Answer
$\Delta V = 6.02 \times 10^{-6}~V$
Work Step by Step
We can find the required speed:
$\lambda = \frac{h}{m~v}$
$v = \frac{h}{m~\lambda}$
$v = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg)(500\times 10^{-9}~m)}$
$v = 1.4548\times 10^{3}~m/s$
We can find the kinetic energy:
$K = \frac{1}{2}mv^2$
$K = (\frac{1}{2})(9.109\times 10^{-31}~kg)(1.4548\times 10^{3}~m/s)^2$
$K = 9.639\times 10^{-25}~J$
We can find the required potential difference:
$\Delta V~\vert q \vert = K$
$\Delta V = \frac{K}{\vert q \vert}$
$\Delta V = \frac{9.639\times 10^{-25}~J}{1.6\times 10^{-19}~C}$
$\Delta V = 6.02 \times 10^{-6}~V$
