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**

Ca^{40}, 96.97%,
^{b}K^{40}

Ca^{42}, 0.64%,
Ar^{40}

Ca^{43}, 0.145%,
Ti^{46}

Ca^{44}, 2.06%,
Ti^{48}

Ca^{46}, 0.0033%
^{a}Ca^{48},
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 Ca^{40} 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 = [E^{2}•M/(2•V)]^{½} = [((16,853V/m)^{2}
(40 • 1.66 x 10^{-27}kg))/(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 Ca^{40}
and Ca^{42 }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^{-27}kg)]

= 1.92 x 10^{-3} m

= 1.92 mm