Answer
(a) $K = 10~keV$
(b) $U = 1.44~MeV$
(c) $K = 1.2\times 10^{19}~eV$
Work Step by Step
(a) 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 ion:
$K = \Delta V~q$
$K = (5000~V)(2)(1.6\times 10^{-19}~C)$
$K = 1.6\times 10^{-15}~J$
$K = (1.6\times 10^{-15}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$K = 10\times 10^3~eV$
$K = 10~keV$
(b) We can find the potential energy:
$U = \frac{k~q^2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)~(1.6\times 10^{-19}~C)^2}{1.0\times 10^{-15}~m}$
$U = 2.304\times 10^{-13}~J$
$U = (2.304\times 10^{-13}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$U = 1.44\times 10^6~eV$
$U = 1.44~MeV$
(c) The kinetic energy just before impact is equal to the initial gravitational potential energy:
$K = U_0 = mgh$
$K = (0.200~kg)(9.8~m/s^2)(1.0~m)$
$K = 1.96~J$
$K = (1.96~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$K = 1.2\times 10^{19}~eV$
