[tex]\begin{gathered} V=\frac{(ma)\cdot d}{q} \\ 463.55kV=9.11\cdot\frac{10^{-31}kg}{1.602\cdot10^{-19}C}\cdot a\cdot d \\ 8.15\cdot\frac{10^{16}m^2}{s^2}=a\cdot d \\ \\ \end{gathered}[/tex][tex]\begin{gathered} vf^2=2\cdot ad \\ vf=\sqrt[2]{2\cdot8.15\cdot10^{16}}=4.03\cdot10^8m/s \end{gathered}[/tex]
First you use the relation between energy and voltage to get the value of axd, then you use kinematics to solve the final speed
