3­20 The vector model
3.12 Exercises
E 3­1
A spectrometer operates with a Larmor frequency of 600 MHz for protons.
For a particular set up the RF field strength, 1/(2) has been determined to
be 25 kHz. Suppose that the transmitter is placed at 5 ppm; what is the offset
(in Hz) of a peak at 10 ppm? Compute the tilt angle, , of a spin with this
offset.
For the normal range of proton shifts (0 ­ 10 ppm), is this 25 kHz field
strong enough to give what could be classed as hard pulses?
E 3­2
In an experiment to determine the pulse length an operator observed a positive
signal for pulse widths of 5 and 10 s; as the pulse was lengthened further the
intensity decreased going through a null at 20.5 s and then turning negative.
Explain what is happening in this experiment and use the data to determine
the RF field strength in Hz and the length of a 90 pulse.
A further null in the signal was seen at 41.0 s; to what do you attribute
this?
E 3­3
Use vector diagrams to describe what happens during a spin echo sequence in
which the 180 pulse is applied about the y axis. Also, draw a phase evolution
diagram appropriate for this pulse sequence.
In what way is the outcome different from the case where the refocusing
pulse is applied about the x axis?
What would the effect of applying the refocusing pulse about the -x axis
be?
E 3­4
The gyromagnetic ratio of phosphorus-31 is 1.08 × 108 rad s-1 T-1. This
nucleus shows a wide range of shifts, covering some 700 ppm.
Estimate the minimum 90 pulse width you would need to excite peaks
in this complete range to within 90% of the their theoretical maximum for a
spectrometer with a B0 field strength of 9.4 T.
E 3­5
A spectrometer operates at a Larmor frequency of 400 MHz for protons and
hence 100 MHz for carbon-13. Suppose that a 90 pulse of length 10 s is
applied to the protons. Does this have a significant effect of the carbon-13
nuclei? Explain your answer carefully.
E 3­6
Referring to the plots of Fig. 3.26 we see that there are some offsets at which
the transverse magnetization goes to zero. Recall that the magnetization is ro-
tating about the effective field, eff; it follows that these nulls in the excitation
3.12 Exercises 3­21
come about when the magnetization executes complete 360 rotations about
the effective field. In such a rotation the magnetization is returned to the z
axis. Make a sketch of a "grapefruit" showing this.
The effective field is given by
eff = 2
1 + 2.
Suppose that we express the offset as a multiple  of the RF field strength:
= 1.
Show that with this values of the effective field is given by:
eff = 1 1 + 2.
(The reason for doing this is to reduce the number of variables.)
Let us assume that on-resonance the pulse flip angle is /2, so the duration
of the pulse, p, is give from
1p = /2 thus p =

21
.
The angle of rotation about the effective field for a pulse of duration p is
(effp). Show that for the effective field given above this angle, eff is given
by
eff =

2
1 + 2.
The null in the excitation will occur when eff is 2 i.e. a complete rota-
tion. Show that this occurs when  =

15 i.e. when ( /1) =

15. Does
this agree with Fig. 3.26?
Predict other values of  at which there will be nulls in the excitation.
E 3­7
When calibrating a pulse by looking for the null produced by a 180 rotation,
why is it important to choose a line which is close to the transmitter frequency
(i.e. one with a small offset)?
E 3­8
Use vector diagrams to predict the outcome of the sequence:
90
- delay  - 90
applied to equilibrium magnetization; both pulses are about the x axis. In
your answer, explain how the x, y and z magnetizations depend on the delay
 and the offset .
E 3­9
Consider the spin echo sequence to which a 90 pulse has been added at the
end:
90
(x) - delay  - 180
(x) - delay  - 90
().
3­22 The vector model
The axis about which the pulse is applied is given in brackets after the flip
angle. Explain in what way the outcome is different depending on whether
the phase  of the pulse is chosen to be x, y, -x or -y.
E 3­10
The so-called "1­1" sequence is:
90
(x) - delay  - 90
(-x)
For a peak which is on resonance the sequence does not excite any observable
magnetization. However, for a peak with an offset such that  = /2 the
sequence results in all of the equilibrium magnetization appearing along the
x axis. Further, if the delay is such that  =  no transverse magnetization
is excited.
Explain these observations and make a sketch graph of the amount of
transverse magnetization generated as a function of the offset for a fixed delay
.
The sequence has been used for suppressing strong solvent signals which
might otherwise overwhelm the spectrum. The solvent is placed on resonance,
and so is not excited;  is chosen so that the peaks of interest are excited. How
does one go about choosing the value for ?
E 3­11
The so-called "1­1" sequence is:
90
(x) - delay  - 90
(y).
Describe the excitation that this sequence produces as a function of offset.
How it could be used for observing spectra in the presence of strong solvent
signals?