# How do electrostatic analysers act as kinetic energy analysers?

Consider a mass spectrometer. The ions produced are accelerated across a potential V.

Now, the kinetic energy of an ion with charge q is given by qV. When it enters the analyser with electric field strength E (orthogonal to the trajectory), the radius of the trajectory is given by

r = 2K/qE, where K is the kinetic energy.

Substituting with qV, we get

r = 2V/E

The mass and charge terms disappear! So how exactly is this a kinetic energy analyser? Because as per this equation, all ions, irrespective of charge, mass and kinetic energy must have the same trajectory. Where is the flaw in my reasoning?

• I think this question may be more appropriate at physics SE. In any case, off the cuff it seems your curvature calculation ignores that the velocity under E is changing ie Fperp = qE so a = dv/dt = (q/m) E while vpar is constant. – Buck Thorn Jan 18 '19 at 18:00
• The electric field in an analyser is orthogonal to the trajectory. – user1089 Jan 18 '19 at 18:02
• @user1089 Related:hyperphysics.phy-astr.gsu.edu/hbase/magnetic/maspec.html#c2 – Tyberius Jan 18 '19 at 18:05
• I know, that's why I used labels perp and par. Label perp refers to motion orthogonal to V. The prep dimension is the resolving dimension. Note the q/m factor – Buck Thorn Jan 18 '19 at 18:06
• @Try Hard I meant that the electric field is orthogonal throught out. It's a radially symmetric field. – user1089 Jan 18 '19 at 18:12

The point is that all ions produced in the ion source will not have a kinetic energy outside of the acceleration region equal to $$q V$$. When produced the ions will most probably be produced at (slightly) different positions in the acceleration field, with a small $$\Delta V$$ potential deviation from the average acceleration potential, and with a thermal kinetic energy distribution ($$E_{kin,thermal}$$).
$$E_{kin,real} = q V + q \Delta V + E_{kin,thermal}$$