PDA

View Full Version : Speed of proton


Biancaraquel
Feb 11, 2007, 11:47 PM
A proton with speed 1.5*10^5 m/s falls through a potential difference of 100 volts, gaining speed. What is the speed reached?

I have the answer he gave us which is 2.04*10^5 m/s, I just don't know what equation I use to find it! Can someone please help me out?

Capuchin
Feb 12, 2007, 03:20 AM
As the proton has a charge of +e, it gains 1eV of energy as it moves through 1V, so moving through 100V it will gain 100eV (this is due to potential energy turning into kinetic energy)
If you want an equation:

PE = QV

This PE is turned completely into KE as the proton moves through the PD.

So if you use

KE = \frac{1}{2}mv^2

Using your values for v and the protonic mass.

You can convert this KE into eV using the electronic charge.

Then you can add 100 to this KE (in eV) then do the process in reverse to find v.

Doing it myself I get the answer you stated.

Let me know if you have trouble making it work!