Acceleration through PD, relativistic momentum

[Stickybees]

When accelerating an electron through a PD of 10^3, that will give a momentum of 1 MeV/c right? Or is there something I'm not taking into account with relativity?

Thanks :D

 

[Meir Achuz]

You probably meant PD=10^6 volts. Then, the NR formula would give p-1 MeV.
But the relativistic result is 1.4 MeV.

 

[Stickybees]

You probably meant PD=10^6 volts. Then, the NR formula would give p-1 MeV.
But the relativistic result is 1.4 MeV.

Ah yes sorry I did, but how would I would out the relativistic result like you did?

 

[Stickybees]

So I've used K=Vq, then K+m(c^2)=E to get the energy and I've got approximately the right answer, but when I divide by c to get units from MeV to MeV/c for momentum I'm of course not even in the same order of magnitude.

 

[Stickybees]

So I've used K=Vq, then K+m(c^2)=E to get the energy and I've got approximately the right answer, but when I divide by c to get units from MeV to MeV/c for momentum I'm of course not even in the same order of magnitude.

Nevermind, I think I must have the right answer I was just thinking that /c had already happened.