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Finding relativistic mass and energy of an electron

  • #1
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Homework Statement



A resting electron was sped up to 0.5 of the speed of light. Find:
A. relativistic mass of the electron,
B. total energy of the electron,
C. kinetic energy of the electron.

Homework Equations


K = mv^2/2
E=mc^2

The Attempt at a Solution


Let’s first find the kinetic energy:
K = m*v^2/2
K = (9.1*10^-31*0.25*9*10^16)/2 = 1.02375*10^-14 J

Now we should find relativistic mass using E=mc^2.
m(rel) = E/c^2 = (1.02375*10^-14)/(9*10^16) = 1.1375*10^-31 kg

Total energy must be = m(rest)*c^2
Then E(total) = 9.1*10^-31*9*10^16 = 8.19*10^-14 J

Is this solution correct?
And I suppose potential energy = 8.19*10^-14 - 1.02375*10^-14 = 7.16625*10^-14 ? :)
 
Last edited:

Answers and Replies

  • #2
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No, that is not correct. Do you have an equation with rest mass and Gamma (which is a relationship that crops up a lot in relativistic formula)?
 
  • #3
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No, that is not correct. Do you have an equation with rest mass and Gamma (which is a relationship that crops up a lot in relativistic formula)?
I'm not sure what equation you mean. I found p = mv*gamma, but I don't understand what gamma means there.
 
  • #4
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This is gamma

##\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}##
 
  • #5
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You were never taught this?
 
  • #7
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You were never taught this?
I have a ton of stuff to figure out by myself in a short period of time, so sorry for stupid questions :( I don't remember anything about relativistic mass from school.
So gamma is also relativistic mass/rest mass?
 
  • #8
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Its not a stupid question, I am just wondering what you know about mass at relativistic speeds. Check out the article I sent, there is a ratio embedded in there that you will find useful. :)
 
  • #9
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Its not a stupid question, I am just wondering what you know about mass at relativistic speeds. Check out the article I sent, there is a ratio embedded in there that you will find useful. :)
Ok, then m(rel) = gamma*m(rest)
gamma = sqrt(1-0.25) = sqrt(0.75)
m(rel) = sqrt(0.75)*9.1*10^-31 = 7.88*10^-31

Is it correct now?
 
  • #10
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Gamma is almost like a way of translating mass/lengths/time from the ordinary or Newtonian way of thinking of life to the relativistic way... It's usually a multiplier of some kind.
 
  • #11
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Ok, then m(rel) = gamma*m(rest)
gamma = sqrt(1-0.25) = sqrt(0.75)
m(rel) = sqrt(0.75)*9.1*10^-31 = 7.88*10^-31

Is it correct now?
You are close. It should be m_electron/sqrt(0.75)
 
  • #12
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You are close. It should be m_electron/sqrt(0.75)
Hm, but why? On wikipedia the equation is m(rel)/m(rest) = gamma, so gamma*m(rest) = m(rel)
 
  • #13
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This is gamma

##\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}##
Mass increases as speed increases
 
  • #14
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Mass increases as speed increases
Right, got it. So then I should plug in relativistic mass as m into K = mv^2/2 and E=mc^2 and get the kinetic and total energies?
Then m = 9.1*10^-31/sqrt(75) = 1.05 * 10^-30
Total energy E = 1.05*10^-30*9*10^16 = 9.45*10^-14
Kinetic energy K = (1.05*10^-30*0.25*9*10^8)/2 = 1.18*10^-22

Or am I wrong again?
 
  • #15
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Oh, looked it up. KE = mc^2 - m0c^2 = 1.05*10^-30*9*10^16 - 9.1*10^-31*9*10^16 = 1.26*10^-14 J
Now I have to figure what "total energy" means.
KE = Total energy - Potential energy, so I suppose m(rel)*c^2 = 9.45*10^-14 is total energy like I wrote in the previous message? I hope it's correct now? :)
 

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