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If I understand correctly, an electron volt is the work done when an electron is moved from a plate with a voltage of 0V to another plate with a voltage of +1V. This is represented by $V = W/Q$, or $W = VQ. W = 1$(volt, change from 0 to +1) * $1.6\cdot 10^{-19}$(coulombs, the charge of an electron), which, of course, is $1.6\cdot 10^{-19} J$. The work, the electron volts eV, is this value of $1.6\cdot 10^{-19}$ J.

My question is: how is this accurate? If we use Einstein's $E = mc^2$, we can arrange this to $m = \frac{E}{c^2}$. The energy is our value ($1.6\cdot 10^{-19}$ J) over the speed of light, squared ($299792458 \frac{m}{s}$). This gives us $1.78\cdot 10^{-36} kg$.

However, a quick Google search tells us that the mass of an electron is 9.10938E−31 kg (quite literally 100% wrong with percent error).

My question is, why does this not work out? Is Wikipedia wrong? Is the math wrong? It does seem rather peculiar that the speed of light would be used here, but that's a simple observation.

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    $\begingroup$ The electron is moving backwards in this scenario: work is actually done when the electron is moved from the +1V plate to the 0V plate. When the electron is moved from 0V to +1V, you can harness its movement to do work elsewhere (but less than 1eV, due to inefficiencies). $\endgroup$ – Dietrich Epp Jun 7 '15 at 12:31
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    $\begingroup$ To put it another way, 1 eV is the energy required to move an electron across a voltage gradient of just 1V. On the other hand, $E=m_e c^2$ computes the energy that would be produced by an electron getting annihilated in a antimatter-matter annihilation. That produces an outrageously larger amount of energy: The entirety of the electron's mass is converted to pure energy. $\endgroup$ – Iwillnotexist Idonotexist Jun 7 '15 at 13:48
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    $\begingroup$ It is not clear at all what the "inconsistency" here is supposed to be. You compute the mass-equivalent of the electron volt and are surprised that it isn't the electron mass, but you do not give any indication why it should be the electron mass. $\endgroup$ – ACuriousMind♦ Jun 7 '15 at 14:40
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    $\begingroup$ @IwillnotexistIdonotexist thank you! That made it very clear to me. $\endgroup$ – Dylan Spano Jun 7 '15 at 18:32
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The problem is that the two calculations have hardly anything to do with one another - so it's no wonder you don't get the same result.

The electron volt, as you say, measures the work you need to move an electron across a potential difference of one volt.

On the other hand, if you want to calculate the mass of an electron using $E=mc^2$, what you need is the energy of the electron in a rest frame. You can measure this to be 511 keV - which explains the difference in your calculation. Why should this be the same as the work you need to move an electron across a potential barrier?

Note also that the electron volt is a unit and your calculation just defines the electron volt in terms of SI-units joule (which is itself defined via other units). If you would redefine the volt, then your eV would also change - the mass of the electron however is independent of the definition of your units except the unit for "mass". Therefore, the only way that the two calculations could give the same result and it would not be pure coincidence would be because the volt was already defined via the electron mass. This is however not the case.

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  • $\begingroup$ Thank you! I was unaware that they were unrelated. This helped. $\endgroup$ – Dylan Spano Jun 6 '15 at 21:22
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    $\begingroup$ The confusion may have arised due to the fact that electron volts are used as units of mass in particle physics. That's because they usually use natural units with $c=1$. What the OP has calculated is that unit of mass in kilograms, which indeed is unrelated to the mass of electron. $\endgroup$ – JiK Jun 7 '15 at 7:00

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