# Speed of Free Electron Moving in Space

by quickclick330
Tags: electron, free, moving, space, speed
 P: 87 The total energy of a free electron moving through empty space is E=1.5mc^2, where m is the mass of the electron and c is the speed of light. What is this electron’s speed? 1. c (the speed of light) 2. 0.7453c 3. 0.8660c 4. 0.9428c 5. 0.9950c 6. 0.9999c I thought since that Etotal = 1.5mc^2 then.... Etotal = KE = .5mv^2 1.5mc^2 = .5mv^2 and solve for v (masses would cancel which is what I was looking for) What is wrong about this approach?
 Mentor P: 41,325 Using KE = .5mv^2 is only good when v << c. Hint: $$E = \gamma m c^2$$
 P: 133 I believe that the proper relativistic kinetic energy equation would be $$E_k = (\gamma - 1)mc^2$$ Or did you make that mistake on purpose, Doc Al?
 P: 87 Speed of Free Electron Moving in Space oh shoot...i forgot about that. Do I need to use the non-approximated version of KE? (I can never remember it.. but i'm looking it up right now)
 Mentor P: 41,325 E is total energy, of course. (There's no need to find kinetic energy.)
 P: 87 okay
 P: 87 so mc^2 would cancel out on both sides and I would be solving 1.5 = gamma right?
Mentor
P: 41,325
 Quote by quickclick330 so mc^2 would cancel out on both sides and I would be solving 1.5 = gamma right?
Yep. That's all you need to do.
 P: 87 ahh...got it, thank you :-)
 P: 133 Oh, I get it... the total energy of the free electron is the sum of its kinetic energy and the energy from the mass-energy equivalence? The potential energy of the electron is zero, since it's free, right? $$E_{total}= (\gamma - 1)mc^2 + mc^2$$ $$\ \ \ = \gamma mc^2$$ Is this so?
Mentor
P: 41,325
 Quote by kudoushinichi88 Oh, I get it... the total energy of the free electron is the sum of its kinetic energy and the energy from the mass-energy equivalence? The potential energy of the electron is zero, since it's free, right? $$E_{total}= (\gamma - 1)mc^2 + mc^2$$ $$\ \ \ = \gamma mc^2$$ Is this so?
Yep.
 P: 133 Ah, thanks...

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