Answer
(a) $K = 71.2~eV$
(b) $U = 14.4~eV$
(c) $K = 5.0~keV$
Work Step by Step
(a) We can find the kinetic energy of the electron:
$K = \frac{1}{2}mv^2$
$K = (\frac{1}{2})~(9.109\times 10^{-31}~kg)(5.0\times 10^6~m/s)^2$
$K = 1.139\times 10^{-17}~J$
$K = (1.139\times 10^{-17}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$K = 71.2~eV$
(b) We can find the potential energy:
$U = \frac{k~q^2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)~(1.6\times 10^{-19}~C)^2}{0.10\times 10^{-9}~m}$
$U = 2.304\times 10^{-18}~J$
$U = (2.304\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$U = 14.4~eV$
(c) The final kinetic energy will be equal to the energy gained by passing through the potential difference. We can find the kinetic energy of the proton:
$K = \Delta V~q$
$K = (5000~V)(1.6\times 10^{-19}~C)$
$K = 8.0\times 10^{-16}~J$
$K = (8.0\times 10^{-16}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$K = 5.0\times 10^3~eV$
$K = 5.0~keV$
