PX156 - E4 - units

E^{2} = (p c)^{2} + (m c^{2})^{2}

- where,
$E =$ total energy
$p =$ momentum
$m =$ rest mass
$c =$ speed of light

here, $p$ and $m$ can be written in terms of energy equivalents
- mass: $e V / c^{2}$
- momentum: $e V / c$
to distinguish momentum from mass, the power of $c$ is tracked
$m_{p} = 1.67 \times 10^{- 27} k g :$ energy equivalent of this mass is $m_{p} c^{2} = 1.5 \times 10^{- 10} J$
$e V$ is preferred
$1 e V \equiv$ energy gained by an electron as it moves through a potential difference of $1 V \equiv 1.602 \times 10^{- 19} J$
in $e V : m_{p} c^{2} = 938 \times 10^{6} e V = 938 M e V ⟹ m_{p} = 938 M e V / c^{2}$