Question

In a certain mass spectrometer, an ion beam passes through a velocity filter consisting of mutually perpendicular fields E and B. The beam then enters a region of another magnetic field B perpendicular to the beam. The radius of curvature of the resulting ion beam is proportional to:

a.EB'/B
b.EB/B'
c.BB'/E
d.B/EB'
e.E/BB'

Answer 1

The radius of curvature of an ion beam in a mass spectrometer velocity filter is proportional to B/EB'. This relation is derived based on the balanced forces in a velocity selector and the circular motion in a subsequent magnetic field.

Step-by-step explanation:

Mass Spectrometer and Velocity Filter

In the velocity selector of a mass spectrometer, ions are subjected to perpendicular electric field (E) and magnetic field (B). For an ion to pass through undeflected, the electric force qE must balance the magnetic force qvB, so v = E/B where q is the charge of the ion, and v is the velocity.

Once through the velocity selector, ions with velocity v enter another region with a magnetic field only, causing them to move in circular paths. These paths have a radius r, which for a moving charged particle in a magnetic field is given by r = mv/qB (m is the mass of the ion). Since all surviving ions have the same velocity from the selector, the radius is therefore proportional to m/qB. Therefore, the correct proportional relationship for the radius of curvature of the ion beam is B/EB' when the magnetic field in the selection region is varied (B').