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Speed of Free Electron Moving in Space 
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#1
Dec1407, 05:12 PM

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? 


#2
Dec1407, 05:29 PM

Mentor
P: 41,471

Using KE = .5mv^2 is only good when v << c.
Hint: [tex]E = \gamma m c^2[/tex] 


#3
Dec1407, 05:40 PM

P: 133

I believe that the proper relativistic kinetic energy equation would be
[tex]E_k = (\gamma  1)mc^2[/tex] Or did you make that mistake on purpose, Doc Al? 


#4
Dec1407, 05:41 PM

P: 87

Speed of Free Electron Moving in Space
oh shoot...i forgot about that. Do I need to use the nonapproximated version of KE? (I can never remember it.. but i'm looking it up right now)



#5
Dec1407, 05:42 PM

Mentor
P: 41,471

E is total energy, of course.
(There's no need to find kinetic energy.) 


#6
Dec1407, 05:44 PM

P: 87

okay



#7
Dec1407, 05:46 PM

P: 87

so mc^2 would cancel out on both sides and I would be solving 1.5 = gamma right?



#9
Dec1407, 05:51 PM

P: 87

ahh...got it, thank you :)



#10
Dec1407, 05:52 PM

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 massenergy equivalence? The potential energy of the electron is zero, since it's free, right?
[tex]E_{total}= (\gamma  1)mc^2 + mc^2[/tex] [tex]\ \ \ = \gamma mc^2[/tex] Is this so? 


#12
Dec1407, 06:00 PM

P: 133

Ah, thanks...



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