1
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ 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. $\endgroup$ – Buck Thorn Jan 18 '19 at 18:00
  • $\begingroup$ The electric field in an analyser is orthogonal to the trajectory. $\endgroup$ – user1089 Jan 18 '19 at 18:02
  • $\begingroup$ @user1089 Related:hyperphysics.phy-astr.gsu.edu/hbase/magnetic/maspec.html#c2 $\endgroup$ – Tyberius Jan 18 '19 at 18:05
  • $\begingroup$ 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 $\endgroup$ – Buck Thorn Jan 18 '19 at 18:06
  • $\begingroup$ @Try Hard I meant that the electric field is orthogonal throught out. It's a radially symmetric field. $\endgroup$ – user1089 Jan 18 '19 at 18:12
1
$\begingroup$

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 ($E_{kin,thermal}$).

In these conditions, the real kinetic energy of the ions as they exit the ion source will be slightly different:

$$E_{kin,real} = q V + q \Delta V + E_{kin,thermal}$$

and the purpose of the electrostatic analyser is to separate the ions depending on their actual kinetic energy and thus compensate for the distribution of kinetic energies of the ions exiting the ion source.

$\endgroup$
1
  • $\begingroup$ Thanks! I had suspected this to be the case, but since neither my textbook nor my professor had mentioned this, I was a little unsure. $\endgroup$ – user1089 Jan 19 '19 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.