When calculating the accelerating voltage to accelerate an electron to a certain speed, we usually use the following equation:

$$e\times U = \frac{1}{2}mv^2$$

But why does this work? On the left side the unit is electronvolt whereas on the right side it's joule. In order to have joule on both sides the equation should be:

$$e\times U \times 1.6022\times10^{-19}\mathrm{\frac{J}{eV}} = \frac{1}{2}mv^2$$

Why do you never see the second formula? Is it wrong?

  • 3
    $\begingroup$ What do you mean "on the left side the unit is electronvolt whereas on the right side it's joule"? There's nothing there telling you to use specific units, and both sides have dimensions of energy. $\endgroup$
    – ACuriousMind
    Commented Nov 13, 2016 at 21:38
  • 1
    $\begingroup$ $e$ is in coulombs, $U$ is in volts. No? $\endgroup$
    – garyp
    Commented Nov 13, 2016 at 21:50
  • 5
    $\begingroup$ I'm voting to close this question as off-topic because there is no evidence of prior research. $\endgroup$
    – garyp
    Commented Nov 13, 2016 at 21:51
  • $\begingroup$ Thanks for your objections. I think I spotted the mistake. I thought when putting in some values on the left side of the equation the result would be in eV. This is not correct as not only 'U' has to be replaced, but also 'e', which means the result will be 'Coulomb*Volt' which can be resolved to Joule. $\endgroup$
    – Duval
    Commented Nov 13, 2016 at 21:54

1 Answer 1


On the left side you have the potential energy of an electron in an electrostatic potential which is $E_{pot}=e·U$. Thus, the energy unit is not $eV$ but joule (J)=coulomb(C)·volts(V). The unit $eV$ is a incoherent energy unit giving the potential energy of a elementary (electron) charge in an electrical potential with a numerical value corresponding to the value of the electrical potential.

The second formula $e × U × 1,6022·10^{-19}$ is wrong because the electron charge $e=1.6022·10^{-19}C$ times potential U (in V) is already joule (J). Therefore multiplication with the electron charge again is wrong.

Note: if you have an energy given in the unit $eV$ (NB: this is not $e×V$), you obtain the the energy value in $J$ by multiplying the value of the energy given in $eV$ with the electron charge in $C$


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