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Potential Difference: Electron Accelerated to Relativistic Speed

  • Thread starter Kung-Fu
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  • #1
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Homework Statement


An electron accelerated from rest through a potential difference V acquires a
speed of 0.8c. Find the value of V.

Homework Equations


E=(gamma)mc^2, E=Vq

The Attempt at a Solution


For this I related the two equations above and chose a value of m=0.5MeV for the electron and found the value of gamma=1.6667, since I did this in electron volts I cancelled out the c^2 and charge of the electron from the resulting equation and got something looking like this: V=(gamma)mc^2/q=1.667*0.5MeV=8.33e5 V

I wanted to get a confirmation if this was the correct way to approach this problem, I do not know if this answer is or isn't correct.

Thank you for your help
 

Answers and Replies

  • #2
Orodruin
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You are computing the total energy of the electron and comparing it to the potential energy that is converted into kinetic energy after acceleration. What about the rest energy of the electron?
 
  • #3
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I'm sorry but I don't see how my equation doesn't take potential energy into consideration. I calculated the total energy of the electron at 0.8c and solved for the potential difference requirement by dividing by the charge on the electron. It is a little convoluted by me cancelling the charge and c^2 so I re-did it like this: E=(gamma)mc^2=1.667*9.11e-31*9e16=1.367e-13 J, this is the total energy. I then related this to E=qV and got V=E/q=1.367e-13/1.6e-19=8.54e5 V; the answer is a little bit different but could be due to rounding errors.

Am I misunderstanding what you are saying?
 
  • #4
Orodruin
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Yes, what I am saying is that the total energy of the electron after acceleration is not going to be equal to the potential energy before acceleration. What is the total energy of an electron at rest?
 
  • #5
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E(int)=9.11e-31*9e16=8.199e-14 J and deltaKE=(1.667-1)9.11e-31*9e16=5.46627e-14 J

With this do I now compare the kinetic energy to the charge and this is the potential difference requirement? Like V=deltaKE/q=5.46627e-14/1.6e-19=3.42e5 V?
 
  • #6
Orodruin
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Yes, although I would still use units of MeV/c^2 for the electron mass as well as MeV for the energy as this will give you significantly easier computations. (Note that your original answer 8.54e5 V is 5.1e5 V larger than your new (and correct) answer, exactly corresponding to the electron mass of 511 keV/c^2 = 5.11e5 eV/c^2 ...)
 
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  • #7
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Alright, I got it down now. Thank you very much for your help!
 
  • #8
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I'm only just getting back into math and physics and am confused by the nomenclature above. For example the calculated potential voltage difference in the problem stated at the beginning is shown as "3.42e5 V". I assume this means 3.42 multiplied by e (2.71828...) raised to the 5th power. But that only comes out to 507.6 volts, which is way too low to impart a relativistic velocity to an electron. So I have to be misinterpreting that expression.
 
  • #9
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No problem, I just realized that 3.42e5 is shorthand for 3.42 times 10 to the 5th power, or 342,000 volts.
 

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