Biomedical Engineering Reference
In-Depth Information
TABLE 16.2
Characteristics of Isotopes Relevant to MRI
Sensitivity Relative to
1
H
Isotopes
Spin State
g
0
(MHz/T)
Natural Isotropic Abundance %
1
H
1/2
42.58
99.99
100
13
C
1/2
10.71
1.11
1.6
19
F
1/2
40.05
100
3.4
23
Na
3/2
11.26
100
9.3
31
P
1/2
17.24
100
6.6
which is known as the Larmor frequency that can be used to calculate the transition
energy. Note that the gyromagnetic constant,
g
, is associated with angular frequency,
and its normalized version,
g
0
, with frequency. This is the important equation that con-
nects frequency to an applied magnetic field. For the hydrogen isotope most often
used in MRI,
g
0
¼
42.58 MHz/T. Fortunately, the values of parameters needed for
MRI fall within reasonable ranges: typical frequencies are under 100 MHz and fields
are 0.1 to 4 T.
The Larmor frequency is a resonance excitation frequency used for externally producing
transitions in spin states. For MRI, only isotopes that have unpaired nuclear spins that are
multiples of ½ are of interest. For example, the spin state for
1
H aligned with the applied
field is assigned the name “½,” and has a dipole moment of
½). The antiparallel
state for hydrogen has a spin designation of “-½” and has a moment of
m
z
¼ g
h
(
þ
m
z
¼ g
h
(-½). In
g
0
are given for different isotopes. In general,
Table 16.2, values of
m
z
¼ g
hm
, in which
m
can
include any of the following values,
m
¼
0,...
(
I
- 1),
I
, where
I
is the spin-state number
in Table 16.2.
How many of these spin states are there? If the number of spins in the low-energy state is
n
-
, then Boltzmann statistics predict the
ratio of spins in either of these two states at any given time
and the number of those in the higher state is
n
þ
n
=
n
þ
¼
exp
ðD
E
=
KT
Þ
ð
16
:
57
Þ
10
23
where the Boltzmann constant is
K
¼
1.3805
J
/
K
,
D
E
is from Eq. (16.55), and
T
is
temperature in
Kelvin (K).
EXAMPLE PROBLEM 16.11
Find the fraction of excess population in the spin-up (
n
þ
) state for a frequency of 20 MHz and a
temperature of 300 K.
Solution
Calculate the change in energy for a given frequency of 20 MHz from Eq. (16.55):
10
34
10
6
10
26
D
E
¼
hv
¼
6
:
626
J
s
20
=
s
¼
1
:
325
J
