Relativistic effects on an electron

AI Thread Summary
An electron accelerated through a potential of 10^9 Volts results in a kinetic energy of 1000 MeV, but the classical kinetic energy formula leads to an impossible speed greater than light. The correct approach involves using the relativistic kinetic energy formula, where the energy in the lab frame is related to the potential difference. The kinetic energy is indeed 10^9 eV, equating to 1000 MeV, and the rest mass energy of the electron must be considered. The calculations yield a Lorentz factor (γ) of approximately 19570.9, indicating the electron is moving at a speed very close to the speed of light. This discussion emphasizes the necessity of applying relativistic equations at high velocities.
khfrekek92
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Homework Statement


An electron is accelerated through a potential of 10^9 Volts. What is the Energy, Kinetic Energy, and Momentum in the lab reference frame?


Homework Equations


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



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??

Thanks in advance!
 
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You're using the classical expression for kinetic energy, which isn't valid at high speeds. You need to use the relativistic formula.
 
So would (γ-1)mc^2=eV now? I know I have to use relativistic kinetic energy, but I just don't know how to relate it to the potential difference. :(
 
Yes, that's right. What does the eV stand for on the righthand side of that equation?
 
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...
 
khfrekek92 said:
eV would be the electron charge times the potential, right?
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.
And also, since it is an electron, wouldn't the Kinetic energy just be 10^9 eV=10000MeV?
1000 MeV, like you said in your first post. You have an extra 0 this time.
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...
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.
 
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