## College Physics (4th Edition)

(a) $E = 0.060~eV$ (b) $\Delta V = 0.060~V$ (c) Photons with a wavelength of 5 nm have a high amount of energy, which is significantly higher than photons of visible light.
(a) We can find the kinetic energy of the electrons: $\lambda = \frac{h}{p}$ $\lambda = \frac{h}{\sqrt{2~m~E}}$ $\sqrt{2~m~E} = \frac{h}{\lambda}$ $2~m~E = \frac{h^2}{\lambda^2}$ $E = \frac{h^2}{2~m~\lambda^2}$ $E = \frac{(6.626\times 10^{-34}~J~s)^2}{(2)(9.1\times 10^{-31}~kg)(5.0\times 10^{-9}~m)^2}$ $E = (9.649\times 10^{-21}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E = 0.060~eV$ (b) We can find the potential difference $\Delta V$: $q~\Delta V = E$ $\Delta V = \frac{E}{q}$ $\Delta V = \frac{9.649\times 10^{-21}~J}{1.6\times 10^{-19}~C}$ $\Delta V = 0.060~V$ (c) Photons with a wavelength of 5 nm have a high amount of energy, which is significantly higher than photons of visible light.