Answer
$K.E=835eV$
Work Step by Step
We know that
$K.E=\frac{1}{2}mv^2$
We plug in the known values in this formula to obtain:
$K.E=\frac{1}{2}(1.67\times10^{-27})(3.9989\times10^5)^2$
$K.E=1.3353\times10^{-16}J$
We know that
$1 eV=1.602\times10^{-19}J$
so $1J=\frac{1}{1.602\times10^{-19}}=6.242\times10^{18}eV$
Thus $K.E=1.3353\times10^{-16}\times6.242\times10^{18}=835eV$
