Name _________________________
Exam 2, 2007:
CHEM/BCMB 4190/6190/8189
1
Exam 2:
CHEM/BCMB 4190/6190/8189 (106 points)
Tuesday, 2 October, 2007
1
).
In the example (right), the effect of a 90
°
(
±
/2) pulse
applied along the “x” axis (90
°
x) is shown for a bulk
magnetization vector (
M
0
) at equilibrium (on the ‘
z
’ axis).
For ‘b’, ‘c’, ‘d’ and ‘f’ below, show the effects of the
indicated pulses by drawing the missing (originating or
resulting) vectors on the coordinate axes.
For ‘a’ and ‘e’,
fill in the blank with the correct pulse that will promote the
indicated movement of the bulk magnetization vector (there
may be more than one correct answer for some of these).
(12 points)
a.
d.
b.
e.
c.
f.
90
°
x
M
0
y
z
x
y
z
x
90
°

x
y
z
x
y
z
x
180
°
y
, 90
°

x
y
z
x
y
z
x
this vector is in
the
x

z
plane
2
±
y
y
z
x
y
z
x
y
z
x
y
z
x
90
°

x
, 90
°

y
90
°
y
y
z
x
y
z
x
this vector is in
the
x

z
plane
y
z
x
y
z
x
135
°
x
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Name _________________________
Exam 2, 2007:
CHEM/BCMB 4190/6190/8189
2
2
).
The inversionrecovery method was
used to estimate the relaxation time
constant (
T
1
) for the H
A
hydrogen of the
compound shown (right).
The signals from
H
A
at various delays (
±
) are shown.
Estimate
T
1
for H
A
.
Shown your work or
otherwise explain your reasoning.
(
5
points
)
In the inversionrecovery sequence, a 180
°
pulse on magnetization initially at
equilibrium (+z) inverts the spin populations to give z magnetization, with M
Z
= M
0
.
The 180
°
pulse is followed by a variable delay (
±
), which is then followed by a 90
°
pulse
to create the observable transverse magnetization.
We know that, following a 180
°
pulse,
the return of bulk magnetization along z to equilibrium (M
Z
= M
0
) is described by the
following first order equation:
For the relaxation data shown, the magnitudes of the signals at the various delay times
are not shown.
However, the value of the delay
±
that produces no observable signal (M
Z
= 0) is reasonably easy to estimate from the data (we’ll call this delay
±
zero
).
For the data
shown,
±
zero
is somewhere between 0.7 and 0.8 s, we’ll call it 0.725 s.
We can then
rearrange the equation to calculate T
1
:
H
A
H
X
RO
CH2R
²
M
z
=
M
0
(1
±
2
e
±
t
/
T
1
)
M
z
=
M
0
(1
±
2
e
±
²
zero
/
T
1
)
0
=
M
0
(1
±
2
e
±
²
zero
/
T
1
)
±
M
0
=
±
M
0
2
e
±
²
zero
/
T
1
1
2
=
e
±
²
zero
/
T
1
ln
1
2
=
±
²
zero
/
T
1
T
1
=
²
zero
ln2
Thus
,
T
1
³
0.725
ln2
³
1.05
s