbers (eg ^{10}B,
^{14}N etc) outside the scope of this particular lecture course.

In
quantum mechanical terms, the nuclear magnetic moment of a nucleus can
align with an externally applied magnetic field of strength **Bo** in
only 2I+1 ways, either re-inforcing or opposing **Bo**. The energetically
preferred orientation has the magnetic moment aligned parallel with the
applied field (spin +1/2) and is often given the notation a,
whereas the higher energy anti-parallel orientation (spin -1/2) is referred
to as b.
The rotational axis of the spinning nucleus cannot be orientated exactly
parallel (or anti-parallel) with the direction of the applied field **Bo**
(defined in our coordinate system as about the *z* axis) but must
precess about this field at an angle (for protons about 54š) with an angular
velocity given by the expression;

w_{o}
= g**Bo****
...(1) (the Larmor frequency, in Hz) **

The
constant g
is called the magnetogyric ratio and relates the magnetic moment m
and the spin number I for any specific nucleus;

g
= 2pm/hI
...(2) (h is Planck's constant)

For a single nucleus with I=1/2 and positive g,
only one transition is possible (D
I=1, a single quantum transition) between the two energy levels;

NMR is all about how to interpret such transitions in terms of chemical
structure. We will first consider the energy of a typical NMR transition.
If angular velocity is related to frequency by w_{o}
= 2¼n,
then

n
= g**Bo****/2¼
...(3) **

It
follows that proton NMR transitions (DI=1)
have the following energy;

hn
= DE
= hg**Bo****./2p
...(4) **

For
a proton g
= 26.75 x 10^{7} rad T^{-1} s^{-1} and **Bo**
~ 2T, DE
= **6** x **10 ^{-26}** J. The relative populations of the
higher (n

n^{2}/n^{1}
= e-^{D}E/kT
~ 0.99999. ...(5)

For
NMR, this means that the probability of observing a transition from n^{1}
to n^{2} is only slightly greater than that for a downward transition,
*ie* the overall probability of observing absorption of energy is
quite small. This relationship also explains why a larger **Bo** favours
sensitivity in NMR measurements, increasing as it does the difference between
the two Boltzmann levels, and why NMR becomes more sensitive at lower temperatures.