CH908, Problem set 6, Mass Analyzers
1. In a MALDI-TOF instrument, calculate the time of flight of singly
charged ions using the following parameters: Ion source potential,
10 kV; drift tube grounded. Drift tube length = 2 meters. A) 100
Da, B) 1000 Da, C) 10 kDa, D) 100 kDa. E) assuming the scan
range is from 0 - 100 kDa, how many scans can be acquired in one
second? F) How much does this affect the signal/noise ratio
compared to a single scan? G) What are the implicit caveats in
question 1F?
A)
t
E
mv 2
2
, thus
2E d 2
 2
m
t
, thus
t
d 2m
2E
2* 2*100*1.67 *1027
 14 s
2*104 *1.6*1019
B) 46 μs
C) 144 μs
D) 456 μs
E) maximum time in the TOF is 456 μs, so lets assume we’ll take
data for 500 μs. This means ~2000 scans can be acquired in one
second.
F) 2000  44.7 , thus a ~45 fold improvement in Signal/Noise ratio is
expected.
G) Signal increases linearly with number of scans, while noise
increases as the n , thus the ratio increase as the n . However,
this requires that the signal is stable (not fluctuating in intensity or
position), and that the noise is random, white noise.
2. At 10kV sourse potential, for two singly charged ions of mass
1000 Da, and 1001 Da, what timing resolution (in microseconds) is
needed to separate them using a timed ion selector (TIS) if the TIS
is placed in the drift tube just after the ion source (10 cm) or just
before the collision cell (1 m)?
Same equation as above in 1A, but now d=10 cm and 100 m.
For 10 cm, 1000 Da = 2.28446 μs, and 1001 Da = 2.28560 μs, so at
least one nanosecond timing resolution in the deflector is required,
preferably 100 picoseconds.
For 1 m, 1000 Da = 22.8446 μs, and 1001 Da = 22.8560 μs, so at
least ten nanosecond timing resolution in the deflector is required,
preferably 1 nanosecond.
3. In MALDI-TOF sample preparation, it is very important to
generate large, flat crystals. Why?
In a MALDI ion source, the sample plate is held at a potential of 10
kV (or so), and the source exit grid is grounded, 5 mm away. Thus,
the electric field is 2e6 volts/meter. A 10 micron difference in height
causes a difference in extraction voltage of 200 Volts, or 0.2%,
which is one of the main causes of the kinetic energy dispersion in
the source. Thus, flat crystals are necessary for generating high
resolution spectra.
Note: delayed extraction and ion mirrors (reflectrons) can partially
correct for an initial kinetic energy distribution, but not completely,
and over a specific window of masses.
4. In a quadrupole ion trap, calculate the low mass cut off for a trap
run at 1000 VRF amplitude, 900 kHz, and with a pole-to-pole inner
diameter of 10 mm, and an z/r aspect ratio of 1.1. What limits the
high m/z range?
Low mass cut off (LMCO) is given by
rearranging, we get
m/ z 
4V
0.908 2 r 2
qz  0.908 
8eV
m (r 2  2 z 2 )
2
,so
, with V = 1000 Volts, ω = 2*pi*900
kHz, and r = 10/2 mm, z = 1.1r. So: LMCO = 154 Da.
High mass range is theoretically unlimited, but in reality limited by
the initial or thermal kinetic energy of the ions which will cause
heavy ions to hit the quadrupole electrodes and be neutralized.
5. QIT-TOF instruments are severely limited in resolution. Why?
What could you, perhaps, do to improve the resolution?
Consider question 3. In a QIT, the ions are contained over a
volume of 1-2 mm diameter, so no matter the extraction voltage,
there will be a substantial kinetic energy distribution caused by
differences in ion position at the time of extraction. Possible ways
to fix this include: cooling the trap and/or squeezing the ion packet
size, drifting the ions into an extraction multipole and performing
orthogonal extraction, using a linac geometry to “bunch” the ions.
6. Estimate the native cyclotron frequency of a 1000 m/z ion in a
12 T FTICR mass spectrometer. If the FTICR uses a 2 cm cubic
ICR cell (geometry constant α = 0.73), how much does the native
cyclotron frequency shift with 1 V, 5 V, 10 V on the trapping plates?
Estimated cyclotron frequency
12
= f  qB 
2 m
2 *(1000*1.6e  19 /1.6e  27)
Adjusted cyclotron frequency =
 
c
   2eV 
  c 
,
2
ma 2
 2 
2
thus
1 V = -48 Hz frequency shift
5 V = -252 Hz frequency shift
10 V = -485 Hz frequency shift.
 199kHz
c  
, c  199kHz * 2*   1.25e  6rad / s
2