Answer
(a) $E = 62~eV$
(b) $E = 0.0038~eV$
(c) $\Delta V = 0.0038~V$
Work Step by Step
(a) We can find the energy of a photon with a wavelength of $20~nm$:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{20\times 10^{-9}~m}$
$E = (9.939\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 62~eV$
(b) We can find the minimum kinetic energy of the electrons:
$\lambda = \frac{h}{p}$
$\lambda = \frac{h}{\sqrt{2~m~E}}$
$\sqrt{2~m~E} = \frac{h}{\lambda}$
$2~m~E = \frac{h^2}{\lambda^2}$
$E = \frac{h^2}{2~m~\lambda^2}$
$E = \frac{(6.626\times 10^{-34}~J~s)^2}{(2)(9.1\times 10^{-31}~kg)(20\times 10^{-9}~m)^2}$
$E = (6.0\times 10^{-22}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 0.0038~eV$
(c) We can find the potential difference $\Delta V$:
$q~\Delta V = E$
$\Delta V = \frac{E}{q}$
$\Delta V = \frac{6.0\times 10^{-22}~J}{1.6\times 10^{-19}~C}$
$\Delta V = 0.0038~V$
