Can you give a few details? where/how was it measured, was the value .511MeV found right away, etc..

Readers who'd like a proper answer to the question may find this link useful.

The value found by this experiment is .537 MeV

I do hope that someone can answer the question from an historical point of view, specifying when this experiment was made first, and if Einstein's conjecture (we all know that his 1905 paper was not even intended as a proof by himself) was verified by any different experiment.

  • $\begingroup$ The energy of an electron can have any value greater than or equal to $mc^2$. Maybe you mean the rest energy? It's not very clear what you mean by this question or what you would consider to be an acceptable answer. $\endgroup$
    – user466
    Dec 28 '18 at 20:28

I suspect what you really are after is the electron's mass, since we can then "easily" get its energy via an obscure :-) equation developed by some fellow named Einstein.

The path thru measuring the charge-to-mass ratio, followed by some famous oil-drop experiments, is discussed pretty well in a physics.SE page.

  • $\begingroup$ are you saying that Einstein knew the energy of an electron was .511 MeV? was it only confirmed later by pair annihilation? How was asking how that valued was concretely measured, was it trhough a calorimeter or what? $\endgroup$
    – user157860
    Sep 20 '18 at 15:23
  • $\begingroup$ @user157860 I was trying to be facetious in reference to $E = mc^2$ . I failed. $\endgroup$ Sep 20 '18 at 15:32
  • $\begingroup$ nevermind, do you know when and how it was actually concretely measured? $\endgroup$
    – user157860
    Sep 20 '18 at 15:36
  • $\begingroup$ I suppose it takes an act of faith to say that the formula you quote established that the energy of rest mass was .511MeV $\endgroup$
    – user157860
    Sep 23 '18 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.