Relativistic Kinetic energy, momentum, speed.

[irre1evantt]

Homework Statement

An Electron (rest mass=9.11*10^-31kg) is accelerated to an energy (mass energy+kinetic energy) of 30*10^6 eV (30 MeV). What is its kinetic energy? Its momentum? Its speed?
(Note: 1 eV = 1.602*10^-19 Joules; c=2.998 * 10^8 ms^-1)

Homework Equations

The Attempt at a Solution

Okay, so this problem was on a test that I missed and I have been trying to understand it/figure it out since then. I solved for Ek (kinetic energy) I believe. (Em= Mass energy; Ek = Kinetic Energy)
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Em + Ek = (3*10^6 eV)* ((1.602*10^-19 J)/(1eV))
=4.806 * 10^-13 J here I converted to Joules.

Then I solved for Em
Em = mc^2
=(9.11*10^-31 kg)(2.998 * 10^8 m*s^-1)^2
=8.188*10^-14 J

So, plugging into Em + Ek = 4.806*10^-13 J and rearranging to solve for Ek i get
Ek= 4.806*10^-13 J - Em
Ek= 4.806*10^-13 J - 8.188*10^-14 J
Ek= 3.9872 * 10^-13 J
I'm not too sure if I found Ek the accurate way.
Now, if I did, I'm a little lost in how to find my momentum and speed.
I know momentum is p = ymv = mv/(sqrt(1-(v/v)^2)) with y being gamma
and I believe after using that to find my momentum I can find velocity by rearranging the Ek equation of
Ek = (1/2)mv^2 to v= sqrt(2Ek/m)

Im just not sure if I'm heading in the right direction. A little guidance would be truly appreciated. Also, I wasn't sure where to post this question.

 

[Simon Bridge]

Not a lot of point converting to Joules: you are just making work for yourself and adding a way to make mistakes.
What is wrong with all energies in keV and speeds as a fraction of the speed of light.

Thus total energy E is given by: ##E = \gamma E_0## ... E0 is the rest mass energy of an electron - which is 0.511 keV or 0.000511MeV and gamma is given by ##\gamma=(1-v^2)^{-1/2}##, with v as the fraction of the speed of light.

Kinetic energy T is then: ##T=E-E_0## and the energy-momentum relation is: ##E^2-E_0^2+p^2##

But otherwise you did OK: KE is the difference between total and rest energy.
Momentum is found from the energy-momentum relation.
You can use the equation above for unified units, for SI units it becomes:
$$E^2=E_0^2+p^2c^2$$

 

[irre1evantt]

Agh, yes I see. Thank you so much!

 

[Simon Bridge]

NO worries, welcome to PF.