Introduction to Magnetic Resonance
All nuclei are characterized by a nuclear spin quantum number, I.

The spin quantum number has multiples of half integral values
ie. I = n/2 where n is an integer.

Even mass number and even atomic number 
Example:

12
C atomic number = 6 and mass number = 12 a.m.u
Odd mass number, even or odd atomic number  I = half-integer values.
Example:
13
C
N
17
O
15

atomic number = 6 and mass number = 13 a.m.u.
atomic number = 7 and mass number = 15 a.m.u.
atomic number = 8 and mass number = 17 a.m.u.
I = ½.
I = ½.
I = 5/2
Even mass number, and odd atomic number  I = Integer values.
Example:
6
Li
B
14
N
10

I=0
atomic number = 3 and mass number = 6 a.m.u.
atomic number = 5 and mass number = 10 a.m.u.
atomic number = 7 and mass number = 14 a.m.u.
I=1
I=3
I=1
All nuclei with spin greater than ½ are termed quadrupolar nuclei, since they posses an electrical
quadrupole moment , which is due to their non-spherical charge distribution.
These nuclei are difficult to observe since the electrical quadrupole moment interacts with local electric
field gradients. In solution state these gradients varying rapidly, causing their signal to relax rapidly
making them very difficult to observe.
Question 1: Determine weather the following nuclei have either, zero, half-integer, or integer spin.
7
Li,
2
H,
19
F,
31
16
P,
O
AN
MN
3
7
1
2
9
19
15
31
8
16
S
3/2
1
½
½
0

Distribution of currents within a nucleus induce a magnetic dipole moment, which is related the spin
quantum number of the nucleus, which is fixed for all nuclei of the same type.

The magnetic dipole moment has symbol 

μ  γI P
where  is known as the gyromagnetic ratio, which has units rad/(s gauss). And P is a vector describing
the nuclear angular momentum, which has components, (Px, Py, Pz).

When placed in a magnetic field the nucleus, the interaction between the nuclear magnetic moment and
the magnetic filed results in a torque experienced by the nucleus as a result it starts to precess along the
field direction. This is known as Larmor precession which is much like a top precesses in the earth’s
gravitational field.
B


The energy of interaction between the magnetic field and magnetic moment is:
E  μ B
Where B is a vector representing the magnetic field, with components (Bx, By, Bz).

Traditionally the applied field is used to define the Z direction, thus B can be written as (0, 0, Bz).
The energy of interaction is thus rewritten as:
E   (Px , Py , Pz )  (0, 0, Bz )   Pz Bz

The z-component of the nuclear angular momentum is related to the z-component of the nuclear spin.
Recall in quantum mechanics that a total angular momentum of L has 2(L+1) components. The same is
true for nuclear angular momentum and hence spin. (this gives rise to the different orbital structures)
A nucleus with spin quantum number I will thus have 2I +1 z-components characterized by an new set
of quantum numbers known as the magnetic quantum number, mI. The magnetic quantum number will
range between –I to I in integer steps.
The interaction energy can thus take on 2I + 1 values given, by the equation:
E    m I Bz
the corresponding angular velocity is
since
Pz   m I
   m I Bz (rad / s) or    m I Bz 2 ( Hz )
Energy
 B/2
increasing B
- B/2
Energy Level diagram for a spin-1/2 nucleus as a function of magnetic field
Question 2:
How many energy levels will 7Li have? Draw the energy level diagram as a function of the magnetic field.
(7Li is a spin 3/2 system, thus it will have 4 energy levels)
Question 3:
Compute the energy levels for 14N in a 11.7 Tesla field. The gyromagnetic ratio for 14N, is 1,934 rad/(sec
gauss). (n.b. there are 104 Gauss to the Tesla and there are 2 radians per revolution, also note that h is
6.62618*10-34 Js.)
14
N is a spin 1 system, thus it has magnetic quantum number –1,0 and 1 and thus has 3 energy levels.
Lets start by computing the rate of presession in each level.
ω  γBm I  1,934 (rad/sG) 11.7T10,000G / Tm I  2.263 *108 m I ( rad / s)
   / 2  36.03mI MHz
note that  and  are both angular velocities,  has units rad/sec, and  has units Hz.
The energy is
E m  mI h  mI (6.62618*1034 Js)(36.03 *106 / s)  mI (2.3875*1026 J)
since mI = 1, 0, -1 the energy levels are, 2.3875, 0 and –2.3875*10-26 J, which have the corresponding
precession frequencies of –36.03, 0, and 36.03 MHz.

In NMR transition between energy levels are observed. In this case a transition from the bottom level to
the middle level involved going from –36.03 MHz to 0 MHz, which is a change of 36.03 MHz.
Therefore a spectrometer working at 11.7 T, operates at 36.03 MHz, for 14N.

The change of frequency experience whilst undergoing a transition at a given field strength is the
operating frequency of that nucleus. The operating frequency of a spectrometer is often regarded as the
frequency of 1H at the given field.
Question 4:
Repeat the above calculation for 1H which has a megnetogyric ratio of 26,753 rad/secG. Calculate the
energy level diagram and precession frequencies for 11.7 and 7 T. What are the operating frequencies,
Larmor frequencies, of these fields?
BmIrad/sec G)][11.7 T][10000 G/T][mI]/[2*3.14] = (498, 000, 000)*mI Hz
Em=h*m = [(498, 000, 000)*mI Hz][6.62618* 10-34 Js] = 3.303*10-25*mI J
1 = 498, 000, 000 Hz * (-½ )= -249, 000, 000 Hz,
E1 = 3.303*10-25 J * (-1/2) = 1.65 *10-25 J
2 = 498, 000, 000 Hz * (½ )= 249, 000, 000 Hz,
E2 = 3.303*10-25 J * (1/2) = 1.65 *10-25 J
The change in frequency is from level 1 to 2 is 498 MHz, the change in energy is 3.303*10-25 J.
exercise: repeat the same problem for 7 T.

The transition energies involve in NMR are very small, and thus the difference in population between
the levels is small as well, as the Boltzmann equation suggests:
N2
 exp( E / kT )
N1
The N’s represent the populations of the energy levels, E is the energy difference between the energy
levels, k is the Boltzmann constant, which is 1.388 *10-23 J/K, and T is the temperature in K. Since the
energy is very small one can approximate this equation with:
N2
 1  E / kT
N1
Question 5:
Let’s use the Boltzmann equation to compute the population difference of between the two energy levels of 1H
at 4 Tesla, given there are 106 protons at 303 K.
Start by computing the energy difference at 4 T for 1H. The gyromagnetic ratio is 26,753 rad/s G thus the
frequency of precession is:
m=m (26,753 rad/s G)(10000 G/T)(4 T)/2 =m*170 MHz.
Therefore bottom level precesses at –85 Mhz, whilst the top level precesses at 85 MHz. The corresponding
energies are –5.65 and 5.65 *10-26 J. The energy difference, E, is thus 1.129 *10-25 J.
E/kT = (1.129*10-25 J)/[(1.388*10-23 J/K)*(303 K)]= 2.685 *10 -5
therefore, N1/N2 = = 1- E/kT = 0.99997315 which implies that N1 = 0.99997315 N2.
Recall that N1 + N2 = 106, therefore 0.99997315 N2 + N2 = 106 implies that N2 = 106/1.99997315 = 500,007
Therefore N1= 106 –500, 007 = 499, 993 thus the difference is difference is 14 spins.
This serves to illustrate that NMR is inherently an insensitive method, since the signal is proportional to
the population difference.
Question 6:
How can the sensitivity be improved? Which gives more signal at a given field, 1H or 14N.
( by increasing  or the magnetic field, 1H gives the most signal since it has the larger 