# Transit time of particle in a particle accelerator question

1. Sep 5, 2009

### bitrex

I have a question about the behavior of a charged particle being accelerated in an electric field. I know that it is possible to find the approximate final velocity of the particle (assuming it doesn't reach too great a fraction of the speed of light) simply by an energy balance, $$1/2mv^2 = qV$$. However, I'm interested in learning how to find the "time in flight" of an electron in the same potential field, and I don't think I can use the above equation to find it out because I need the average velocity, not the final velocity. Particularly I'm interested in the case of knowing the flight time of an electron in a small electrostatically-deflected CRT. I know there should be a straightforward way to do this, but I'm blocking on it. Thanks for any ideas.

2. Sep 5, 2009

### clem

Do you want to know the TOF during the acceleration of the electron up to speed, or the time spent being electrostatically deflected?

3. Sep 5, 2009

### bitrex

I'm interested in figuring out the TOF from the cathode to the CRT screen, assuming the electrons in the beam aren't being deflected sideways by the electrostatic deflection plates, for now. So yes, the first one.

4. Sep 5, 2009

### Bob S

Here is a description of both TV and oscilloscope cathode ray tubes. As I recall, the accelerating voltages range from about 5,000 volts to 15,000 volts.
http://en.wikipedia.org/wiki/Cathode_ray_tube
The acceleration occurs in the first few cm, and the electron drifts the rest of the way at constant velocity. Your foumula, 1/2 mv2 = qV is adequate.

5. Sep 5, 2009

### bitrex

Thanks, Bob. I wasn't sure if the electron was quickly accelerated to its final velocity, or whether it underwent a constant acceleration all the way down the barrel. I was reading about a neat experimental way to determine the mass of a charged particle (in this case an electron) by putting the CRT in a longitudinal magnetic field and giving the electron a sideways deflection as it leaves the gun. The magnetic field causes the particle to rotate (I guess because the magnetic field is always providing a force tangential to the particle's velocity component that's perpendicular to the longitudinal field) with an angular frequency of $$w = \frac{q}{m}B$$. Like a cyclotron essentially. So if you can get the electron to complete one complete rotation, and you know the travel time down the tube, you know the angular frequency, and if you know the magnetic field strength you can calculate the mass of the electron.