## College Physics (4th Edition)

(a) $E = 62~eV$ (b) $E = 0.0038~eV$ (c) $\Delta V = 0.0038~V$
(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 $\Delta V$: $q~\Delta V = E$ $\Delta V = \frac{E}{q}$ $\Delta V = \frac{6.0\times 10^{-22}~J}{1.6\times 10^{-19}~C}$ $\Delta V = 0.0038~V$