Velocity Filter Dispersion

The capability of the filter is determined by the following formula:

D ~ [laE (DM)]/[4VM], where

where
D  =  the dispersion between the masses M and (M - DM), and

M = 2 q V [B/E]2

where
M  = the mass (mass number) which is passing through the filter undeflected and
          collected on the target
a   =  the length of the filter
l    = the drift distance from the center of the filter to target (drift distance)
E   = the electric field strength in V/m
V   = the acceleration voltage or ion energy
B   = the magnetic field strength
q   =  the ionic charge
M/DM  =  maximum mass resolution; DM= the full width at half maximum
                  intensity
As an example, suppose that calcium metal is used as the charge material in the Colutron ion source. After an arc discharge is established in the source, the ions are extracted at 5keV and focused on a target 30 cm from the center of the velocity filter. Calcium has the following naturally occurring isotopes:
CALCIUM
Natural Abundance, Stable Isobars
Ca40, 96.97%, bK40
Ca42, 0.64%, Ar40
Ca43, 0.145%, Ti46
Ca44, 2.06%, Ti48
Ca46, 0.0033%
aCa48, 0.185%

NOTE:
See the online charge preparation database at http://www.colutron.com/products/chargeprep.html
for information on other ion source charge materials.

Suppose also that Ca40 is required at the target (undeflected) with maximum dispersion. If the filter is operated at maximum electric field (300V), the magnetic field required can be calculated using equation 3. Velocity filter parameters and fundamental constants are shown in Tables 1 and 2.

The electric field with 300V on velocity filter is
V = 300V/(0.0178m) = 16,853 V/m

Therefore, the required magnetic field will be

B = [E2•M/(2•V)]½ = [((16,853V/m)2 (40 • 1.66 x 10-27kg))/(2•(5000eV•1.602x10-19 J/eV)]½

   = 0.1085 T = 1,085 Gauss

From Figure 1 below, the magnet power supply current required for a magnetic field  of 1,085 Gauss is roughly  5.5 - 5.75 amperes.

The approximate dispersion at the target between Ca40 and Ca42 can now be calculated from equation 2, and is found to be

D = [(0.3m)(0.152m)(16,853 V/m)(2 • 1.66 x 10-27 kg)]/[4(5000 eV)(40(1.66 x 10-27kg)]
    =  1.92 x 10-3 m
    =  1.92 mm

 

Table 1. Velocity Filter Specifications and Parameters
 
Table 2. Useful fundamental constants

 Figure 1. Velocity Filter Magnetic Field