16.17MRI is a non-invasive technique for showing the presence of fluids.
For an accident, MRI would show the build-up of fluids in the brain
at the trauma point.16.18 (a)=11.5 × 109 s−^1 =11.5 GHz(b)=8.4 × 109 s−^1 =8.4 GHz16.19
16.20
=11 GHz16.21 The splitting arises from the unpaired electron interacting with a
nucleus ofI=1.16.22The EPR spectrum will split into two lines that are separated by the
hyperfine coupling constant.16.23 The oxidation state determines whether there are five or six elec-
trons in the d orbitals. The five electronic levels are divided in energy
into a lower group of three and an upper group of two depending
upon coordination. The relative energies of these five levels deter-
mines the distributions of the electrons.16.24 There are six electrons in five electronic levels that are divided in
energy into a lower group of three and upper group of two. For high
spin, the spacing between the two groups is small, so one electron
is in each level except for the lowest, which has two. This results
in a net spin of 2. The low-spin value is zero.16.25 The low-spin value is 1/2. There are five electrons in five electronic
levels, which are divided in energy into a lower group of three and
an upper group of two. For low spin, the spacing between the two
groups is large so all electrons are in the lowest three levels with each
level having two electrons except for the highest, which has 1. The
presence of the single unpaired electron results in a net spin of 1/2.16.26 The Fe^2 +of the heme will give rise to a spectrum around g=2 for
low spin and g=6 for high spin.ννμ
EPR NMR μB
NEPR
NMRB
B
.
=×× = 200 ×
16
MHz7710
91 10
03
10
27.^31
×.
×
×
−
−kg
kgT
T
g
h
BB(. )( )
(.
==
ν ××−
μ662 10 9 10
92
(^34) Js (^9) Hz
7710 03
24 1 214
)(. )
.
×
−− =
JT T
ν
μ (. )(. )(.
==
gB × −−
heB 2 0023 9 27 10 0 3(^24) JT (^1) TT
Js
)
662 10. × −^34
ν
μ (. )(. )(.
==
gB × −−
heB 2 0023 9 27 10 0 4(^24) JT (^1) TT
Js
)
662 10. × −^34
478 ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 478