Velocity Selectors & Mass Spectrometers

You’ll be able to

Determine the speed selected by crossed electric and magnetic fields
Use the radius of a circular path to find a charge-to-mass ratio
Explain why the selector stage is independent of charge and mass

Crossed fields select one speed

A velocity selector places an electric field and a magnetic field perpendicular to each other and to the beam, arranged so their forces oppose. A particle passes straight through only when the two balance: qE = qvB, so v = E/B. Notice what cancels — the charge q drops out entirely, so the selected speed is the same for every particle regardless of charge or mass. That independence is what makes the device useful as a first stage.

The two stages

selector: v = E/B · separator: r = mv/(qB) → m/q = rB/v

Stage one fixes the speed; stage two sorts by mass-to-charge ratio. Combining them gives m/q = rBB'/E when the two magnetic fields differ.

How the mass spectrometer sorts

Particles leaving the selector all share a speed, so when they enter a second magnetic field their radius r = mv/(qB) depends only on m/q. Heavier particles curve less and land further out; more highly charged ones curve more. Measuring where each lands therefore measures its mass-to-charge ratio, which is how isotopes are separated and how unknown molecules are identified.

What the instrument actually measures

Note carefully that the radius gives m/q, not m alone. Two particles with the same ratio — a singly charged atom and a doubly charged atom of twice the mass — land in the same place and cannot be distinguished by radius alone. Real instruments resolve this by controlling the ionization stage so the charge state is known. An exam answer that claims the device measures mass directly has skipped the step where the charge is pinned down.

Worked example

A velocity selector uses E = 3.0 × 10⁴ V/m and B = 0.10 T. The emerging ions then enter a 0.25 T field and follow a circle of radius 0.12 m. Find the selected speed and the mass-to-charge ratio.

1.Selected speed: v = E/B = (3.0 × 10⁴)/0.10 = 3.0 × 10⁵ m/s.
2.In the second field, qvB' = mv²/r rearranges to m/q = rB'/v.
3.m/q = (0.12)(0.25)/(3.0 × 10⁵).
4.m/q = 0.030/(3.0 × 10⁵) = 1.0 × 10⁻⁷ kg/C.

Answer: v = 3.0 × 10⁵ m/s and m/q = 1.0 × 10⁻⁷ kg/C. For a singly charged ion that implies a mass of about 1.6 × 10⁻²⁶ kg, but only once the charge state is known independently.

Watch out

In the selector the electric and magnetic forces must be antiparallel, which requires the two fields to be mutually perpendicular and correctly oriented. If the geometry has them assisting rather than opposing, no speed passes through undeflected and the device does not work at all.

Checkpoint

A velocity selector uses E = 2.0 × 10⁴ V/m and B = 0.050 T. The speed that passes through undeflected is:

Checkpoint

Two ions with the same charge but different masses enter the same magnetic field at the same speed. The heavier ion follows a path of:

Checkpoint

The radius of an ion's path in a mass spectrometer determines its:

On the exam

Treat the two stages separately and in order. The selector fixes v with no reference to the particle; only then does the second field sort by m/q. Trying to combine them in one equation before establishing v is where these problems go wrong.

Answer the 3 checkpoints as you read.