# If we are using [eV] as a unit for energy, what, then, should the unit for mass and distance be?

This seems obvious but it is confusing since I know that in MKS (i.e. SI) system, we use Joule with Meter with KG.

But if we're using electron volt (1.6e-19 J) for energy, should we change the mass and distance units too?

• You may. It's not required. – Michael Brown Mar 15 '13 at 2:09
• Wadda ya mean "if"? //particle physicist who knows that $c = 1$ so that mass is in units of $\text{eV}$ and length is in units of $\text{eV}^{-1}$ like God and Leon Lederman intended – dmckee --- ex-moderator kitten Mar 15 '13 at 4:34

It depends on what you mean by "should." You can, if you'd like, quote all of your energies in $\mathrm{eV}$, all of your masses in $\mathrm{kg}$, and all of your distances in $\mathrm{m}$ without any inconsistency.
However, you're probably referring to the fact that joules are naturally suited to kilograms, meters, and seconds in the sense that $$\mathrm J = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}$$ without any weird dimensionless prefactors on either side of the equation, and you want to know what mass and length units, lets call them $\mathrm M_\mathrm{eV}$ and $\mathrm L_\mathrm{eV}$ give you a relationship similar to the relationship for $\mathrm{J}$; \begin{align} \mathrm{eV} = \frac{\mathrm M_\mathrm{eV}\cdot \mathrm L_\mathrm{eV}^2}{s^2}\qquad ?\tag{$\star$} \end{align} If so, then note that there isn't a unique pair that does this. As you noted, there is a dimensionless constant $a$ relating Joules and electron volts; $$a\cdot \mathrm{eV} = \mathrm J$$ This means that we can use the first relation above to write $$a \cdot\mathrm{eV} = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}$$ Now we notice that both of the following pairs of choices will lead to an expression like $(\star)$: for choice 1 take $$\mathrm M_\mathrm{ev} = a^{-1}\cdot \mathrm{kg}, \qquad \mathrm L_\mathrm{eV} = \mathrm m$$ and for choice 2 take $$\mathrm M_\mathrm{ev} = \mathrm{kg}, \qquad \mathrm L_\mathrm{eV} = a^{-1/2}\cdot \mathrm m$$
• What about choosing units such that $c=1$, so both mass and energy would be measured in $\mathrm{eV}$? I'm no particle physicist, but that would seem the most sensible way to me. – Nathaniel Mar 15 '13 at 2:34
• @Nathaniel that's exactly what we do in particle physics. Though it comes with a danger of forgetting that mass and energy aren't the same thing. (Just look around and see how many websites quote the mass of the Higgs boson as 126 GeV, with no factor of $1/c^2$!) – David Z Mar 15 '13 at 5:07