Natural units

Definitions of speed of light, reduced Planck's constant, and electronvolt in terms of kilograms, meters, seconds:

c=299792458\ {m \over s}

\hbar =1.054571817\times 10^{-34}\ kg{m^{2} \over s}

eV=1.602176463\times 10^{-19}\ kg{m^{2} \over s^{2}}

Then solve for second:

{\hbar  \over eV}=6.582120268\times 10^{-16}s

s=1.519267287\times 10^{15}{\hbar  \over eV}

Solve for kilogram:

c^{2}=8.987551787\times 10^{16}\ {m^{2} \over s^{2}}

{eV \over c^{2}}=1.782661731\times 10^{-36}\ kg

kg=5.609589202\times 10^{35}\ {eV \over c^{2}}

Solve for meter:

{\hbar  \over c}=3.51767294\times 10^{-43}\ kg\cdot m

{\hbar  \over c}=1.973270014\times 10^{-7}\ {eV \over c^{2}}m

5067730.179\ {\hbar  \over c}{c^{2} \over eV}=m

m=5067730.179\ {\hbar c \over eV}

Summarize results:

kg=5.609589202\times 10^{35}\ {eV \over c^{2}}

m=5067730.179\ {\hbar c \over eV}

s=1.519267287\times 10^{15}{\hbar  \over eV}

Dimensional conversion:

kg^{x}m^{y}s^{z}\sim c^{y-2x}\hbar ^{y+z}eV^{x-y-z}

Now omit c and ℏ:

kg=5.609589202\times 10^{35}\ eV

m=5,067,730.179\ {1 \over eV}

s=1.519267287\times 10^{15}{1 \over eV}

Dimensional conversion:

kg^{x}m^{y}s^{z}\sim eV^{x-y-z}