J.J.Thompson used a cathode ray instrument to measure the charge-to-mass ratio for the electron.

Question

Consider a beam of particles travelling in the  direction through electrostatic plates of  between which the electrons are deflected (down) in the  direction by an electric field of magnitude  . A phosphorous screen is placed  away from the end of the electrostatic plates and a displacement of 35.0 cm is observed.  The magnetic field is turned on and the beam is restored to its non-displaced position when B is measured to be .  Evaluate the charge-to-mass ratio.

Answer 1

Some (but very little) of the drop of 0.35 m will take place while the beam is transiting the 0.1 horizontal distance through the plates. I will ignore this small amount in the interest of simplicity, but also, because the result for the charge to mass ratio (q/m) comes out to be very close to the accepted value for the electron. I will also use a non-relativistic analysis. The key item here is the horizontal velocity of the beam. which turns out to be only about 1% of the speed of light.

As the beam transits the plates, the horizontal component of the velocity, vx, does not change, but the vertical component, vy, changes from from zero to - 300 q/m tof (where tof is the time of flight through the plates - which is equal to 0.1/ vx). Thus on exiting the plates, vy = - 300 (q/m) 0.1 /vx.

This means that the velocity vector makes an angle of θ below the horizontal given by

tan(θ) = 300 (q/m) 0.1 /( vx²)

By simple trig, we then have 0.35 = 0.9 tan(θ) . With some algebra this can be rearranged as

vx² = 300 (q/m) (0.1) (0.9) /.35

However, for the crossed fields geometry here, we also will have

vx² = (E/B)² to get back to the zero deflection condition with the B field turned on. Equating the two expressions for vx² gives (after some algebra)

q/m = [ 300 / 81E-6 ]² ,35 / [ 300 (0.1 ) (0.9) ] = 1.78 x 10¹¹

This is close to the accepted value for the electron of 1.76 x 10¹¹