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Illustration 27.3: A Mass Spectrometer

Multiple Masses Demonstration | For a single particle:

initial velocity = m/s | electric field = N/C | magnetic field = T

mass = x 10-3kg | charge = x 10-3 C

Please wait for the animation to completely load.

Begin this animation by selecting the Multiple Masses Demonstration. This shows five particles passing through a model of a mass spectrometer. The particles have different masses but are otherwise identical. Notice how the particles are separated based on their mass. Restart.

You can enter values for the initial conditions and then press the "register values and play" button to see a single particle pass through the mass spectrometer. The particle initially enters a region with an electric field directed downward and a magnetic field directed into the screen. Since the particle is negatively charged, the electric field exerts an upward force (F = q E; see, for example, Illustration 23.4) and the magnetic field initially exerts a downward force (F = q v x B). Try setting the magnetic or the electric field to zero to see the effect of just one field on the particle. For certain values of the magnetic and electric fields, the magnetic and electric forces on the particle will exactly cancel and the particle will pass through the first region. This region is called a velocity selector, since only particles with a certain initial velocity will pass through for given values of the magnetic and electric fields. Other Explorations and Problems from this chapter will require you to formulate a mathematical relationship between the initial values for particles that pass through the velocity selector.

If a particle is able to pass through the first region, it enters a region where only the magnetic field is present. Since the magnetic field exerts a force perpendicular to the direction of the velocity (F = q v x B), the particle follows a circular path (since v and B are constant and v and B are perpendicular). The radius of this path depends on the mass. From Newton's second law for uniform circular motion, |F| = mv2/R = q |v x B| = qvB, since v and B are perpendicular. By measuring where the particle strikes one of the walls, you can determine the mass of the particle.

Illustration authored by Melissa Dancy and Wolfgang Christian.

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