A mass spectrometer separates ions according to their ratio of charge to mass. Such devices are widely used in silence and engineering to analyze unknown mixtures and to separate isotopes of chemical elements. Figure 1.24 shows ions of charge q and mass m first being accelerated from rest through a potential difference V and then entering a region of uniform magnetic field B pointing out of the page. Only the magnetic force acts on the ions in this region, so they undergo circular motion and, after half an orbit, land on a detector. Find an expression for the horizontal distancexxfrom the entrance slit to the point where an ion lands on the detector.

Explanation

This problem is about charged particle undergoing circular motion in a uniform magnetic field. The distance we are asked for is the diameter of the particle's circular path.
Equation, `r=mv//qB`, shows that the path radius depends on the field and on the particle's mass, change and speed. We know everything but the speed, so this becomes a two-step problem in which we will first find the speed, We are given the potential difference -energy per unit charge-so we can use energy conservation to find the kinetic energy and hence the ion's speed in the magnetic field region, Then we will use equation to find the radius of the ion's circular path.
A charge q gains kinetic energy qV in ''faling'' through a potential difference V, so an ion's kinetic energy once it enters the magnetic field is
`1/2mv^2=qV`
Solving for v gives `v=sqrt(2qV//m)`.
Our answer, the path diameter x, is then twice the radius gives in equation.
`x=2r=(2mv)/(qB)=(2msqrt(2qV//m))/(qB)=2/Bsqrt((2mv)/q)`.