# Relativistic effects on an electron

1. Feb 6, 2012

### khfrekek92

1. The problem statement, all variables and given/known data
An electron is accelerated through a potential of 10^9 Volts. What is the Energy, Kinetic Energy, and Momentum in the lab reference frame?

2. Relevant equations
(1/2)mv^2=eV=1.6E-10 J=1000 MeV

3. The attempt at a solution
Solving for the kinetic energy gives 1000 MeV, which then solving for v gives 2E10 m/s, which is greater than the speed of light.. In the rest frame of the electron, E=mc^2=8.2E-14 J, then I need to multiply that by γ to get the energy in the lab reference frame, but I can't solve for v, because I got v>c. What am I doing wrong??

2. Feb 6, 2012

### vela

Staff Emeritus
You're using the classical expression for kinetic energy, which isn't valid at high speeds. You need to use the relativistic formula.

3. Feb 6, 2012

### khfrekek92

So would (γ-1)mc^2=eV now? I know I have to use relativistic kinetic energy, but I just dont know how to relate it to the potential difference. :(

4. Feb 6, 2012

### vela

Staff Emeritus
Yes, that's right. What does the eV stand for on the righthand side of that equation?

5. Feb 6, 2012

### khfrekek92

eV would be the electron charge times the potential, right? And also, since it is an electron, wouldn't the Kinetic energy just be 10^9 eV=10000MeV? Then I can use this and E=γmc^2 where mc^2 is the rest mass and E=K+mc^2? Then solving for gamme, I get γ=19570.9, making β=.999999999. This seems a little too high, but 10^9 V is a lot too...

6. Feb 6, 2012

### vela

Staff Emeritus
Yup, I just asked because you said you didn't know how to work in the potential difference but it was already in your equation.
1000 MeV, like you said in your first post. You have an extra 0 this time.
Yes, that's exactly how you solve it, but your γ is off by a factor of 10. The mass of the electron, 0.511 MeV, is much smaller compared to its energy E, so it's very relativistic. You should expect a speed very close to c.