Converting Electron Mass from kg to MeV/c^2: A Quick Guide

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Homework Help Overview

The discussion centers around converting the mass of an electron from kilograms to MeV/c², utilizing concepts from relativistic physics, specifically Einstein's equation E=mc².

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Einstein's equation for the conversion process and raise questions about the calculations involved in changing units from kg to MeV/c².

Discussion Status

Some participants have provided guidance on the conversion process, while others express confusion and request further clarification on the calculations. Multiple interpretations of the conversion method are being explored.

Contextual Notes

There is mention of specific values for the mass of the electron and the speed of light, as well as the conversion factor for joules to electronvolts, which may be relevant to the calculations discussed.

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how do you change the measurement unit for an electron mass from kg to MeV/c^2?
 
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use Einstein's famous equation
[tex]E=mc^2[/tex]
and use the conversion for joules per electron volt.
 
thanks! :)
 
It is a pleasure.
 
@@a I'm still having trouble doing the calculations~
could you write out the whole equation to transfer the mass of an electron from kg to Mev?
 
You know the mass of the electron and you know the speed of light. Just plug those into [tex]E=mc^2[/tex]. You will have an energy value in terms of joules. Since you also know that one electronvolt is equivalent to [tex]1.602*10^-^1^9[/tex] joules, you can find the energy value in terms of electronvolts instead. Rearrange the original expression and you can reexpress the mass in terms of [tex]MeV/c^2[/tex].
 
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It is amazing how much confusion such a deceptively simple formula can generate. First calculate the rest mass of an electron:
[tex]\begin{split*} E_e = m_ec^2 \\<br /> \ = 9.11 \times 10^{-31} \times (3.0 \times 10^8)^2 \\<br /> \ = 8.2 \times 10^{-14} \ joule \\<br /> \ = 512 \ keV \\<br /> \ = 0.512 \ MeV \\<br /> \ m_ec^2=0.512 \ MeV\end{split*}[/tex]
Now get the mass in the "strange" new units:
[tex]m_e=0.512 \ MeV/{c^2}[/tex]
 
Last edited:
thank you very much! :)
 

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