Physics Help Forum Electron Velocity

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 Apr 14th 2008, 10:02 PM #1 Junior Member   Join Date: Apr 2008 Posts: 21 Electron Velocity Is there a way to calculate this in a Non-Calculus way? I saw one of my friends try to use a relativistic view, specifically: $\displaystyle v = \sqrt{\frac{E^2 - m_0^2c^4}{m^2c^2}}$ I know that this can be simplified even more, but he chose to leave it as such... I don't even know if he was right or not to use this equation... That leaves me to my second question... If you cannot use a non-calc method to do this, then what is the BEST way to do it?
Apr 15th 2008, 05:03 AM   #2

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 Originally Posted by Aryth Is there a way to calculate this in a Non-Calculus way? I saw one of my friends try to use a relativistic view, specifically: $\displaystyle v = \sqrt{\frac{E^2 - m_0^2c^4}{m^2c^2}}$ I know that this can be simplified even more, but he chose to leave it as such... I don't even know if he was right or not to use this equation... That leaves me to my second question... If you cannot use a non-calc method to do this, then what is the BEST way to do it?
Ummmm....What problem are you trying to solve? It looks like you are trying to find the speed of the moving frame such that the electron has a specific energy?

-Dan
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 Apr 15th 2008, 08:40 AM #3 Junior Member   Join Date: Apr 2008 Posts: 21 Well, it deals with Linear Accelerators. We're dealing with energies between 6 and 25 MeV. I just want to know the best way to find the range of electron velocities for the given energy range.
Apr 15th 2008, 09:09 AM   #4

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 Originally Posted by Aryth Well, it deals with Linear Accelerators. We're dealing with energies between 6 and 25 MeV. I just want to know the best way to find the range of electron velocities for the given energy range.
Oh okay. The energy given is going to be the kinetic energy of the electrons, so
$\displaystyle K = (\gamma - 1)m_0c^2$

$\displaystyle v = \left ( 1 - \frac{m^2c^4}{(K + mc^2)^2} \right ) c$

-Dan
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 Apr 15th 2008, 05:06 PM #5 Junior Member   Join Date: Apr 2008 Posts: 21 If I may ask... What is the $\displaystyle \gamma$ for?
 Apr 16th 2008, 07:58 PM #6 Junior Member   Join Date: Apr 2008 Posts: 21 Would the gamma be the Lorentz factor? $\displaystyle \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$
Apr 17th 2008, 05:03 AM   #7

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 Originally Posted by Aryth Would the gamma be the Lorentz factor? $\displaystyle \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$
Yes, sorry I didn't specify that earlier.

-Dan
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