# How do we calculate the relativistic energy equivalent contained in one proton in "MeV"?

Feb 23, 2017

Consider that an electron acts as a wave. It has a spin of $\pm \frac{1}{2}$. Similarly, a proton has a spin of $\pm \frac{1}{2}$. Either particle can be treated using

$E = m {c}^{2}$,

since both have significant wave characteristics (though they are less prominent in protons).

The rest mass of a proton is $1.672621898 \times {10}^{- 27}$ $\text{kg}$. Calculating its energy is a simple conversion:

$E = {\left(1.672621898 \times {10}^{- 27} \text{kg")(2.99792458xx10^8 "m/s}\right)}^{2}$

$= 1.50327759 \times {10}^{- 10} \text{J}$

Now that we're in units of energy, we can convert from $\text{J}$ to $\text{eV}$.

1.50327759 xx 10^(-10) cancel"J" xx cancel"1 eV"/(1.602176565 xx 10^(-19) cancel"J") xx ("1 MeV")/(10^6 cancel"eV")

$=$ $\textcolor{b l u e}{\text{938.272 MeV}}$

This website gives $938.28$ $\text{MeV}$. Pretty close.